3.1.16 \(\int \frac {1+x^4}{1-3 x^4+x^8} \, dx\)

Optimal. Leaf size=131 \[ \frac {\tan ^{-1}\left (\sqrt {\frac {2}{\sqrt {5}-1}} x\right )}{\sqrt {2 \left (\sqrt {5}-1\right )}}-\frac {\tan ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {2 \left (1+\sqrt {5}\right )}}+\frac {\tanh ^{-1}\left (\sqrt {\frac {2}{\sqrt {5}-1}} x\right )}{\sqrt {2 \left (\sqrt {5}-1\right )}}-\frac {\tanh ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {2 \left (1+\sqrt {5}\right )}} \]

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Rubi [A]  time = 0.09, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1419, 1093, 203, 207} \begin {gather*} \frac {\tan ^{-1}\left (\sqrt {\frac {2}{\sqrt {5}-1}} x\right )}{\sqrt {2 \left (\sqrt {5}-1\right )}}-\frac {\tan ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {2 \left (1+\sqrt {5}\right )}}+\frac {\tanh ^{-1}\left (\sqrt {\frac {2}{\sqrt {5}-1}} x\right )}{\sqrt {2 \left (\sqrt {5}-1\right )}}-\frac {\tanh ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {2 \left (1+\sqrt {5}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^4)/(1 - 3*x^4 + x^8),x]

[Out]

ArcTan[Sqrt[2/(-1 + Sqrt[5])]*x]/Sqrt[2*(-1 + Sqrt[5])] - ArcTan[Sqrt[2/(1 + Sqrt[5])]*x]/Sqrt[2*(1 + Sqrt[5])
] + ArcTanh[Sqrt[2/(-1 + Sqrt[5])]*x]/Sqrt[2*(-1 + Sqrt[5])] - ArcTanh[Sqrt[2/(1 + Sqrt[5])]*x]/Sqrt[2*(1 + Sq
rt[5])]

Rule 203

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

Rule 207

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

Rule 1093

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/(b/
2 - q/2 + c*x^2), x], x] - Dist[c/q, Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*
a*c, 0] && PosQ[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}{1-3 x^4+x^8} \, dx &=\frac {1}{2} \int \frac {1}{1-\sqrt {5} x^2+x^4} \, dx+\frac {1}{2} \int \frac {1}{1+\sqrt {5} x^2+x^4} \, dx\\ &=\frac {1}{2} \int \frac {1}{-\frac {1}{2}-\frac {\sqrt {5}}{2}+x^2} \, dx-\frac {1}{2} \int \frac {1}{\frac {1}{2}-\frac {\sqrt {5}}{2}+x^2} \, dx+\frac {1}{2} \int \frac {1}{-\frac {1}{2}+\frac {\sqrt {5}}{2}+x^2} \, dx-\frac {1}{2} \int \frac {1}{\frac {1}{2}+\frac {\sqrt {5}}{2}+x^2} \, dx\\ &=\frac {\tan ^{-1}\left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {2 \left (-1+\sqrt {5}\right )}}-\frac {\tan ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {2 \left (1+\sqrt {5}\right )}}+\frac {\tanh ^{-1}\left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {2 \left (-1+\sqrt {5}\right )}}-\frac {\tanh ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {2 \left (1+\sqrt {5}\right )}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 131, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\sqrt {\frac {2}{\sqrt {5}-1}} x\right )}{\sqrt {2 \left (\sqrt {5}-1\right )}}-\frac {\tan ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {2 \left (1+\sqrt {5}\right )}}+\frac {\tanh ^{-1}\left (\sqrt {\frac {2}{\sqrt {5}-1}} x\right )}{\sqrt {2 \left (\sqrt {5}-1\right )}}-\frac {\tanh ^{-1}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {2 \left (1+\sqrt {5}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^4)/(1 - 3*x^4 + x^8),x]

[Out]

ArcTan[Sqrt[2/(-1 + Sqrt[5])]*x]/Sqrt[2*(-1 + Sqrt[5])] - ArcTan[Sqrt[2/(1 + Sqrt[5])]*x]/Sqrt[2*(1 + Sqrt[5])
] + ArcTanh[Sqrt[2/(-1 + Sqrt[5])]*x]/Sqrt[2*(-1 + Sqrt[5])] - ArcTanh[Sqrt[2/(1 + Sqrt[5])]*x]/Sqrt[2*(1 + Sq
rt[5])]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1+x^4}{1-3 x^4+x^8} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(1 + x^4)/(1 - 3*x^4 + x^8),x]

[Out]

IntegrateAlgebraic[(1 + x^4)/(1 - 3*x^4 + x^8), x]

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fricas [B]  time = 1.57, size = 247, normalized size = 1.89 \begin {gather*} -\frac {1}{2} \, \sqrt {2} \sqrt {\sqrt {5} + 1} \arctan \left (-\frac {1}{2} \, \sqrt {2} x \sqrt {\sqrt {5} + 1} + \frac {1}{2} \, \sqrt {2 \, x^{2} + \sqrt {5} - 1} \sqrt {\sqrt {5} + 1}\right ) + \frac {1}{2} \, \sqrt {2} \sqrt {\sqrt {5} - 1} \arctan \left (-\frac {1}{2} \, \sqrt {2} x \sqrt {\sqrt {5} - 1} + \frac {1}{2} \, \sqrt {2 \, x^{2} + \sqrt {5} + 1} \sqrt {\sqrt {5} - 1}\right ) + \frac {1}{8} \, \sqrt {2} \sqrt {\sqrt {5} + 1} \log \left ({\left (\sqrt {5} \sqrt {2} - \sqrt {2}\right )} \sqrt {\sqrt {5} + 1} + 4 \, x\right ) - \frac {1}{8} \, \sqrt {2} \sqrt {\sqrt {5} + 1} \log \left (-{\left (\sqrt {5} \sqrt {2} - \sqrt {2}\right )} \sqrt {\sqrt {5} + 1} + 4 \, x\right ) - \frac {1}{8} \, \sqrt {2} \sqrt {\sqrt {5} - 1} \log \left ({\left (\sqrt {5} \sqrt {2} + \sqrt {2}\right )} \sqrt {\sqrt {5} - 1} + 4 \, x\right ) + \frac {1}{8} \, \sqrt {2} \sqrt {\sqrt {5} - 1} \log \left (-{\left (\sqrt {5} \sqrt {2} + \sqrt {2}\right )} \sqrt {\sqrt {5} - 1} + 4 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+1)/(x^8-3*x^4+1),x, algorithm="fricas")

[Out]

-1/2*sqrt(2)*sqrt(sqrt(5) + 1)*arctan(-1/2*sqrt(2)*x*sqrt(sqrt(5) + 1) + 1/2*sqrt(2*x^2 + sqrt(5) - 1)*sqrt(sq
rt(5) + 1)) + 1/2*sqrt(2)*sqrt(sqrt(5) - 1)*arctan(-1/2*sqrt(2)*x*sqrt(sqrt(5) - 1) + 1/2*sqrt(2*x^2 + sqrt(5)
+ 1)*sqrt(sqrt(5) - 1)) + 1/8*sqrt(2)*sqrt(sqrt(5) + 1)*log((sqrt(5)*sqrt(2) - sqrt(2))*sqrt(sqrt(5) + 1) + 4
*x) - 1/8*sqrt(2)*sqrt(sqrt(5) + 1)*log(-(sqrt(5)*sqrt(2) - sqrt(2))*sqrt(sqrt(5) + 1) + 4*x) - 1/8*sqrt(2)*sq
rt(sqrt(5) - 1)*log((sqrt(5)*sqrt(2) + sqrt(2))*sqrt(sqrt(5) - 1) + 4*x) + 1/8*sqrt(2)*sqrt(sqrt(5) - 1)*log(-
(sqrt(5)*sqrt(2) + sqrt(2))*sqrt(sqrt(5) - 1) + 4*x)

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giac [A]  time = 0.96, size = 147, normalized size = 1.12 \begin {gather*} -\frac {1}{4} \, \sqrt {2 \, \sqrt {5} - 2} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}}}\right ) + \frac {1}{4} \, \sqrt {2 \, \sqrt {5} + 2} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}}}\right ) - \frac {1}{8} \, \sqrt {2 \, \sqrt {5} - 2} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) + \frac {1}{8} \, \sqrt {2 \, \sqrt {5} - 2} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) + \frac {1}{8} \, \sqrt {2 \, \sqrt {5} + 2} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) - \frac {1}{8} \, \sqrt {2 \, \sqrt {5} + 2} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+1)/(x^8-3*x^4+1),x, algorithm="giac")

[Out]

-1/4*sqrt(2*sqrt(5) - 2)*arctan(x/sqrt(1/2*sqrt(5) + 1/2)) + 1/4*sqrt(2*sqrt(5) + 2)*arctan(x/sqrt(1/2*sqrt(5)
- 1/2)) - 1/8*sqrt(2*sqrt(5) - 2)*log(abs(x + sqrt(1/2*sqrt(5) + 1/2))) + 1/8*sqrt(2*sqrt(5) - 2)*log(abs(x -
sqrt(1/2*sqrt(5) + 1/2))) + 1/8*sqrt(2*sqrt(5) + 2)*log(abs(x + sqrt(1/2*sqrt(5) - 1/2))) - 1/8*sqrt(2*sqrt(5
) + 2)*log(abs(x - sqrt(1/2*sqrt(5) - 1/2)))

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maple [A]  time = 0.04, size = 96, normalized size = 0.73 \begin {gather*} \frac {\arctanh \left (\frac {2 x}{\sqrt {-2+2 \sqrt {5}}}\right )}{\sqrt {-2+2 \sqrt {5}}}-\frac {\arctanh \left (\frac {2 x}{\sqrt {2+2 \sqrt {5}}}\right )}{\sqrt {2+2 \sqrt {5}}}+\frac {\arctan \left (\frac {2 x}{\sqrt {-2+2 \sqrt {5}}}\right )}{\sqrt {-2+2 \sqrt {5}}}-\frac {\arctan \left (\frac {2 x}{\sqrt {2+2 \sqrt {5}}}\right )}{\sqrt {2+2 \sqrt {5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+1)/(x^8-3*x^4+1),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} + 1}{x^{8} - 3 \, x^{4} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+1)/(x^8-3*x^4+1),x, algorithm="maxima")

[Out]

integrate((x^4 + 1)/(x^8 - 3*x^4 + 1), x)

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mupad [B]  time = 0.20, size = 269, normalized size = 2.05 \begin {gather*} -\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x\,\sqrt {\sqrt {5}-1}\,1875{}\mathrm {i}}{2\,\left (875\,\sqrt {5}-1875\right )}-\frac {\sqrt {2}\,\sqrt {5}\,x\,\sqrt {\sqrt {5}-1}\,875{}\mathrm {i}}{2\,\left (875\,\sqrt {5}-1875\right )}\right )\,\sqrt {\sqrt {5}-1}\,1{}\mathrm {i}}{4}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x\,\sqrt {\sqrt {5}+1}\,1875{}\mathrm {i}}{2\,\left (875\,\sqrt {5}+1875\right )}+\frac {\sqrt {2}\,\sqrt {5}\,x\,\sqrt {\sqrt {5}+1}\,875{}\mathrm {i}}{2\,\left (875\,\sqrt {5}+1875\right )}\right )\,\sqrt {\sqrt {5}+1}\,1{}\mathrm {i}}{4}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x\,\sqrt {1-\sqrt {5}}\,1875{}\mathrm {i}}{2\,\left (875\,\sqrt {5}-1875\right )}-\frac {\sqrt {2}\,\sqrt {5}\,x\,\sqrt {1-\sqrt {5}}\,875{}\mathrm {i}}{2\,\left (875\,\sqrt {5}-1875\right )}\right )\,\sqrt {1-\sqrt {5}}\,1{}\mathrm {i}}{4}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x\,\sqrt {-\sqrt {5}-1}\,1875{}\mathrm {i}}{2\,\left (875\,\sqrt {5}+1875\right )}+\frac {\sqrt {2}\,\sqrt {5}\,x\,\sqrt {-\sqrt {5}-1}\,875{}\mathrm {i}}{2\,\left (875\,\sqrt {5}+1875\right )}\right )\,\sqrt {-\sqrt {5}-1}\,1{}\mathrm {i}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 + 1)/(x^8 - 3*x^4 + 1),x)

[Out]

(2^(1/2)*atan((2^(1/2)*x*(1 - 5^(1/2))^(1/2)*1875i)/(2*(875*5^(1/2) - 1875)) - (2^(1/2)*5^(1/2)*x*(1 - 5^(1/2)
)^(1/2)*875i)/(2*(875*5^(1/2) - 1875)))*(1 - 5^(1/2))^(1/2)*1i)/4 - (2^(1/2)*atan((2^(1/2)*x*(5^(1/2) + 1)^(1/
2)*1875i)/(2*(875*5^(1/2) + 1875)) + (2^(1/2)*5^(1/2)*x*(5^(1/2) + 1)^(1/2)*875i)/(2*(875*5^(1/2) + 1875)))*(5
^(1/2) + 1)^(1/2)*1i)/4 - (2^(1/2)*atan((2^(1/2)*x*(5^(1/2) - 1)^(1/2)*1875i)/(2*(875*5^(1/2) - 1875)) - (2^(1
/2)*5^(1/2)*x*(5^(1/2) - 1)^(1/2)*875i)/(2*(875*5^(1/2) - 1875)))*(5^(1/2) - 1)^(1/2)*1i)/4 + (2^(1/2)*atan((2
^(1/2)*x*(- 5^(1/2) - 1)^(1/2)*1875i)/(2*(875*5^(1/2) + 1875)) + (2^(1/2)*5^(1/2)*x*(- 5^(1/2) - 1)^(1/2)*875i
)/(2*(875*5^(1/2) + 1875)))*(- 5^(1/2) - 1)^(1/2)*1i)/4

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sympy [A]  time = 1.19, size = 49, normalized size = 0.37 \begin {gather*} \operatorname {RootSum} {\left (256 t^{4} - 16 t^{2} - 1, \left (t \mapsto t \log {\left (1024 t^{5} - 8 t + x \right )} \right )\right )} + \operatorname {RootSum} {\left (256 t^{4} + 16 t^{2} - 1, \left (t \mapsto t \log {\left (1024 t^{5} - 8 t + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4+1)/(x**8-3*x**4+1),x)

[Out]

RootSum(256*_t**4 - 16*_t**2 - 1, Lambda(_t, _t*log(1024*_t**5 - 8*_t + x))) + RootSum(256*_t**4 + 16*_t**2 -
1, Lambda(_t, _t*log(1024*_t**5 - 8*_t + x)))

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