Integrand size = 13, antiderivative size = 25 \[ \int x^3 \left (-1+x^2\right )^{2/3} \, dx=\frac {3}{80} \left (-1+x^2\right )^{2/3} \left (-3-2 x^2+5 x^4\right ) \]
[Out]
Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int x^3 \left (-1+x^2\right )^{2/3} \, dx=\frac {3}{16} \left (x^2-1\right )^{8/3}+\frac {3}{10} \left (x^2-1\right )^{5/3} \]
[In]
[Out]
Rule 45
Rule 272
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int (-1+x)^{2/3} x \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left ((-1+x)^{2/3}+(-1+x)^{5/3}\right ) \, dx,x,x^2\right ) \\ & = \frac {3}{10} \left (-1+x^2\right )^{5/3}+\frac {3}{16} \left (-1+x^2\right )^{8/3} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int x^3 \left (-1+x^2\right )^{2/3} \, dx=\frac {3}{80} \left (-1+x^2\right )^{5/3} \left (3+5 x^2\right ) \]
[In]
[Out]
Time = 0.84 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68
method | result | size |
pseudoelliptic | \(\frac {3 \left (x^{2}-1\right )^{\frac {5}{3}} \left (5 x^{2}+3\right )}{80}\) | \(17\) |
trager | \(\left (\frac {3}{16} x^{4}-\frac {3}{40} x^{2}-\frac {9}{80}\right ) \left (x^{2}-1\right )^{\frac {2}{3}}\) | \(21\) |
risch | \(\frac {3 \left (x^{2}-1\right )^{\frac {2}{3}} \left (5 x^{4}-2 x^{2}-3\right )}{80}\) | \(22\) |
gosper | \(\frac {3 \left (x -1\right ) \left (1+x \right ) \left (5 x^{2}+3\right ) \left (x^{2}-1\right )^{\frac {2}{3}}}{80}\) | \(23\) |
meijerg | \(\frac {\operatorname {signum}\left (x^{2}-1\right )^{\frac {2}{3}} x^{4} \operatorname {hypergeom}\left (\left [-\frac {2}{3}, 2\right ], \left [3\right ], x^{2}\right )}{4 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {2}{3}}}\) | \(33\) |
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int x^3 \left (-1+x^2\right )^{2/3} \, dx=\frac {3}{80} \, {\left (5 \, x^{4} - 2 \, x^{2} - 3\right )} {\left (x^{2} - 1\right )}^{\frac {2}{3}} \]
[In]
[Out]
Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int x^3 \left (-1+x^2\right )^{2/3} \, dx=\frac {3 x^{4} \left (x^{2} - 1\right )^{\frac {2}{3}}}{16} - \frac {3 x^{2} \left (x^{2} - 1\right )^{\frac {2}{3}}}{40} - \frac {9 \left (x^{2} - 1\right )^{\frac {2}{3}}}{80} \]
[In]
[Out]
none
Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x^3 \left (-1+x^2\right )^{2/3} \, dx=\frac {3}{16} \, {\left (x^{2} - 1\right )}^{\frac {8}{3}} + \frac {3}{10} \, {\left (x^{2} - 1\right )}^{\frac {5}{3}} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x^3 \left (-1+x^2\right )^{2/3} \, dx=\frac {3}{16} \, {\left (x^{2} - 1\right )}^{\frac {8}{3}} + \frac {3}{10} \, {\left (x^{2} - 1\right )}^{\frac {5}{3}} \]
[In]
[Out]
Time = 5.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int x^3 \left (-1+x^2\right )^{2/3} \, dx=-{\left (x^2-1\right )}^{2/3}\,\left (-\frac {3\,x^4}{16}+\frac {3\,x^2}{40}+\frac {9}{80}\right ) \]
[In]
[Out]