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

Optimal. Leaf size=82 \[ -\frac{\log \left (x^4-\sqrt{3} x^2+1\right )}{8 \sqrt{3}}+\frac{\log \left (x^4+\sqrt{3} x^2+1\right )}{8 \sqrt{3}}-\frac{1}{4} \tan ^{-1}\left (\sqrt{3}-2 x^2\right )+\frac{1}{4} \tan ^{-1}\left (2 x^2+\sqrt{3}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0571404, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {1359, 1094, 634, 618, 204, 628} \[ -\frac{\log \left (x^4-\sqrt{3} x^2+1\right )}{8 \sqrt{3}}+\frac{\log \left (x^4+\sqrt{3} x^2+1\right )}{8 \sqrt{3}}-\frac{1}{4} \tan ^{-1}\left (\sqrt{3}-2 x^2\right )+\frac{1}{4} \tan ^{-1}\left (2 x^2+\sqrt{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x/(1 - x^4 + x^8),x]

[Out]

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

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]

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 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.0555856, size = 83, normalized size = 1.01 \[ \frac{i \left (\sqrt{-1-i \sqrt{3}} \tan ^{-1}\left (\frac{1}{2} \left (1-i \sqrt{3}\right ) x^2\right )-\sqrt{-1+i \sqrt{3}} \tan ^{-1}\left (\frac{1}{2} \left (1+i \sqrt{3}\right ) x^2\right )\right )}{2 \sqrt{6}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x/(1 - x^4 + x^8),x]

[Out]

((I/2)*(Sqrt[-1 - I*Sqrt[3]]*ArcTan[((1 - I*Sqrt[3])*x^2)/2] - Sqrt[-1 + I*Sqrt[3]]*ArcTan[((1 + I*Sqrt[3])*x^
2)/2]))/Sqrt[6]

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Maple [A]  time = 0.008, size = 65, normalized size = 0.8 \begin{align*}{\frac{\arctan \left ( 2\,{x}^{2}-\sqrt{3} \right ) }{4}}+{\frac{\arctan \left ( 2\,{x}^{2}+\sqrt{3} \right ) }{4}}-{\frac{\ln \left ( 1+{x}^{4}-{x}^{2}\sqrt{3} \right ) \sqrt{3}}{24}}+{\frac{\ln \left ( 1+{x}^{4}+{x}^{2}\sqrt{3} \right ) \sqrt{3}}{24}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*arctan(2*x^2-3^(1/2))+1/4*arctan(2*x^2+3^(1/2))-1/24*ln(1+x^4-x^2*3^(1/2))*3^(1/2)+1/24*ln(1+x^4+x^2*3^(1/
2))*3^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(x^8 - x^4 + 1), x)

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Fricas [B]  time = 1.59651, size = 545, normalized size = 6.65 \begin{align*} -\frac{1}{12} \, \sqrt{6} \sqrt{3} \sqrt{2} \arctan \left (-\frac{1}{3} \, \sqrt{6} \sqrt{3} \sqrt{2} x^{2} + \frac{1}{3} \, \sqrt{6} \sqrt{3} \sqrt{2 \, x^{4} + \sqrt{6} \sqrt{2} x^{2} + 2} - \sqrt{3}\right ) - \frac{1}{12} \, \sqrt{6} \sqrt{3} \sqrt{2} \arctan \left (-\frac{1}{3} \, \sqrt{6} \sqrt{3} \sqrt{2} x^{2} + \frac{1}{3} \, \sqrt{6} \sqrt{3} \sqrt{2 \, x^{4} - \sqrt{6} \sqrt{2} x^{2} + 2} + \sqrt{3}\right ) + \frac{1}{48} \, \sqrt{6} \sqrt{2} \log \left (2 \, x^{4} + \sqrt{6} \sqrt{2} x^{2} + 2\right ) - \frac{1}{48} \, \sqrt{6} \sqrt{2} \log \left (2 \, x^{4} - \sqrt{6} \sqrt{2} x^{2} + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*sqrt(6)*sqrt(3)*sqrt(2)*arctan(-1/3*sqrt(6)*sqrt(3)*sqrt(2)*x^2 + 1/3*sqrt(6)*sqrt(3)*sqrt(2*x^4 + sqrt(
6)*sqrt(2)*x^2 + 2) - sqrt(3)) - 1/12*sqrt(6)*sqrt(3)*sqrt(2)*arctan(-1/3*sqrt(6)*sqrt(3)*sqrt(2)*x^2 + 1/3*sq
rt(6)*sqrt(3)*sqrt(2*x^4 - sqrt(6)*sqrt(2)*x^2 + 2) + sqrt(3)) + 1/48*sqrt(6)*sqrt(2)*log(2*x^4 + sqrt(6)*sqrt
(2)*x^2 + 2) - 1/48*sqrt(6)*sqrt(2)*log(2*x^4 - sqrt(6)*sqrt(2)*x^2 + 2)

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Sympy [A]  time = 0.195369, size = 70, normalized size = 0.85 \begin{align*} - \frac{\sqrt{3} \log{\left (x^{4} - \sqrt{3} x^{2} + 1 \right )}}{24} + \frac{\sqrt{3} \log{\left (x^{4} + \sqrt{3} x^{2} + 1 \right )}}{24} + \frac{\operatorname{atan}{\left (2 x^{2} - \sqrt{3} \right )}}{4} + \frac{\operatorname{atan}{\left (2 x^{2} + \sqrt{3} \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(3)*log(x**4 - sqrt(3)*x**2 + 1)/24 + sqrt(3)*log(x**4 + sqrt(3)*x**2 + 1)/24 + atan(2*x**2 - sqrt(3))/4
+ atan(2*x**2 + sqrt(3))/4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(x^8 - x^4 + 1), x)