\(\int \frac {1+x^2}{(-1+x^2) \sqrt [3]{-x^2+x^4}} \, dx\) [262]

Optimal result

Integrand size = 26, antiderivative size = 25 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=-\frac {3 \left (-x^2+x^4\right )^{2/3}}{x \left (-1+x^2\right )} \]

[Out]

-3*(x^4-x^2)^(2/3)/x/(x^2-1)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2081, 460} \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=-\frac {3 x}{\sqrt [3]{x^4-x^2}} \]

[In]

Int[(1 + x^2)/((-1 + x^2)*(-x^2 + x^4)^(1/3)),x]

[Out]

(-3*x)/(-x^2 + x^4)^(1/3)

Rule 460

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[
a*d*(m + 1) - b*c*(m + n*(p + 1) + 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 (x^{2/3} \sqrt [3]{-1+x^2}\right ) \int \frac {1+x^2}{x^{2/3} \left (-1+x^2\right )^{4/3}} \, dx}{\sqrt [3]{-x^2+x^4}} \\ & = -\frac {3 x}{\sqrt [3]{-x^2+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=-\frac {3 x}{\sqrt [3]{x^2 \left (-1+x^2\right )}} \]

[In]

Integrate[(1 + x^2)/((-1 + x^2)*(-x^2 + x^4)^(1/3)),x]

[Out]

(-3*x)/(x^2*(-1 + x^2))^(1/3)

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.60

method	result	size
gosper	\(-\frac {3 x}{\left (x^{4}-x^{2}\right )^{\frac {1}{3}}}\)	\(15\)
risch	\(-\frac {3 x}{\left (x^{2} \left (x^{2}-1\right )\right )^{\frac {1}{3}}}\)	\(15\)
pseudoelliptic	\(-\frac {3 x}{\left (x^{4}-x^{2}\right )^{\frac {1}{3}}}\)	\(15\)
trager	\(-\frac {3 \left (x^{4}-x^{2}\right )^{\frac {2}{3}}}{x \left (x^{2}-1\right )}\)	\(24\)
meijerg	\(-\frac {3 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{3}} x^{\frac {1}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{6}, \frac {4}{3}\right ], \left [\frac {7}{6}\right ], x^{2}\right )}{\operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{3}}}-\frac {3 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{3}} x^{\frac {7}{3}} \operatorname {hypergeom}\left (\left [\frac {7}{6}, \frac {4}{3}\right ], \left [\frac {13}{6}\right ], x^{2}\right )}{7 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{3}}}\)	\(66\)

[In]

int((x^2+1)/(x^2-1)/(x^4-x^2)^(1/3),x,method=_RETURNVERBOSE)

[Out]

-3*x/(x^4-x^2)^(1/3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=-\frac {3 \, {\left (x^{4} - x^{2}\right )}^{\frac {2}{3}}}{x^{3} - x} \]

[In]

integrate((x^2+1)/(x^2-1)/(x^4-x^2)^(1/3),x, algorithm="fricas")

[Out]

-3*(x^4 - x^2)^(2/3)/(x^3 - x)

Sympy [F]

\[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=\int \frac {x^{2} + 1}{\sqrt [3]{x^{2} \left (x - 1\right ) \left (x + 1\right )} \left (x - 1\right ) \left (x + 1\right )}\, dx \]

[In]

integrate((x**2+1)/(x**2-1)/(x**4-x**2)**(1/3),x)

[Out]

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

Maxima [F]

\[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=\int { \frac {x^{2} + 1}{{\left (x^{4} - x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )}} \,d x } \]

[In]

integrate((x^2+1)/(x^2-1)/(x^4-x^2)^(1/3),x, algorithm="maxima")

[Out]

integrate((x^2 + 1)/((x^4 - x^2)^(1/3)*(x^2 - 1)), x)

Giac [F]

\[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=\int { \frac {x^{2} + 1}{{\left (x^{4} - x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )}} \,d x } \]

[In]

integrate((x^2+1)/(x^2-1)/(x^4-x^2)^(1/3),x, algorithm="giac")

[Out]

integrate((x^2 + 1)/((x^4 - x^2)^(1/3)*(x^2 - 1)), x)

Mupad [B] (verification not implemented)

Time = 5.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=-\frac {3\,{\left (x^4-x^2\right )}^{2/3}}{x\,\left (x^2-1\right )} \]

[In]

int((x^2 + 1)/((x^2 - 1)*(x^4 - x^2)^(1/3)),x)

[Out]

-(3*(x^4 - x^2)^(2/3))/(x*(x^2 - 1))