\(\int \frac {1}{(-1+x^2) \sqrt [3]{-x+x^3}} \, dx\) [204]

Optimal result

Integrand size = 19, antiderivative size = 22 \[ \int \frac {1}{\left (-1+x^2\right ) \sqrt [3]{-x+x^3}} \, dx=-\frac {3 \left (-x+x^3\right )^{2/3}}{2 \left (-1+x^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2081, 270} \[ \int \frac {1}{\left (-1+x^2\right ) \sqrt [3]{-x+x^3}} \, dx=-\frac {3 x}{2 \sqrt [3]{x^3-x}} \]

[In]

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

[Out]

(-3*x)/(2*(-x + x^3)^(1/3))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 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]{x} \sqrt [3]{-1+x^2}\right ) \int \frac {1}{\sqrt [3]{x} \left (-1+x^2\right )^{4/3}} \, dx}{\sqrt [3]{-x+x^3}} \\ & = -\frac {3 x}{2 \sqrt [3]{-x+x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\left (-1+x^2\right ) \sqrt [3]{-x+x^3}} \, dx=-\frac {3 x}{2 \sqrt [3]{x \left (-1+x^2\right )}} \]

[In]

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

[Out]

(-3*x)/(2*(x*(-1 + x^2))^(1/3))

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59

method	result	size
gosper	\(-\frac {3 x}{2 \left (x^{3}-x \right )^{\frac {1}{3}}}\)	\(13\)
risch	\(-\frac {3 x}{2 {\left (x \left (x^{2}-1\right )\right )}^{\frac {1}{3}}}\)	\(13\)
pseudoelliptic	\(-\frac {3 x}{2 \left (x^{3}-x \right )^{\frac {1}{3}}}\)	\(13\)
trager	\(-\frac {3 \left (x^{3}-x \right )^{\frac {2}{3}}}{2 \left (x^{2}-1\right )}\)	\(19\)
meijerg	\(-\frac {3 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{3}} x^{\frac {2}{3}}}{2 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}}\)	\(33\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (-1+x^2\right ) \sqrt [3]{-x+x^3}} \, dx=-\frac {3 \, {\left (x^{3} - x\right )}^{\frac {2}{3}}}{2 \, {\left (x^{2} - 1\right )}} \]

[In]

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

[Out]

-3/2*(x^3 - x)^(2/3)/(x^2 - 1)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\left (-1+x^2\right ) \sqrt [3]{-x+x^3}} \, dx=-\frac {3}{2 \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}} \]

[In]

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

[Out]

-3/2/(-1/x^2 + 1)^(1/3)

Mupad [B] (verification not implemented)

Time = 5.39 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (-1+x^2\right ) \sqrt [3]{-x+x^3}} \, dx=-\frac {3\,{\left (x^3-x\right )}^{2/3}}{2\,\left (x^2-1\right )} \]

[In]

int(1/((x^3 - x)^(1/3)*(x^2 - 1)),x)

[Out]

-(3*(x^3 - x)^(2/3))/(2*(x^2 - 1))