\(\int \frac {1}{\sqrt [3]{-x^2+x^3}} \, dx\) [1252]

Optimal result

Integrand size = 13, antiderivative size = 91 \[ \int \frac {1}{\sqrt [3]{-x^2+x^3}} \, dx=\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-x^2+x^3}}\right )-\log \left (-x+\sqrt [3]{-x^2+x^3}\right )+\frac {1}{2} \log \left (x^2+x \sqrt [3]{-x^2+x^3}+\left (-x^2+x^3\right )^{2/3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.48, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2036, 61} \[ \int \frac {1}{\sqrt [3]{-x^2+x^3}} \, dx=-\frac {\sqrt {3} \sqrt [3]{x-1} x^{2/3} \arctan \left (\frac {2 \sqrt [3]{x-1}}{\sqrt {3} \sqrt [3]{x}}+\frac {1}{\sqrt {3}}\right )}{\sqrt [3]{x^3-x^2}}-\frac {3 \sqrt [3]{x-1} x^{2/3} \log \left (\frac {\sqrt [3]{x-1}}{\sqrt [3]{x}}-1\right )}{2 \sqrt [3]{x^3-x^2}}-\frac {\sqrt [3]{x-1} x^{2/3} \log (x)}{2 \sqrt [3]{x^3-x^2}} \]

[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 61

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

Rule 2036

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*} \text {integral}& = \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} \arctan \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*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.20 \[ \int \frac {1}{\sqrt [3]{-x^2+x^3}} \, dx=\frac {\sqrt [3]{-1+x} x^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{2 \sqrt [3]{-1+x}+\sqrt [3]{x}}\right )-2 \log \left (\sqrt [3]{-1+x}-\sqrt [3]{x}\right )+\log \left ((-1+x)^{2/3}+\sqrt [3]{-1+x} \sqrt [3]{x}+x^{2/3}\right )\right )}{2 \sqrt [3]{(-1+x) x^2}} \]

[In]

Integrate[(-x^2 + x^3)^(-1/3),x]

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 3.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.30

method	result	size
meijerg	\(\frac {3 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{3}} x^{\frac {1}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], x\right )}{\operatorname {signum}\left (x -1\right )^{\frac {1}{3}}}\)	\(27\)
pseudoelliptic	\(-\ln \left (\frac {\left (\left (x -1\right ) x^{2}\right )^{\frac {1}{3}}-x}{x}\right )+\frac {\ln \left (\frac {\left (\left (x -1\right ) x^{2}\right )^{\frac {2}{3}}+\left (\left (x -1\right ) x^{2}\right )^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\left (2 \left (\left (x -1\right ) x^{2}\right )^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )\)	\(80\)
trager	\(-\ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x -48 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+30 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {1}{3}} x +16 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}-18 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x +36 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}-96 x \left (x^{3}-x^{2}\right )^{\frac {1}{3}}+64 x^{2}-16 x}{x}\right )+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x +24 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {2}{3}}-9 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{3}-x^{2}\right )^{\frac {1}{3}} x -19 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}+10 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x -30 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}+48 x \left (x^{3}-x^{2}\right )^{\frac {1}{3}}-10 x^{2}+6 x}{x}\right )}{2}\)	\(300\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\sqrt [3]{-x^2+x^3}} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) - \log \left (-\frac {x - {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} x + {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]

[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)

Sympy [F]

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

[In]

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

[Out]

Integral((x**3 - x**2)**(-1/3), x)

Maxima [F]

\[ \int \frac {1}{\sqrt [3]{-x^2+x^3}} \, dx=\int { \frac {1}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}} \,d x } \]

[In]

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

[Out]

integrate((x^3 - x^2)^(-1/3), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\sqrt [3]{-x^2+x^3}} \, dx=-\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{2} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) - \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 6.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.30 \[ \int \frac {1}{\sqrt [3]{-x^2+x^3}} \, dx=\frac {3\,x\,{\left (1-x\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ x\right )}{{\left (x^3-x^2\right )}^{1/3}} \]

[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)