\(\int \frac {1}{-x^3+x^6} \, dx\) [5]

Optimal result

Integrand size = 11, antiderivative size = 48 \[ \int \frac {1}{-x^3+x^6} \, dx=\frac {1}{2 x^2}-\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log (1-x)-\frac {1}{6} \log \left (1+x+x^2\right ) \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {1607, 331, 206, 31, 648, 632, 210, 642} \[ \int \frac {1}{-x^3+x^6} \, dx=-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2 x^2}-\frac {1}{6} \log \left (x^2+x+1\right )+\frac {1}{3} \log (1-x) \]

[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 206

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 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 331

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 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]

Rule 1607

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*} \text {integral}& = \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 )+\text {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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {1}{-x^3+x^6} \, dx=\frac {1}{2 x^2}-\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log (1-x)-\frac {1}{6} \log \left (1+x+x^2\right ) \]

[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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {1}{2 x^{2}}+\frac {\ln \left (x -1\right )}{3}-\frac {\ln \left (x^{2}+x +1\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x +\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{3}\)	\(36\)
default	\(\frac {1}{2 x^{2}}-\frac {\ln \left (x^{2}+x +1\right )}{6}-\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}+\frac {\ln \left (x -1\right )}{3}\)	\(38\)
meijerg	\(-\frac {\left (-1\right )^{\frac {2}{3}} \left (\frac {3 \left (-1\right )^{\frac {1}{3}}}{2 x^{2}}+\frac {x \left (-1\right )^{\frac {1}{3}} \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3}\)	\(78\)

[In]

int(1/(x^6-x^3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96 \[ \int \frac {1}{-x^3+x^6} \, dx=-\frac {2 \, \sqrt {3} x^{2} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + x^{2} \log \left (x^{2} + x + 1\right ) - 2 \, x^{2} \log \left (x - 1\right ) - 3}{6 \, x^{2}} \]

[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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {1}{-x^3+x^6} \, dx=\frac {\log {\left (x - 1 \right )}}{3} - \frac {\log {\left (x^{2} + x + 1 \right )}}{6} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{3} + \frac {1}{2 x^{2}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.77 \[ \int \frac {1}{-x^3+x^6} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{2 \, x^{2}} - \frac {1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.79 \[ \int \frac {1}{-x^3+x^6} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{2 \, x^{2}} - \frac {1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{3} \, \log \left ({\left | x - 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06 \[ \int \frac {1}{-x^3+x^6} \, dx=\frac {\ln \left (x-1\right )}{3}+\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\frac {1}{2\,x^2} \]

[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)