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

Optimal. Leaf size=91 \[ -\log \left (\sqrt [3]{x^3-x^2}-x\right )+\frac {1}{2} \log \left (x^2+\sqrt [3]{x^3-x^2} x+\left (x^3-x^2\right )^{2/3}\right )+\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-x^2}+x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 135, normalized size of antiderivative = 1.48, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2011, 59} \begin {gather*} -\frac {3 \sqrt [3]{x-1} x^{2/3} \log \left (\frac {\sqrt [3]{x-1}}{\sqrt [3]{x}}-1\right )}{2 \sqrt [3]{x^3-x^2}}-\frac {\sqrt [3]{x-1} x^{2/3} \log (x)}{2 \sqrt [3]{x^3-x^2}}-\frac {\sqrt {3} \sqrt [3]{x-1} x^{2/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x-1}}{\sqrt {3} \sqrt [3]{x}}+\frac {1}{\sqrt {3}}\right )}{\sqrt [3]{x^3-x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x^2 + x^3)^(-1/3),x]

[Out]

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

Rule 59

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, -Simp[(Sqrt
[3]*q*ArcTan[(2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/3)) + 1/Sqrt[3]])/d, x] + (-Simp[(3*q*Log[(q*(a + b*x
)^(1/3))/(c + d*x)^(1/3) - 1])/(2*d), x] - Simp[(q*Log[c + d*x])/(2*d), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[
b*c - a*d, 0] && PosQ[d/b]

Rule 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.38 \begin {gather*} \frac {3 \left ((x-1) x^2\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};1-x\right )}{2 x^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^2 + x^3)^(-1/3),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.16, size = 91, normalized size = 1.00 \begin {gather*} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-x^2+x^3}}\right )-\log \left (-x+\sqrt [3]{-x^2+x^3}\right )+\frac {1}{2} \log \left (x^2+x \sqrt [3]{-x^2+x^3}+\left (-x^2+x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-x^2 + x^3)^(-1/3),x]

[Out]

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

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fricas [A]  time = 0.49, size = 92, normalized size = 1.01 \begin {gather*} -\sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) - \log \left (-\frac {x - {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} x + {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 63, normalized size = 0.69 \begin {gather*} -\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{2} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) - \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.61, size = 27, normalized size = 0.30

method	result	size
meijerg	\(\frac {3 \left (-\mathrm {signum}\left (-1+x \right )\right )^{\frac {1}{3}} x^{\frac {1}{3}} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], x\right )}{\mathrm {signum}\left (-1+x \right )^{\frac {1}{3}}}\)	\(27\)
trager	\(-\ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{2}+48 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {2}{3}}-30 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {1}{3}} x +2 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x -16 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}-36 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+96 x \left (x^{3}-x^{2}\right )^{\frac {1}{3}}+18 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x -64 x^{2}+16 x}{x}\right )+\frac {\RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \ln \left (-\frac {2 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{2}+24 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {2}{3}}-9 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {1}{3}} x -4 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x -19 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}-30 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+48 x \left (x^{3}-x^{2}\right )^{\frac {1}{3}}+10 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x -10 x^{2}+6 x}{x}\right )}{2}\)	\(300\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3-x^2)^(1/3),x,method=_RETURNVERBOSE)

[Out]

3/signum(-1+x)^(1/3)*(-signum(-1+x))^(1/3)*x^(1/3)*hypergeom([1/3,1/3],[4/3],x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^3 - x^2)^(-1/3), x)

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mupad [B]  time = 0.98, size = 27, normalized size = 0.30 \begin {gather*} \frac {3\,x\,{\left (1-x\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ x\right )}{{\left (x^3-x^2\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3 - x^2)^(1/3),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [3]{x^{3} - x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**3 - x**2)**(-1/3), x)

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