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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 56, normalized size of antiderivative = 1.00, 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 31

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

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.85, size = 52, normalized size = 0.93 \[ \frac {2 \, \sqrt {3} x^{4} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + x^{4} \log \left (x^{4} + x^{2} + 1\right ) - 2 \, x^{4} \log \left (x^{2} - 1\right ) - 3}{12 \, x^{4}} \]

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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giac [A]  time = 0.16, size = 44, normalized size = 0.79 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) - \frac {1}{4 \, x^{4}} + \frac {1}{12} \, \log \left (x^{4} + x^{2} + 1\right ) - \frac {1}{6} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]

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))

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maple [A]  time = 0.01, size = 71, normalized size = 1.27 \[ \frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}-\frac {\ln \left (x -1\right )}{6}-\frac {\ln \left (x +1\right )}{6}+\frac {\ln \left (x^{2}-x +1\right )}{12}+\frac {\ln \left (x^{2}+x +1\right )}{12}-\frac {1}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*ln(x-1)-1/4/x^4+1/12*ln(x^2-x+1)+1/6*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))-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 = 2.22, size = 43, normalized size = 0.77 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) - \frac {1}{4 \, x^{4}} + \frac {1}{12} \, \log \left (x^{4} + x^{2} + 1\right ) - \frac {1}{6} \, \log \left (x^{2} - 1\right ) \]

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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mupad [B]  time = 0.08, size = 57, normalized size = 1.02 \[ -\frac {\ln \left (x^2-1\right )}{6}-\frac {1}{4\,x^4}-\ln \left (x^2-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}+\frac {1}{2}\right )\,\left (-\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\ln \left (x^2+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}+\frac {1}{2}\right )\,\left (\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(x^5*(x^6 - 1)),x)

[Out]

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

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sympy [A]  time = 0.21, size = 53, normalized size = 0.95 \[ - \frac {\log {\left (x^{2} - 1 \right )}}{6} + \frac {\log {\left (x^{4} + x^{2} + 1 \right )}}{12} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} + \frac {\sqrt {3}}{3} \right )}}{6} - \frac {1}{4 x^{4}} \]

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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