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

Optimal. Leaf size=73 \[ \frac{1}{3} \sqrt [3]{1-x^3} x^2-\frac{1}{6} \log \left (-\sqrt [3]{1-x^3}-x\right )-\frac{\tan ^{-1}\left (\frac{1-\frac{2 x}{\sqrt [3]{1-x^3}}}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

[Out]

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

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Rubi [A]  time = 0.0435833, antiderivative size = 107, normalized size of antiderivative = 1.47, number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {279, 331, 292, 31, 634, 618, 204, 628} \[ \frac{1}{3} \sqrt [3]{1-x^3} x^2+\frac{1}{18} \log \left (\frac{x^2}{\left (1-x^3\right )^{2/3}}-\frac{x}{\sqrt [3]{1-x^3}}+1\right )-\frac{1}{9} \log \left (\frac{x}{\sqrt [3]{1-x^3}}+1\right )-\frac{\tan ^{-1}\left (\frac{1-\frac{2 x}{\sqrt [3]{1-x^3}}}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[x*(1 - x^3)^(1/3),x]

[Out]

(x^2*(1 - x^3)^(1/3))/3 - ArcTan[(1 - (2*x)/(1 - x^3)^(1/3))/Sqrt[3]]/(3*Sqrt[3]) + Log[1 + x^2/(1 - x^3)^(2/3
) - x/(1 - x^3)^(1/3)]/18 - Log[1 + x/(1 - x^3)^(1/3)]/9

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, 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 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 31

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

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]

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 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 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]

Rubi steps

\begin{align*} \int x \sqrt [3]{1-x^3} \, dx &=\frac{1}{3} x^2 \sqrt [3]{1-x^3}+\frac{1}{3} \int \frac{x}{\left (1-x^3\right )^{2/3}} \, dx\\ &=\frac{1}{3} x^2 \sqrt [3]{1-x^3}+\frac{1}{3} \operatorname{Subst}\left (\int \frac{x}{1+x^3} \, dx,x,\frac{x}{\sqrt [3]{1-x^3}}\right )\\ &=\frac{1}{3} x^2 \sqrt [3]{1-x^3}-\frac{1}{9} \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\frac{x}{\sqrt [3]{1-x^3}}\right )+\frac{1}{9} \operatorname{Subst}\left (\int \frac{1+x}{1-x+x^2} \, dx,x,\frac{x}{\sqrt [3]{1-x^3}}\right )\\ &=\frac{1}{3} x^2 \sqrt [3]{1-x^3}-\frac{1}{9} \log \left (1+\frac{x}{\sqrt [3]{1-x^3}}\right )+\frac{1}{18} \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,\frac{x}{\sqrt [3]{1-x^3}}\right )+\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,\frac{x}{\sqrt [3]{1-x^3}}\right )\\ &=\frac{1}{3} x^2 \sqrt [3]{1-x^3}+\frac{1}{18} \log \left (1+\frac{x^2}{\left (1-x^3\right )^{2/3}}-\frac{x}{\sqrt [3]{1-x^3}}\right )-\frac{1}{9} \log \left (1+\frac{x}{\sqrt [3]{1-x^3}}\right )-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+\frac{2 x}{\sqrt [3]{1-x^3}}\right )\\ &=\frac{1}{3} x^2 \sqrt [3]{1-x^3}-\frac{\tan ^{-1}\left (\frac{1-\frac{2 x}{\sqrt [3]{1-x^3}}}{\sqrt{3}}\right )}{3 \sqrt{3}}+\frac{1}{18} \log \left (1+\frac{x^2}{\left (1-x^3\right )^{2/3}}-\frac{x}{\sqrt [3]{1-x^3}}\right )-\frac{1}{9} \log \left (1+\frac{x}{\sqrt [3]{1-x^3}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0026731, size = 20, normalized size = 0.27 \[ \frac{1}{2} x^2 \, _2F_1\left (-\frac{1}{3},\frac{2}{3};\frac{5}{3};x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*(1 - x^3)^(1/3),x]

[Out]

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

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Maple [C]  time = 0.046, size = 69, normalized size = 1. \begin{align*} -{\frac{{x}^{2} \left ({x}^{3}-1 \right ) }{3} \left ( -{x}^{3}+1 \right ) ^{-{\frac{2}{3}}}}+{\frac{{x}^{2}}{6} \left ({x}^{3}-1 \right ) ^{{\frac{2}{3}}} \left ( -{\it signum} \left ({x}^{3}-1 \right ) \right ) ^{{\frac{2}{3}}}{\mbox{$_2$F$_1$}({\frac{2}{3}},{\frac{2}{3}};\,{\frac{5}{3}};\,{x}^{3})} \left ({\it signum} \left ({x}^{3}-1 \right ) \right ) ^{-{\frac{2}{3}}} \left ( -{x}^{3}+1 \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^2*(x^3-1)/(-x^3+1)^(2/3)+1/6*(x^3-1)^(2/3)/signum(x^3-1)^(2/3)*(-signum(x^3-1))^(2/3)*x^2*hypergeom([2/
3,2/3],[5/3],x^3)/(-x^3+1)^(2/3)

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Maxima [A]  time = 1.42811, size = 142, normalized size = 1.95 \begin{align*} -\frac{1}{9} \, \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{{\left (-x^{3} + 1\right )}^{\frac{1}{3}}}{3 \, x{\left (\frac{x^{3} - 1}{x^{3}} - 1\right )}} - \frac{1}{9} \, \log \left (\frac{{\left (-x^{3} + 1\right )}^{\frac{1}{3}}}{x} + 1\right ) + \frac{1}{18} \, \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-x^3 + 1)^(1/3)/x - 1)) - 1/3*(-x^3 + 1)^(1/3)/(x*((x^3 - 1)/x^3 - 1)) - 1
/9*log((-x^3 + 1)^(1/3)/x + 1) + 1/18*log(-(-x^3 + 1)^(1/3)/x + (-x^3 + 1)^(2/3)/x^2 + 1)

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Fricas [A]  time = 2.05284, size = 262, normalized size = 3.59 \begin{align*} \frac{1}{3} \,{\left (-x^{3} + 1\right )}^{\frac{1}{3}} x^{2} - \frac{1}{9} \, \sqrt{3} \arctan \left (-\frac{\sqrt{3} x - 2 \, \sqrt{3}{\left (-x^{3} + 1\right )}^{\frac{1}{3}}}{3 \, x}\right ) - \frac{1}{9} \, \log \left (\frac{x +{\left (-x^{3} + 1\right )}^{\frac{1}{3}}}{x}\right ) + \frac{1}{18} \, \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-x^3 + 1)^(1/3)*x^2 - 1/9*sqrt(3)*arctan(-1/3*(sqrt(3)*x - 2*sqrt(3)*(-x^3 + 1)^(1/3))/x) - 1/9*log((x +
(-x^3 + 1)^(1/3))/x) + 1/18*log((x^2 - (-x^3 + 1)^(1/3)*x + (-x^3 + 1)^(2/3))/x^2)

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Sympy [C]  time = 1.10618, size = 32, normalized size = 0.44 \begin{align*} \frac{x^{2} \Gamma \left (\frac{2}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{3}, \frac{2}{3} \\ \frac{5}{3} \end{matrix}\middle |{x^{3} e^{2 i \pi }} \right )}}{3 \Gamma \left (\frac{5}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*gamma(2/3)*hyper((-1/3, 2/3), (5/3,), x**3*exp_polar(2*I*pi))/(3*gamma(5/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-x^{3} + 1\right )}^{\frac{1}{3}} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-x^3 + 1)^(1/3)*x, x)