3.18.70 \(\int \frac {-1+x^{10}}{\sqrt {1+x^4} (1+x^{10})} \, dx\)

Optimal. Leaf size=119 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{5 \sqrt {2}}-\frac {1}{5} \sqrt {2 \left (1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}} x}{\sqrt {x^4+1}}\right )-\frac {1}{5} \sqrt {2 \left (\sqrt {5}-1\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt {x^4+1}}\right ) \]

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Rubi [F]  time = 1.21, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-1+x^{10}}{\sqrt {1+x^4} \left (1+x^{10}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 + x^10)/(Sqrt[1 + x^4]*(1 + x^10)),x]

[Out]

-1/5*ArcTan[(Sqrt[2]*x)/Sqrt[1 + x^4]]/Sqrt[2] + (2*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x
], 1/2])/(5*Sqrt[1 + x^4]) - (8*Defer[Int][1/(Sqrt[1 + x^4]*(1 - x^2 + x^4 - x^6 + x^8)), x])/5 + (6*Defer[Int
][x^2/(Sqrt[1 + x^4]*(1 - x^2 + x^4 - x^6 + x^8)), x])/5 - (4*Defer[Int][x^4/(Sqrt[1 + x^4]*(1 - x^2 + x^4 - x
^6 + x^8)), x])/5 + (2*Defer[Int][x^6/(Sqrt[1 + x^4]*(1 - x^2 + x^4 - x^6 + x^8)), x])/5

Rubi steps

\begin {align*} \int \frac {-1+x^{10}}{\sqrt {1+x^4} \left (1+x^{10}\right )} \, dx &=\int \left (\frac {1}{\sqrt {1+x^4}}-\frac {2}{\sqrt {1+x^4} \left (1+x^{10}\right )}\right ) \, dx\\ &=-\left (2 \int \frac {1}{\sqrt {1+x^4} \left (1+x^{10}\right )} \, dx\right )+\int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-2 \int \left (\frac {1}{5 \left (1+x^2\right ) \sqrt {1+x^4}}+\frac {4-3 x^2+2 x^4-x^6}{5 \sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )}\right ) \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\frac {2}{5} \int \frac {1}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx-\frac {2}{5} \int \frac {4-3 x^2+2 x^4-x^6}{\sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )} \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\frac {1}{5} \int \frac {1}{\sqrt {1+x^4}} \, dx-\frac {1}{5} \int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx-\frac {2}{5} \int \left (\frac {4}{\sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )}-\frac {3 x^2}{\sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )}+\frac {2 x^4}{\sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )}-\frac {x^6}{\sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )}\right ) \, dx\\ &=\frac {2 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{5 \sqrt {1+x^4}}-\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )+\frac {2}{5} \int \frac {x^6}{\sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )} \, dx-\frac {4}{5} \int \frac {x^4}{\sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )} \, dx+\frac {6}{5} \int \frac {x^2}{\sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )} \, dx-\frac {8}{5} \int \frac {1}{\sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )} \, dx\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{5 \sqrt {2}}+\frac {2 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{5 \sqrt {1+x^4}}+\frac {2}{5} \int \frac {x^6}{\sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )} \, dx-\frac {4}{5} \int \frac {x^4}{\sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )} \, dx+\frac {6}{5} \int \frac {x^2}{\sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )} \, dx-\frac {8}{5} \int \frac {1}{\sqrt {1+x^4} \left (1-x^2+x^4-x^6+x^8\right )} \, dx\\ \end {align*}

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Mathematica [C]  time = 0.76, size = 375, normalized size = 3.15 \begin {gather*} \frac {\sqrt [4]{-1} \left (\left (-4-\sqrt [5]{-1}+(-1)^{2/5}-(-1)^{3/5}+(-1)^{4/5}\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+2 \left (\Pi \left (-i;\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\left (2-\sqrt [5]{-1}+(-1)^{2/5}-(-1)^{3/5}+(-1)^{4/5}\right ) \Pi \left (-\sqrt [10]{-1};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )-(-1)^{4/5} \Pi \left ((-1)^{3/10};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+(-1)^{3/5} \Pi \left ((-1)^{3/10};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )-(-1)^{2/5} \Pi \left ((-1)^{3/10};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\sqrt [5]{-1} \Pi \left ((-1)^{3/10};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\Pi \left ((-1)^{7/10};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )-(-1)^{4/5} \Pi \left (-(-1)^{9/10};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+(-1)^{3/5} \Pi \left (-(-1)^{9/10};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )-(-1)^{2/5} \Pi \left (-(-1)^{9/10};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\sqrt [5]{-1} \Pi \left (-(-1)^{9/10};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )\right )\right )}{4+\sqrt [5]{-1}-(-1)^{2/5}+(-1)^{3/5}-(-1)^{4/5}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(-1 + x^10)/(Sqrt[1 + x^4]*(1 + x^10)),x]

[Out]

((-1)^(1/4)*((-4 - (-1)^(1/5) + (-1)^(2/5) - (-1)^(3/5) + (-1)^(4/5))*EllipticF[I*ArcSinh[(-1)^(1/4)*x], -1] +
2*(EllipticPi[-I, I*ArcSinh[(-1)^(1/4)*x], -1] + (2 - (-1)^(1/5) + (-1)^(2/5) - (-1)^(3/5) + (-1)^(4/5))*Elli
pticPi[-(-1)^(1/10), I*ArcSinh[(-1)^(1/4)*x], -1] + (-1)^(1/5)*EllipticPi[(-1)^(3/10), I*ArcSinh[(-1)^(1/4)*x]
, -1] - (-1)^(2/5)*EllipticPi[(-1)^(3/10), I*ArcSinh[(-1)^(1/4)*x], -1] + (-1)^(3/5)*EllipticPi[(-1)^(3/10), I
*ArcSinh[(-1)^(1/4)*x], -1] - (-1)^(4/5)*EllipticPi[(-1)^(3/10), I*ArcSinh[(-1)^(1/4)*x], -1] + EllipticPi[(-1
)^(7/10), I*ArcSinh[(-1)^(1/4)*x], -1] + (-1)^(1/5)*EllipticPi[-(-1)^(9/10), I*ArcSinh[(-1)^(1/4)*x], -1] - (-
1)^(2/5)*EllipticPi[-(-1)^(9/10), I*ArcSinh[(-1)^(1/4)*x], -1] + (-1)^(3/5)*EllipticPi[-(-1)^(9/10), I*ArcSinh
[(-1)^(1/4)*x], -1] - (-1)^(4/5)*EllipticPi[-(-1)^(9/10), I*ArcSinh[(-1)^(1/4)*x], -1])))/(4 + (-1)^(1/5) - (-
1)^(2/5) + (-1)^(3/5) - (-1)^(4/5))

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IntegrateAlgebraic [A]  time = 2.45, size = 119, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{5 \sqrt {2}}-\frac {1}{5} \sqrt {2 \left (1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt {1+x^4}}\right )-\frac {1}{5} \sqrt {2 \left (-1+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt {1+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-1 + x^10)/(Sqrt[1 + x^4]*(1 + x^10)),x]

[Out]

-1/5*ArcTan[(Sqrt[2]*x)/Sqrt[1 + x^4]]/Sqrt[2] - (Sqrt[2*(1 + Sqrt[5])]*ArcTan[(Sqrt[-1/2 + Sqrt[5]/2]*x)/Sqrt
[1 + x^4]])/5 - (Sqrt[2*(-1 + Sqrt[5])]*ArcTanh[(Sqrt[1/2 + Sqrt[5]/2]*x)/Sqrt[1 + x^4]])/5

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fricas [B]  time = 0.60, size = 395, normalized size = 3.32 \begin {gather*} -\frac {1}{10} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) - \frac {1}{5} \, \sqrt {2 \, \sqrt {5} + 2} \arctan \left (-\frac {2 \, {\left (x^{5} - 2 \, x^{3} - \sqrt {5} {\left (x^{5} + x\right )} + x\right )} \sqrt {x^{4} + 1} \sqrt {2 \, \sqrt {5} + 2} - {\left (x^{8} + 5 \, x^{6} + 3 \, x^{4} + 5 \, x^{2} + \sqrt {5} {\left (x^{8} + x^{6} + 3 \, x^{4} + x^{2} + 1\right )} + 1\right )} \sqrt {2 \, \sqrt {5} + 2} \sqrt {\sqrt {5} - 2}}{4 \, {\left (x^{8} + x^{6} + x^{4} + x^{2} + 1\right )}}\right ) - \frac {1}{20} \, \sqrt {2 \, \sqrt {5} - 2} \log \left (-\frac {4 \, {\left (3 \, x^{5} + x^{3} + \sqrt {5} {\left (x^{5} + x^{3} + x\right )} + 3 \, x\right )} \sqrt {x^{4} + 1} + {\left (3 \, x^{8} + 5 \, x^{6} + 9 \, x^{4} + 5 \, x^{2} + \sqrt {5} {\left (x^{8} + 3 \, x^{6} + 3 \, x^{4} + 3 \, x^{2} + 1\right )} + 3\right )} \sqrt {2 \, \sqrt {5} - 2}}{x^{8} - x^{6} + x^{4} - x^{2} + 1}\right ) + \frac {1}{20} \, \sqrt {2 \, \sqrt {5} - 2} \log \left (-\frac {4 \, {\left (3 \, x^{5} + x^{3} + \sqrt {5} {\left (x^{5} + x^{3} + x\right )} + 3 \, x\right )} \sqrt {x^{4} + 1} - {\left (3 \, x^{8} + 5 \, x^{6} + 9 \, x^{4} + 5 \, x^{2} + \sqrt {5} {\left (x^{8} + 3 \, x^{6} + 3 \, x^{4} + 3 \, x^{2} + 1\right )} + 3\right )} \sqrt {2 \, \sqrt {5} - 2}}{x^{8} - x^{6} + x^{4} - x^{2} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^10-1)/(x^4+1)^(1/2)/(x^10+1),x, algorithm="fricas")

[Out]

-1/10*sqrt(2)*arctan(sqrt(2)*x/sqrt(x^4 + 1)) - 1/5*sqrt(2*sqrt(5) + 2)*arctan(-1/4*(2*(x^5 - 2*x^3 - sqrt(5)*
(x^5 + x) + x)*sqrt(x^4 + 1)*sqrt(2*sqrt(5) + 2) - (x^8 + 5*x^6 + 3*x^4 + 5*x^2 + sqrt(5)*(x^8 + x^6 + 3*x^4 +
x^2 + 1) + 1)*sqrt(2*sqrt(5) + 2)*sqrt(sqrt(5) - 2))/(x^8 + x^6 + x^4 + x^2 + 1)) - 1/20*sqrt(2*sqrt(5) - 2)*
log(-(4*(3*x^5 + x^3 + sqrt(5)*(x^5 + x^3 + x) + 3*x)*sqrt(x^4 + 1) + (3*x^8 + 5*x^6 + 9*x^4 + 5*x^2 + sqrt(5)
*(x^8 + 3*x^6 + 3*x^4 + 3*x^2 + 1) + 3)*sqrt(2*sqrt(5) - 2))/(x^8 - x^6 + x^4 - x^2 + 1)) + 1/20*sqrt(2*sqrt(5
) - 2)*log(-(4*(3*x^5 + x^3 + sqrt(5)*(x^5 + x^3 + x) + 3*x)*sqrt(x^4 + 1) - (3*x^8 + 5*x^6 + 9*x^4 + 5*x^2 +
sqrt(5)*(x^8 + 3*x^6 + 3*x^4 + 3*x^2 + 1) + 3)*sqrt(2*sqrt(5) - 2))/(x^8 - x^6 + x^4 - x^2 + 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{10} - 1}{{\left (x^{10} + 1\right )} \sqrt {x^{4} + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^10-1)/(x^4+1)^(1/2)/(x^10+1),x, algorithm="giac")

[Out]

integrate((x^10 - 1)/((x^10 + 1)*sqrt(x^4 + 1)), x)

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maple [A]  time = 2.54, size = 87, normalized size = 0.73

method	result	size
elliptic	\(\frac {\left (\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{5}+\frac {4 \arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x \sqrt {\sqrt {5}-1}}\right )}{5 \sqrt {\sqrt {5}-1}}-\frac {4 \arctanh \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x \sqrt {\sqrt {5}+1}}\right )}{5 \sqrt {\sqrt {5}+1}}\right ) \sqrt {2}}{2}\)	\(87\)
default	\(\frac {\sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {2 \left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{5 \sqrt {x^{4}+1}}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{8}-\textit {\_Z}^{6}+\textit {\_Z}^{4}-\textit {\_Z}^{2}+1\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {\arctanh \left (\frac {\underline {\hspace {1.25 ex}}\alpha ^{2} \left (-\underline {\hspace {1.25 ex}}\alpha ^{6}+\underline {\hspace {1.25 ex}}\alpha ^{4}-\underline {\hspace {1.25 ex}}\alpha ^{2}+x^{2}+1\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}\, \sqrt {x^{4}+1}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}}+\frac {2 \left (-1\right )^{\frac {3}{4}} \left (-\underline {\hspace {1.25 ex}}\alpha ^{7}+\underline {\hspace {1.25 ex}}\alpha ^{5}-\underline {\hspace {1.25 ex}}\alpha ^{3}+\underline {\hspace {1.25 ex}}\alpha \right ) \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i \underline {\hspace {1.25 ex}}\alpha ^{6}-i \underline {\hspace {1.25 ex}}\alpha ^{4}+i \underline {\hspace {1.25 ex}}\alpha ^{2}-i, i\right )}{\sqrt {x^{4}+1}}\right )\right )}{10}\)	\(260\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) x +\sqrt {x^{4}+1}}{x^{2}+1}\right )}{10}-\frac {\RootOf \left (\textit {\_Z}^{2}+25 \RootOf \left (625 \textit {\_Z}^{4}+25 \textit {\_Z}^{2}-1\right )^{2}+1\right ) \ln \left (-\frac {25 \RootOf \left (625 \textit {\_Z}^{4}+25 \textit {\_Z}^{2}-1\right )^{2} \RootOf \left (\textit {\_Z}^{2}+25 \RootOf \left (625 \textit {\_Z}^{4}+25 \textit {\_Z}^{2}-1\right )^{2}+1\right ) x^{2}-\RootOf \left (\textit {\_Z}^{2}+25 \RootOf \left (625 \textit {\_Z}^{4}+25 \textit {\_Z}^{2}-1\right )^{2}+1\right ) x^{4}-2 \sqrt {x^{4}+1}\, x -\RootOf \left (\textit {\_Z}^{2}+25 \RootOf \left (625 \textit {\_Z}^{4}+25 \textit {\_Z}^{2}-1\right )^{2}+1\right )}{25 \RootOf \left (625 \textit {\_Z}^{4}+25 \textit {\_Z}^{2}-1\right )^{2} x^{2}+x^{4}+1}\right )}{5}-\RootOf \left (625 \textit {\_Z}^{4}+25 \textit {\_Z}^{2}-1\right ) \ln \left (-\frac {125 \RootOf \left (625 \textit {\_Z}^{4}+25 \textit {\_Z}^{2}-1\right )^{3} x^{2}+5 \RootOf \left (625 \textit {\_Z}^{4}+25 \textit {\_Z}^{2}-1\right ) x^{4}+5 \RootOf \left (625 \textit {\_Z}^{4}+25 \textit {\_Z}^{2}-1\right ) x^{2}+2 \sqrt {x^{4}+1}\, x +5 \RootOf \left (625 \textit {\_Z}^{4}+25 \textit {\_Z}^{2}-1\right )}{25 \RootOf \left (625 \textit {\_Z}^{4}+25 \textit {\_Z}^{2}-1\right )^{2} x^{2}-x^{4}+x^{2}-1}\right )\)	\(329\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^10-1)/(x^4+1)^(1/2)/(x^10+1),x,method=_RETURNVERBOSE)

[Out]

1/2*(1/5*arctan(1/2*2^(1/2)/x*(x^4+1)^(1/2))+4/5/(5^(1/2)-1)^(1/2)*arctan(2^(1/2)/x*(x^4+1)^(1/2)/(5^(1/2)-1)^
(1/2))-4/5/(5^(1/2)+1)^(1/2)*arctanh(2^(1/2)/x*(x^4+1)^(1/2)/(5^(1/2)+1)^(1/2)))*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{10} - 1}{{\left (x^{10} + 1\right )} \sqrt {x^{4} + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^10-1)/(x^4+1)^(1/2)/(x^10+1),x, algorithm="maxima")

[Out]

integrate((x^10 - 1)/((x^10 + 1)*sqrt(x^4 + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{10}-1}{\sqrt {x^4+1}\,\left (x^{10}+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^10 - 1)/((x^4 + 1)^(1/2)*(x^10 + 1)),x)

[Out]

int((x^10 - 1)/((x^4 + 1)^(1/2)*(x^10 + 1)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**10-1)/(x**4+1)**(1/2)/(x**10+1),x)

[Out]

Timed out

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