\(\int \frac {1+x^6}{-1+x^6} \, dx\) [425]

Optimal result

Integrand size = 13, antiderivative size = 69 \[ \int \frac {1+x^6}{-1+x^6} \, dx=x+\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2 \text {arctanh}(x)}{3}+\frac {1}{6} \log \left (1-x+x^2\right )-\frac {1}{6} \log \left (1+x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {396, 216, 648, 632, 210, 642, 212} \[ \int \frac {1+x^6}{-1+x^6} \, dx=\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2 \text {arctanh}(x)}{3}+\frac {1}{6} \log \left (x^2-x+1\right )-\frac {1}{6} \log \left (x^2+x+1\right )+x \]

[In]

Int[(1 + x^6)/(-1 + x^6),x]

[Out]

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

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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.13 \[ \int \frac {1+x^6}{-1+x^6} \, dx=\frac {1}{6} \left (6 x-2 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )-2 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )+2 \log (1-x)-2 \log (1+x)+\log \left (1-x+x^2\right )-\log \left (1+x+x^2\right )\right ) \]

[In]

Integrate[(1 + x^6)/(-1 + x^6),x]

[Out]

(6*x - 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])/6

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97

method	result	size
default	\(x -\frac {\ln \left (x^{2}+x +1\right )}{6}-\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}+\frac {\ln \left (x -1\right )}{3}-\frac {\ln \left (x +1\right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}\)	\(67\)
risch	\(x -\frac {\ln \left (4 x^{2}+4 x +4\right )}{6}-\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}+\frac {\ln \left (4 x^{2}-4 x +4\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}-\frac {\ln \left (x +1\right )}{3}+\frac {\ln \left (x -1\right )}{3}\)	\(73\)
meijerg	\(\frac {x \left (\ln \left (1-\left (x^{6}\right )^{\frac {1}{6}}\right )-\ln \left (1+\left (x^{6}\right )^{\frac {1}{6}}\right )+\frac {\ln \left (1-\left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{2-\left (x^{6}\right )^{\frac {1}{6}}}\right )-\frac {\ln \left (1+\left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{2+\left (x^{6}\right )^{\frac {1}{6}}}\right )\right )}{6 \left (x^{6}\right )^{\frac {1}{6}}}-\frac {\left (-1\right )^{\frac {5}{6}} \left (6 x \left (-1\right )^{\frac {1}{6}}+\frac {x \left (-1\right )^{\frac {1}{6}} \left (\ln \left (1-\left (x^{6}\right )^{\frac {1}{6}}\right )-\ln \left (1+\left (x^{6}\right )^{\frac {1}{6}}\right )+\frac {\ln \left (1-\left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{2-\left (x^{6}\right )^{\frac {1}{6}}}\right )-\frac {\ln \left (1+\left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{2+\left (x^{6}\right )^{\frac {1}{6}}}\right )\right )}{\left (x^{6}\right )^{\frac {1}{6}}}\right )}{6}\)	\(246\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {1+x^6}{-1+x^6} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + x - \frac {1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{3} \, \log \left (x + 1\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.23 \[ \int \frac {1+x^6}{-1+x^6} \, dx=x + \frac {\log {\left (x - 1 \right )}}{3} - \frac {\log {\left (x + 1 \right )}}{3} + \frac {\log {\left (x^{2} - x + 1 \right )}}{6} - \frac {\log {\left (x^{2} + x + 1 \right )}}{6} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{3} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {1+x^6}{-1+x^6} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + x - \frac {1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{3} \, \log \left (x + 1\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99 \[ \int \frac {1+x^6}{-1+x^6} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + x - \frac {1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{3} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{3} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.36 \[ \int \frac {1+x^6}{-1+x^6} \, dx=x+\frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{3}-\mathrm {atan}\left (\frac {x\,32{}\mathrm {i}}{-32+\sqrt {3}\,32{}\mathrm {i}}-\frac {32\,\sqrt {3}\,x}{-32+\sqrt {3}\,32{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{3}-\frac {1}{3}{}\mathrm {i}\right )-\mathrm {atan}\left (\frac {x\,32{}\mathrm {i}}{32+\sqrt {3}\,32{}\mathrm {i}}+\frac {32\,\sqrt {3}\,x}{32+\sqrt {3}\,32{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{3}+\frac {1}{3}{}\mathrm {i}\right ) \]

[In]

int((x^6 + 1)/(x^6 - 1),x)

[Out]

x + (atan(x*1i)*2i)/3 - atan((x*32i)/(3^(1/2)*32i - 32) - (32*3^(1/2)*x)/(3^(1/2)*32i - 32))*(3^(1/2)/3 - 1i/3
) - atan((x*32i)/(3^(1/2)*32i + 32) + (32*3^(1/2)*x)/(3^(1/2)*32i + 32))*(3^(1/2)/3 + 1i/3)