3.175 \(\int \frac {1}{-x^3+x^6} \, dx\)

Optimal. Leaf size=48 \[ \frac {1}{2 x^2}-\frac {1}{6} \log \left (x^2+x+1\right )+\frac {1}{3} \log (1-x)-\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {1593, 325, 200, 31, 634, 618, 204, 628} \[ \frac {1}{2 x^2}-\frac {1}{6} \log \left (x^2+x+1\right )+\frac {1}{3} \log (1-x)-\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(-x^3 + x^6)^(-1),x]

[Out]

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

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 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]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 48, normalized size = 1.00 \[ \frac {1}{2 x^2}-\frac {1}{6} \log \left (x^2+x+1\right )+\frac {1}{3} \log (1-x)-\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-x^3 + x^6)^(-1),x]

[Out]

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

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fricas [A]  time = 0.40, size = 46, normalized size = 0.96 \[ -\frac {2 \, \sqrt {3} x^{2} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + x^{2} \log \left (x^{2} + x + 1\right ) - 2 \, x^{2} \log \left (x - 1\right ) - 3}{6 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^6-x^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.97, size = 38, normalized size = 0.79 \[ -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{2 \, x^{2}} - \frac {1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{3} \, \log \left ({\left | x - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^6-x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 38, normalized size = 0.79 \[ -\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{3}+\frac {\ln \left (x -1\right )}{3}-\frac {\ln \left (x^{2}+x +1\right )}{6}+\frac {1}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6-x^3),x)

[Out]

1/2/x^2+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))

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maxima [A]  time = 1.27, size = 37, normalized size = 0.77 \[ -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{2 \, x^{2}} - \frac {1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^6-x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.11, size = 51, normalized size = 1.06 \[ \frac {\ln \left (x-1\right )}{3}+\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\frac {1}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(x^3 - x^6),x)

[Out]

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

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sympy [A]  time = 0.16, size = 48, normalized size = 1.00 \[ \frac {\log {\left (x - 1 \right )}}{3} - \frac {\log {\left (x^{2} + x + 1 \right )}}{6} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{3} + \frac {1}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**6-x**3),x)

[Out]

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

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