3.1348 \(\int \frac{x^4}{1-x^6} \, dx\)

Optimal. Leaf size=47 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{3} x}{1-x^2}\right )}{2 \sqrt{3}}+\frac{1}{6} \tanh ^{-1}\left (\frac{x}{x^2+1}\right )+\frac{1}{3} \tanh ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.148602, antiderivative size = 73, normalized size of antiderivative = 1.55, number of steps used = 10, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {296, 634, 618, 204, 628, 206} \[ -\frac{1}{12} \log \left (x^2-x+1\right )+\frac{1}{12} \log \left (x^2+x+1\right )+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{2 \sqrt{3}}-\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{2 \sqrt{3}}+\frac{1}{3} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x^4/(1 - x^6),x]

[Out]

ArcTan[(1 - 2*x)/Sqrt[3]]/(2*Sqrt[3]) - ArcTan[(1 + 2*x)/Sqrt[3]]/(2*Sqrt[3]) + ArcTanh[x]/3 - Log[1 - x + x^2
]/12 + Log[1 + x + x^2]/12

Rule 296

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt
[-(a/b), n]], k, u}, Simp[u = Int[(r*Cos[(2*k*m*Pi)/n] - s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi
)/n]*x + s^2*x^2), x] + Int[(r*Cos[(2*k*m*Pi)/n] + s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x
+ s^2*x^2), x]; (2*r^(m + 2)*Int[1/(r^2 - s^2*x^2), x])/(a*n*s^m) + Dist[(2*r^(m + 1))/(a*n*s^m), Sum[u, {k,
1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && NegQ[a/b]

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]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{x^4}{1-x^6} \, dx &=\frac{1}{3} \int \frac{-\frac{1}{2}-\frac{x}{2}}{1-x+x^2} \, dx+\frac{1}{3} \int \frac{-\frac{1}{2}+\frac{x}{2}}{1+x+x^2} \, dx+\frac{1}{3} \int \frac{1}{1-x^2} \, dx\\ &=\frac{1}{3} \tanh ^{-1}(x)-\frac{1}{12} \int \frac{-1+2 x}{1-x+x^2} \, dx+\frac{1}{12} \int \frac{1+2 x}{1+x+x^2} \, dx-\frac{1}{4} \int \frac{1}{1-x+x^2} \, dx-\frac{1}{4} \int \frac{1}{1+x+x^2} \, dx\\ &=\frac{1}{3} \tanh ^{-1}(x)-\frac{1}{12} \log \left (1-x+x^2\right )+\frac{1}{12} \log \left (1+x+x^2\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{2 \sqrt{3}}-\frac{\tan ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )}{2 \sqrt{3}}+\frac{1}{3} \tanh ^{-1}(x)-\frac{1}{12} \log \left (1-x+x^2\right )+\frac{1}{12} \log \left (1+x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0101258, size = 75, normalized size = 1.6 \[ \frac{1}{12} \left (-\log \left (x^2-x+1\right )+\log \left (x^2+x+1\right )-2 \log (1-x)+2 \log (x+1)-2 \sqrt{3} \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )-2 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(1 - x^6),x]

[Out]

(-2*Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]] - 2*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] - 2*Log[1 - x] + 2*Log[1 + x] - L
og[1 - x + x^2] + Log[1 + x + x^2])/12

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Maple [A]  time = 0.007, size = 66, normalized size = 1.4 \begin{align*} -{\frac{\ln \left ( -1+x \right ) }{6}}+{\frac{\ln \left ({x}^{2}+x+1 \right ) }{12}}-{\frac{\sqrt{3}}{6}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) }-{\frac{\ln \left ({x}^{2}-x+1 \right ) }{12}}-{\frac{\sqrt{3}}{6}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }+{\frac{\ln \left ( 1+x \right ) }{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(-x^6+1),x)

[Out]

-1/6*ln(-1+x)+1/12*ln(x^2+x+1)-1/6*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)-1/12*ln(x^2-x+1)-1/6*3^(1/2)*arctan(1/3
*(2*x-1)*3^(1/2))+1/6*ln(1+x)

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Maxima [A]  time = 1.50281, size = 88, normalized size = 1.87 \begin{align*} -\frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{1}{12} \, \log \left (x^{2} + x + 1\right ) - \frac{1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac{1}{6} \, \log \left (x + 1\right ) - \frac{1}{6} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(-x^6+1),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.46658, size = 230, normalized size = 4.89 \begin{align*} -\frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{1}{12} \, \log \left (x^{2} + x + 1\right ) - \frac{1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac{1}{6} \, \log \left (x + 1\right ) - \frac{1}{6} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(-x^6+1),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 0.226036, size = 83, normalized size = 1.77 \begin{align*} - \frac{\log{\left (x - 1 \right )}}{6} + \frac{\log{\left (x + 1 \right )}}{6} - \frac{\log{\left (x^{2} - x + 1 \right )}}{12} + \frac{\log{\left (x^{2} + x + 1 \right )}}{12} - \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} - \frac{\sqrt{3}}{3} \right )}}{6} - \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} + \frac{\sqrt{3}}{3} \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(-x**6+1),x)

[Out]

-log(x - 1)/6 + log(x + 1)/6 - log(x**2 - x + 1)/12 + log(x**2 + x + 1)/12 - sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt
(3)/3)/6 - sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/6

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Giac [A]  time = 1.15246, size = 90, normalized size = 1.91 \begin{align*} -\frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{1}{12} \, \log \left (x^{2} + x + 1\right ) - \frac{1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac{1}{6} \, \log \left ({\left | x + 1 \right |}\right ) - \frac{1}{6} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(-x^6+1),x, algorithm="giac")

[Out]

-1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/12*log(x^2 + x + 1)
- 1/12*log(x^2 - x + 1) + 1/6*log(abs(x + 1)) - 1/6*log(abs(x - 1))