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

Optimal. Leaf size=331 \[ -\frac {\log \left (x^2-\sqrt {2-\sqrt {3}} x+1\right )}{8 \sqrt {2-\sqrt {3}}}+\frac {\log \left (x^2+\sqrt {2-\sqrt {3}} x+1\right )}{8 \sqrt {2-\sqrt {3}}}-\frac {\log \left (x^2-\sqrt {2+\sqrt {3}} x+1\right )}{8 \sqrt {2+\sqrt {3}}}+\frac {\log \left (x^2+\sqrt {2+\sqrt {3}} x+1\right )}{8 \sqrt {2+\sqrt {3}}}-\frac {1}{4} \sqrt {2-\sqrt {3}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {3}}-2 x}{\sqrt {2+\sqrt {3}}}\right )-\frac {1}{4} \sqrt {2+\sqrt {3}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {3}}-2 x}{\sqrt {2-\sqrt {3}}}\right )+\frac {1}{4} \sqrt {2-\sqrt {3}} \tan ^{-1}\left (\frac {2 x+\sqrt {2-\sqrt {3}}}{\sqrt {2+\sqrt {3}}}\right )+\frac {1}{4} \sqrt {2+\sqrt {3}} \tan ^{-1}\left (\frac {2 x+\sqrt {2+\sqrt {3}}}{\sqrt {2-\sqrt {3}}}\right ) \]

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Rubi [A]  time = 0.23, antiderivative size = 331, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1419, 1094, 634, 618, 204, 628} \begin {gather*} -\frac {\log \left (x^2-\sqrt {2-\sqrt {3}} x+1\right )}{8 \sqrt {2-\sqrt {3}}}+\frac {\log \left (x^2+\sqrt {2-\sqrt {3}} x+1\right )}{8 \sqrt {2-\sqrt {3}}}-\frac {\log \left (x^2-\sqrt {2+\sqrt {3}} x+1\right )}{8 \sqrt {2+\sqrt {3}}}+\frac {\log \left (x^2+\sqrt {2+\sqrt {3}} x+1\right )}{8 \sqrt {2+\sqrt {3}}}-\frac {1}{4} \sqrt {2-\sqrt {3}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {3}}-2 x}{\sqrt {2+\sqrt {3}}}\right )-\frac {1}{4} \sqrt {2+\sqrt {3}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {3}}-2 x}{\sqrt {2-\sqrt {3}}}\right )+\frac {1}{4} \sqrt {2-\sqrt {3}} \tan ^{-1}\left (\frac {2 x+\sqrt {2-\sqrt {3}}}{\sqrt {2+\sqrt {3}}}\right )+\frac {1}{4} \sqrt {2+\sqrt {3}} \tan ^{-1}\left (\frac {2 x+\sqrt {2+\sqrt {3}}}{\sqrt {2-\sqrt {3}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^4)/(1 - x^4 + x^8),x]

[Out]

-(Sqrt[2 - Sqrt[3]]*ArcTan[(Sqrt[2 - Sqrt[3]] - 2*x)/Sqrt[2 + Sqrt[3]]])/4 - (Sqrt[2 + Sqrt[3]]*ArcTan[(Sqrt[2
+ Sqrt[3]] - 2*x)/Sqrt[2 - Sqrt[3]]])/4 + (Sqrt[2 - Sqrt[3]]*ArcTan[(Sqrt[2 - Sqrt[3]] + 2*x)/Sqrt[2 + Sqrt[3
]]])/4 + (Sqrt[2 + Sqrt[3]]*ArcTan[(Sqrt[2 + Sqrt[3]] + 2*x)/Sqrt[2 - Sqrt[3]]])/4 - Log[1 - Sqrt[2 - Sqrt[3]]
*x + x^2]/(8*Sqrt[2 - Sqrt[3]]) + Log[1 + Sqrt[2 - Sqrt[3]]*x + x^2]/(8*Sqrt[2 - Sqrt[3]]) - Log[1 - Sqrt[2 +
Sqrt[3]]*x + x^2]/(8*Sqrt[2 + Sqrt[3]]) + Log[1 + Sqrt[2 + Sqrt[3]]*x + x^2]/(8*Sqrt[2 + Sqrt[3]])

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

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

Rule 1094

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

Rule 1419

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[(2*d)/e -
b/c, 2]}, Dist[e/(2*c), Int[1/Simp[d/e + q*x^(n/2) + x^n, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x^(n/2
) + x^n, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2,
0] && IGtQ[n/2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !LtQ[(2*d)/e - b/c, 0] && EqQ[d, e*Rt[a/c, 2]]))

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 55, normalized size = 0.17 \begin {gather*} \frac {1}{4} \text {RootSum}\left [\text {$\#$1}^8-\text {$\#$1}^4+1\&,\frac {\text {$\#$1}^4 \log (x-\text {$\#$1})+\log (x-\text {$\#$1})}{2 \text {$\#$1}^7-\text {$\#$1}^3}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

RootSum[1 - #1^4 + #1^8 & , (Log[x - #1] + Log[x - #1]*#1^4)/(-#1^3 + 2*#1^7) & ]/4

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1+x^4}{1-x^4+x^8} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.32, size = 377, normalized size = 1.14 \begin {gather*} -\frac {1}{8} \, \sqrt {\sqrt {3} + 2} {\left (\sqrt {3} - 2\right )} \log \left (2 \, x^{2} + 2 \, x \sqrt {\sqrt {3} + 2} + 2\right ) + \frac {1}{8} \, \sqrt {\sqrt {3} + 2} {\left (\sqrt {3} - 2\right )} \log \left (2 \, x^{2} - 2 \, x \sqrt {\sqrt {3} + 2} + 2\right ) + \frac {1}{16} \, {\left (\sqrt {3} + 2\right )} \sqrt {-4 \, \sqrt {3} + 8} \log \left (2 \, x^{2} + x \sqrt {-4 \, \sqrt {3} + 8} + 2\right ) - \frac {1}{16} \, {\left (\sqrt {3} + 2\right )} \sqrt {-4 \, \sqrt {3} + 8} \log \left (2 \, x^{2} - x \sqrt {-4 \, \sqrt {3} + 8} + 2\right ) - \frac {1}{2} \, \sqrt {\sqrt {3} + 2} \arctan \left (\sqrt {2} \sqrt {2 \, x^{2} + 2 \, x \sqrt {\sqrt {3} + 2} + 2} \sqrt {\sqrt {3} + 2} - 2 \, x \sqrt {\sqrt {3} + 2} - \sqrt {3} - 2\right ) - \frac {1}{2} \, \sqrt {\sqrt {3} + 2} \arctan \left (\sqrt {2} \sqrt {2 \, x^{2} - 2 \, x \sqrt {\sqrt {3} + 2} + 2} \sqrt {\sqrt {3} + 2} - 2 \, x \sqrt {\sqrt {3} + 2} + \sqrt {3} + 2\right ) - \frac {1}{4} \, \sqrt {-4 \, \sqrt {3} + 8} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {2 \, x^{2} + x \sqrt {-4 \, \sqrt {3} + 8} + 2} \sqrt {-4 \, \sqrt {3} + 8} - x \sqrt {-4 \, \sqrt {3} + 8} + \sqrt {3} - 2\right ) - \frac {1}{4} \, \sqrt {-4 \, \sqrt {3} + 8} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {2 \, x^{2} - x \sqrt {-4 \, \sqrt {3} + 8} + 2} \sqrt {-4 \, \sqrt {3} + 8} - x \sqrt {-4 \, \sqrt {3} + 8} - \sqrt {3} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(sqrt(3) + 2)*(sqrt(3) - 2)*log(2*x^2 + 2*x*sqrt(sqrt(3) + 2) + 2) + 1/8*sqrt(sqrt(3) + 2)*(sqrt(3) -
2)*log(2*x^2 - 2*x*sqrt(sqrt(3) + 2) + 2) + 1/16*(sqrt(3) + 2)*sqrt(-4*sqrt(3) + 8)*log(2*x^2 + x*sqrt(-4*sqr
t(3) + 8) + 2) - 1/16*(sqrt(3) + 2)*sqrt(-4*sqrt(3) + 8)*log(2*x^2 - x*sqrt(-4*sqrt(3) + 8) + 2) - 1/2*sqrt(sq
rt(3) + 2)*arctan(sqrt(2)*sqrt(2*x^2 + 2*x*sqrt(sqrt(3) + 2) + 2)*sqrt(sqrt(3) + 2) - 2*x*sqrt(sqrt(3) + 2) -
sqrt(3) - 2) - 1/2*sqrt(sqrt(3) + 2)*arctan(sqrt(2)*sqrt(2*x^2 - 2*x*sqrt(sqrt(3) + 2) + 2)*sqrt(sqrt(3) + 2)
- 2*x*sqrt(sqrt(3) + 2) + sqrt(3) + 2) - 1/4*sqrt(-4*sqrt(3) + 8)*arctan(1/2*sqrt(2)*sqrt(2*x^2 + x*sqrt(-4*sq
rt(3) + 8) + 2)*sqrt(-4*sqrt(3) + 8) - x*sqrt(-4*sqrt(3) + 8) + sqrt(3) - 2) - 1/4*sqrt(-4*sqrt(3) + 8)*arctan
(1/2*sqrt(2)*sqrt(2*x^2 - x*sqrt(-4*sqrt(3) + 8) + 2)*sqrt(-4*sqrt(3) + 8) - x*sqrt(-4*sqrt(3) + 8) - sqrt(3)
+ 2)

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giac [A]  time = 0.50, size = 245, normalized size = 0.74 \begin {gather*} \frac {1}{8} \, {\left (\sqrt {6} - \sqrt {2}\right )} \arctan \left (\frac {4 \, x + \sqrt {6} - \sqrt {2}}{\sqrt {6} + \sqrt {2}}\right ) + \frac {1}{8} \, {\left (\sqrt {6} - \sqrt {2}\right )} \arctan \left (\frac {4 \, x - \sqrt {6} + \sqrt {2}}{\sqrt {6} + \sqrt {2}}\right ) + \frac {1}{8} \, {\left (\sqrt {6} + \sqrt {2}\right )} \arctan \left (\frac {4 \, x + \sqrt {6} + \sqrt {2}}{\sqrt {6} - \sqrt {2}}\right ) + \frac {1}{8} \, {\left (\sqrt {6} + \sqrt {2}\right )} \arctan \left (\frac {4 \, x - \sqrt {6} - \sqrt {2}}{\sqrt {6} - \sqrt {2}}\right ) + \frac {1}{16} \, {\left (\sqrt {6} - \sqrt {2}\right )} \log \left (x^{2} + \frac {1}{2} \, x {\left (\sqrt {6} + \sqrt {2}\right )} + 1\right ) - \frac {1}{16} \, {\left (\sqrt {6} - \sqrt {2}\right )} \log \left (x^{2} - \frac {1}{2} \, x {\left (\sqrt {6} + \sqrt {2}\right )} + 1\right ) + \frac {1}{16} \, {\left (\sqrt {6} + \sqrt {2}\right )} \log \left (x^{2} + \frac {1}{2} \, x {\left (\sqrt {6} - \sqrt {2}\right )} + 1\right ) - \frac {1}{16} \, {\left (\sqrt {6} + \sqrt {2}\right )} \log \left (x^{2} - \frac {1}{2} \, x {\left (\sqrt {6} - \sqrt {2}\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(sqrt(6) - sqrt(2))*arctan((4*x + sqrt(6) - sqrt(2))/(sqrt(6) + sqrt(2))) + 1/8*(sqrt(6) - sqrt(2))*arctan
((4*x - sqrt(6) + sqrt(2))/(sqrt(6) + sqrt(2))) + 1/8*(sqrt(6) + sqrt(2))*arctan((4*x + sqrt(6) + sqrt(2))/(sq
rt(6) - sqrt(2))) + 1/8*(sqrt(6) + sqrt(2))*arctan((4*x - sqrt(6) - sqrt(2))/(sqrt(6) - sqrt(2))) + 1/16*(sqrt
(6) - sqrt(2))*log(x^2 + 1/2*x*(sqrt(6) + sqrt(2)) + 1) - 1/16*(sqrt(6) - sqrt(2))*log(x^2 - 1/2*x*(sqrt(6) +
sqrt(2)) + 1) + 1/16*(sqrt(6) + sqrt(2))*log(x^2 + 1/2*x*(sqrt(6) - sqrt(2)) + 1) - 1/16*(sqrt(6) + sqrt(2))*l
og(x^2 - 1/2*x*(sqrt(6) - sqrt(2)) + 1)

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maple [C]  time = 0.01, size = 42, normalized size = 0.13 \begin {gather*} \frac {\left (\RootOf \left (\textit {\_Z}^{8}-\textit {\_Z}^{4}+1\right )^{4}+1\right ) \ln \left (-\RootOf \left (\textit {\_Z}^{8}-\textit {\_Z}^{4}+1\right )+x \right )}{8 \RootOf \left (\textit {\_Z}^{8}-\textit {\_Z}^{4}+1\right )^{7}-4 \RootOf \left (\textit {\_Z}^{8}-\textit {\_Z}^{4}+1\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sum((_R^4+1)/(2*_R^7-_R^3)*ln(-_R+x),_R=RootOf(_Z^8-_Z^4+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.22, size = 145, normalized size = 0.44 \begin {gather*} -\mathrm {atan}\left (\frac {\sqrt {6}\,x\,\left (27-27{}\mathrm {i}\right )}{27\,\sqrt {3}-81{}\mathrm {i}}\right )\,\left (\sqrt {2}\,\left (\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+\sqrt {6}\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )\right )-\mathrm {atan}\left (\frac {\sqrt {6}\,x\,\left (27+27{}\mathrm {i}\right )}{27\,\sqrt {3}-81{}\mathrm {i}}\right )\,\left (\sqrt {2}\,\left (\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )+\sqrt {6}\,\left (\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )\right )-\mathrm {atan}\left (\frac {\sqrt {6}\,x\,\left (27-27{}\mathrm {i}\right )}{27\,\sqrt {3}+81{}\mathrm {i}}\right )\,\left (\sqrt {2}\,\left (\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+\sqrt {6}\,\left (\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )\right )-\mathrm {atan}\left (\frac {\sqrt {6}\,x\,\left (27+27{}\mathrm {i}\right )}{27\,\sqrt {3}+81{}\mathrm {i}}\right )\,\left (\sqrt {2}\,\left (\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )+\sqrt {6}\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 + 1)/(x^8 - x^4 + 1),x)

[Out]

- atan((6^(1/2)*x*(27 - 27i))/(27*3^(1/2) - 81i))*(2^(1/2)*(1/8 + 1i/8) - 6^(1/2)*(1/8 - 1i/8)) - atan((6^(1/2
)*x*(27 + 27i))/(27*3^(1/2) - 81i))*(2^(1/2)*(1/8 - 1i/8) + 6^(1/2)*(1/8 + 1i/8)) - atan((6^(1/2)*x*(27 - 27i)
)/(27*3^(1/2) + 81i))*(2^(1/2)*(1/8 + 1i/8) + 6^(1/2)*(1/8 - 1i/8)) - atan((6^(1/2)*x*(27 + 27i))/(27*3^(1/2)
+ 81i))*(2^(1/2)*(1/8 - 1i/8) - 6^(1/2)*(1/8 + 1i/8))

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sympy [A]  time = 3.10, size = 20, normalized size = 0.06 \begin {gather*} \operatorname {RootSum} {\left (65536 t^{8} - 256 t^{4} + 1, \left (t \mapsto t \log {\left (1024 t^{5} + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(65536*_t**8 - 256*_t**4 + 1, Lambda(_t, _t*log(1024*_t**5 + x)))

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