\(\int \frac {1}{1-x^6} \, dx\) [1352]

Optimal result

Integrand size = 9, antiderivative size = 47 \[ \int \frac {1}{1-x^6} \, dx=\frac {\arctan \left (\frac {\sqrt {3} x}{1-x^2}\right )}{2 \sqrt {3}}+\frac {\text {arctanh}(x)}{3}+\frac {1}{6} \text {arctanh}\left (\frac {x}{1+x^2}\right ) \]

[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 = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.55, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {216, 648, 632, 210, 642, 212} \[ \int \frac {1}{1-x^6} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\text {arctanh}(x)}{3}-\frac {1}{12} \log \left (x^2-x+1\right )+\frac {1}{12} \log \left (x^2+x+1\right ) \]

[In]

Int[(1 - x^6)^(-1),x]

[Out]

-1/2*ArcTan[(1 - 2*x)/Sqrt[3]]/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 216

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[-a/b, n]], s = Denominator[Rt[-a/b, n
]], k, u}, Simp[u = Int[(r - s*Cos[(2*k*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x] + Int[(r + s*C
os[(2*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; 2*(r^2/(a*n))*Int[1/(r^2 - s^2*x^2), x] + Dis
t[2*(r/(a*n)), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && 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 {1-\frac {x}{2}}{1-x+x^2} \, dx+\frac {1}{3} \int \frac {1+\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 \[ \int \frac {1}{1-x^6} \, dx=\frac {1}{12} \left (2 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )+2 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-2 \log (1-x)+2 \log (1+x)-\log \left (1-x+x^2\right )+\log \left (1+x+x^2\right )\right ) \]

[In]

Integrate[(1 - x^6)^(-1),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] - Lo
g[1 - x + x^2] + Log[1 + x + x^2])/12

Maple [A] (verified)

Time = 4.40 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.40

method	result	size
default	\(-\frac {\ln \left (-1+x \right )}{6}+\frac {\ln \left (x^{2}+x +1\right )}{12}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}+\frac {\ln \left (1+x \right )}{6}-\frac {\ln \left (x^{2}-x +1\right )}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{6}\)	\(66\)
risch	\(\frac {\ln \left (4 x^{2}+4 x +4\right )}{12}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {\ln \left (4 x^{2}-4 x +4\right )}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{6}+\frac {\ln \left (1+x \right )}{6}-\frac {\ln \left (-1+x \right )}{6}\)	\(72\)
meijerg	\(-\frac {x \left (\ln \left (1-\left (x^{6}\right )^{\frac {1}{6}}\right )-\ln \left (1+\left (x^{6}\right )^{\frac {1}{6}}\right )+\frac {\ln \left (1-\left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{2-\left (x^{6}\right )^{\frac {1}{6}}}\right )-\frac {\ln \left (1+\left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{2+\left (x^{6}\right )^{\frac {1}{6}}}\right )\right )}{6 \left (x^{6}\right )^{\frac {1}{6}}}\)	\(116\)

[In]

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

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

Fricas [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.38 \[ \int \frac {1}{1-x^6} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{12} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{6} \, \log \left (x + 1\right ) - \frac {1}{6} \, \log \left (x - 1\right ) \]

[In]

integrate(1/(-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 \[ \int \frac {1}{1-x^6} \, dx=- \frac {\log {\left (x - 1 \right )}}{6} + \frac {\log {\left (x + 1 \right )}}{6} - \frac {\log {\left (x^{2} - x + 1 \right )}}{12} + \frac {\log {\left (x^{2} + x + 1 \right )}}{12} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{6} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{6} \]

[In]

integrate(1/(-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 \[ \int \frac {1}{1-x^6} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{12} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{6} \, \log \left (x + 1\right ) - \frac {1}{6} \, \log \left (x - 1\right ) \]

[In]

integrate(1/(-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 = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.43 \[ \int \frac {1}{1-x^6} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{12} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{6} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{6} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

integrate(1/(-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 = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.83 \[ \int \frac {1}{1-x^6} \, dx=\frac {\mathrm {atanh}\left (x\right )}{3}+\mathrm {atan}\left (\frac {x\,1{}\mathrm {i}}{1+\sqrt {3}\,1{}\mathrm {i}}+\frac {\sqrt {3}\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{6}+\frac {1}{6}{}\mathrm {i}\right )+\mathrm {atan}\left (\frac {x\,1{}\mathrm {i}}{-1+\sqrt {3}\,1{}\mathrm {i}}-\frac {\sqrt {3}\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{6}-\frac {1}{6}{}\mathrm {i}\right ) \]

[In]

int(-1/(x^6 - 1),x)

[Out]

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