\(\int \frac {1}{x \sqrt [3]{-x^2+x^3}} \, dx\) [169]

Optimal result

Integrand size = 17, antiderivative size = 20 \[ \int \frac {1}{x \sqrt [3]{-x^2+x^3}} \, dx=\frac {3 \left (-x^2+x^3\right )^{2/3}}{2 x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2039} \[ \int \frac {1}{x \sqrt [3]{-x^2+x^3}} \, dx=\frac {3 \left (x^3-x^2\right )^{2/3}}{2 x^2} \]

[In]

Int[1/(x*(-x^2 + x^3)^(1/3)),x]

[Out]

(3*(-x^2 + x^3)^(2/3))/(2*x^2)

Rule 2039

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

Rubi steps \begin{align*} \text {integral}& = \frac {3 \left (-x^2+x^3\right )^{2/3}}{2 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x \sqrt [3]{-x^2+x^3}} \, dx=\frac {3 (-1+x)}{2 \sqrt [3]{(-1+x) x^2}} \]

[In]

Integrate[1/(x*(-x^2 + x^3)^(1/3)),x]

[Out]

(3*(-1 + x))/(2*((-1 + x)*x^2)^(1/3))

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {\frac {3 x}{2}-\frac {3}{2}}{\left (\left (x -1\right ) x^{2}\right )^{\frac {1}{3}}}\)	\(15\)
pseudoelliptic	\(\frac {3 \left (\left (x -1\right ) x^{2}\right )^{\frac {2}{3}}}{2 x^{2}}\)	\(15\)
gosper	\(\frac {\frac {3 x}{2}-\frac {3}{2}}{\left (x^{3}-x^{2}\right )^{\frac {1}{3}}}\)	\(17\)
trager	\(\frac {3 \left (x^{3}-x^{2}\right )^{\frac {2}{3}}}{2 x^{2}}\)	\(17\)
meijerg	\(-\frac {3 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{3}} \left (1-x \right )^{\frac {2}{3}}}{2 \operatorname {signum}\left (x -1\right )^{\frac {1}{3}} x^{\frac {2}{3}}}\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x \sqrt [3]{-x^2+x^3}} \, dx=\frac {3 \, {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{2 \, x^{2}} \]

[In]

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

[Out]

3/2*(x^3 - x^2)^(2/3)/x^2

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate(1/((x^3 - x^2)^(1/3)*x), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55 \[ \int \frac {1}{x \sqrt [3]{-x^2+x^3}} \, dx=\frac {3}{2} \, {\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}} \]

[In]

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

[Out]

3/2*(-1/x + 1)^(2/3)

Mupad [B] (verification not implemented)

Time = 5.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x \sqrt [3]{-x^2+x^3}} \, dx=\frac {3\,{\left (x^3-x^2\right )}^{2/3}}{2\,x^2} \]

[In]

int(1/(x*(x^3 - x^2)^(1/3)),x)

[Out]

(3*(x^3 - x^2)^(2/3))/(2*x^2)