3.1354 \(\int \frac {1}{x^2 (1-x^6)} \, dx\)

Optimal. Leaf size=52 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{1-x^2}\right )}{2 \sqrt {3}}+\frac {1}{6} \tanh ^{-1}\left (\frac {x}{x^2+1}\right )-\frac {1}{x}+\frac {1}{3} \tanh ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 78, normalized size of antiderivative = 1.50, number of steps used = 11, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {325, 296, 634, 618, 204, 628, 206} \[ -\frac {1}{12} \log \left (x^2-x+1\right )+\frac {1}{12} \log \left (x^2+x+1\right )-\frac {1}{x}+\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {1}{3} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(1 - x^6)),x]

[Out]

-x^(-1) + 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 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 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])

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 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 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]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (1-x^6\right )} \, dx &=-\frac {1}{x}+\int \frac {x^4}{1-x^6} \, dx\\ &=-\frac {1}{x}+\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}{x}+\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}{x}+\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 {1}{x}+\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*}

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Mathematica [A]  time = 0.02, size = 86, normalized size = 1.65 \[ -\frac {x \log \left (x^2-x+1\right )-x \log \left (x^2+x+1\right )+2 x \log (1-x)-2 x \log (x+1)+2 \sqrt {3} x \tan ^{-1}\left (\frac {2 x-1}{\sqrt {3}}\right )+2 \sqrt {3} x \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )+12}{12 x} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(1 - x^6)),x]

[Out]

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

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fricas [A]  time = 0.68, size = 76, normalized size = 1.46 \[ -\frac {2 \, \sqrt {3} x \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + 2 \, \sqrt {3} x \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - x \log \left (x^{2} + x + 1\right ) + x \log \left (x^{2} - x + 1\right ) - 2 \, x \log \left (x + 1\right ) + 2 \, x \log \left (x - 1\right ) + 12}{12 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-x^6+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 72, normalized size = 1.38 \[ -\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{x} + \frac {1}{12} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{6} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{6} \, \log \left ({\left | x - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-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/x + 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))

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maple [A]  time = 0.01, size = 71, normalized size = 1.37 \[ -\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}-\frac {\ln \left (x -1\right )}{6}+\frac {\ln \left (x +1\right )}{6}-\frac {\ln \left (x^{2}-x +1\right )}{12}+\frac {\ln \left (x^{2}+x +1\right )}{12}-\frac {1}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(-x^6+1),x)

[Out]

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

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maxima [A]  time = 2.41, size = 70, normalized size = 1.35 \[ -\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{x} + \frac {1}{12} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{6} \, \log \left (x + 1\right ) - \frac {1}{6} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-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/x + 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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mupad [B]  time = 0.06, size = 64, normalized size = 1.23 \[ -\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{6}-\frac {1}{6}{}\mathrm {i}\right )-\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{6}+\frac {1}{6}{}\mathrm {i}\right )-\frac {1}{x}-\frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(x^2*(x^6 - 1)),x)

[Out]

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

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sympy [B]  time = 0.55, size = 87, normalized size = 1.67 \[ - \frac {\log {\left (x - 1 \right )}}{6} + \frac {\log {\left (x + 1 \right )}}{6} - \frac {\log {\left (x^{2} - x + 1 \right )}}{12} + \frac {\log {\left (x^{2} + x + 1 \right )}}{12} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{6} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{6} - \frac {1}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(-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 - 1/x

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