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

Optimal. Leaf size=40 \[ -\frac {1}{6} \log \left (x^2+x+1\right )+\frac {1}{3} \log (1-x)+\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

1/3*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 = 40, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {292, 31, 634, 618, 204, 628} \[ -\frac {1}{6} \log \left (x^2+x+1\right )+\frac {1}{3} \log (1-x)+\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[x/(-1 + x^3),x]

[Out]

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 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 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(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 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 {x}{-1+x^3} \, dx &=\frac {1}{3} \int \frac {1}{-1+x} \, dx-\frac {1}{3} \int \frac {-1+x}{1+x+x^2} \, dx\\ &=\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}{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 {\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 = 40, normalized size = 1.00 \[ -\frac {1}{6} \log \left (x^2+x+1\right )+\frac {1}{3} \log (1-x)+\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(-1 + x^3),x]

[Out]

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 = 0.43, size = 32, normalized size = 0.80 \[ \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.07, size = 33, normalized size = 0.82 \[ \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{3} \, \log \left ({\left | x - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 33, normalized size = 0.82 \[ \frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{3}+\frac {\ln \left (x -1\right )}{3}-\frac {\ln \left (x^{2}+x +1\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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 = 0.96, size = 32, normalized size = 0.80 \[ \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.11, size = 46, normalized size = 1.15 \[ \frac {\ln \left (x-1\right )}{3}-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^3 - 1),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)

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sympy [A]  time = 0.13, size = 41, normalized size = 1.02 \[ \frac {\log {\left (x - 1 \right )}}{3} - \frac {\log {\left (x^{2} + x + 1 \right )}}{6} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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