Integrand size = 26, antiderivative size = 26 \[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=-\frac {4 \sqrt [4]{-1+x^3} \left (-1+x^3+5 x^4\right )}{5 x^5} \]
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Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1849, 1600, 460} \[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=-\frac {4 \sqrt [4]{x^3-1}}{x}+\frac {4 \sqrt [4]{x^3-1}}{5 x^5}-\frac {4 \sqrt [4]{x^3-1}}{5 x^2} \]
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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^2-40 x^3+10 x^5+10 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-40 x^2+10 x^4+10 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 {-16-40 x+10 x^3+10 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 {-160+40 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 {4 \sqrt [4]{-1+x^3}}{x} \\ \end{align*}
Time = 3.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=-\frac {4 \sqrt [4]{-1+x^3} \left (-1+x^3+5 x^4\right )}{5 x^5} \]
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Time = 0.84 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88
method | result | size |
trager | \(-\frac {4 \left (x^{3}-1\right )^{\frac {1}{4}} \left (5 x^{4}+x^{3}-1\right )}{5 x^{5}}\) | \(23\) |
gosper | \(-\frac {4 \left (x -1\right ) \left (x^{2}+x +1\right ) \left (5 x^{4}+x^{3}-1\right )}{5 x^{5} \left (x^{3}-1\right )^{\frac {3}{4}}}\) | \(32\) |
risch | \(-\frac {4 \left (5 x^{7}+x^{6}-5 x^{4}-2 x^{3}+1\right )}{5 \left (x^{3}-1\right )^{\frac {3}{4}} x^{5}}\) | \(33\) |
meijerg | \(\frac {{\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {3}{4}} x^{2} \operatorname {hypergeom}\left (\left [\frac {2}{3}, \frac {3}{4}\right ], \left [\frac {5}{3}\right ], x^{3}\right )}{2 \operatorname {signum}\left (x^{3}-1\right )^{\frac {3}{4}}}+\frac {{\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {3}{4}} x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {3}{4}\right ], \left [\frac {4}{3}\right ], x^{3}\right )}{\operatorname {signum}\left (x^{3}-1\right )^{\frac {3}{4}}}+\frac {5 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [-\frac {2}{3}, \frac {3}{4}\right ], \left [\frac {1}{3}\right ], x^{3}\right )}{2 \operatorname {signum}\left (x^{3}-1\right )^{\frac {3}{4}} x^{2}}-\frac {4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [-\frac {5}{3}, \frac {3}{4}\right ], \left [-\frac {2}{3}\right ], x^{3}\right )}{5 \operatorname {signum}\left (x^{3}-1\right )^{\frac {3}{4}} x^{5}}+\frac {4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {3}{4}\right ], \left [\frac {2}{3}\right ], x^{3}\right )}{\operatorname {signum}\left (x^{3}-1\right )^{\frac {3}{4}} x}\) | \(159\) |
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=-\frac {4 \, {\left (5 \, x^{4} + x^{3} - 1\right )} {\left (x^{3} - 1\right )}^{\frac {1}{4}}}{5 \, x^{5}} \]
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Result contains complex when optimal does not.
Time = 2.03 (sec) , antiderivative size = 178, normalized size of antiderivative = 6.85 \[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=\frac {x^{2} e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {3}{4} \\ \frac {5}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} + \frac {x e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {3}{4} \\ \frac {4}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {4 e^{\frac {i \pi }{4}} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{4} \\ \frac {2}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} + \frac {5 e^{\frac {i \pi }{4}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {3}{4} \\ \frac {1}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} - \frac {4 e^{\frac {i \pi }{4}} \Gamma \left (- \frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{3}, \frac {3}{4} \\ - \frac {2}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 x^{5} \Gamma \left (- \frac {2}{3}\right )} \]
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Time = 0.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=-\frac {4 \, {\left (5 \, x^{7} + x^{6} - 5 \, x^{4} - 2 \, x^{3} + 1\right )}}{5 \, {\left (x^{2} + x + 1\right )}^{\frac {3}{4}} {\left (x - 1\right )}^{\frac {3}{4}} x^{5}} \]
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\[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=\int { \frac {{\left (x^{4} + x^{3} - 1\right )} {\left (x^{3} - 4\right )}}{{\left (x^{3} - 1\right )}^{\frac {3}{4}} x^{6}} \,d x } \]
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Time = 5.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=-\frac {4\,x^3\,{\left (x^3-1\right )}^{1/4}-4\,{\left (x^3-1\right )}^{1/4}+20\,x^4\,{\left (x^3-1\right )}^{1/4}}{5\,x^5} \]
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