\(\int \frac {1}{(-1+x) \sqrt [3]{-x^2+x^3}} \, dx\) [211]

Optimal result

Integrand size = 19, antiderivative size = 23 \[ \int \frac {1}{(-1+x) \sqrt [3]{-x^2+x^3}} \, dx=-\frac {3 \left (-x^2+x^3\right )^{2/3}}{(-1+x) x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2081, 37} \[ \int \frac {1}{(-1+x) \sqrt [3]{-x^2+x^3}} \, dx=-\frac {3 x}{\sqrt [3]{x^3-x^2}} \]

[In]

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

[Out]

(-3*x)/(-x^2 + x^3)^(1/3)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [3]{-1+x} x^{2/3}\right ) \int \frac {1}{(-1+x)^{4/3} x^{2/3}} \, dx}{\sqrt [3]{-x^2+x^3}} \\ & = -\frac {3 x}{\sqrt [3]{-x^2+x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(-1+x) \sqrt [3]{-x^2+x^3}} \, dx=-\frac {3 x}{\sqrt [3]{(-1+x) x^2}} \]

[In]

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

[Out]

(-3*x)/((-1 + x)*x^2)^(1/3)

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57

method	result	size
risch	\(-\frac {3 x}{\left (\left (x -1\right ) x^{2}\right )^{\frac {1}{3}}}\)	\(13\)
pseudoelliptic	\(-\frac {3 x}{\left (\left (x -1\right ) x^{2}\right )^{\frac {1}{3}}}\)	\(13\)
gosper	\(-\frac {3 x}{\left (x^{3}-x^{2}\right )^{\frac {1}{3}}}\)	\(15\)
trager	\(-\frac {3 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}}{\left (x -1\right ) x}\)	\(22\)
meijerg	\(-\frac {3 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{3}} x^{\frac {1}{3}}}{\operatorname {signum}\left (x -1\right )^{\frac {1}{3}} \left (1-x \right )^{\frac {1}{3}}}\)	\(27\)

[In]

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

[Out]

-3*x/((x-1)*x^2)^(1/3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(-1+x) \sqrt [3]{-x^2+x^3}} \, dx=-\frac {3 \, {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{x^{2} - x} \]

[In]

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

[Out]

-3*(x^3 - x^2)^(2/3)/(x^2 - x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.48 \[ \int \frac {1}{(-1+x) \sqrt [3]{-x^2+x^3}} \, dx=-\frac {3}{{\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}}} \]

[In]

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

[Out]

-3/(-1/x + 1)^(1/3)

Mupad [B] (verification not implemented)

Time = 5.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(-1+x) \sqrt [3]{-x^2+x^3}} \, dx=-\frac {3\,{\left (x^3-x^2\right )}^{2/3}}{x\,\left (x-1\right )} \]

[In]

int(1/((x^3 - x^2)^(1/3)*(x - 1)),x)

[Out]

-(3*(x^3 - x^2)^(2/3))/(x*(x - 1))