Integrand size = 28, antiderivative size = 38 \[ \int \frac {\left (1+x^5\right )^{2/3} \left (1+x^3+x^5\right ) \left (-3+2 x^5\right )}{x^9} \, dx=\frac {3 \left (1+x^5\right )^{2/3} \left (5+8 x^3+10 x^5+8 x^8+5 x^{10}\right )}{40 x^8} \]
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Leaf count is larger than twice the leaf count of optimal. \(89\) vs. \(2(38)=76\).
Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.34, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1840, 1849, 1600, 1598, 460} \[ \int \frac {\left (1+x^5\right )^{2/3} \left (1+x^3+x^5\right ) \left (-3+2 x^5\right )}{x^9} \, dx=-\frac {6 \left (x^5+1\right )^{2/3}}{5 x^5}-\frac {15 \left (x^5+1\right )^{2/3}}{56 x^8}+\frac {15 \left (x^5+1\right )^{2/3}}{4 x^3}+\frac {3 \left (x^5+1\right )^{2/3} \left (35 x^{11}+56 x^9-280 x^6+168 x^4+60 x\right )}{280 x^9} \]
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Rule 460
Rule 1598
Rule 1600
Rule 1840
Rule 1849
Rubi steps \begin{align*} \text {integral}& = \frac {3 \left (1+x^5\right )^{2/3} \left (60 x+168 x^4-280 x^6+56 x^9+35 x^{11}\right )}{280 x^9}+\frac {10}{3} \int \frac {\frac {9}{14}+\frac {9 x^3}{5}-3 x^5+\frac {3 x^8}{5}+\frac {3 x^{10}}{8}}{x^9 \sqrt [3]{1+x^5}} \, dx \\ & = -\frac {15 \left (1+x^5\right )^{2/3}}{56 x^8}+\frac {3 \left (1+x^5\right )^{2/3} \left (60 x+168 x^4-280 x^6+56 x^9+35 x^{11}\right )}{280 x^9}-\frac {5}{24} \int \frac {-\frac {144 x^2}{5}+54 x^4-\frac {48 x^7}{5}-6 x^9}{x^8 \sqrt [3]{1+x^5}} \, dx \\ & = -\frac {15 \left (1+x^5\right )^{2/3}}{56 x^8}+\frac {3 \left (1+x^5\right )^{2/3} \left (60 x+168 x^4-280 x^6+56 x^9+35 x^{11}\right )}{280 x^9}-\frac {5}{24} \int \frac {-\frac {144 x}{5}+54 x^3-\frac {48 x^6}{5}-6 x^8}{x^7 \sqrt [3]{1+x^5}} \, dx \\ & = -\frac {15 \left (1+x^5\right )^{2/3}}{56 x^8}+\frac {3 \left (1+x^5\right )^{2/3} \left (60 x+168 x^4-280 x^6+56 x^9+35 x^{11}\right )}{280 x^9}-\frac {5}{24} \int \frac {-\frac {144}{5}+54 x^2-\frac {48 x^5}{5}-6 x^7}{x^6 \sqrt [3]{1+x^5}} \, dx \\ & = -\frac {15 \left (1+x^5\right )^{2/3}}{56 x^8}-\frac {6 \left (1+x^5\right )^{2/3}}{5 x^5}+\frac {3 \left (1+x^5\right )^{2/3} \left (60 x+168 x^4-280 x^6+56 x^9+35 x^{11}\right )}{280 x^9}+\frac {1}{48} \int \frac {-540 x+60 x^6}{x^5 \sqrt [3]{1+x^5}} \, dx \\ & = -\frac {15 \left (1+x^5\right )^{2/3}}{56 x^8}-\frac {6 \left (1+x^5\right )^{2/3}}{5 x^5}+\frac {3 \left (1+x^5\right )^{2/3} \left (60 x+168 x^4-280 x^6+56 x^9+35 x^{11}\right )}{280 x^9}+\frac {1}{48} \int \frac {-540+60 x^5}{x^4 \sqrt [3]{1+x^5}} \, dx \\ & = -\frac {15 \left (1+x^5\right )^{2/3}}{56 x^8}-\frac {6 \left (1+x^5\right )^{2/3}}{5 x^5}+\frac {15 \left (1+x^5\right )^{2/3}}{4 x^3}+\frac {3 \left (1+x^5\right )^{2/3} \left (60 x+168 x^4-280 x^6+56 x^9+35 x^{11}\right )}{280 x^9} \\ \end{align*}
Time = 0.81 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {\left (1+x^5\right )^{2/3} \left (1+x^3+x^5\right ) \left (-3+2 x^5\right )}{x^9} \, dx=\frac {3 \left (1+x^5\right )^{5/3} \left (5+8 x^3+5 x^5\right )}{40 x^8} \]
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Time = 1.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(\frac {3 \left (x^{5}+1\right )^{\frac {5}{3}} \left (5 x^{5}+8 x^{3}+5\right )}{40 x^{8}}\) | \(25\) |
trager | \(\frac {3 \left (x^{5}+1\right )^{\frac {2}{3}} \left (5 x^{10}+8 x^{8}+10 x^{5}+8 x^{3}+5\right )}{40 x^{8}}\) | \(35\) |
gosper | \(\frac {3 \left (x^{4}-x^{3}+x^{2}-x +1\right ) \left (1+x \right ) \left (5 x^{5}+8 x^{3}+5\right ) \left (x^{5}+1\right )^{\frac {2}{3}}}{40 x^{8}}\) | \(44\) |
risch | \(\frac {\frac {6}{5} x^{8}+\frac {3}{5} x^{3}+\frac {9}{8} x^{10}+\frac {9}{8} x^{5}+\frac {3}{8}+\frac {3}{8} x^{15}+\frac {3}{5} x^{13}}{x^{8} \left (x^{5}+1\right )^{\frac {1}{3}}}\) | \(45\) |
meijerg | \(x^{2} \operatorname {hypergeom}\left (\left [-\frac {2}{3}, \frac {2}{5}\right ], \left [\frac {7}{5}\right ], -x^{5}\right )+\frac {\operatorname {hypergeom}\left (\left [-\frac {2}{3}, -\frac {3}{5}\right ], \left [\frac {2}{5}\right ], -x^{5}\right )}{3 x^{3}}-\frac {2 \sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (-\frac {2 \pi \sqrt {3}\, x^{5} \operatorname {hypergeom}\left (\left [\frac {1}{3}, 1, 1\right ], \left [2, 2\right ], -x^{5}\right )}{3 \Gamma \left (\frac {2}{3}\right )}-\frac {\left (\frac {3}{2}-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+5 \ln \left (x \right )\right ) \pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right )}\right )}{15 \pi }+\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (\frac {\pi \sqrt {3}\, x^{5} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 3\right ], -x^{5}\right )}{9 \Gamma \left (\frac {2}{3}\right )}-\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}-1+5 \ln \left (x \right )\right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}+\frac {\pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right ) x^{5}}\right )}{5 \pi }+\frac {3 \operatorname {hypergeom}\left (\left [-\frac {8}{5}, -\frac {2}{3}\right ], \left [-\frac {3}{5}\right ], -x^{5}\right )}{8 x^{8}}\) | \(187\) |
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Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {\left (1+x^5\right )^{2/3} \left (1+x^3+x^5\right ) \left (-3+2 x^5\right )}{x^9} \, dx=\frac {3 \, {\left (5 \, x^{10} + 8 \, x^{8} + 10 \, x^{5} + 8 \, x^{3} + 5\right )} {\left (x^{5} + 1\right )}^{\frac {2}{3}}}{40 \, x^{8}} \]
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Result contains complex when optimal does not.
Time = 2.81 (sec) , antiderivative size = 184, normalized size of antiderivative = 4.84 \[ \int \frac {\left (1+x^5\right )^{2/3} \left (1+x^3+x^5\right ) \left (-3+2 x^5\right )}{x^9} \, dx=- \frac {2 x^{\frac {10}{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{5}}} \right )}}{5 \Gamma \left (\frac {1}{3}\right )} + \frac {2 x^{2} \Gamma \left (\frac {2}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {2}{5} \\ \frac {7}{5} \end {matrix}\middle | {x^{5} e^{i \pi }} \right )}}{5 \Gamma \left (\frac {7}{5}\right )} - \frac {\Gamma \left (- \frac {3}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {3}{5} \\ \frac {2}{5} \end {matrix}\middle | {x^{5} e^{i \pi }} \right )}}{5 x^{3} \Gamma \left (\frac {2}{5}\right )} - \frac {3 \Gamma \left (- \frac {8}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {8}{5}, - \frac {2}{3} \\ - \frac {3}{5} \end {matrix}\middle | {x^{5} e^{i \pi }} \right )}}{5 x^{8} \Gamma \left (- \frac {3}{5}\right )} + \frac {3 \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{5}}} \right )}}{5 x^{\frac {5}{3}} \Gamma \left (\frac {4}{3}\right )} \]
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Time = 0.32 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.32 \[ \int \frac {\left (1+x^5\right )^{2/3} \left (1+x^3+x^5\right ) \left (-3+2 x^5\right )}{x^9} \, dx=\frac {3 \, {\left (5 \, x^{10} + 8 \, x^{8} + 10 \, x^{5} + 8 \, x^{3} + 5\right )} {\left (x^{4} - x^{3} + x^{2} - x + 1\right )}^{\frac {2}{3}} {\left (x + 1\right )}^{\frac {2}{3}}}{40 \, x^{8}} \]
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\[ \int \frac {\left (1+x^5\right )^{2/3} \left (1+x^3+x^5\right ) \left (-3+2 x^5\right )}{x^9} \, dx=\int { \frac {{\left (2 \, x^{5} - 3\right )} {\left (x^{5} + x^{3} + 1\right )} {\left (x^{5} + 1\right )}^{\frac {2}{3}}}{x^{9}} \,d x } \]
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Time = 5.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.37 \[ \int \frac {\left (1+x^5\right )^{2/3} \left (1+x^3+x^5\right ) \left (-3+2 x^5\right )}{x^9} \, dx={\left (x^5+1\right )}^{2/3}\,\left (\frac {3\,x^2}{8}+\frac {3}{5}\right )+\frac {3\,{\left (x^5+1\right )}^{2/3}}{4\,x^3}+\frac {3\,{\left (x^5+1\right )}^{2/3}}{5\,x^5}+\frac {3\,{\left (x^5+1\right )}^{2/3}}{8\,x^8} \]
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