Integrand size = 24, antiderivative size = 20 \[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\frac {3 \left (-x^2+x^5\right )^{4/3}}{4 x^4} \]
[Out]
Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {1604} \[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\frac {3 \left (x^5-x^2\right )^{4/3}}{4 x^4} \]
[In]
[Out]
Rule 1604
Rubi steps \begin{align*} \text {integral}& = \frac {3 \left (-x^2+x^5\right )^{4/3}}{4 x^4} \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\frac {3 \left (x^2 \left (-1+x^3\right )\right )^{4/3}}{4 x^4} \]
[In]
[Out]
Time = 0.84 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10
method | result | size |
trager | \(\frac {3 \left (x^{3}-1\right ) \left (x^{5}-x^{2}\right )^{\frac {1}{3}}}{4 x^{2}}\) | \(22\) |
pseudoelliptic | \(\frac {3 \left (x^{3}-1\right ) \left (x^{5}-x^{2}\right )^{\frac {1}{3}}}{4 x^{2}}\) | \(22\) |
gosper | \(\frac {3 \left (x -1\right ) \left (x^{2}+x +1\right ) \left (x^{5}-x^{2}\right )^{\frac {1}{3}}}{4 x^{2}}\) | \(26\) |
risch | \(\frac {3 \left (x^{2} \left (x^{3}-1\right )\right )^{\frac {1}{3}} \left (x^{6}-2 x^{3}+1\right )}{4 x^{2} \left (x^{3}-1\right )}\) | \(34\) |
meijerg | \(-\frac {3 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} \operatorname {hypergeom}\left (\left [-\frac {4}{9}, -\frac {1}{3}\right ], \left [\frac {5}{9}\right ], x^{3}\right )}{4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}} x^{\frac {4}{3}}}+\frac {6 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} x^{\frac {5}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {5}{9}\right ], \left [\frac {14}{9}\right ], x^{3}\right )}{5 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}}}\) | \(66\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\frac {3 \, {\left (x^{5} - x^{2}\right )}^{\frac {1}{3}} {\left (x^{3} - 1\right )}}{4 \, x^{2}} \]
[In]
[Out]
\[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\int \frac {\sqrt [3]{x^{2} \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (2 x^{3} + 1\right )}{x^{3}}\, dx \]
[In]
[Out]
\[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\int { \frac {{\left (x^{5} - x^{2}\right )}^{\frac {1}{3}} {\left (2 \, x^{3} + 1\right )}}{x^{3}} \,d x } \]
[In]
[Out]
\[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\int { \frac {{\left (x^{5} - x^{2}\right )}^{\frac {1}{3}} {\left (2 \, x^{3} + 1\right )}}{x^{3}} \,d x } \]
[In]
[Out]
Time = 5.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\frac {3\,\left (x^3-1\right )\,{\left (x^5-x^2\right )}^{1/3}}{4\,x^2} \]
[In]
[Out]