3.7.83 \(\int \frac {-1+x^8}{\sqrt {1+x^4} (1+x^8)} \, dx\)

Optimal. Leaf size=53 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}} \]

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Rubi [C]  time = 0.54, antiderivative size = 314, normalized size of antiderivative = 5.92, number of steps used = 21, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1586, 6725, 406, 220, 409, 1217, 1707} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}}+\frac {i \left (\sqrt {2}+(1+i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{8 \sqrt {x^4+1}}+\frac {i \left (\sqrt {2}+(-1+i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{8 \sqrt {x^4+1}}+\frac {\left ((-1-i)-i \sqrt {2}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{8 \sqrt {x^4+1}}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \left (1+(-1)^{3/4}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {x^4+1}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {x^4+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x^8)/(Sqrt[1 + x^4]*(1 + x^8)),x]

[Out]

-1/2*ArcTan[(2^(1/4)*x)/Sqrt[1 + x^4]]/2^(1/4) - ArcTanh[(2^(1/4)*x)/Sqrt[1 + x^4]]/(2*2^(1/4)) + ((1 + x^2)*S
qrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(2*Sqrt[1 + x^4]) - ((1/8 - I/8)*(1 + (-1)^(3/4))*(1 +
x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/Sqrt[1 + x^4] + (((-1 - I) - I*Sqrt[2])*(1 + x^
2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(8*Sqrt[1 + x^4]) + ((I/8)*((-1 + I) + Sqrt[2])*(1
+ x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/Sqrt[1 + x^4] + ((I/8)*((1 + I) + Sqrt[2])*(1
+ x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/Sqrt[1 + x^4]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 406

Int[Sqrt[(a_) + (b_.)*(x_)^4]/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[b/d, Int[1/Sqrt[a + b*x^4], x], x] - Di
st[(b*c - a*d)/d, Int[1/(Sqrt[a + b*x^4]*(c + d*x^4)), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 409

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1
- Rt[-(d/c), 2]*x^2)), x], x] + Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-(d/c), 2]*x^2)), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1217

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(c*d + a*e*q
)/(c*d^2 - a*e^2), Int[1/Sqrt[a + c*x^4], x], x] - Dist[(a*e*(e + d*q))/(c*d^2 - a*e^2), Int[(1 + q*x^2)/((d +
e*x^2)*Sqrt[a + c*x^4]), x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0]
&& PosQ[c/a]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1707

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]
}, -Simp[((B*d - A*e)*ArcTan[(Rt[(c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + c*x^4]])/(2*d*e*Rt[(c*d)/e + (a*e)/d, 2]),
x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + c*x^4))/(a*(A + B*x^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2
/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2])/(4*d*e*A*q*Sqrt[a + c*x^4]), x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c
*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {-1+x^8}{\sqrt {1+x^4} \left (1+x^8\right )} \, dx &=\int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+x^8} \, dx\\ &=\int \left (-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {1+x^4}}{i-x^4}+\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {1+x^4}}{i+x^4}\right ) \, dx\\ &=\left (-\frac {1}{2}-\frac {i}{2}\right ) \int \frac {\sqrt {1+x^4}}{i-x^4} \, dx+\left (\frac {1}{2}-\frac {i}{2}\right ) \int \frac {\sqrt {1+x^4}}{i+x^4} \, dx\\ &=-\left (i \int \frac {1}{\left (i-x^4\right ) \sqrt {1+x^4}} \, dx\right )-i \int \frac {1}{\left (i+x^4\right ) \sqrt {1+x^4}} \, dx+\left (\frac {1}{2}-\frac {i}{2}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx+\left (\frac {1}{2}+\frac {i}{2}\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}}-\frac {1}{2} \int \frac {1}{\left (1-\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1-(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+(-1)^{3/4} x^2\right ) \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}}-\left (\left (\frac {1}{4}+\frac {i}{4}\right ) \left (1+\sqrt [4]{-1}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\left (\left (-\frac {1}{4}-\frac {i}{4}\right ) \sqrt [4]{-1} \left (1+\sqrt [4]{-1}\right )\right ) \int \frac {1+x^2}{\left (1+\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\left (\left (\frac {1}{4}-\frac {i}{4}\right ) (-1)^{3/4} \left (1-(-1)^{3/4}\right )\right ) \int \frac {1+x^2}{\left (1-(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx-\left (\left (\frac {1}{4}-\frac {i}{4}\right ) \left (1+(-1)^{3/4}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\left (\left (-\frac {1}{4}+\frac {i}{4}\right ) (-1)^{3/4} \left (1+(-1)^{3/4}\right )\right ) \int \frac {1+x^2}{\left (1+(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{4} \left ((1+i)-i \sqrt {2}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\frac {1}{4} \left (i \left ((-1-i)+\sqrt {2}\right )\right ) \int \frac {1+x^2}{\left (1-\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{4} \left (i \left ((1+i)+\sqrt {2}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt [4]{2}}+\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 {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (1+\sqrt [4]{-1}\right ) \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 )}{\sqrt {1+x^4}}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \left (1+(-1)^{3/4}\right ) \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 )}{\sqrt {1+x^4}}-\frac {\left ((1+i)-i \sqrt {2}\right ) \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 )}{8 \sqrt {1+x^4}}+\frac {i \left ((1+i)+\sqrt {2}\right ) \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 )}{8 \sqrt {1+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.33, size = 106, normalized size = 2.00 \begin {gather*} \frac {1}{2} \sqrt [4]{-1} \left (-2 F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\Pi \left (-\sqrt [4]{-1};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\Pi \left (\sqrt [4]{-1};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\Pi \left (-(-1)^{3/4};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\Pi \left ((-1)^{3/4};\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x^8)/(Sqrt[1 + x^4]*(1 + x^8)),x]

[Out]

((-1)^(1/4)*(-2*EllipticF[I*ArcSinh[(-1)^(1/4)*x], -1] + EllipticPi[-(-1)^(1/4), I*ArcSinh[(-1)^(1/4)*x], -1]
+ EllipticPi[(-1)^(1/4), I*ArcSinh[(-1)^(1/4)*x], -1] + EllipticPi[-(-1)^(3/4), I*ArcSinh[(-1)^(1/4)*x], -1] +
EllipticPi[(-1)^(3/4), I*ArcSinh[(-1)^(1/4)*x], -1]))/2

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IntegrateAlgebraic [A]  time = 0.27, size = 53, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-1 + x^8)/(Sqrt[1 + x^4]*(1 + x^8)),x]

[Out]

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

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fricas [B]  time = 0.62, size = 199, normalized size = 3.75 \begin {gather*} -\frac {1}{4} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )} + 2^{\frac {1}{4}} {\left (x^{8} + 4 \, x^{4} + 1\right )}\right )} + 4 \, \sqrt {x^{4} + 1} {\left (2^{\frac {3}{4}} x^{3} + 2^{\frac {1}{4}} {\left (x^{5} + x\right )}\right )}}{2 \, {\left (x^{8} + 1\right )}}\right ) - \frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (-\frac {2^{\frac {3}{4}} {\left (x^{8} + 4 \, x^{4} + 1\right )} + 4 \, {\left (x^{5} + \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}}{x^{8} + 1}\right ) + \frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (\frac {2^{\frac {3}{4}} {\left (x^{8} + 4 \, x^{4} + 1\right )} - 4 \, {\left (x^{5} + \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}}{x^{8} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^8-1)/(x^4+1)^(1/2)/(x^8+1),x, algorithm="fricas")

[Out]

-1/4*2^(3/4)*arctan(1/2*(2^(3/4)*(2*2^(3/4)*(x^6 + x^2) + 2^(1/4)*(x^8 + 4*x^4 + 1)) + 4*sqrt(x^4 + 1)*(2^(3/4
)*x^3 + 2^(1/4)*(x^5 + x)))/(x^8 + 1)) - 1/16*2^(3/4)*log(-(2^(3/4)*(x^8 + 4*x^4 + 1) + 4*(x^5 + sqrt(2)*x^3 +
x)*sqrt(x^4 + 1) + 4*2^(1/4)*(x^6 + x^2))/(x^8 + 1)) + 1/16*2^(3/4)*log((2^(3/4)*(x^8 + 4*x^4 + 1) - 4*(x^5 +
sqrt(2)*x^3 + x)*sqrt(x^4 + 1) + 4*2^(1/4)*(x^6 + x^2))/(x^8 + 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8} - 1}{{\left (x^{8} + 1\right )} \sqrt {x^{4} + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^8-1)/(x^4+1)^(1/2)/(x^8+1),x, algorithm="giac")

[Out]

integrate((x^8 - 1)/((x^8 + 1)*sqrt(x^4 + 1)), x)

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maple [B]  time = 0.94, size = 79, normalized size = 1.49

method	result	size
elliptic	\(\frac {\left (\frac {2^{\frac {1}{4}} \arctan \left (\frac {2^{\frac {3}{4}} \sqrt {x^{4}+1}}{2 x}\right )}{2}-\frac {2^{\frac {1}{4}} \ln \left (\frac {\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}+\frac {2^{\frac {3}{4}}}{2}}{\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}-\frac {2^{\frac {3}{4}}}{2}}\right )}{4}\right ) \sqrt {2}}{2}\)	\(79\)
default	\(\frac {\sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{8}+1\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {\arctanh \left (\frac {\underline {\hspace {1.25 ex}}\alpha ^{2} \left (-\underline {\hspace {1.25 ex}}\alpha ^{6}+x^{2}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}\, \sqrt {x^{4}+1}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{4}+1}}-\frac {2 \left (-1\right )^{\frac {3}{4}} \underline {\hspace {1.25 ex}}\alpha ^{7} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i \underline {\hspace {1.25 ex}}\alpha ^{6}, i\right )}{\sqrt {x^{4}+1}}\right )\right )}{8}\)	\(161\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {-\RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x^{2}-2 x^{4} \RootOf \left (\textit {\_Z}^{4}-8\right )+8 \sqrt {x^{4}+1}\, x -2 \RootOf \left (\textit {\_Z}^{4}-8\right )}{\RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-2 x^{4}-2}\right )}{8}-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{2} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2}-2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}-8 \sqrt {x^{4}+1}\, x -2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right )}{2 x^{4}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+2}\right )}{8}\)	\(184\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8-1)/(x^4+1)^(1/2)/(x^8+1),x,method=_RETURNVERBOSE)

[Out]

1/2*(1/2*2^(1/4)*arctan(1/2*2^(3/4)/x*(x^4+1)^(1/2))-1/4*2^(1/4)*ln((1/2*2^(1/2)/x*(x^4+1)^(1/2)+1/2*2^(3/4))/
(1/2*2^(1/2)/x*(x^4+1)^(1/2)-1/2*2^(3/4))))*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8} - 1}{{\left (x^{8} + 1\right )} \sqrt {x^{4} + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^8-1)/(x^4+1)^(1/2)/(x^8+1),x, algorithm="maxima")

[Out]

integrate((x^8 - 1)/((x^8 + 1)*sqrt(x^4 + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^8-1}{\sqrt {x^4+1}\,\left (x^8+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8 - 1)/((x^4 + 1)^(1/2)*(x^8 + 1)),x)

[Out]

int((x^8 - 1)/((x^4 + 1)^(1/2)*(x^8 + 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \sqrt {x^{4} + 1}}{x^{8} + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**8-1)/(x**4+1)**(1/2)/(x**8+1),x)

[Out]

Integral((x - 1)*(x + 1)*(x**2 + 1)*sqrt(x**4 + 1)/(x**8 + 1), x)

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