Integrand size = 30, antiderivative size = 28 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+2 x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4 \sqrt [4]{1+x^3} \left (-1-x^3+10 x^4\right )}{5 x^5} \]
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Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.68, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1849, 1600, 460} \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+2 x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=-\frac {8 \sqrt [4]{x^3+1}}{x}+\frac {4 \sqrt [4]{x^3+1}}{5 x^5}+\frac {4 \sqrt [4]{x^3+1}}{5 x^2} \]
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Rule 460
Rule 1600
Rule 1849
Rubi steps \begin{align*} \text {integral}& = \frac {4 \sqrt [4]{1+x^3}}{5 x^5}-\frac {1}{10} \int \frac {16 x^2-80 x^3+10 x^5-20 x^6}{x^5 \left (1+x^3\right )^{3/4}} \, dx \\ & = \frac {4 \sqrt [4]{1+x^3}}{5 x^5}-\frac {1}{10} \int \frac {16 x-80 x^2+10 x^4-20 x^5}{x^4 \left (1+x^3\right )^{3/4}} \, dx \\ & = \frac {4 \sqrt [4]{1+x^3}}{5 x^5}-\frac {1}{10} \int \frac {16-80 x+10 x^3-20 x^4}{x^3 \left (1+x^3\right )^{3/4}} \, dx \\ & = \frac {4 \sqrt [4]{1+x^3}}{5 x^5}+\frac {4 \sqrt [4]{1+x^3}}{5 x^2}+\frac {1}{40} \int \frac {320+80 x^3}{x^2 \left (1+x^3\right )^{3/4}} \, dx \\ & = \frac {4 \sqrt [4]{1+x^3}}{5 x^5}+\frac {4 \sqrt [4]{1+x^3}}{5 x^2}-\frac {8 \sqrt [4]{1+x^3}}{x} \\ \end{align*}
Time = 3.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+2 x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4 \sqrt [4]{1+x^3} \left (-1-x^3+10 x^4\right )}{5 x^5} \]
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Time = 0.78 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89
method | result | size |
trager | \(-\frac {4 \left (x^{3}+1\right )^{\frac {1}{4}} \left (10 x^{4}-x^{3}-1\right )}{5 x^{5}}\) | \(25\) |
risch | \(-\frac {4 \left (10 x^{7}-x^{6}+10 x^{4}-2 x^{3}-1\right )}{5 \left (x^{3}+1\right )^{\frac {3}{4}} x^{5}}\) | \(35\) |
gosper | \(-\frac {4 \left (10 x^{4}-x^{3}-1\right ) \left (1+x \right ) \left (x^{2}-x +1\right )}{5 \left (x^{3}+1\right )^{\frac {3}{4}} x^{5}}\) | \(36\) |
meijerg | \(x^{2} \operatorname {hypergeom}\left (\left [\frac {2}{3}, \frac {3}{4}\right ], \left [\frac {5}{3}\right ], -x^{3}\right )-x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {3}{4}\right ], \left [\frac {4}{3}\right ], -x^{3}\right )+\frac {5 \operatorname {hypergeom}\left (\left [-\frac {2}{3}, \frac {3}{4}\right ], \left [\frac {1}{3}\right ], -x^{3}\right )}{2 x^{2}}+\frac {4 \operatorname {hypergeom}\left (\left [-\frac {5}{3}, \frac {3}{4}\right ], \left [-\frac {2}{3}\right ], -x^{3}\right )}{5 x^{5}}-\frac {8 \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {3}{4}\right ], \left [\frac {2}{3}\right ], -x^{3}\right )}{x}\) | \(79\) |
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Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+2 x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4 \, {\left (10 \, x^{4} - x^{3} - 1\right )} {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{5 \, x^{5}} \]
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Result contains complex when optimal does not.
Time = 2.04 (sec) , antiderivative size = 168, normalized size of antiderivative = 6.00 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+2 x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=\frac {2 x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {3}{4} \\ \frac {5}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} - \frac {x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {3}{4} \\ \frac {4}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {8 \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{4} \\ \frac {2}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} - \frac {5 \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {3}{4} \\ \frac {1}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} - \frac {4 \Gamma \left (- \frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{3}, \frac {3}{4} \\ - \frac {2}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{5} \Gamma \left (- \frac {2}{3}\right )} \]
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Time = 0.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+2 x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4 \, {\left (10 \, x^{7} - x^{6} + 10 \, x^{4} - 2 \, x^{3} - 1\right )}}{5 \, {\left (x^{2} - x + 1\right )}^{\frac {3}{4}} {\left (x + 1\right )}^{\frac {3}{4}} x^{5}} \]
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\[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+2 x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=\int { \frac {{\left (2 \, x^{4} - x^{3} - 1\right )} {\left (x^{3} + 4\right )}}{{\left (x^{3} + 1\right )}^{\frac {3}{4}} x^{6}} \,d x } \]
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Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+2 x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=\frac {4\,{\left (x^3+1\right )}^{1/4}+4\,x^3\,{\left (x^3+1\right )}^{1/4}-40\,x^4\,{\left (x^3+1\right )}^{1/4}}{5\,x^5} \]
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