Integrand size = 22, antiderivative size = 20 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}}{x^3} \, dx=\frac {3 \left (-x^2+x^4\right )^{4/3}}{4 x^4} \]
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Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {1604} \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}}{x^3} \, dx=\frac {3 \left (x^4-x^2\right )^{4/3}}{4 x^4} \]
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Rule 1604
Rubi steps \begin{align*} \text {integral}& = \frac {3 \left (-x^2+x^4\right )^{4/3}}{4 x^4} \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}}{x^3} \, dx=\frac {3 \left (x^2 \left (-1+x^2\right )\right )^{4/3}}{4 x^4} \]
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Time = 0.84 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10
method | result | size |
trager | \(\frac {3 \left (x^{2}-1\right ) \left (x^{4}-x^{2}\right )^{\frac {1}{3}}}{4 x^{2}}\) | \(22\) |
pseudoelliptic | \(\frac {3 \left (x^{2}-1\right ) \left (x^{4}-x^{2}\right )^{\frac {1}{3}}}{4 x^{2}}\) | \(22\) |
gosper | \(\frac {3 \left (1+x \right ) \left (x -1\right ) \left (x^{4}-x^{2}\right )^{\frac {1}{3}}}{4 x^{2}}\) | \(23\) |
risch | \(\frac {3 \left (x^{2} \left (x^{2}-1\right )\right )^{\frac {1}{3}} \left (x^{4}-2 x^{2}+1\right )}{4 x^{2} \left (x^{2}-1\right )}\) | \(34\) |
meijerg | \(-\frac {3 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{3}} \operatorname {hypergeom}\left (\left [-\frac {2}{3}, -\frac {1}{3}\right ], \left [\frac {1}{3}\right ], x^{2}\right )}{4 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{3}} x^{\frac {4}{3}}}+\frac {3 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{3}} x^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], x^{2}\right )}{2 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{3}}}\) | \(66\) |
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none
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}}{x^3} \, dx=\frac {3 \, {\left (x^{4} - x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )}}{4 \, x^{2}} \]
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\[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}}{x^3} \, dx=\int \frac {\sqrt [3]{x^{2} \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 1\right )}{x^{3}}\, dx \]
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\[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}}{x^3} \, dx=\int { \frac {{\left (x^{4} - x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )}}{x^{3}} \,d x } \]
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\[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}}{x^3} \, dx=\int { \frac {{\left (x^{4} - x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )}}{x^{3}} \,d x } \]
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Time = 4.99 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}}{x^3} \, dx=\frac {3\,\left (x^2-1\right )\,{\left (x^4-x^2\right )}^{1/3}}{4\,x^2} \]
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