3.581 \(\int \frac {x}{(1-x^3)^{2/3}} \, dx\)

Optimal. Leaf size=53 \[ -\frac {1}{2} \log \left (-\sqrt [3]{1-x^3}-x\right )-\frac {\tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 87, normalized size of antiderivative = 1.64, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {331, 292, 31, 634, 618, 204, 628} \[ \frac {1}{6} \log \left (\frac {x^2}{\left (1-x^3\right )^{2/3}}-\frac {x}{\sqrt [3]{1-x^3}}+1\right )-\frac {1}{3} \log \left (\frac {x}{\sqrt [3]{1-x^3}}+1\right )-\frac {\tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[x/(1 - x^3)^(2/3),x]

[Out]

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

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (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]

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

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Mathematica [C]  time = 0.01, size = 37, normalized size = 0.70 \[ \frac {x^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {x^3}{x^3-1}\right )}{2 \left (1-x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(1 - x^3)^(2/3),x]

[Out]

(x^2*Hypergeometric2F1[2/3, 1, 5/3, x^3/(-1 + x^3)])/(2*(1 - x^3)^(2/3))

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fricas [A]  time = 0.71, size = 82, normalized size = 1.55 \[ -\frac {1}{3} \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} x - 2 \, \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{3 \, x}\right ) - \frac {1}{3} \, \log \left (\frac {x + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x}\right ) + \frac {1}{6} \, \log \left (\frac {x^{2} - {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{{\left (-x^{3} + 1\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(-x^3 + 1)^(2/3), x)

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maple [C]  time = 0.11, size = 15, normalized size = 0.28 \[ \frac {x^{2} \hypergeom \left (\left [\frac {2}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x^{3}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(-x^3+1)^(2/3),x)

[Out]

1/2*x^2*hypergeom([2/3,2/3],[5/3],x^3)

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maxima [A]  time = 2.89, size = 78, normalized size = 1.47 \[ -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x} - 1\right )}\right ) - \frac {1}{3} \, \log \left (\frac {{\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x} + 1\right ) + \frac {1}{6} \, \log \left (-\frac {{\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x}{{\left (1-x^3\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(1 - x^3)^(2/3),x)

[Out]

int(x/(1 - x^3)^(2/3), x)

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sympy [C]  time = 2.01, size = 31, normalized size = 0.58 \[ \frac {x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {x^{3} e^{2 i \pi }} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*gamma(2/3)*hyper((2/3, 2/3), (5/3,), x**3*exp_polar(2*I*pi))/(3*gamma(5/3))

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