\(\int \frac {3+x}{(-1+x)^2 \sqrt [3]{-1+x^2}} \, dx\) [121]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {801} \[ \int \frac {3+x}{(-1+x)^2 \sqrt [3]{-1+x^2}} \, dx=-\frac {3 \left (x^2-1\right )^{2/3}}{2 (1-x)^2} \]

[In]

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

[Out]

(-3*(-1 + x^2)^(2/3))/(2*(1 - x)^2)

Rule 801

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*
((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&
EqQ[m*(d*g + e*f) + 2*e*f*(p + 1), 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 \left (-1+x^2\right )^{2/3}}{2 (1-x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {3+x}{(-1+x)^2 \sqrt [3]{-1+x^2}} \, dx=-\frac {3 \left (-1+x^2\right )^{2/3}}{2 (-1+x)^2} \]

[In]

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

[Out]

(-3*(-1 + x^2)^(2/3))/(2*(-1 + x)^2)

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
trager	\(-\frac {3 \left (x^{2}-1\right )^{\frac {2}{3}}}{2 \left (x -1\right )^{2}}\)	\(15\)
gosper	\(-\frac {3 \left (1+x \right )}{2 \left (x -1\right ) \left (x^{2}-1\right )^{\frac {1}{3}}}\)	\(18\)
risch	\(-\frac {3 \left (1+x \right )}{2 \left (x -1\right ) \left (x^{2}-1\right )^{\frac {1}{3}}}\)	\(18\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {3+x}{(-1+x)^2 \sqrt [3]{-1+x^2}} \, dx=-\frac {3 \, {\left (x^{2} - 1\right )}^{\frac {2}{3}}}{2 \, {\left (x^{2} - 2 \, x + 1\right )}} \]

[In]

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

[Out]

-3/2*(x^2 - 1)^(2/3)/(x^2 - 2*x + 1)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {3+x}{(-1+x)^2 \sqrt [3]{-1+x^2}} \, dx=-\frac {3 \, {\left (x + 1\right )}^{\frac {2}{3}}}{2 \, {\left (x - 1\right )}^{\frac {4}{3}}} \]

[In]

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

[Out]

-3/2*(x + 1)^(2/3)/(x - 1)^(4/3)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

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

[In]

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

[Out]

-(3*(x^2 - 1)^(2/3))/(2*(x - 1)^2)