\(\int \frac {1}{x^2 \sqrt [3]{-x+x^3}} \, dx\) [123]

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int \frac {1}{x^2 \sqrt [3]{-x+x^3}} \, dx=\frac {3 \left (-x+x^3\right )^{2/3}}{4 x^2} \]

[Out]

3/4*(x^3-x)^(2/3)/x^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2039} \[ \int \frac {1}{x^2 \sqrt [3]{-x+x^3}} \, dx=\frac {3 \left (x^3-x\right )^{2/3}}{4 x^2} \]

[In]

Int[1/(x^2*(-x + x^3)^(1/3)),x]

[Out]

(3*(-x + x^3)^(2/3))/(4*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+x^3\right )^{2/3}}{4 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \sqrt [3]{-x+x^3}} \, dx=\frac {3 \left (x \left (-1+x^2\right )\right )^{2/3}}{4 x^2} \]

[In]

Integrate[1/(x^2*(-x + x^3)^(1/3)),x]

[Out]

(3*(x*(-1 + x^2))^(2/3))/(4*x^2)

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
trager	\(\frac {3 \left (x^{3}-x \right )^{\frac {2}{3}}}{4 x^{2}}\)	\(15\)
pseudoelliptic	\(\frac {3 \left (x^{3}-x \right )^{\frac {2}{3}}}{4 x^{2}}\)	\(15\)
risch	\(\frac {\frac {3 x^{2}}{4}-\frac {3}{4}}{x {\left (x \left (x^{2}-1\right )\right )}^{\frac {1}{3}}}\)	\(20\)
gosper	\(\frac {3 \left (1+x \right ) \left (x -1\right )}{4 x \left (x^{3}-x \right )^{\frac {1}{3}}}\)	\(21\)
meijerg	\(-\frac {3 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{3}} \left (-x^{2}+1\right )^{\frac {2}{3}}}{4 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{3}} x^{\frac {4}{3}}}\)	\(33\)

[In]

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

[Out]

3/4*(x^3-x)^(2/3)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^2 \sqrt [3]{-x+x^3}} \, dx=\frac {3 \, {\left (x^{3} - x\right )}^{\frac {2}{3}}}{4 \, x^{2}} \]

[In]

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

[Out]

3/4*(x^3 - x)^(2/3)/x^2

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^2 \sqrt [3]{-x+x^3}} \, dx=\frac {3 \, {\left (x^{3} - x\right )}}{4 \, {\left (x + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )}^{\frac {1}{3}} x^{\frac {7}{3}}} \]

[In]

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

[Out]

3/4*(x^3 - x)/((x + 1)^(1/3)*(x - 1)^(1/3)*x^(7/3))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x^2 \sqrt [3]{-x+x^3}} \, dx=\frac {3}{4} \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} \]

[In]

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

[Out]

3/4*(-1/x^2 + 1)^(2/3)

Mupad [B] (verification not implemented)

Time = 4.95 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^2 \sqrt [3]{-x+x^3}} \, dx=\frac {3\,{\left (x^3-x\right )}^{2/3}}{4\,x^2} \]

[In]

int(1/(x^2*(x^3 - x)^(1/3)),x)

[Out]

(3*(x^3 - x)^(2/3))/(4*x^2)