\(\int \frac {x^4}{1-x^6} \, dx\) [1348]

Optimal result

Integrand size = 13, antiderivative size = 47 \[ \int \frac {x^4}{1-x^6} \, dx=-\frac {\arctan \left (\frac {\sqrt {3} x}{1-x^2}\right )}{2 \sqrt {3}}+\frac {\text {arctanh}(x)}{3}+\frac {1}{6} \text {arctanh}\left (\frac {x}{1+x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.55, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {302, 648, 632, 210, 642, 212} \[ \int \frac {x^4}{1-x^6} \, dx=\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\text {arctanh}(x)}{3}-\frac {1}{12} \log \left (x^2-x+1\right )+\frac {1}{12} \log \left (x^2+x+1\right ) \]

[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 210

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

Rule 212

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

Rule 302

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)/(a*n*s^m))*Int[1/(r^2 - s^2*x^2), x] + 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 632

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 642

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 648

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]

Rubi steps \begin{align*} \text {integral}& = \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} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )+\frac {1}{2} \text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.60 \[ \int \frac {x^4}{1-x^6} \, dx=\frac {1}{12} \left (-2 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )-2 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-2 \log (1-x)+2 \log (1+x)-\log \left (1-x+x^2\right )+\log \left (1+x+x^2\right )\right ) \]

[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

Maple [A] (verified)

Time = 4.41 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.32

method	result	size
risch	\(\frac {\ln \left (1+x \right )}{6}+\frac {\ln \left (x^{2}+x +1\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x +\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{6}-\frac {\ln \left (-1+x \right )}{6}-\frac {\ln \left (x^{2}-x +1\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{6}\)	\(62\)
default	\(-\frac {\ln \left (-1+x \right )}{6}+\frac {\ln \left (x^{2}+x +1\right )}{12}-\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}+\frac {\ln \left (1+x \right )}{6}-\frac {\ln \left (x^{2}-x +1\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{6}\)	\(66\)
meijerg	\(-\frac {x^{5} \left (\ln \left (1-\left (x^{6}\right )^{\frac {1}{6}}\right )-\ln \left (1+\left (x^{6}\right )^{\frac {1}{6}}\right )+\frac {\ln \left (1-\left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{2-\left (x^{6}\right )^{\frac {1}{6}}}\right )-\frac {\ln \left (1+\left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{2+\left (x^{6}\right )^{\frac {1}{6}}}\right )\right )}{6 \left (x^{6}\right )^{\frac {5}{6}}}\)	\(116\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.38 \[ \int \frac {x^4}{1-x^6} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{12} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{6} \, \log \left (x + 1\right ) - \frac {1}{6} \, \log \left (x - 1\right ) \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (34) = 68\).

Time = 0.13 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.77 \[ \int \frac {x^4}{1-x^6} \, dx=- \frac {\log {\left (x - 1 \right )}}{6} + \frac {\log {\left (x + 1 \right )}}{6} - \frac {\log {\left (x^{2} - x + 1 \right )}}{12} + \frac {\log {\left (x^{2} + x + 1 \right )}}{12} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{6} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{6} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.38 \[ \int \frac {x^4}{1-x^6} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{12} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{6} \, \log \left (x + 1\right ) - \frac {1}{6} \, \log \left (x - 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.43 \[ \int \frac {x^4}{1-x^6} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{12} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{6} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{6} \, \log \left ({\left | x - 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 5.49 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.13 \[ \int \frac {x^4}{1-x^6} \, dx=\frac {\mathrm {atanh}\left (x\right )}{3}-\mathrm {atanh}\left (\frac {2\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\mathrm {atanh}\left (\frac {2\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \]

[In]

int(-x^4/(x^6 - 1),x)

[Out]

atanh(x)/3 - atanh((2*x)/(3^(1/2)*1i - 1))*((3^(1/2)*1i)/6 + 1/6) - atanh((2*x)/(3^(1/2)*1i + 1))*((3^(1/2)*1i
)/6 - 1/6)