On Tuesday, March 1, 2016 at 4:00:34 AM UTC+11, clicl...@freenet.de wrote:

Waldek Hebisch schrieb:

In developement version I now get:

(3) -> integrate(1/(x*(x^2 - 3*x + 2)^(1/3)), x)

(3)
-
3+-+2
\|2
*
log
+-----------+2
2 3+-+2 3| 2
(12x - 72x + 72)\|2 \|x - 3x + 2
+
+-----------+
3 2 3| 2
(- 6x + 84x - 216x + 144)\|x - 3x + 2
+
4 3 2 3+-+
(x - 36x + 180x - 288x + 144)\|2
/
4
x
+
+-----------+2 +-----------+
3| 2 3+-+3| 2 2 3+-+2
3+-+2 12\|x - 3x + 2 + (6x - 12)\|2 \|x - 3x + 2 + x \|2
2\|2 log(---------------------------------------------------------)
2
x
+
+-----+
+-+ | 3+-+
2\|2 \|3\|2
*
atan
4 3 2 +-+3+-+2
(48x - 1008x + 4464x - 6912x + 3456)\|2 \|2
*
+-----------+2
3| 2
\|x - 3x + 2
+
5 4 3 2 +-+
(- 12x - 120x + 2160x - 7200x + 8640x - 3456)\|2
*
+-----------+
3| 2
\|x - 3x + 2
+
6 5 4 3 2 +-+3+-+
(x - 108x + 972x - 3456x + 6048x - 5184x + 1728)\|2 \|2
*
+-----+
| 3+-+
\|3\|2
/
6 5 4 3 2
2x + 648x - 11016x + 51840x - 103680x + 93312x - 31104
/
24
Type: Union(Expression(Integer),...)
Time: 0.01 (IN) + 4.49 (EV) + 0.02 (OT) = 4.52 sec

This is somewhat specific to the example: problematic part of
Trager resultant is of form z^3 - a, and I hardcoded knowledge
of splitting field of such polynomials. This seem to cure all
Goursat-like examples (however, some take a very long time).
I wonder if Goursat always gives residues which are cube roots
or may produce something more complicated?

Congratulations! A compact version of the antiderivative is:

INT(1/(x*(x^2 - 3*x + 2)^(1/3)), x)
= - 1/(4*2^(1/3))*LN(4*(x^2 - 3*x + 2) + (x - 2)^3)
+ 3/(4*2^(1/3))*LN(2^(2/3)*(x^2 - 3*x + 2)^(1/3) + (x - 2))
- SQRT(3)/(2*2^(1/3))
*ATAN(1/SQRT(3) - 2^(1/3)*(x - 2)/(SQRT(3)*(x^2 - 3*x + 2)^(1/3)))

How long is "very long" in minutes, say? If this could be speeded up,
any rewrite rules specific to cube-root integrands could probably be
removed from the FriCAS integrator. And, maybe, other common and simple,
yet problematic factors of Trager resultants could be hard-coded too? To
me this looks like a better place for hard-coding than rewriting at the integrand level.

Collected examples of Goursat pseudo-elliptics involving cube roots have been added to "my" integration problems on the Rubi website, according
to Albert Rich. Other examples are numbers 99, 107, 109 and 112 (!) from Chapter 4 of Timofeev's 1948 textbook. Indeed, I believe that all
classical examples of transcendental elementary algebraic integrals involving a cube root of a non-degenerate quadratic or cubic polynomial belong to this class of pseudo-elliptics.

I don't know the answer to your question but could mail Derive code that tests if algebraics of the forms R(x)/((a + b*x + c*x^2 + d*x^3))^(1/3)
and R(x)/((a + b*x + c*x^2 + d*x^3))^(2/3) are Goursat pseudo-elliptic;
its inspection might enable you to answer the question yourself. My code checks the symmetry properties of R(x) under Möbius transformations that map the radicand roots onto each other, the mappings being a realization
of the symmetric group S_3 (equivalently dihedral D_3) with subgroups
C_2, C_2, C_2 and A_3 (alternating A_3 is isomorphic to cyclic C_3). Alternatively, you could post suspect candidates of cube-root algebraics
for testing by me.

Martin.

Hi Martin,

I know I am a bit late (6+ years) to this conversation, but did you ever publish your code/algorithmic extensions for Goursat's cube-root type integrals?

Cheers,

Sam

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Sam Blake schrieb:

On Tuesday, March 1, 2016 at 4:00:34 AM UTC+11, clicl...@freenet.de wrote:

Waldek Hebisch schrieb:

In developement version I now get:

(3) -> integrate(1/(x*(x^2 - 3*x + 2)^(1/3)), x)

(3)
-
3+-+2
\|2
*
log
+-----------+2
2 3+-+2 3| 2
(12x - 72x + 72)\|2 \|x - 3x + 2
+
+-----------+
3 2 3| 2
(- 6x + 84x - 216x + 144)\|x - 3x + 2
+
4 3 2 3+-+
(x - 36x + 180x - 288x + 144)\|2
/
4
x
+
+-----------+2 +-----------+
3| 2 3+-+3| 2 2 3+-+2
3+-+2 12\|x - 3x + 2 + (6x - 12)\|2 \|x - 3x + 2 + x \|2
2\|2 log(---------------------------------------------------------)
2
x
+
+-----+
+-+ | 3+-+
2\|2 \|3\|2
*
atan
4 3 2 +-+3+-+2
(48x - 1008x + 4464x - 6912x + 3456)\|2 \|2
*
+-----------+2
3| 2
\|x - 3x + 2
+
5 4 3 2 +-+
(- 12x - 120x + 2160x - 7200x + 8640x - 3456)\|2
*
+-----------+
3| 2
\|x - 3x + 2
+
6 5 4 3 2 +-+3+-+
(x - 108x + 972x - 3456x + 6048x - 5184x + 1728)\|2 \|2
*
+-----+
| 3+-+
\|3\|2
/
6 5 4 3 2
2x + 648x - 11016x + 51840x - 103680x + 93312x - 31104
/
24
Type: Union(Expression(Integer),...)
Time: 0.01 (IN) + 4.49 (EV) + 0.02 (OT) = 4.52 sec

This is somewhat specific to the example: problematic part of
Trager resultant is of form z^3 - a, and I hardcoded knowledge
of splitting field of such polynomials. This seem to cure all Goursat-like examples (however, some take a very long time).
I wonder if Goursat always gives residues which are cube roots
or may produce something more complicated?

Congratulations! A compact version of the antiderivative is:

INT(1/(x*(x^2 - 3*x + 2)^(1/3)), x)
= - 1/(4*2^(1/3))*LN(4*(x^2 - 3*x + 2) + (x - 2)^3)
+ 3/(4*2^(1/3))*LN(2^(2/3)*(x^2 - 3*x + 2)^(1/3) + (x - 2))
- SQRT(3)/(2*2^(1/3))
*ATAN(1/SQRT(3) - 2^(1/3)*(x - 2)/(SQRT(3)*(x^2 - 3*x + 2)^(1/3)))

How long is "very long" in minutes, say? If this could be speeded up,
any rewrite rules specific to cube-root integrands could probably be removed from the FriCAS integrator. And, maybe, other common and simple, yet problematic factors of Trager resultants could be hard-coded too? To
me this looks like a better place for hard-coding than rewriting at the integrand level.

Collected examples of Goursat pseudo-elliptics involving cube roots have been added to "my" integration problems on the Rubi website, according
to Albert Rich. Other examples are numbers 99, 107, 109 and 112 (!) from Chapter 4 of Timofeev's 1948 textbook. Indeed, I believe that all
classical examples of transcendental elementary algebraic integrals involving a cube root of a non-degenerate quadratic or cubic polynomial belong to this class of pseudo-elliptics.

I don't know the answer to your question but could mail Derive code that tests if algebraics of the forms R(x)/((a + b*x + c*x^2 + d*x^3))^(1/3)
and R(x)/((a + b*x + c*x^2 + d*x^3))^(2/3) are Goursat pseudo-elliptic;
its inspection might enable you to answer the question yourself. My code checks the symmetry properties of R(x) under Möbius transformations that map the radicand roots onto each other, the mappings being a realization
of the symmetric group S_3 (equivalently dihedral D_3) with subgroups
C_2, C_2, C_2 and A_3 (alternating A_3 is isomorphic to cyclic C_3). Alternatively, you could post suspect candidates of cube-root algebraics for testing by me.

Martin.

Hi Martin,

I know I am a bit late (6+ years) to this conversation, but did you
ever publish your code/algorithmic extensions for Goursat's cube-root
type integrals?

Cheers,

Sam

It took me a week to find the time for this context switch. No, I
haven't written anything about my cube root integrals of Goursat type,
apart from explanatory remarks in this (January to February 2016) and
many other <sci.math.symbolic> threads. Posts under "Announce: FriCAS
1.2.4 has been released" (November 2014 to February 2015) in particular
come to my mind.

Since hardcoding of the splitting field for factors of the form z^3 - a
in the Trager resultant has greatly improved the performance of FriCAS
on cube-root integrands, implementing an analogue for factors of the
form z^4 - a might have a similar effect with fourth-root integrands.
There are cases by Euler that FriCAS still fails to solve.

Martin.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Monday, December 12, 2022 at 4:12:09 AM UTC+11, nob...@nowhere.invalid wrote:

Sam Blake schrieb:

On Tuesday, March 1, 2016 at 4:00:34 AM UTC+11, clicl...@freenet.de wrote:

Waldek Hebisch schrieb:

In developement version I now get:

(3) -> integrate(1/(x*(x^2 - 3*x + 2)^(1/3)), x)

(3)
-
3+-+2
\|2
*
log
+-----------+2
2 3+-+2 3| 2
(12x - 72x + 72)\|2 \|x - 3x + 2
+
+-----------+
3 2 3| 2
(- 6x + 84x - 216x + 144)\|x - 3x + 2
+
4 3 2 3+-+
(x - 36x + 180x - 288x + 144)\|2
/
4
x
+
+-----------+2 +-----------+
3| 2 3+-+3| 2 2 3+-+2
3+-+2 12\|x - 3x + 2 + (6x - 12)\|2 \|x - 3x + 2 + x \|2
2\|2 log(---------------------------------------------------------)
2
x
+
+-----+
+-+ | 3+-+
2\|2 \|3\|2
*
atan
4 3 2 +-+3+-+2
(48x - 1008x + 4464x - 6912x + 3456)\|2 \|2
*
+-----------+2
3| 2
\|x - 3x + 2
+
5 4 3 2 +-+
(- 12x - 120x + 2160x - 7200x + 8640x - 3456)\|2
*
+-----------+
3| 2
\|x - 3x + 2
+
6 5 4 3 2 +-+3+-+
(x - 108x + 972x - 3456x + 6048x - 5184x + 1728)\|2 \|2
*
+-----+
| 3+-+
\|3\|2
/
6 5 4 3 2
2x + 648x - 11016x + 51840x - 103680x + 93312x - 31104
/
24
Type: Union(Expression(Integer),...)
Time: 0.01 (IN) + 4.49 (EV) + 0.02 (OT) = 4.52 sec

This is somewhat specific to the example: problematic part of
Trager resultant is of form z^3 - a, and I hardcoded knowledge
of splitting field of such polynomials. This seem to cure all Goursat-like examples (however, some take a very long time).
I wonder if Goursat always gives residues which are cube roots
or may produce something more complicated?

Congratulations! A compact version of the antiderivative is:

INT(1/(x*(x^2 - 3*x + 2)^(1/3)), x)
= - 1/(4*2^(1/3))*LN(4*(x^2 - 3*x + 2) + (x - 2)^3)
+ 3/(4*2^(1/3))*LN(2^(2/3)*(x^2 - 3*x + 2)^(1/3) + (x - 2))
- SQRT(3)/(2*2^(1/3))
*ATAN(1/SQRT(3) - 2^(1/3)*(x - 2)/(SQRT(3)*(x^2 - 3*x + 2)^(1/3)))

How long is "very long" in minutes, say? If this could be speeded up, any rewrite rules specific to cube-root integrands could probably be removed from the FriCAS integrator. And, maybe, other common and simple, yet problematic factors of Trager resultants could be hard-coded too? To me this looks like a better place for hard-coding than rewriting at the integrand level.

Collected examples of Goursat pseudo-elliptics involving cube roots have been added to "my" integration problems on the Rubi website, according to Albert Rich. Other examples are numbers 99, 107, 109 and 112 (!) from Chapter 4 of Timofeev's 1948 textbook. Indeed, I believe that all classical examples of transcendental elementary algebraic integrals involving a cube root of a non-degenerate quadratic or cubic polynomial belong to this class of pseudo-elliptics.

I don't know the answer to your question but could mail Derive code that tests if algebraics of the forms R(x)/((a + b*x + c*x^2 + d*x^3))^(1/3) and R(x)/((a + b*x + c*x^2 + d*x^3))^(2/3) are Goursat pseudo-elliptic; its inspection might enable you to answer the question yourself. My code checks the symmetry properties of R(x) under Möbius transformations that
map the radicand roots onto each other, the mappings being a realization of the symmetric group S_3 (equivalently dihedral D_3) with subgroups C_2, C_2, C_2 and A_3 (alternating A_3 is isomorphic to cyclic C_3). Alternatively, you could post suspect candidates of cube-root algebraics for testing by me.

Martin.

Hi Martin,

I know I am a bit late (6+ years) to this conversation, but did you
ever publish your code/algorithmic extensions for Goursat's cube-root
type integrals?

Cheers,

Sam

It took me a week to find the time for this context switch. No, I
haven't written anything about my cube root integrals of Goursat type,
apart from explanatory remarks in this (January to February 2016) and
many other <sci.math.symbolic> threads. Posts under "Announce: FriCAS
1.2.4 has been released" (November 2014 to February 2015) in particular
come to my mind.

Since hardcoding of the splitting field for factors of the form z^3 - a
in the Trager resultant has greatly improved the performance of FriCAS
on cube-root integrands, implementing an analogue for factors of the
form z^4 - a might have a similar effect with fourth-root integrands.
There are cases by Euler that FriCAS still fails to solve.

Martin.

Here are some nice cube root-type integrals which may or may not be integrable using your Goursat adaption. I am guessing that as Rubi cannot integrate these that they may not be Goursat type integrals?

(1 + x)/(x*(-1 + x^3)^(1/3))
(-1 + x^2)/(x*(-1 + x^3)^(2/3))
((1 + x)*(-1 + x^3)^(1/3))/((-1 + x)^2*x)
((-1 + x)^2*(1 + x))/(x*(1 + x + x^2)*(-1 + x^3)^(1/3))
((1 + x)*(-1 + x^3)^(2/3))/((-1 + x)^3*x)
(x*(-2 + x^3))/((-1 + x^3)^(1/3)*(-1 + x^3 + x^6))
(x^3*(2 + x^3))/((1 + x^3)^(2/3)*(1 + x^3 + x^6))
((1 - x^3)^(1/3)*(-2 + x^3))/(x^3*(-1 + x^3 + x^6))
((2 + x + x^2)*(-1 + x^3)^(1/3))/(x*(-1 + x^2)^2*(-3 - 2*x + x^2 + x^3))
(1 + 3*x)/((-1 + 3*x)*(-x + x^3)^(1/3))
((-1 + x)*(1 + 3*x))/((-1 + 3*x)*(-x + x^3)^(2/3))
((-1 + x)^2*(1 + 3*x))/(x*(1 + x)*(-1 + 3*x)*(-x + x^3)^(1/3))
(x*(1 + x)*(1 + 3*x))/((-1 + x)*(-1 + 3*x)*(-x + x^3)^(2/3))
(x*(1 + x)*(1 + 3*x))/((-1 + x)^2*(-1 + 3*x)*(-x + x^3)^(1/3))

Sam

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Sam Blake schrieb:

On Monday, December 12, 2022 at 4:12:09 AM UTC+11, nob...@nowhere.invalid wrote:

Sam Blake schrieb:

On Tuesday, March 1, 2016 at 4:00:34 AM UTC+11, clicl...@freenet.de wrote:

Waldek Hebisch schrieb:

In developement version I now get:

(3) -> integrate(1/(x*(x^2 - 3*x + 2)^(1/3)), x)

(3)
-
3+-+2
\|2
*
log
+-----------+2
2 3+-+2 3| 2
(12x - 72x + 72)\|2 \|x - 3x + 2
+
+-----------+
3 2 3| 2
(- 6x + 84x - 216x + 144)\|x - 3x + 2
+
4 3 2 3+-+
(x - 36x + 180x - 288x + 144)\|2
/
4
x
+
+-----------+2 +-----------+
3| 2 3+-+3| 2 2 3+-+2
3+-+2 12\|x - 3x + 2 + (6x - 12)\|2 \|x - 3x + 2 + x \|2
2\|2 log(---------------------------------------------------------)
2
x
+
+-----+
+-+ | 3+-+
2\|2 \|3\|2
*
atan
4 3 2 +-+3+-+2
(48x - 1008x + 4464x - 6912x + 3456)\|2 \|2
*
+-----------+2
3| 2
\|x - 3x + 2
+
5 4 3 2 +-+
(- 12x - 120x + 2160x - 7200x + 8640x - 3456)\|2
*
+-----------+
3| 2
\|x - 3x + 2
+
6 5 4 3 2 +-+3+-+
(x - 108x + 972x - 3456x + 6048x - 5184x + 1728)\|2 \|2
*
+-----+
| 3+-+
\|3\|2
/
6 5 4 3 2
2x + 648x - 11016x + 51840x - 103680x + 93312x - 31104
/
24
Type: Union(Expression(Integer),...)
Time: 0.01 (IN) + 4.49 (EV) + 0.02 (OT) = 4.52 sec

This is somewhat specific to the example: problematic part of
Trager resultant is of form z^3 - a, and I hardcoded knowledge
of splitting field of such polynomials. This seem to cure all Goursat-like examples (however, some take a very long time).
I wonder if Goursat always gives residues which are cube roots
or may produce something more complicated?

Congratulations! A compact version of the antiderivative is:

INT(1/(x*(x^2 - 3*x + 2)^(1/3)), x)
= - 1/(4*2^(1/3))*LN(4*(x^2 - 3*x + 2) + (x - 2)^3)
+ 3/(4*2^(1/3))*LN(2^(2/3)*(x^2 - 3*x + 2)^(1/3) + (x - 2))
- SQRT(3)/(2*2^(1/3))
*ATAN(1/SQRT(3) - 2^(1/3)*(x - 2)/(SQRT(3)*(x^2 - 3*x + 2)^(1/3)))

How long is "very long" in minutes, say? If this could be speeded up, any rewrite rules specific to cube-root integrands could probably be removed from the FriCAS integrator. And, maybe, other common and simple,
yet problematic factors of Trager resultants could be hard-coded too? To
me this looks like a better place for hard-coding than rewriting at the integrand level.

Collected examples of Goursat pseudo-elliptics involving cube roots have
been added to "my" integration problems on the Rubi website, according to Albert Rich. Other examples are numbers 99, 107, 109 and 112 (!) from
Chapter 4 of Timofeev's 1948 textbook. Indeed, I believe that all classical examples of transcendental elementary algebraic integrals involving a cube root of a non-degenerate quadratic or cubic polynomial belong to this class of pseudo-elliptics.

I don't know the answer to your question but could mail Derive code that
tests if algebraics of the forms R(x)/((a + b*x + c*x^2 + d*x^3))^(1/3) and R(x)/((a + b*x + c*x^2 + d*x^3))^(2/3) are Goursat pseudo-elliptic; its inspection might enable you to answer the question yourself. My code
checks the symmetry properties of R(x) under Möbius transformations that
map the radicand roots onto each other, the mappings being a realization
of the symmetric group S_3 (equivalently dihedral D_3) with subgroups C_2, C_2, C_2 and A_3 (alternating A_3 is isomorphic to cyclic C_3). Alternatively, you could post suspect candidates of cube-root algebraics
for testing by me.

Martin.

Hi Martin,

I know I am a bit late (6+ years) to this conversation, but did you
ever publish your code/algorithmic extensions for Goursat's cube-root type integrals?

Cheers,

Sam

It took me a week to find the time for this context switch. No, I
haven't written anything about my cube root integrals of Goursat type, apart from explanatory remarks in this (January to February 2016) and
many other <sci.math.symbolic> threads. Posts under "Announce: FriCAS
1.2.4 has been released" (November 2014 to February 2015) in particular come to my mind.

Since hardcoding of the splitting field for factors of the form z^3 - a
in the Trager resultant has greatly improved the performance of FriCAS
on cube-root integrands, implementing an analogue for factors of the
form z^4 - a might have a similar effect with fourth-root integrands.
There are cases by Euler that FriCAS still fails to solve.

Martin.

Here are some nice cube root-type integrals which may or may not be integrable using your Goursat adaption. I am guessing that as Rubi cannot integrate these that they may not be Goursat type integrals?

(1 + x)/(x*(-1 + x^3)^(1/3))
(-1 + x^2)/(x*(-1 + x^3)^(2/3))
((1 + x)*(-1 + x^3)^(1/3))/((-1 + x)^2*x)
((-1 + x)^2*(1 + x))/(x*(1 + x + x^2)*(-1 + x^3)^(1/3))
((1 + x)*(-1 + x^3)^(2/3))/((-1 + x)^3*x)
(x*(-2 + x^3))/((-1 + x^3)^(1/3)*(-1 + x^3 + x^6))
(x^3*(2 + x^3))/((1 + x^3)^(2/3)*(1 + x^3 + x^6))
((1 - x^3)^(1/3)*(-2 + x^3))/(x^3*(-1 + x^3 + x^6))
((2 + x + x^2)*(-1 + x^3)^(1/3))/(x*(-1 + x^2)^2*(-3 - 2*x + x^2 + x^3))
(1 + 3*x)/((-1 + 3*x)*(-x + x^3)^(1/3))
((-1 + x)*(1 + 3*x))/((-1 + 3*x)*(-x + x^3)^(2/3))
((-1 + x)^2*(1 + 3*x))/(x*(1 + x)*(-1 + 3*x)*(-x + x^3)^(1/3))
(x*(1 + x)*(1 + 3*x))/((-1 + x)*(-1 + 3*x)*(-x + x^3)^(2/3))
(x*(1 + x)*(1 + 3*x))/((-1 + x)^2*(-1 + 3*x)*(-x + x^3)^(1/3))

There are 14 integrands here. The first two are readily integrated by
Derive 6.10 and should therefore be doable by standard recipes from
books like G&R and Timofeev:

INT((x + 1)/(x*(x^3 - 1)^(1/3)), x)
INT((x^2 - 1)/(x*(x^3 - 1)^(2/3)), x)

So Rubi could and should be tought to do these as well.

The subsequent twelve integrands, however, fail in Derive 6.10:

INT((x + 1)*(x^3 - 1)^(1/3)/(x*(x - 1)^2), x)
INT((x + 1)*(x - 1)^2/(x*(x^2 + x + 1)*(x^3 - 1)^(1/3)), x)
INT((x + 1)*(x^3 - 1)^(2/3)/(x*(x - 1)^3), x)
INT(x*(x^3 - 2)/((x^3 - 1)^(1/3)*(x^6 + x^3 - 1)), x)
INT(x^3*(x^3 + 2)/((x^3 + 1)^(2/3)*(x^6 + x^3 + 1)), x)
INT((1 - x^3)^(1/3)*(x^3 - 2)/(x^3*(x^6 + x^3 - 1)), x)
INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)),
x)
INT((3*x + 1)/((3*x - 1)*(x^3 - x)^(1/3)), x)
INT((x - 1)*(3*x + 1)/((3*x - 1)*(x^3 - x)^(2/3)), x)
INT((x - 1)^2*(3*x + 1)/(x*(x + 1)*(3*x - 1)*(x^3 - x)^(1/3)), x)
INT(x*(x + 1)*(3*x + 1)/((x - 1)*(3*x - 1)*(x^3 - x)^(2/3)), x)
INT(x*(x + 1)*(3*x + 1)/((x - 1)^2*(3*x - 1)*(x^3 - x)^(1/3)), x)

My Goursat tests for cube-root integrands determine all but the four
integrals from #6 to #9 to be directly of Goursat type, which means
that the integrands become rational under one of the characteristic substitutions. Of the other four, the first three become Goursat
integrable under the fairly obvious substitution t = x^3.

A more involved substitution must have been used to scramble integrand
#9; even FriCAS fails on this one, but succeeds once t = x^3 has been
applied. The substitution doesn't make the integrand look nicer though,
and while the change cannot be considered momentous either, for FriCAS
it evidently is. The integrand simplifies a bit when an algebraic part
of the antiderivative is removed:

INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/
(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)), x) =
(x^3 - 1)^(1/3)/(x^2 - 1) +
INT((x^2 - 1)*(x^2 + x + 2)/
(x*(x^3 - 1)^(2/3)*(x^3 + x^2 - 2*x - 3)), x)

There is no reason to assume that Rubi can handle Goursat type
integrals in general.

Martin.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

"clicliclic@freenet.de" schrieb:

Sam Blake schrieb:

Here are some nice cube root-type integrals which may or may not be integrable using your Goursat adaption. I am guessing that as Rubi
cannot integrate these that they may not be Goursat type integrals?

(1 + x)/(x*(-1 + x^3)^(1/3))
(-1 + x^2)/(x*(-1 + x^3)^(2/3))
((1 + x)*(-1 + x^3)^(1/3))/((-1 + x)^2*x)
((-1 + x)^2*(1 + x))/(x*(1 + x + x^2)*(-1 + x^3)^(1/3))
((1 + x)*(-1 + x^3)^(2/3))/((-1 + x)^3*x)
(x*(-2 + x^3))/((-1 + x^3)^(1/3)*(-1 + x^3 + x^6))
(x^3*(2 + x^3))/((1 + x^3)^(2/3)*(1 + x^3 + x^6))
((1 - x^3)^(1/3)*(-2 + x^3))/(x^3*(-1 + x^3 + x^6))
((2 + x + x^2)*(-1 + x^3)^(1/3))/(x*(-1 + x^2)^2*(-3 - 2*x + x^2 + x^3))
(1 + 3*x)/((-1 + 3*x)*(-x + x^3)^(1/3))
((-1 + x)*(1 + 3*x))/((-1 + 3*x)*(-x + x^3)^(2/3))
((-1 + x)^2*(1 + 3*x))/(x*(1 + x)*(-1 + 3*x)*(-x + x^3)^(1/3))
(x*(1 + x)*(1 + 3*x))/((-1 + x)*(-1 + 3*x)*(-x + x^3)^(2/3))
(x*(1 + x)*(1 + 3*x))/((-1 + x)^2*(-1 + 3*x)*(-x + x^3)^(1/3))

There are 14 integrands here. The first two are readily integrated by
Derive 6.10 and should therefore be doable by standard recipes from
books like G&R and Timofeev:

INT((x + 1)/(x*(x^3 - 1)^(1/3)), x)
INT((x^2 - 1)/(x*(x^3 - 1)^(2/3)), x)

So Rubi could and should be tought to do these as well.

The subsequent twelve integrands, however, fail in Derive 6.10:

INT((x + 1)*(x^3 - 1)^(1/3)/(x*(x - 1)^2), x)
INT((x + 1)*(x - 1)^2/(x*(x^2 + x + 1)*(x^3 - 1)^(1/3)), x)
INT((x + 1)*(x^3 - 1)^(2/3)/(x*(x - 1)^3), x)
INT(x*(x^3 - 2)/((x^3 - 1)^(1/3)*(x^6 + x^3 - 1)), x)
INT(x^3*(x^3 + 2)/((x^3 + 1)^(2/3)*(x^6 + x^3 + 1)), x)
INT((1 - x^3)^(1/3)*(x^3 - 2)/(x^3*(x^6 + x^3 - 1)), x)
INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)),
x)
INT((3*x + 1)/((3*x - 1)*(x^3 - x)^(1/3)), x)
INT((x - 1)*(3*x + 1)/((3*x - 1)*(x^3 - x)^(2/3)), x)
INT((x - 1)^2*(3*x + 1)/(x*(x + 1)*(3*x - 1)*(x^3 - x)^(1/3)), x)
INT(x*(x + 1)*(3*x + 1)/((x - 1)*(3*x - 1)*(x^3 - x)^(2/3)), x)
INT(x*(x + 1)*(3*x + 1)/((x - 1)^2*(3*x - 1)*(x^3 - x)^(1/3)), x)

My Goursat tests for cube-root integrands determine all but the four integrals from #6 to #9 to be directly of Goursat type, which means
that the integrands become rational under one of the characteristic substitutions. Of the other four, the first three become Goursat
integrable under the fairly obvious substitution t = x^3.

A more involved substitution must have been used to scramble integrand
#9; even FriCAS fails on this one, but succeeds once t = x^3 has been applied. The substitution doesn't make the integrand look nicer
though, and while the change cannot be considered momentous either,
for FriCAS it evidently is. The integrand simplifies a bit when an
algebraic part of the antiderivative is removed:

INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/
(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)), x) =
(x^3 - 1)^(1/3)/(x^2 - 1) +
INT((x^2 - 1)*(x^2 + x + 2)/
(x*(x^3 - 1)^(2/3)*(x^3 + x^2 - 2*x - 3)), x)

There is no reason to assume that Rubi can handle Goursat type
integrals in general.

For the record, and as an "optimal" reference solution for Nasser's
integrator tests, a compact antiderivative of the non-Goursat integral
#9:

INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)),
x)

is given by a sum of the following five components:

- INT(x*(x^2 + x + 2)/((x + 1)*(x^2 - 1)*(x^3 - 1)^(2/3)), x) =
(x^3 - 1)^(1/3)/(x^2 - 1)

INT(2/(3*x*(x^3 - 1)^(2/3)), x) =
- 2*SQRT(3)/9*ATAN(SQRT(3)/3*(1 - 2*(x^3 - 1)^(1/3)))
+ 1/3*LN((x^3 - 1)^(1/3) + 1) - 1/9*LN(x^3)

- INT(x^2*(2*x^6 - 19*x^3 - 10)
/(3*(x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
SQRT(3)/9*ATAN(SQRT(3)/3*(1 - 2*(2*(x^3 - 1)^(2/3) + x^3 - 4)
/((x^3 - 1)^(1/3)*(2*(x^3 - 1)^(1/3) + 3))))
+ 1/6*LN(- 2*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) + x^3 - 4)
- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

INT(x*(x^3 - 1)^(1/3)*(3*x^3 - 1)/(x^9 - 2*x^6 + x^3 - 27), x) = SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x*(x^3 - 1)^(2/3))/3))
+ 1/6*LN(x*(x^3 - 1)^(2/3) - 3) - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

INT((x^9 - 3*x^6 + 8*x^3 + 3)
/((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3) /(3*((x^3 - 1)^(1/3) + x^2))))
+ 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

All but the first and last of these turn into classical cases of
elementary integrals under the substitutions t = x^3 or t = 1/x^3.

Of course, the three -1/18*LN(x^9 - 2*x^6 + x^3 - 27) terms should be
merged into one. Is the FriCAS antiderivative suboptimal because its
leaf count exceeds twice that of this optimal result?

Martin.

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On 10/30/2023 2:02 PM, clicliclic@freenet.de wrote:

Of course, the three -1/18*LN(x^9 - 2*x^6 + x^3 - 27) terms should be
merged into one. Is the FriCAS antiderivative suboptimal because its
leaf count exceeds twice that of this optimal result?

Martin.

fyi, I've added your 14 integrals as separate new test file
to CAS integration tests. Maybe this will make it easier to reference things.

Created a new test file (#213) of them, and converted them to all
the other CAS formats and added them to
CAS integration tests summer 2023 version.

So now you can see the full report for all 8 CAS systems. Link at end.

Result % solved is
-------------------
1. Maple 100.00% 14/14
2. Fricas 92.86% 13/14
3. Mathematica 78.57% 11/14
4. Rubi 71.43% 10/14
5. Maxima 14.29% 2/14
6. Sympy 14.29% 2/14
7. Mupad 7.14% 1/14
8. Giac 0.00% 0/14

In terms of grading:

A B C F
Fricas 85.714 7.143 0.000 7.143
Mathematica 78.571 0.000 0.000 21.429
Rubi 35.714 0.000 35.714 28.571
Maxima 14.286 0.000 0.000 85.714
Maple 7.143 0.000 92.857 0.000
Mupad 0.000 7.143 0.000 92.857
Giac 0.000 0.000 0.000 100.000
Sympy 0.000 0.000 14.286 85.714

For optimal anti, I picked the smallest one by trying each integral
on different CAS's. So it is possible a better optimal exists that
I missed.

Only Maple was able to solve them all but the grading was not the best.
FriCAS got the best A grading and was in second place in terms of number solved. Maxima, mupad, giac and sympy did not do well on these hard
integrals.

Link to the new test report is

<https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/test_cases/213_Goursat/report.htm>

Which you can also access from the main web page

<https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/index.htm>

by going to the "links to individual test reports" page and scroll all
the way down.

--Nasser

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"Nasser M. Abbasi" schrieb:

On 10/30/2023 2:02 PM, clicliclic@freenet.de wrote:

Of course, the three -1/18*LN(x^9 - 2*x^6 + x^3 - 27) terms should
be merged into one. Is the FriCAS antiderivative suboptimal because
its leaf count exceeds twice that of this optimal result?

fyi, I've added your 14 integrals as separate new test file
to CAS integration tests. Maybe this will make it easier to reference
things.

Just to avoid misunderstandings: these 14 integrals are by Sam Blake
who posted them on 11 December 2022; they may even be duplicated from
his test suite of algebraic integrals - I simply don't know.

Created a new test file (#213) of them, and converted them to all
the other CAS formats and added them to
CAS integration tests summer 2023 version.

So now you can see the full report for all 8 CAS systems. Link at end.

Result % solved is
-------------------
1. Maple 100.00% 14/14
2. Fricas 92.86% 13/14
3. Mathematica 78.57% 11/14
4. Rubi 71.43% 10/14
5. Maxima 14.29% 2/14
6. Sympy 14.29% 2/14
7. Mupad 7.14% 1/14
8. Giac 0.00% 0/14

Rubi's success rate on these 14 cube root integrals is impressive even
though many solutions appear to involve functions higher than
elementary. And I notice that my antiderivative for integral #9 was
very far from optimal! The optimal solution is just something like:

INT((x^2 + x + 2)*(x^3 - 1)^(1/3)
/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)), x) =
(x^3 - 1)^(1/3)/(x^2 - 1)
+ SQRT(3)/3*ATAN(SQRT(3)/3*(1 + 2*(x^2 - 1)/(x^3 - 1)^(1/3)))
+ 1/2*LN((x^3 - 1)^(1/3) - x^2 + 1)
- 1/6*LN(x^2*(x^4 - 3*x^2 - x + 3))

Wow,

Martin.

In terms of grading:

A B C F
Fricas 85.714 7.143 0.000 7.143
Mathematica 78.571 0.000 0.000 21.429
Rubi 35.714 0.000 35.714 28.571
Maxima 14.286 0.000 0.000 85.714
Maple 7.143 0.000 92.857 0.000
Mupad 0.000 7.143 0.000 92.857
Giac 0.000 0.000 0.000 100.000
Sympy 0.000 0.000 14.286 85.714

For optimal anti, I picked the smallest one by trying each integral
on different CAS's. So it is possible a better optimal exists that
I missed.

Only Maple was able to solve them all but the grading was not the
best. FriCAS got the best A grading and was in second place in terms
of number solved. Maxima, mupad, giac and sympy did not do well on
these hard integrals.

Link to the new test report is

<https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/test_cases/213_Goursat/report.htm>

Which you can also access from the main web page

<https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/index.htm>

by going to the "links to individual test reports" page and scroll all
the way down.

PS: Can any of the systems solve the fifth component integral:

INT((x^9 - 3*x^6 + 8*x^3 + 3)
/((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3) /(3*((x^3 - 1)^(1/3) + x^2))))
+ 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

in elementary terms?

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On Tuesday, October 31, 2023 at 5:59:01 AM UTC+11, nob...@nowhere.invalid wrote:

"clicl...@freenet.de" schrieb:

Sam Blake schrieb:

Here are some nice cube root-type integrals which may or may not be integrable using your Goursat adaption. I am guessing that as Rubi cannot integrate these that they may not be Goursat type integrals?

(1 + x)/(x*(-1 + x^3)^(1/3))
(-1 + x^2)/(x*(-1 + x^3)^(2/3))
((1 + x)*(-1 + x^3)^(1/3))/((-1 + x)^2*x)
((-1 + x)^2*(1 + x))/(x*(1 + x + x^2)*(-1 + x^3)^(1/3))
((1 + x)*(-1 + x^3)^(2/3))/((-1 + x)^3*x)
(x*(-2 + x^3))/((-1 + x^3)^(1/3)*(-1 + x^3 + x^6))
(x^3*(2 + x^3))/((1 + x^3)^(2/3)*(1 + x^3 + x^6))
((1 - x^3)^(1/3)*(-2 + x^3))/(x^3*(-1 + x^3 + x^6))
((2 + x + x^2)*(-1 + x^3)^(1/3))/(x*(-1 + x^2)^2*(-3 - 2*x + x^2 + x^3)) (1 + 3*x)/((-1 + 3*x)*(-x + x^3)^(1/3))
((-1 + x)*(1 + 3*x))/((-1 + 3*x)*(-x + x^3)^(2/3))
((-1 + x)^2*(1 + 3*x))/(x*(1 + x)*(-1 + 3*x)*(-x + x^3)^(1/3))
(x*(1 + x)*(1 + 3*x))/((-1 + x)*(-1 + 3*x)*(-x + x^3)^(2/3))
(x*(1 + x)*(1 + 3*x))/((-1 + x)^2*(-1 + 3*x)*(-x + x^3)^(1/3))

There are 14 integrands here. The first two are readily integrated by Derive 6.10 and should therefore be doable by standard recipes from
books like G&R and Timofeev:

INT((x + 1)/(x*(x^3 - 1)^(1/3)), x)
INT((x^2 - 1)/(x*(x^3 - 1)^(2/3)), x)

So Rubi could and should be tought to do these as well.

The subsequent twelve integrands, however, fail in Derive 6.10:

INT((x + 1)*(x^3 - 1)^(1/3)/(x*(x - 1)^2), x)
INT((x + 1)*(x - 1)^2/(x*(x^2 + x + 1)*(x^3 - 1)^(1/3)), x)
INT((x + 1)*(x^3 - 1)^(2/3)/(x*(x - 1)^3), x)
INT(x*(x^3 - 2)/((x^3 - 1)^(1/3)*(x^6 + x^3 - 1)), x)
INT(x^3*(x^3 + 2)/((x^3 + 1)^(2/3)*(x^6 + x^3 + 1)), x)
INT((1 - x^3)^(1/3)*(x^3 - 2)/(x^3*(x^6 + x^3 - 1)), x)
INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)), x)
INT((3*x + 1)/((3*x - 1)*(x^3 - x)^(1/3)), x)
INT((x - 1)*(3*x + 1)/((3*x - 1)*(x^3 - x)^(2/3)), x)
INT((x - 1)^2*(3*x + 1)/(x*(x + 1)*(3*x - 1)*(x^3 - x)^(1/3)), x)
INT(x*(x + 1)*(3*x + 1)/((x - 1)*(3*x - 1)*(x^3 - x)^(2/3)), x)
INT(x*(x + 1)*(3*x + 1)/((x - 1)^2*(3*x - 1)*(x^3 - x)^(1/3)), x)

My Goursat tests for cube-root integrands determine all but the four integrals from #6 to #9 to be directly of Goursat type, which means
that the integrands become rational under one of the characteristic substitutions. Of the other four, the first three become Goursat integrable under the fairly obvious substitution t = x^3.

A more involved substitution must have been used to scramble integrand
#9; even FriCAS fails on this one, but succeeds once t = x^3 has been applied. The substitution doesn't make the integrand look nicer
though, and while the change cannot be considered momentous either,
for FriCAS it evidently is. The integrand simplifies a bit when an algebraic part of the antiderivative is removed:

INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/
(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)), x) =
(x^3 - 1)^(1/3)/(x^2 - 1) +
INT((x^2 - 1)*(x^2 + x + 2)/
(x*(x^3 - 1)^(2/3)*(x^3 + x^2 - 2*x - 3)), x)

There is no reason to assume that Rubi can handle Goursat type
integrals in general.

For the record, and as an "optimal" reference solution for Nasser's integrator tests, a compact antiderivative of the non-Goursat integral
#9:
INT((x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)),
x)
is given by a sum of the following five components:

- INT(x*(x^2 + x + 2)/((x + 1)*(x^2 - 1)*(x^3 - 1)^(2/3)), x) =
(x^3 - 1)^(1/3)/(x^2 - 1)
INT(2/(3*x*(x^3 - 1)^(2/3)), x) =
- 2*SQRT(3)/9*ATAN(SQRT(3)/3*(1 - 2*(x^3 - 1)^(1/3)))
+ 1/3*LN((x^3 - 1)^(1/3) + 1) - 1/9*LN(x^3)

- INT(x^2*(2*x^6 - 19*x^3 - 10)
/(3*(x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) = SQRT(3)/9*ATAN(SQRT(3)/3*(1 - 2*(2*(x^3 - 1)^(2/3) + x^3 - 4)
/((x^3 - 1)^(1/3)*(2*(x^3 - 1)^(1/3) + 3))))
+ 1/6*LN(- 2*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) + x^3 - 4)
- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

INT(x*(x^3 - 1)^(1/3)*(3*x^3 - 1)/(x^9 - 2*x^6 + x^3 - 27), x) = SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x*(x^3 - 1)^(2/3))/3))
+ 1/6*LN(x*(x^3 - 1)^(2/3) - 3) - 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

INT((x^9 - 3*x^6 + 8*x^3 + 3)
/((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) = SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3) /(3*((x^3 - 1)^(1/3) + x^2))))
+ 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

All but the first and last of these turn into classical cases of
elementary integrals under the substitutions t = x^3 or t = 1/x^3.

Of course, the three -1/18*LN(x^9 - 2*x^6 + x^3 - 27) terms should be
merged into one. Is the FriCAS antiderivative suboptimal because its
leaf count exceeds twice that of this optimal result?

Martin.

Hey Martin,

Here's my solution for this integral:

In[431]:= IntegrateAlgebraic[((x^2 + x + 2) (x^3 - 1)^(1/3))/(x (x^2 - 1)^2 (x^3 + x^2 - 2 x - 3)), x]

Out[431]= (-1 + x^3)^(1/3)/(-1 + x^2) - ArcTan[(Sqrt[3] (-1 + x^3)^(1/3))/(-2 + 2 x^2 + (-1 + x^3)^(1/3))]/Sqrt[3] +
1/3 Log[1 - x^2 + (-1 + x^3)^(1/3)] - 1/6 Log[1 - 2 x^2 + x^4 + (-1 + x^2) (-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)]

Where the substitution u == (1-x^2)/(-1+x^3)^(1/3) reduces the integral to 1/(u^2 (1+u^3)).

Cheers,

Sam

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On 10/31/2023 12:46 PM, clicliclic@freenet.de wrote:

PS: Can any of the systems solve the fifth component integral:

INT((x^9 - 3*x^6 + 8*x^3 + 3)
/((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3) /(3*((x^3 - 1)^(1/3) + x^2))))
+ 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

in elementary terms?

I've added this as integral 15 and updated the report at

<https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/test_cases/213_Goursat/report.htm>

it shows that only Fricas and Maple can solve it. But Fricas only one that gives solution in terms of elementary functions:

1/9*3^(1/2)*arctan((258*3^(1/2)*(2*x^5-59*x^2)*(x^3-1)^(2/3)-258*3^(1/2)*(13*x^7+56*x^4-24*x)*(x^3-1)^(1/3)-3^
(1/2)*(169*x^9+7789*x^6+7135*x^3-10368))/(2197*x^9+13021*x^6-25667*x^3+13824))+1/18*ln((x^9-83*x^6+82*x^3+18*(
2*x^5-5*x^2)*(x^3-1)^(2/3)-9*(x^7-13*x^4+3*x)*(x^3-1)^(1/3)-27)/(x^9-2*x^6+x^3-27))

I see now that Sam's integrator also solved this giving

(-1 + x^3)^(1/3)/(-1 + x^2) - ArcTan[(Sqrt[3] (-1 + x^3)^(1/3))/(-2 + 2 x^2 + (-1 + x^3)^(1/3))]/Sqrt[3] +
1/3 Log[1 - x^2 + (-1 + x^3)^(1/3)] - 1/6 Log[1 - 2 x^2 + x^4 + (-1 + x^2) (-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)]

Which is smaller than Fricas result. Next time I build the pages will
use Sam's result as the optimal for this problem as I only saw it now.

So file #213 now has 15 integrals not 14. I also updated the optimal for 3 integrals
to better ones I found later. Currently A grade result is

Fricas 86.667
Mathematica 66.667
Rubi 20.000
Maxima 13.333
Maple 0.000
Giac 0.000
Mupad 0.000
Sympy 0.000

--Nasser

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On 10/31/2023 6:19 PM, Nasser M. Abbasi wrote:

On 10/31/2023 12:46 PM, clicliclic@freenet.de wrote:

PS: Can any of the systems solve the fifth component integral:

INT((x^9 - 3*x^6 + 8*x^3 + 3)
/((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) =
SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3)
/(3*((x^3 - 1)^(1/3) + x^2))))
+ 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

in elementary terms?

I've added this as integral 15 and updated the report at

<https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/test_cases/213_Goursat/report.htm>

it shows that only Fricas and Maple can solve it. But Fricas only one that gives solution in terms of elementary functions:

1/9*3^(1/2)*arctan((258*3^(1/2)*(2*x^5-59*x^2)*(x^3-1)^(2/3)-258*3^(1/2)*(13*x^7+56*x^4-24*x)*(x^3-1)^(1/3)-3^
(1/2)*(169*x^9+7789*x^6+7135*x^3-10368))/(2197*x^9+13021*x^6-25667*x^3+13824))+1/18*ln((x^9-83*x^6+82*x^3+18*(
2*x^5-5*x^2)*(x^3-1)^(2/3)-9*(x^7-13*x^4+3*x)*(x^3-1)^(1/3)-27)/(x^9-2*x^6+x^3-27))

I see now that Sam's integrator also solved this giving

Opps, sorry, I see now Sam was answering about #9 and not #15.
So one can ignore this below. There is nothing I need to change or update.

(-1 + x^3)^(1/3)/(-1 + x^2) - ArcTan[(Sqrt[3] (-1 + x^3)^(1/3))/(-2 + 2 x^2 + (-1 + x^3)^(1/3))]/Sqrt[3] +
1/3 Log[1 - x^2 + (-1 + x^3)^(1/3)] - 1/6 Log[1 - 2 x^2 + x^4 + (-1 + x^2) (-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)]

Which is smaller than Fricas result. Next time I build the pages will
use Sam's result as the optimal for this problem as I only saw it now.

--Nasser

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On Wednesday, November 1, 2023 at 4:42:47 AM UTC+11, nob...@nowhere.invalid wrote:

"Nasser M. Abbasi" schrieb:

On 10/30/2023 2:02 PM, clicl...@freenet.de wrote:

Of course, the three -1/18*LN(x^9 - 2*x^6 + x^3 - 27) terms should
be merged into one. Is the FriCAS antiderivative suboptimal because
its leaf count exceeds twice that of this optimal result?

fyi, I've added your 14 integrals as separate new test file
to CAS integration tests. Maybe this will make it easier to reference things.

Just to avoid misunderstandings: these 14 integrals are by Sam Blake
who posted them on 11 December 2022; they may even be duplicated from
his test suite of algebraic integrals - I simply don't know.

Created a new test file (#213) of them, and converted them to all
the other CAS formats and added them to
CAS integration tests summer 2023 version.

So now you can see the full report for all 8 CAS systems. Link at end.

Result % solved is
-------------------
1. Maple 100.00% 14/14
2. Fricas 92.86% 13/14
3. Mathematica 78.57% 11/14
4. Rubi 71.43% 10/14
5. Maxima 14.29% 2/14
6. Sympy 14.29% 2/14
7. Mupad 7.14% 1/14
8. Giac 0.00% 0/14

Rubi's success rate on these 14 cube root integrals is impressive even though many solutions appear to involve functions higher than
elementary. And I notice that my antiderivative for integral #9 was
very far from optimal! The optimal solution is just something like:
INT((x^2 + x + 2)*(x^3 - 1)^(1/3)
/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)), x) =
(x^3 - 1)^(1/3)/(x^2 - 1)
+ SQRT(3)/3*ATAN(SQRT(3)/3*(1 + 2*(x^2 - 1)/(x^3 - 1)^(1/3)))
+ 1/2*LN((x^3 - 1)^(1/3) - x^2 + 1)
- 1/6*LN(x^2*(x^4 - 3*x^2 - x + 3))

Wow,

Martin.

In terms of grading:

A B C F
Fricas 85.714 7.143 0.000 7.143
Mathematica 78.571 0.000 0.000 21.429
Rubi 35.714 0.000 35.714 28.571
Maxima 14.286 0.000 0.000 85.714
Maple 7.143 0.000 92.857 0.000
Mupad 0.000 7.143 0.000 92.857
Giac 0.000 0.000 0.000 100.000
Sympy 0.000 0.000 14.286 85.714

For optimal anti, I picked the smallest one by trying each integral
on different CAS's. So it is possible a better optimal exists that
I missed.

Only Maple was able to solve them all but the grading was not the
best. FriCAS got the best A grading and was in second place in terms
of number solved. Maxima, mupad, giac and sympy did not do well on
these hard integrals.

Link to the new test report is

<https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/test_cases/213_Goursat/report.htm>

Which you can also access from the main web page

<https://12000.org/my_notes/CAS_integration_tests/reports/summer_2023/index.htm>

by going to the "links to individual test reports" page and scroll all
the way down.

PS: Can any of the systems solve the fifth component integral:
INT((x^9 - 3*x^6 + 8*x^3 + 3)
/((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) = SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3) /(3*((x^3 - 1)^(1/3) + x^2))))
+ 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

in elementary terms?

Hi Martin,

Regarding the integral of

(x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3))

Seeing the term in your answer

- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

makes me think that this integrand must be split into a rational term + an algebraic term(or terms). In general, IntegrateAlgebraic will not compute such an integral unless the splitting of the rational part and the algebraic part can be achieved with
Apart. Which in this case it cannot.

Cheers,

Sam

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Sam Blake schrieb:

On Wednesday, November 1, 2023 at 4:42:47 AM UTC+11, nob...@nowhere.invalid wrote:

[...]

Rubi's success rate on these 14 cube root integrals is impressive
even though many solutions appear to involve functions higher than elementary. And I notice that my antiderivative for integral #9 was
very far from optimal! The optimal solution is just something like:

INT((x^2 + x + 2)*(x^3 - 1)^(1/3)
/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3)), x) =
(x^3 - 1)^(1/3)/(x^2 - 1)
+ SQRT(3)/3*ATAN(SQRT(3)/3*(1 + 2*(x^2 - 1)/(x^3 - 1)^(1/3)))
+ 1/2*LN((x^3 - 1)^(1/3) - x^2 + 1)
- 1/6*LN(x^2*(x^4 - 3*x^2 - x + 3))

[...]

PS: Can any of the systems solve the fifth component integral:
INT((x^9 - 3*x^6 + 8*x^3 + 3)
/((x^3 - 1)^(2/3)*(x^9 - 2*x^6 + x^3 - 27)), x) = SQRT(3)/9*ATAN(SQRT(3)/3*(1 + 2*(x^3 - 1)^(1/3)*(x*(x^3 - 1)^(1/3) - 3) /(3*((x^3 - 1)^(1/3) + x^2))))
+ 1/6*LN(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

in elementary terms?

Regarding the integral of

(x^2 + x + 2)*(x^3 - 1)^(1/3)/(x*(x^2 - 1)^2*(x^3 + x^2 - 2*x - 3))

Seeing the term in your answer

- 1/18*LN(x^9 - 2*x^6 + x^3 - 27)

makes me think that this integrand must be split into a rational term
+ an algebraic term(or terms). In general, IntegrateAlgebraic will
not compute such an integral unless the splitting of the rational
part and the algebraic part can be achieved with Apart. Which in this
case it cannot.

In the antiderivatives of cube-root integrands, one regularly finds
paired logarithms that correspond with algebraic factorizations of the integrand's denominator in terms of the radical.

For the denominator of your integral one has:

x^2*(x - 1)*(x^3 + x^2 - 2*x - 3) = (x^2 - 1 - (x^3 - 1)^(1/3))
*((x^2 - 1)^2 + (x^2 - 1)*(x^3 - 1)^(1/3) + (x^3 - 1)^(2/3))

Similarly for the denominator of the third and fifth components of my multi-component antiderivative:

x^9 - 2*x^6 + x^3 - 27 =
(- 2*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) + x^3 - 4)
*((x^3 - 1)^(1/3)*((x^3 - 1)^(1/3)*(2*x^3 + 1) + x^3 + 8)
+ x^6 - 2*x^3 + 10)

x^9 - 2*x^6 + x^3 - 27 =
(x*(x^3 - 1)^(2/3) + 3*(x^3 - 1)^(1/3) - 3*x^2)
*((x^3 - 1)^(1/3)*(3*(x^3 + 3)*(x^3 - 1)^(1/3) + x^2*(x^3 + 8))
+ 3*x*(2*x^3 + 1))

which incidently shows that algebraic factorization is not unique. More
compact antiderivatives result when they are expressed in terms of the
smaller factor and the full denominator polynominal.

For the component integrals, I just simplified the antiderivative
produced by FriCAS for t = x^3 substituted in your integral - this was
the only solution I had access to. Unfortunately, I didn't try to lump
its logarithms and its arc tangents - I should have been surprised.

I have no idea how to recocognize that the fifth component integral is elementary. FriCAS can solve this one "as is".

Martin.

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