Integrand size = 79, antiderivative size = 24 \[ \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx=e^{e^3 \left (5-\frac {3 \left (\log (3)+\log \left (3 x^2\right )\right )}{-1+x}\right )} \]
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\[ \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx=\int \frac {\exp \left (\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}\right ) \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}\right ) \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x \left (1-2 x+x^2\right )} \, dx \\ & = \int \frac {\exp \left (\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}\right ) \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{(-1+x)^2 x} \, dx \\ & = \int \frac {3 \exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \left (2-2 x (1-\log (3))+x \log \left (x^2\right )\right )}{(1-x)^2 x} \, dx \\ & = 3 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \left (2-2 x (1-\log (3))+x \log \left (x^2\right )\right )}{(1-x)^2 x} \, dx \\ & = 3 \int \left (\frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) (2-x (2-\log (9)))}{(1-x)^2 x}+\frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \log \left (x^2\right )}{(-1+x)^2}\right ) \, dx \\ & = 3 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) (2-x (2-\log (9)))}{(1-x)^2 x} \, dx+3 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \log \left (x^2\right )}{(-1+x)^2} \, dx \\ & = 3 \int \left (-\frac {2 \exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right )}{-1+x}+\frac {2 \exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right )}{x}+\frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \log (9)}{(-1+x)^2}\right ) \, dx+3 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \log \left (x^2\right )}{(-1+x)^2} \, dx \\ & = 3 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \log \left (x^2\right )}{(-1+x)^2} \, dx-6 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right )}{-1+x} \, dx+6 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right )}{x} \, dx+(3 \log (9)) \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right )}{(-1+x)^2} \, dx \\ \end{align*}
Time = 0.49 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx=729^{\frac {e^3}{1-x}} e^{5 e^3} \left (x^2\right )^{-\frac {3 e^3}{-1+x}} \]
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Time = 0.94 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17
method | result | size |
risch | \({\mathrm e}^{-\frac {{\mathrm e}^{3} \left (3 \ln \left (3 x^{2}\right )+3 \ln \left (3\right )-5 x +5\right )}{-1+x}}\) | \(28\) |
parallelrisch | \({\mathrm e}^{-\frac {{\mathrm e}^{3} \left (3 \ln \left (3 x^{2}\right )+3 \ln \left (3\right )-5 x +5\right )}{-1+x}}\) | \(28\) |
default | \(\frac {x \,{\mathrm e}^{\frac {-3 \,{\mathrm e}^{3} \ln \left (3 x^{2}\right )-3 \,{\mathrm e}^{3} \ln \left (3\right )+\left (5 x -5\right ) {\mathrm e}^{3}}{-1+x}}-{\mathrm e}^{\frac {-3 \,{\mathrm e}^{3} \ln \left (3 x^{2}\right )-3 \,{\mathrm e}^{3} \ln \left (3\right )+\left (5 x -5\right ) {\mathrm e}^{3}}{-1+x}}}{-1+x}\) | \(76\) |
norman | \(\frac {x \,{\mathrm e}^{\frac {-3 \,{\mathrm e}^{3} \ln \left (3 x^{2}\right )-3 \,{\mathrm e}^{3} \ln \left (3\right )+\left (5 x -5\right ) {\mathrm e}^{3}}{-1+x}}-{\mathrm e}^{\frac {-3 \,{\mathrm e}^{3} \ln \left (3 x^{2}\right )-3 \,{\mathrm e}^{3} \ln \left (3\right )+\left (5 x -5\right ) {\mathrm e}^{3}}{-1+x}}}{-1+x}\) | \(76\) |
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Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx=e^{\left (\frac {5 \, {\left (x - 1\right )} e^{3} - 3 \, e^{3} \log \left (3\right ) - 3 \, e^{3} \log \left (3 \, x^{2}\right )}{x - 1}\right )} \]
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Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx=e^{\frac {\left (5 x - 5\right ) e^{3} - 3 e^{3} \log {\left (3 x^{2} \right )} - 3 e^{3} \log {\left (3 \right )}}{x - 1}} \]
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Time = 0.43 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx=e^{\left (-\frac {6 \, e^{3} \log \left (3\right )}{x - 1} - \frac {6 \, e^{3} \log \left (x\right )}{x - 1} + 5 \, e^{3}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx=e^{\left (\frac {5 \, x e^{3}}{x - 1} - \frac {3 \, e^{3} \log \left (3\right )}{x - 1} - \frac {3 \, e^{3} \log \left (3 \, x^{2}\right )}{x - 1} - \frac {5 \, e^{3}}{x - 1}\right )} \]
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Time = 18.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx={\mathrm {e}}^{-\frac {5\,{\mathrm {e}}^3}{x-1}}\,{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^3}{x-1}}\,{\left (\frac {1}{729\,x^6}\right )}^{\frac {{\mathrm {e}}^3}{x-1}} \]
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