\(\int x (-1+x^2)^{2/3} \, dx\) [8]

Optimal result

Integrand size = 11, antiderivative size = 13 \[ \int x \left (-1+x^2\right )^{2/3} \, dx=\frac {3}{10} \left (-1+x^2\right )^{5/3} \]

[Out]

3/10*(x^2-1)^(5/3)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {267} \[ \int x \left (-1+x^2\right )^{2/3} \, dx=\frac {3}{10} \left (x^2-1\right )^{5/3} \]

[In]

Int[x*(-1 + x^2)^(2/3),x]

[Out]

(3*(-1 + x^2)^(5/3))/10

Rule 267

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

Rubi steps \begin{align*} \text {integral}& = \frac {3}{10} \left (-1+x^2\right )^{5/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int x \left (-1+x^2\right )^{2/3} \, dx=\frac {3}{10} \left (-1+x^2\right )^{5/3} \]

[In]

Integrate[x*(-1 + x^2)^(2/3),x]

[Out]

(3*(-1 + x^2)^(5/3))/10

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {3 \left (x^{2}-1\right )^{\frac {5}{3}}}{10}\)	\(10\)
default	\(\frac {3 \left (x^{2}-1\right )^{\frac {5}{3}}}{10}\)	\(10\)
risch	\(\frac {3 \left (x^{2}-1\right )^{\frac {5}{3}}}{10}\)	\(10\)
pseudoelliptic	\(\frac {3 \left (x^{2}-1\right )^{\frac {5}{3}}}{10}\)	\(10\)
gosper	\(\frac {3 \left (x -1\right ) \left (1+x \right ) \left (x^{2}-1\right )^{\frac {2}{3}}}{10}\)	\(16\)
trager	\(\left (\frac {3 x^{2}}{10}-\frac {3}{10}\right ) \left (x^{2}-1\right )^{\frac {2}{3}}\)	\(16\)
meijerg	\(\frac {\operatorname {signum}\left (x^{2}-1\right )^{\frac {2}{3}} x^{2} \operatorname {hypergeom}\left (\left [-\frac {2}{3}, 1\right ], \left [2\right ], x^{2}\right )}{2 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {2}{3}}}\)	\(33\)

[In]

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

[Out]

3/10*(x^2-1)^(5/3)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x \left (-1+x^2\right )^{2/3} \, dx=\frac {3}{10} \, {\left (x^{2} - 1\right )}^{\frac {5}{3}} \]

[In]

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

[Out]

3/10*(x^2 - 1)^(5/3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (10) = 20\).

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.00 \[ \int x \left (-1+x^2\right )^{2/3} \, dx=\frac {3 x^{2} \left (x^{2} - 1\right )^{\frac {2}{3}}}{10} - \frac {3 \left (x^{2} - 1\right )^{\frac {2}{3}}}{10} \]

[In]

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

[Out]

3*x**2*(x**2 - 1)**(2/3)/10 - 3*(x**2 - 1)**(2/3)/10

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x \left (-1+x^2\right )^{2/3} \, dx=\frac {3}{10} \, {\left (x^{2} - 1\right )}^{\frac {5}{3}} \]

[In]

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

[Out]

3/10*(x^2 - 1)^(5/3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x \left (-1+x^2\right )^{2/3} \, dx=\frac {3}{10} \, {\left (x^{2} - 1\right )}^{\frac {5}{3}} \]

[In]

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

[Out]

3/10*(x^2 - 1)^(5/3)

Mupad [B] (verification not implemented)

Time = 5.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x \left (-1+x^2\right )^{2/3} \, dx=\frac {3\,{\left (x^2-1\right )}^{5/3}}{10} \]

[In]

int(x*(x^2 - 1)^(2/3),x)

[Out]

(3*(x^2 - 1)^(5/3))/10