3.2.51 \(\int \frac {1}{(-1+x^3)^2} \, dx\) [151]

Optimal. Leaf size=57 \[ \frac {x}{3 \left (1-x^3\right )}+\frac {2 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {2}{9} \log (1-x)+\frac {1}{9} \log \left (1+x+x^2\right ) \]

[Out]

1/3*x/(-x^3+1)-2/9*ln(1-x)+1/9*ln(x^2+x+1)+2/9*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {205, 206, 31, 648, 632, 210, 642} \begin {gather*} \frac {2 \text {ArcTan}\left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {x}{3 \left (1-x^3\right )}+\frac {1}{9} \log \left (x^2+x+1\right )-\frac {2}{9} \log (1-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x^3)^(-2),x]

[Out]

x/(3*(1 - x^3)) + (2*ArcTan[(1 + 2*x)/Sqrt[3]])/(3*Sqrt[3]) - (2*Log[1 - x])/9 + Log[1 + x + x^2]/9

Rule 31

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

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 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]

Rubi steps

\begin {align*} \int \frac {1}{\left (-1+x^3\right )^2} \, dx &=\frac {x}{3 \left (1-x^3\right )}-\frac {2}{3} \int \frac {1}{-1+x^3} \, dx\\ &=\frac {x}{3 \left (1-x^3\right )}-\frac {2}{9} \int \frac {1}{-1+x} \, dx-\frac {2}{9} \int \frac {-2-x}{1+x+x^2} \, dx\\ &=\frac {x}{3 \left (1-x^3\right )}-\frac {2}{9} \log (1-x)+\frac {1}{9} \int \frac {1+2 x}{1+x+x^2} \, dx+\frac {1}{3} \int \frac {1}{1+x+x^2} \, dx\\ &=\frac {x}{3 \left (1-x^3\right )}-\frac {2}{9} \log (1-x)+\frac {1}{9} \log \left (1+x+x^2\right )-\frac {2}{3} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=\frac {x}{3 \left (1-x^3\right )}+\frac {2 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {2}{9} \log (1-x)+\frac {1}{9} \log \left (1+x+x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 49, normalized size = 0.86 \begin {gather*} \frac {1}{9} \left (-\frac {3 x}{-1+x^3}+2 \sqrt {3} \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )-2 \log (1-x)+\log \left (1+x+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x^3)^(-2),x]

[Out]

((-3*x)/(-1 + x^3) + 2*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] - 2*Log[1 - x] + Log[1 + x + x^2])/9

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Maple [A]
time = 0.06, size = 53, normalized size = 0.93

method	result	size
risch	\(-\frac {x}{3 \left (x^{3}-1\right )}+\frac {\ln \left (4 x^{2}+4 x +4\right )}{9}+\frac {2 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}-\frac {2 \ln \left (-1+x \right )}{9}\)	\(47\)
default	\(-\frac {1}{9 \left (-1+x \right )}-\frac {2 \ln \left (-1+x \right )}{9}+\frac {-1+x}{9 x^{2}+9 x +9}+\frac {\ln \left (x^{2}+x +1\right )}{9}+\frac {2 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}\)	\(53\)
meijerg	\(-\frac {\left (-1\right )^{\frac {2}{3}} \left (\frac {3 x \left (-1\right )^{\frac {1}{3}}}{-3 x^{3}+3}-\frac {2 x \left (-1\right )^{\frac {1}{3}} \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}\right )}{3}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3-1)^2,x,method=_RETURNVERBOSE)

[Out]

-1/9/(-1+x)-2/9*ln(-1+x)+1/9*(-1+x)/(x^2+x+1)+1/9*ln(x^2+x+1)+2/9*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)

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Maxima [A]
time = 3.92, size = 42, normalized size = 0.74 \begin {gather*} \frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {x}{3 \, {\left (x^{3} - 1\right )}} + \frac {1}{9} \, \log \left (x^{2} + x + 1\right ) - \frac {2}{9} \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-1)^2,x, algorithm="maxima")

[Out]

2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/3*x/(x^3 - 1) + 1/9*log(x^2 + x + 1) - 2/9*log(x - 1)

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Fricas [A]
time = 0.48, size = 58, normalized size = 1.02 \begin {gather*} \frac {2 \, \sqrt {3} {\left (x^{3} - 1\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + {\left (x^{3} - 1\right )} \log \left (x^{2} + x + 1\right ) - 2 \, {\left (x^{3} - 1\right )} \log \left (x - 1\right ) - 3 \, x}{9 \, {\left (x^{3} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-1)^2,x, algorithm="fricas")

[Out]

1/9*(2*sqrt(3)*(x^3 - 1)*arctan(1/3*sqrt(3)*(2*x + 1)) + (x^3 - 1)*log(x^2 + x + 1) - 2*(x^3 - 1)*log(x - 1) -
3*x)/(x^3 - 1)

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Sympy [A]
time = 0.06, size = 53, normalized size = 0.93 \begin {gather*} - \frac {x}{3 x^{3} - 3} - \frac {2 \log {\left (x - 1 \right )}}{9} + \frac {\log {\left (x^{2} + x + 1 \right )}}{9} + \frac {2 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**3-1)**2,x)

[Out]

-x/(3*x**3 - 3) - 2*log(x - 1)/9 + log(x**2 + x + 1)/9 + 2*sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/9

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Giac [A]
time = 0.44, size = 43, normalized size = 0.75 \begin {gather*} \frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {x}{3 \, {\left (x^{3} - 1\right )}} + \frac {1}{9} \, \log \left (x^{2} + x + 1\right ) - \frac {2}{9} \, \log \left ({\left | x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-1)^2,x, algorithm="giac")

[Out]

2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/3*x/(x^3 - 1) + 1/9*log(x^2 + x + 1) - 2/9*log(abs(x - 1))

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Mupad [B]
time = 0.08, size = 60, normalized size = 1.05 \begin {gather*} -\frac {2\,\ln \left (x-1\right )}{9}-\frac {x}{3\,\left (x^3-1\right )}-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )+\ln \left (2\,x+1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3 - 1)^2,x)

[Out]

log(2*x + 3^(1/2)*1i + 1)*((3^(1/2)*1i)/9 + 1/9) - x/(3*(x^3 - 1)) - log(x - (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*1
i)/9 - 1/9) - (2*log(x - 1))/9

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