Integrand size = 22, antiderivative size = 20 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^6}}{x^4} \, dx=\frac {2 \left (-x^2+x^6\right )^{5/4}}{5 x^5} \]
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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} \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^6}}{x^4} \, dx=\frac {2 \left (x^6-x^2\right )^{5/4}}{5 x^5} \]
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Rule 1604
Rubi steps \begin{align*} \text {integral}& = \frac {2 \left (-x^2+x^6\right )^{5/4}}{5 x^5} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^6}}{x^4} \, dx=\frac {2 \left (x^2 \left (-1+x^4\right )\right )^{5/4}}{5 x^5} \]
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Time = 0.86 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10
method | result | size |
trager | \(\frac {2 \left (x^{4}-1\right ) \left (x^{6}-x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}\) | \(22\) |
pseudoelliptic | \(\frac {2 \left (x^{4}-1\right ) \left (x^{6}-x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}\) | \(22\) |
gosper | \(\frac {2 \left (x^{2}+1\right ) \left (x -1\right ) \left (1+x \right ) \left (x^{6}-x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}\) | \(28\) |
risch | \(\frac {2 \left (x^{2} \left (x^{4}-1\right )\right )^{\frac {1}{4}} \left (x^{8}-2 x^{4}+1\right )}{5 x^{3} \left (x^{4}-1\right )}\) | \(34\) |
meijerg | \(-\frac {2 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {5}{8}, -\frac {1}{4}\right ], \left [\frac {3}{8}\right ], x^{4}\right )}{5 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} x^{\frac {5}{2}}}+\frac {2 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{\frac {3}{2}} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {3}{8}\right ], \left [\frac {11}{8}\right ], x^{4}\right )}{3 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}}}\) | \(66\) |
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^6}}{x^4} \, dx=\frac {2 \, {\left (x^{6} - x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} - 1\right )}}{5 \, x^{3}} \]
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\[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^6}}{x^4} \, dx=\int \frac {\sqrt [4]{x^{2} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )} \left (x^{4} + 1\right )}{x^{4}}\, dx \]
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\[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^6}}{x^4} \, dx=\int { \frac {{\left (x^{6} - x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )}}{x^{4}} \,d x } \]
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\[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^6}}{x^4} \, dx=\int { \frac {{\left (x^{6} - x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )}}{x^{4}} \,d x } \]
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Time = 5.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^6}}{x^4} \, dx=\frac {2\,\left (x^4-1\right )\,{\left (x^6-x^2\right )}^{1/4}}{5\,x^3} \]
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