3.1357 \(\int \frac{1}{x^5 (1-x^6)} \, dx\)

Optimal. Leaf size=56 \[ -\frac{1}{4 x^4}-\frac{1}{6} \log \left (1-x^2\right )+\frac{1}{12} \log \left (x^4+x^2+1\right )+\frac{\tan ^{-1}\left (\frac{2 x^2+1}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

[Out]

-1/(4*x^4) + ArcTan[(1 + 2*x^2)/Sqrt[3]]/(2*Sqrt[3]) - Log[1 - x^2]/6 + Log[1 + x^2 + x^4]/12

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Rubi [A]  time = 0.0381724, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {275, 325, 200, 31, 634, 618, 204, 628} \[ -\frac{1}{4 x^4}-\frac{1}{6} \log \left (1-x^2\right )+\frac{1}{12} \log \left (x^4+x^2+1\right )+\frac{\tan ^{-1}\left (\frac{2 x^2+1}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(1 - x^6)),x]

[Out]

-1/(4*x^4) + ArcTan[(1 + 2*x^2)/Sqrt[3]]/(2*Sqrt[3]) - Log[1 - x^2]/6 + Log[1 + x^2 + x^4]/12

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{x^5 \left (1-x^6\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^3 \left (1-x^3\right )} \, dx,x,x^2\right )\\ &=-\frac{1}{4 x^4}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1-x^3} \, dx,x,x^2\right )\\ &=-\frac{1}{4 x^4}+\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,x^2\right )+\frac{1}{6} \operatorname{Subst}\left (\int \frac{2+x}{1+x+x^2} \, dx,x,x^2\right )\\ &=-\frac{1}{4 x^4}-\frac{1}{6} \log \left (1-x^2\right )+\frac{1}{12} \operatorname{Subst}\left (\int \frac{1+2 x}{1+x+x^2} \, dx,x,x^2\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,x^2\right )\\ &=-\frac{1}{4 x^4}-\frac{1}{6} \log \left (1-x^2\right )+\frac{1}{12} \log \left (1+x^2+x^4\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x^2\right )\\ &=-\frac{1}{4 x^4}+\frac{\tan ^{-1}\left (\frac{1+2 x^2}{\sqrt{3}}\right )}{2 \sqrt{3}}-\frac{1}{6} \log \left (1-x^2\right )+\frac{1}{12} \log \left (1+x^2+x^4\right )\\ \end{align*}

Mathematica [A]  time = 0.0223837, size = 78, normalized size = 1.39 \[ \frac{1}{12} \left (-\frac{3}{x^4}+\log \left (x^2-x+1\right )+\log \left (x^2+x+1\right )-2 \log (1-x)-2 \log (x+1)+2 \sqrt{3} \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )-2 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(1 - x^6)),x]

[Out]

(-3/x^4 + 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

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Maple [A]  time = 0.008, size = 71, normalized size = 1.3 \begin{align*}{\frac{\ln \left ({x}^{2}-x+1 \right ) }{12}}+{\frac{\sqrt{3}}{6}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }-{\frac{\ln \left ( 1+x \right ) }{6}}-{\frac{\ln \left ( -1+x \right ) }{6}}+{\frac{\ln \left ({x}^{2}+x+1 \right ) }{12}}-{\frac{\sqrt{3}}{6}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) }-{\frac{1}{4\,{x}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(-x^6+1),x)

[Out]

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)-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)-1/4/x^4

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Maxima [A]  time = 1.47288, size = 58, normalized size = 1.04 \begin{align*} \frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} + 1\right )}\right ) - \frac{1}{4 \, x^{4}} + \frac{1}{12} \, \log \left (x^{4} + x^{2} + 1\right ) - \frac{1}{6} \, \log \left (x^{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-x^6+1),x, algorithm="maxima")

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) - 1/4/x^4 + 1/12*log(x^4 + x^2 + 1) - 1/6*log(x^2 - 1)

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Fricas [A]  time = 1.59949, size = 146, normalized size = 2.61 \begin{align*} \frac{2 \, \sqrt{3} x^{4} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} + 1\right )}\right ) + x^{4} \log \left (x^{4} + x^{2} + 1\right ) - 2 \, x^{4} \log \left (x^{2} - 1\right ) - 3}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-x^6+1),x, algorithm="fricas")

[Out]

1/12*(2*sqrt(3)*x^4*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + x^4*log(x^4 + x^2 + 1) - 2*x^4*log(x^2 - 1) - 3)/x^4

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Sympy [A]  time = 0.158901, size = 53, normalized size = 0.95 \begin{align*} - \frac{\log{\left (x^{2} - 1 \right )}}{6} + \frac{\log{\left (x^{4} + x^{2} + 1 \right )}}{12} + \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x^{2}}{3} + \frac{\sqrt{3}}{3} \right )}}{6} - \frac{1}{4 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(-x**6+1),x)

[Out]

-log(x**2 - 1)/6 + log(x**4 + x**2 + 1)/12 + sqrt(3)*atan(2*sqrt(3)*x**2/3 + sqrt(3)/3)/6 - 1/(4*x**4)

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Giac [A]  time = 1.13112, size = 59, normalized size = 1.05 \begin{align*} \frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} + 1\right )}\right ) - \frac{1}{4 \, x^{4}} + \frac{1}{12} \, \log \left (x^{4} + x^{2} + 1\right ) - \frac{1}{6} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-x^6+1),x, algorithm="giac")

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) - 1/4/x^4 + 1/12*log(x^4 + x^2 + 1) - 1/6*log(abs(x^2 - 1))