Integrand size = 7, antiderivative size = 47 \[ \int \frac {1}{-1+x^6} \, dx=-\frac {\arctan \left (\frac {\sqrt {3} x}{1-x^2}\right )}{2 \sqrt {3}}-\frac {\text {arctanh}(x)}{3}-\frac {1}{6} \text {arctanh}\left (\frac {x}{1+x^2}\right ) \]
Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.60 \[ \int \frac {1}{-1+x^6} \, dx=\frac {1}{12} \left (-2 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )-2 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )+2 \log (1-x)-2 \log (1+x)+\log \left (1-x+x^2\right )-\log \left (1+x+x^2\right )\right ) \]
(-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] + Log[1 - x + x^2] - Log[1 + x + x^2])/12
Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.68, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.143, Rules used = {754, 27, 219, 1142, 25, 1083, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x^6-1} \, dx\) |
\(\Big \downarrow \) 754 |
\(\displaystyle -\frac {1}{3} \int \frac {1}{1-x^2}dx-\frac {1}{3} \int \frac {2-x}{2 \left (x^2-x+1\right )}dx-\frac {1}{3} \int \frac {x+2}{2 \left (x^2+x+1\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{3} \int \frac {1}{1-x^2}dx-\frac {1}{6} \int \frac {2-x}{x^2-x+1}dx-\frac {1}{6} \int \frac {x+2}{x^2+x+1}dx\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {1}{6} \int \frac {2-x}{x^2-x+1}dx-\frac {1}{6} \int \frac {x+2}{x^2+x+1}dx-\frac {\text {arctanh}(x)}{3}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{6} \left (\frac {1}{2} \int -\frac {1-2 x}{x^2-x+1}dx-\frac {3}{2} \int \frac {1}{x^2-x+1}dx\right )+\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{x^2+x+1}dx-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )-\frac {\text {arctanh}(x)}{3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{x^2-x+1}dx-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{x^2+x+1}dx-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )-\frac {\text {arctanh}(x)}{3}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{6} \left (3 \int \frac {1}{-(2 x-1)^2-3}d(2 x-1)-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{6} \left (3 \int \frac {1}{-(2 x+1)^2-3}d(2 x+1)-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )-\frac {\text {arctanh}(x)}{3}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{6} \left (-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx-\sqrt {3} \arctan \left (\frac {2 x-1}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx-\sqrt {3} \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )\right )-\frac {\text {arctanh}(x)}{3}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{6} \left (\frac {1}{2} \log \left (x^2-x+1\right )-\sqrt {3} \arctan \left (\frac {2 x-1}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (-\sqrt {3} \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (x^2+x+1\right )\right )-\frac {\text {arctanh}(x)}{3}\) |
-1/3*ArcTanh[x] + (-(Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]]) + Log[1 - x + x^2 ]/2)/6 + (-(Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]]) - Log[1 + x + x^2]/2)/6
3.1.50.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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] && (Gt Q[a, 0] || LtQ[b, 0])
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*Cos[(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] + 2*(r/(a*n)) Sum[u, {k, 1, (n - 2)/4}], x]] / ; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.40
method | result | size |
default | \(\frac {\ln \left (-1+x \right )}{6}-\frac {\ln \left (x^{2}+x +1\right )}{12}-\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}+\frac {\ln \left (x^{2}-x +1\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}-\frac {\ln \left (1+x \right )}{6}\) | \(66\) |
risch | \(\frac {\ln \left (4 x^{2}-4 x +4\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}-\frac {\ln \left (1+x \right )}{6}-\frac {\ln \left (x^{2}+x +1\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{6}+\frac {\ln \left (-1+x \right )}{6}\) | \(66\) |
meijerg | \(\frac {x \left (\ln \left (1-\left (x^{6}\right )^{\frac {1}{6}}\right )-\ln \left (1+\left (x^{6}\right )^{\frac {1}{6}}\right )+\frac {\ln \left (1-\left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{2-\left (x^{6}\right )^{\frac {1}{6}}}\right )-\frac {\ln \left (1+\left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{2+\left (x^{6}\right )^{\frac {1}{6}}}\right )\right )}{6 \left (x^{6}\right )^{\frac {1}{6}}}\) | \(116\) |
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)
Time = 0.24 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.38 \[ \int \frac {1}{-1+x^6} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{12} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{6} \, \log \left (x + 1\right ) + \frac {1}{6} \, \log \left (x - 1\right ) \]
-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)
Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (36) = 72\).
Time = 0.13 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.77 \[ \int \frac {1}{-1+x^6} \, dx=\frac {\log {\left (x - 1 \right )}}{6} - \frac {\log {\left (x + 1 \right )}}{6} + \frac {\log {\left (x^{2} - x + 1 \right )}}{12} - \frac {\log {\left (x^{2} + x + 1 \right )}}{12} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{6} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{6} \]
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
Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.38 \[ \int \frac {1}{-1+x^6} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{12} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{6} \, \log \left (x + 1\right ) + \frac {1}{6} \, \log \left (x - 1\right ) \]
-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)
Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.43 \[ \int \frac {1}{-1+x^6} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{12} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{6} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{6} \, \log \left ({\left | x - 1 \right |}\right ) \]
-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))
Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.87 \[ \int \frac {1}{-1+x^6} \, dx=-\frac {\mathrm {atanh}\left (x\right )}{3}-\mathrm {atan}\left (\frac {x\,1{}\mathrm {i}}{1+\sqrt {3}\,1{}\mathrm {i}}+\frac {\sqrt {3}\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{6}+\frac {1}{6}{}\mathrm {i}\right )-\mathrm {atan}\left (\frac {x\,1{}\mathrm {i}}{-1+\sqrt {3}\,1{}\mathrm {i}}-\frac {\sqrt {3}\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{6}-\frac {1}{6}{}\mathrm {i}\right ) \]