3.306 \(\int \sqrt [3]{x (1-x^2)} \, dx\)

Optimal. Leaf size=93 \[ \frac{1}{2} \sqrt [3]{x \left (1-x^2\right )} x-\frac{1}{4} \log \left (\sqrt [3]{x \left (1-x^2\right )}+x\right )+\frac{\tan ^{-1}\left (\frac{2 x-\sqrt [3]{x \left (1-x^2\right )}}{\sqrt{3} \sqrt [3]{x \left (1-x^2\right )}}\right )}{2 \sqrt{3}}+\frac{\log (x)}{12} \]

[Out]

(x*(x*(1 - x^2))^(1/3))/2 + ArcTan[(2*x - (x*(1 - x^2))^(1/3))/(Sqrt[3]*(x*(1 - x^2))^(1/3))]/(2*Sqrt[3]) + Lo
g[x]/12 - Log[x + (x*(1 - x^2))^(1/3)]/4

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Rubi [B]  time = 0.143999, antiderivative size = 200, normalized size of antiderivative = 2.15, number of steps used = 12, number of rules used = 12, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.923, Rules used = {1979, 2004, 2032, 329, 275, 331, 292, 31, 634, 618, 204, 628} \[ \frac{1}{2} \sqrt [3]{x-x^3} x+\frac{\left (1-x^2\right )^{2/3} x^{2/3} \log \left (\frac{x^{4/3}}{\left (1-x^2\right )^{2/3}}-\frac{x^{2/3}}{\sqrt [3]{1-x^2}}+1\right )}{12 \left (x-x^3\right )^{2/3}}-\frac{\left (1-x^2\right )^{2/3} x^{2/3} \log \left (\frac{x^{2/3}}{\sqrt [3]{1-x^2}}+1\right )}{6 \left (x-x^3\right )^{2/3}}-\frac{\left (1-x^2\right )^{2/3} x^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 x^{2/3}}{\sqrt [3]{1-x^2}}}{\sqrt{3}}\right )}{2 \sqrt{3} \left (x-x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(x*(1 - x^2))^(1/3),x]

[Out]

(x*(x - x^3)^(1/3))/2 - (x^(2/3)*(1 - x^2)^(2/3)*ArcTan[(1 - (2*x^(2/3))/(1 - x^2)^(1/3))/Sqrt[3]])/(2*Sqrt[3]
*(x - x^3)^(2/3)) + (x^(2/3)*(1 - x^2)^(2/3)*Log[1 + x^(4/3)/(1 - x^2)^(2/3) - x^(2/3)/(1 - x^2)^(1/3)])/(12*(
x - x^3)^(2/3)) - (x^(2/3)*(1 - x^2)^(2/3)*Log[1 + x^(2/3)/(1 - x^2)^(1/3)])/(6*(x - x^3)^(2/3))

Rule 1979

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedBinomialQ[u, x] &&  !Gene
ralizedBinomialMatchQ[u, x]

Rule 2004

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(x*(a*x^j + b*x^n)^p)/(n*p + 1), x] + Dist[(
a*(n - j)*p)/(n*p + 1), Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0,
j, n] && GtQ[p, 0] && NeQ[n*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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 \sqrt [3]{x \left (1-x^2\right )} \, dx &=\int \sqrt [3]{x-x^3} \, dx\\ &=\frac{1}{2} x \sqrt [3]{x-x^3}+\frac{1}{3} \int \frac{x}{\left (x-x^3\right )^{2/3}} \, dx\\ &=\frac{1}{2} x \sqrt [3]{x-x^3}+\frac{\left (x^{2/3} \left (1-x^2\right )^{2/3}\right ) \int \frac{\sqrt [3]{x}}{\left (1-x^2\right )^{2/3}} \, dx}{3 \left (x-x^3\right )^{2/3}}\\ &=\frac{1}{2} x \sqrt [3]{x-x^3}+\frac{\left (x^{2/3} \left (1-x^2\right )^{2/3}\right ) \operatorname{Subst}\left (\int \frac{x^3}{\left (1-x^6\right )^{2/3}} \, dx,x,\sqrt [3]{x}\right )}{\left (x-x^3\right )^{2/3}}\\ &=\frac{1}{2} x \sqrt [3]{x-x^3}+\frac{\left (x^{2/3} \left (1-x^2\right )^{2/3}\right ) \operatorname{Subst}\left (\int \frac{x}{\left (1-x^3\right )^{2/3}} \, dx,x,x^{2/3}\right )}{2 \left (x-x^3\right )^{2/3}}\\ &=\frac{1}{2} x \sqrt [3]{x-x^3}+\frac{\left (x^{2/3} \left (1-x^2\right )^{2/3}\right ) \operatorname{Subst}\left (\int \frac{x}{1+x^3} \, dx,x,\frac{x^{2/3}}{\sqrt [3]{1-x^2}}\right )}{2 \left (x-x^3\right )^{2/3}}\\ &=\frac{1}{2} x \sqrt [3]{x-x^3}-\frac{\left (x^{2/3} \left (1-x^2\right )^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\frac{x^{2/3}}{\sqrt [3]{1-x^2}}\right )}{6 \left (x-x^3\right )^{2/3}}+\frac{\left (x^{2/3} \left (1-x^2\right )^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1+x}{1-x+x^2} \, dx,x,\frac{x^{2/3}}{\sqrt [3]{1-x^2}}\right )}{6 \left (x-x^3\right )^{2/3}}\\ &=\frac{1}{2} x \sqrt [3]{x-x^3}-\frac{x^{2/3} \left (1-x^2\right )^{2/3} \log \left (1+\frac{x^{2/3}}{\sqrt [3]{1-x^2}}\right )}{6 \left (x-x^3\right )^{2/3}}+\frac{\left (x^{2/3} \left (1-x^2\right )^{2/3}\right ) \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,\frac{x^{2/3}}{\sqrt [3]{1-x^2}}\right )}{12 \left (x-x^3\right )^{2/3}}+\frac{\left (x^{2/3} \left (1-x^2\right )^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,\frac{x^{2/3}}{\sqrt [3]{1-x^2}}\right )}{4 \left (x-x^3\right )^{2/3}}\\ &=\frac{1}{2} x \sqrt [3]{x-x^3}+\frac{x^{2/3} \left (1-x^2\right )^{2/3} \log \left (1+\frac{x^{4/3}}{\left (1-x^2\right )^{2/3}}-\frac{x^{2/3}}{\sqrt [3]{1-x^2}}\right )}{12 \left (x-x^3\right )^{2/3}}-\frac{x^{2/3} \left (1-x^2\right )^{2/3} \log \left (1+\frac{x^{2/3}}{\sqrt [3]{1-x^2}}\right )}{6 \left (x-x^3\right )^{2/3}}-\frac{\left (x^{2/3} \left (1-x^2\right )^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+\frac{2 x^{2/3}}{\sqrt [3]{1-x^2}}\right )}{2 \left (x-x^3\right )^{2/3}}\\ &=\frac{1}{2} x \sqrt [3]{x-x^3}-\frac{x^{2/3} \left (1-x^2\right )^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 x^{2/3}}{\sqrt [3]{1-x^2}}}{\sqrt{3}}\right )}{2 \sqrt{3} \left (x-x^3\right )^{2/3}}+\frac{x^{2/3} \left (1-x^2\right )^{2/3} \log \left (1+\frac{x^{4/3}}{\left (1-x^2\right )^{2/3}}-\frac{x^{2/3}}{\sqrt [3]{1-x^2}}\right )}{12 \left (x-x^3\right )^{2/3}}-\frac{x^{2/3} \left (1-x^2\right )^{2/3} \log \left (1+\frac{x^{2/3}}{\sqrt [3]{1-x^2}}\right )}{6 \left (x-x^3\right )^{2/3}}\\ \end{align*}

Mathematica [C]  time = 0.0088829, size = 40, normalized size = 0.43 \[ \frac{3 x \sqrt [3]{x-x^3} \, _2F_1\left (-\frac{1}{3},\frac{2}{3};\frac{5}{3};x^2\right )}{4 \sqrt [3]{1-x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(1 - x^2))^(1/3),x]

[Out]

(3*x*(x - x^3)^(1/3)*Hypergeometric2F1[-1/3, 2/3, 5/3, x^2])/(4*(1 - x^2)^(1/3))

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Maple [C]  time = 0.026, size = 15, normalized size = 0.2 \begin{align*}{\frac{3}{4}{x}^{{\frac{4}{3}}}{\mbox{$_2$F$_1$}(-{\frac{1}{3}},{\frac{2}{3}};\,{\frac{5}{3}};\,{x}^{2})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*x^(4/3)*hypergeom([-1/3,2/3],[5/3],x^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (-{\left (x^{2} - 1\right )} x\right )^{\frac{1}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-(x^2 - 1)*x)^(1/3), x)

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Fricas [A]  time = 3.46897, size = 347, normalized size = 3.73 \begin{align*} -\frac{1}{6} \, \sqrt{3} \arctan \left (\frac{44032959556 \, \sqrt{3}{\left (-x^{3} + x\right )}^{\frac{1}{3}} x - \sqrt{3}{\left (16754327161 \, x^{2} - 2707204793\right )} + 10524305234 \, \sqrt{3}{\left (-x^{3} + x\right )}^{\frac{2}{3}}}{81835897185 \, x^{2} - 1102302937}\right ) + \frac{1}{2} \,{\left (-x^{3} + x\right )}^{\frac{1}{3}} x - \frac{1}{12} \, \log \left (3 \,{\left (-x^{3} + x\right )}^{\frac{1}{3}} x + 3 \,{\left (-x^{3} + x\right )}^{\frac{2}{3}} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3)*arctan((44032959556*sqrt(3)*(-x^3 + x)^(1/3)*x - sqrt(3)*(16754327161*x^2 - 2707204793) + 1052430
5234*sqrt(3)*(-x^3 + x)^(2/3))/(81835897185*x^2 - 1102302937)) + 1/2*(-x^3 + x)^(1/3)*x - 1/12*log(3*(-x^3 + x
)^(1/3)*x + 3*(-x^3 + x)^(2/3) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{x \left (1 - x^{2}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x*(1 - x**2))**(1/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (-{\left (x^{2} - 1\right )} x\right )^{\frac{1}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-(x^2 - 1)*x)^(1/3), x)