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

Optimal. Leaf size=75 \[ \frac {1}{2} \sqrt {\frac {1}{10} \left (3+\sqrt {5}\right )} \tanh ^{-1}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )-\frac {\tanh ^{-1}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )}{\sqrt {10 \left (3+\sqrt {5}\right )}} \]

[Out]

1/2*arctanh(x^2*(1/2+1/2*5^(1/2)))*(1/2+1/10*5^(1/2))-arctanh(x^2*2^(1/2)/(3+5^(1/2))^(1/2))/(5+5^(1/2))

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Rubi [A]  time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1359, 1093, 207} \[ \frac {1}{2} \sqrt {\frac {1}{10} \left (3+\sqrt {5}\right )} \tanh ^{-1}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )-\frac {\tanh ^{-1}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )}{\sqrt {10 \left (3+\sqrt {5}\right )}} \]

Antiderivative was successfully verified.

[In]

Int[x/(1 - 3*x^4 + x^8),x]

[Out]

-(ArcTanh[Sqrt[2/(3 + Sqrt[5])]*x^2]/Sqrt[10*(3 + Sqrt[5])]) + (Sqrt[(3 + Sqrt[5])/10]*ArcTanh[Sqrt[(3 + Sqrt[
5])/2]*x^2])/2

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 1359

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[
1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k) + c*x^((2*n)/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b,
c, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 91, normalized size = 1.21 \[ \frac {1}{40} \left (-\left (\left (5+\sqrt {5}\right ) \log \left (-2 x^2+\sqrt {5}-1\right )\right )-\left (\sqrt {5}-5\right ) \log \left (-2 x^2+\sqrt {5}+1\right )+\left (5+\sqrt {5}\right ) \log \left (2 x^2+\sqrt {5}-1\right )+\left (\sqrt {5}-5\right ) \log \left (2 x^2+\sqrt {5}+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/(1 - 3*x^4 + x^8),x]

[Out]

(-((5 + Sqrt[5])*Log[-1 + Sqrt[5] - 2*x^2]) - (-5 + Sqrt[5])*Log[1 + Sqrt[5] - 2*x^2] + (5 + Sqrt[5])*Log[-1 +
Sqrt[5] + 2*x^2] + (-5 + Sqrt[5])*Log[1 + Sqrt[5] + 2*x^2])/40

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fricas [B]  time = 0.87, size = 107, normalized size = 1.43 \[ \frac {1}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{4} + 2 \, x^{2} + \sqrt {5} {\left (2 \, x^{2} + 1\right )} + 3}{x^{4} + x^{2} - 1}\right ) + \frac {1}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{4} - 2 \, x^{2} + \sqrt {5} {\left (2 \, x^{2} - 1\right )} + 3}{x^{4} - x^{2} - 1}\right ) - \frac {1}{8} \, \log \left (x^{4} + x^{2} - 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*sqrt(5)*log((2*x^4 + 2*x^2 + sqrt(5)*(2*x^2 + 1) + 3)/(x^4 + x^2 - 1)) + 1/40*sqrt(5)*log((2*x^4 - 2*x^2
+ sqrt(5)*(2*x^2 - 1) + 3)/(x^4 - x^2 - 1)) - 1/8*log(x^4 + x^2 - 1) + 1/8*log(x^4 - x^2 - 1)

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giac [B]  time = 0.43, size = 92, normalized size = 1.23 \[ -\frac {1}{40} \, \sqrt {5} \log \left (\frac {{\left | 2 \, x^{2} - \sqrt {5} + 1 \right |}}{2 \, x^{2} + \sqrt {5} + 1}\right ) - \frac {1}{40} \, \sqrt {5} \log \left (\frac {{\left | 2 \, x^{2} - \sqrt {5} - 1 \right |}}{{\left | 2 \, x^{2} + \sqrt {5} - 1 \right |}}\right ) - \frac {1}{8} \, \log \left ({\left | x^{4} + x^{2} - 1 \right |}\right ) + \frac {1}{8} \, \log \left ({\left | x^{4} - x^{2} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/40*sqrt(5)*log(abs(2*x^2 - sqrt(5) + 1)/(2*x^2 + sqrt(5) + 1)) - 1/40*sqrt(5)*log(abs(2*x^2 - sqrt(5) - 1)/
abs(2*x^2 + sqrt(5) - 1)) - 1/8*log(abs(x^4 + x^2 - 1)) + 1/8*log(abs(x^4 - x^2 - 1))

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maple [A]  time = 0.00, size = 62, normalized size = 0.83 \[ \frac {\sqrt {5}\, \arctanh \left (\frac {\left (2 x^{2}-1\right ) \sqrt {5}}{5}\right )}{20}+\frac {\sqrt {5}\, \arctanh \left (\frac {\left (2 x^{2}+1\right ) \sqrt {5}}{5}\right )}{20}+\frac {\ln \left (x^{4}-x^{2}-1\right )}{8}-\frac {\ln \left (x^{4}+x^{2}-1\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*ln(x^4+x^2-1)+1/20*5^(1/2)*arctanh(1/5*(2*x^2+1)*5^(1/2))+1/8*ln(x^4-x^2-1)+1/20*5^(1/2)*arctanh(1/5*(2*x
^2-1)*5^(1/2))

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maxima [B]  time = 1.45, size = 87, normalized size = 1.16 \[ -\frac {1}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{2} - \sqrt {5} + 1}{2 \, x^{2} + \sqrt {5} + 1}\right ) - \frac {1}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{2} - \sqrt {5} - 1}{2 \, x^{2} + \sqrt {5} - 1}\right ) - \frac {1}{8} \, \log \left (x^{4} + x^{2} - 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/40*sqrt(5)*log((2*x^2 - sqrt(5) + 1)/(2*x^2 + sqrt(5) + 1)) - 1/40*sqrt(5)*log((2*x^2 - sqrt(5) - 1)/(2*x^2
+ sqrt(5) - 1)) - 1/8*log(x^4 + x^2 - 1) + 1/8*log(x^4 - x^2 - 1)

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mupad [B]  time = 1.30, size = 83, normalized size = 1.11 \[ \mathrm {atanh}\left (\frac {29\,x^2}{8\,\sqrt {5}-18}-\frac {13\,\sqrt {5}\,x^2}{8\,\sqrt {5}-18}\right )\,\left (\frac {\sqrt {5}}{20}-\frac {1}{4}\right )+\mathrm {atanh}\left (\frac {29\,x^2}{8\,\sqrt {5}+18}+\frac {13\,\sqrt {5}\,x^2}{8\,\sqrt {5}+18}\right )\,\left (\frac {\sqrt {5}}{20}+\frac {1}{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^8 - 3*x^4 + 1),x)

[Out]

atanh((29*x^2)/(8*5^(1/2) - 18) - (13*5^(1/2)*x^2)/(8*5^(1/2) - 18))*(5^(1/2)/20 - 1/4) + atanh((29*x^2)/(8*5^
(1/2) + 18) + (13*5^(1/2)*x^2)/(8*5^(1/2) + 18))*(5^(1/2)/20 + 1/4)

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sympy [B]  time = 0.37, size = 165, normalized size = 2.20 \[ \left (\frac {\sqrt {5}}{40} + \frac {1}{8}\right ) \log {\left (x^{2} - \frac {7}{2} - \frac {7 \sqrt {5}}{10} + 960 \left (\frac {\sqrt {5}}{40} + \frac {1}{8}\right )^{3} \right )} + \left (\frac {1}{8} - \frac {\sqrt {5}}{40}\right ) \log {\left (x^{2} - \frac {7}{2} + 960 \left (\frac {1}{8} - \frac {\sqrt {5}}{40}\right )^{3} + \frac {7 \sqrt {5}}{10} \right )} + \left (- \frac {1}{8} + \frac {\sqrt {5}}{40}\right ) \log {\left (x^{2} - \frac {7 \sqrt {5}}{10} + 960 \left (- \frac {1}{8} + \frac {\sqrt {5}}{40}\right )^{3} + \frac {7}{2} \right )} + \left (- \frac {1}{8} - \frac {\sqrt {5}}{40}\right ) \log {\left (x^{2} + 960 \left (- \frac {1}{8} - \frac {\sqrt {5}}{40}\right )^{3} + \frac {7 \sqrt {5}}{10} + \frac {7}{2} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(5)/40 + 1/8)*log(x**2 - 7/2 - 7*sqrt(5)/10 + 960*(sqrt(5)/40 + 1/8)**3) + (1/8 - sqrt(5)/40)*log(x**2 -
7/2 + 960*(1/8 - sqrt(5)/40)**3 + 7*sqrt(5)/10) + (-1/8 + sqrt(5)/40)*log(x**2 - 7*sqrt(5)/10 + 960*(-1/8 + sq
rt(5)/40)**3 + 7/2) + (-1/8 - sqrt(5)/40)*log(x**2 + 960*(-1/8 - sqrt(5)/40)**3 + 7*sqrt(5)/10 + 7/2)

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