3.36 \(\int \frac {1}{x \sqrt [3]{1-x^3}} \, dx\)

Optimal. Leaf size=55 \[ \frac {1}{2} \log \left (1-\sqrt [3]{1-x^3}\right )+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\log (x)}{2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {266, 55, 618, 204, 31} \[ \frac {1}{2} \log \left (1-\sqrt [3]{1-x^3}\right )+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\log (x)}{2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(1 - x^3)^(1/3)),x]

[Out]

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

Rule 31

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

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 55, normalized size = 1.00 \[ \frac {1}{2} \log \left (1-\sqrt [3]{1-x^3}\right )+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\log (x)}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 - x^3)^(1/3)),x]

[Out]

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

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fricas [A]  time = 1.11, size = 64, normalized size = 1.16 \[ \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{6} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.82, size = 63, normalized size = 1.15 \[ \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{6} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left | {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.10, size = 65, normalized size = 1.18 \[ \frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (\frac {2 \pi \sqrt {3}\, x^{3} \hypergeom \left (\left [1, 1, \frac {4}{3}\right ], \left [2, 2\right ], x^{3}\right )}{9 \Gamma \left (\frac {2}{3}\right )}+\frac {2 \left (3 \ln \relax (x )-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+i \pi \right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}\right )}{6 \pi } \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/Pi*3^(1/2)*GAMMA(2/3)*(2/9*Pi*3^(1/2)/GAMMA(2/3)*x^3*hypergeom([1,1,4/3],[2,2],x^3)+2/3*(-1/6*Pi*3^(1/2)-3
/2*ln(3)+3*ln(x)+I*Pi)*Pi*3^(1/2)/GAMMA(2/3))

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maxima [A]  time = 1.21, size = 62, normalized size = 1.13 \[ \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{6} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.51, size = 80, normalized size = 1.45 \[ \frac {\ln \left ({\left (1-x^3\right )}^{1/3}-1\right )}{3}+\ln \left ({\left (1-x^3\right )}^{1/3}-9\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left ({\left (1-x^3\right )}^{1/3}-9\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(1 - x^3)^(1/3)),x)

[Out]

log((1 - x^3)^(1/3) - 1)/3 + log((1 - x^3)^(1/3) - 9*((3^(1/2)*1i)/6 - 1/6)^2)*((3^(1/2)*1i)/6 - 1/6) - log((1
- x^3)^(1/3) - 9*((3^(1/2)*1i)/6 + 1/6)^2)*((3^(1/2)*1i)/6 + 1/6)

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sympy [C]  time = 0.92, size = 32, normalized size = 0.58 \[ - \frac {e^{- \frac {i \pi }{3}} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {1}{x^{3}}} \right )}}{3 x \Gamma \left (\frac {4}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(-I*pi/3)*gamma(1/3)*hyper((1/3, 1/3), (4/3,), x**(-3))/(3*x*gamma(4/3))

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