An interesting rational integral in Mathematica

[Sam Blake]

Some of you may find this result interesting...

In[1047]:= $Version

Out[1047]= "13.2.0 for Mac OS X x86 (64-bit) (November 18, 2022)"

In[1048]:= Integrate[(2 - Sqrt[7] x^2 + 3 x^4)/(2 + Sqrt[7] x^2 - x^4)^2, x]

Out[1048]= 1/60 (30 x AppellF1[1/2, 2, 2, 3/2, (2 x^2)/(Sqrt[7] + Sqrt[15]), (2 x^2)/(Sqrt[7] - Sqrt[15])] -
5 Sqrt[7] x^3 AppellF1[3/2, 2, 2, 5/2, (2 x^2)/(Sqrt[7] + Sqrt[15]), (2 x^2)/(Sqrt[7] - Sqrt[15])] +
9 x^5 AppellF1[5/2, 2, 2, 7/2, (2 x^2)/(Sqrt[7] + Sqrt[15]), (2 x^2)/(Sqrt[7] - Sqrt[15])])

Cheers,

Sam

[Sam Blake]

... and here's an example where Rubi outperforms both Mathematica and my implementation of Mack's linear Hermite reduction (as given in Symbolic Integration I by Manuel Bronstein)

In[1151]:= (2 - 9 x^4 + 2 Sqrt[2] x^4 - 12 x^6 - 3 x^8)/(Sqrt[2] - 3 x^2 - x^4)^3;
HermiteReduce[%, x]
Integrate[%%, x]
Int[%%%, x]

Out[1152]= {-((37549921350591017222596 (102743992 x +
72655641 Sqrt[2] x))/((-8 + 9 Sqrt[2]) (-113 +
72 Sqrt[2])^4 (2200 + 1593 Sqrt[2])^2 (-32544 +
23137 Sqrt[2]) (3352883803641 +
2370846461804 Sqrt[2]) (-3186 - 2200 Sqrt[2] + 6600 x^2 +
4779 Sqrt[2] x^2 + 2200 x^4 + 1593 Sqrt[2] x^4))), 0}

Out[1153]= -(1/(
105 (-2 + 3 Sqrt[2] x^2 + Sqrt[2] x^4)^3))(Sqrt[2] - 3 x^2 -
x^4)^3 (210 x AppellF1[1/2, 3, 3, 3/2, (
2 x^2)/(-3 + Sqrt[9 + 4 Sqrt[2]]), -((2 x^2)/(
3 + Sqrt[9 + 4 Sqrt[2]]))] +
21 (-9 + 2 Sqrt[2]) x^5 AppellF1[5/2, 3, 3, 7/2, (
2 x^2)/(-3 + Sqrt[9 + 4 Sqrt[2]]), -((2 x^2)/(
3 + Sqrt[9 + 4 Sqrt[2]]))] -
5 (36 x^7 AppellF1[7/2, 3, 3, 9/2, (
2 x^2)/(-3 + Sqrt[9 + 4 Sqrt[2]]), -((2 x^2)/(
3 + Sqrt[9 + 4 Sqrt[2]]))] +
7 x^9 AppellF1[9/2, 3, 3, 11/2, (
2 x^2)/(-3 + Sqrt[9 + 4 Sqrt[2]]), -((2 x^2)/(
3 + Sqrt[9 + 4 Sqrt[2]]))]))

Out[1154]= x/(Sqrt[2] - 3 x^2 - x^4)

[Sam Blake]

For fairness, Rubi makes a mess of this one

In[1578]:= Int[(-2 - 2*Sqrt[3]*x - 3*x^2 + 4*Sqrt[3]*x^3 + 7*x^4 + 6*Sqrt[3]*x^5 + 10*x^6 + 2*Sqrt[3]*x^7 + 6*x^8 + x^10)/(1 + 3*x^2 + Sqrt[3]*x^3 + 3*x^4 + x^6)^2, x]

Out[1578]= -((3*3^(1/6) + 4*Sqrt[3] -
5*3^(5/6) + (3 - 4*3^(2/3))*
x)/(3*(3 - 4*3^(2/3))*(1 + 3^(1/6)*x + x^2))) -
(3^(1/3)*(6*I - 3*(-1)^(2/3)*3^(1/6) + 2*Sqrt[3] +
5*3^(5/6) + (-1)^(1/3)*(3 + 3^(2/3)*(2 + 2*I*Sqrt[3]))*x))/
((1 + (-1)^(1/3))^4*(4 - (-1)^(2/3)*3^(1/3))*(1 - (-1)^(1/
3)*3^(1/6)*x + x^2)) -
(2*(-1)^(1/3)*(3*(-1)^(2/3)*3^(1/6) + 4*Sqrt[3] +
5*(-1)^(1/3)*3^(5/6) + (3 - 4*(-3)^(2/3))*x))/
(3*3^(2/3)*(8 + 3^(1/3) + I*3^(5/6))*(1 + (-1)^(2/3)*3^(1/6)*x +
x^2)) +
(2*ArcTan[((-1)^(1/3)*3^(1/6) - 2*x)/
Sqrt[4 - (-1)^(2/3)*3^(1/3)]])/(3*
Sqrt[(1/2)*(8 + 3^(1/3) - I*3^(5/6))]) -
(2*ArcTan[((-1)^(2/3)*3^(1/6) + 2*x)/Sqrt[4 + (-3)^(1/3)]])/(3*
Sqrt[(1/2)*(8 + 3^(1/3) + I*3^(5/6))]) +
(2*I*ArcTanh[(3^(1/6)*(I - Sqrt[3]) - 4*I*x)/
Sqrt[2*(8 + 3^(1/3) - I*3^(5/6))]])/
(3*Sqrt[(1/2)*(8 + 3^(1/3) - I*3^(5/6))]) +
(2*I*ArcTanh[(3^(1/6)*(I + Sqrt[3]) - 4*I*x)/
Sqrt[2*(8 + 3^(1/3) + I*3^(5/6))]])/
(3*Sqrt[(1/2)*(8 + 3^(1/3) + I*3^(5/6))])

[nob...@nowhere.invalid]

Sam Blake schrieb:

Acording to Derive 6.10, the antiderivative is simply:

INT((2 - SQRT(7)*x^2 + 3*x^4)/(2 + SQRT(7)*x^2 - x^4)^2, x)

- x/(x^4 - SQRT(7)*x^2 - 2)

Martin.

[nob...@nowhere.invalid]

Sam Blake schrieb:

... with the same result being obtained by Derive 6.10:

INT((2 - 9*x^4 + 2*SQRT(2)*x^4 - 12*x^6 - 3*x^8)
/(SQRT(2) - 3*x^2 - x^4)^3, x)

- x/(x^4 + 3*x^2 - SQRT(2))

Martin.

[nob...@nowhere.invalid]

Sam Blake schrieb:

Derive 6.10 handles this in one step by one rule:

" INT(F'(x)/F(x)^n,x) -> -1/((n-1)*F(x)^(n-1)) "

Perhaps Rubi should be taught this rule as well?

Martin.

[Sam Blake]

Here's a similar example where Hermite reduction should return the integral without any logarithmic terms:

In[1793]:= (x^3 (2 + x^2))/(-2 + Sqrt[3] + (-2 + Sqrt[3]) x^2 - x^4)^2;
Integrate[%, x]
D[%, x] - %% // Simplify

Out[1794]= (((-2 - I) + Sqrt[3] - 2 x^2) ((1 + 2 I) - I Sqrt[3] +
2 I x^2) (x^4 (-2 I + ((2 - I) + Sqrt[3]) x^2) +
2 (-52 +
30 Sqrt[3] + ((-78 - 7 I) + (45 + 4 I) Sqrt[
3]) x^2 + ((-40 - 7 I) + (23 + 4 I) Sqrt[
3]) x^4 + ((-7 - 2 I) + (4 + I) Sqrt[3]) x^6) Log[(
2 (-2 + Sqrt[3]) + ((-2 - I) + Sqrt[3]) x^2)/(
2 (-2 + Sqrt[3]) + ((-2 + I) + Sqrt[3]) x^2)] +
2 (-52 +
30 Sqrt[3] + ((-78 - 7 I) + (45 + 4 I) Sqrt[
3]) x^2 + ((-40 - 7 I) + (23 + 4 I) Sqrt[
3]) x^4 + ((-7 - 2 I) + (4 + I) Sqrt[3]) x^6) Log[(
2 (-2 + Sqrt[3]) + ((-2 + I) + Sqrt[3]) x^2)/(
2 (-2 + Sqrt[3]) + ((-2 - I) + Sqrt[3]) x^2)]))/(8 (2 (-2 +
Sqrt[3]) + ((-2 - I) + Sqrt[3]) x^2) (-2 + Sqrt[
3] + (-2 + Sqrt[3]) x^2 - x^4)^2)

Out[1795]= 0

[Sam Blake]

On Saturday, January 13, 2024 at 8:29:08 PM UTC+11, nob...@nowhere.invalid wrote:

Rubi gets the general form

In[1796]:= Int[f'[x]/f[x]^n, x]

Out[1796]= f[x]^(1 - n)/(1 - n)

Sam