Integrand size = 145, antiderivative size = 28 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=-2+x-\frac {5 e^2}{5+x-\log \left (-2+x^2 \log \left (e^{2 x}\right )\right )} \]
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\[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=\int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-(5+x)^2 \left (-1+x^3\right )-5 e^2 \left (-1-3 x^2+x^3\right )+2 \left (-5-x+5 x^3+x^4\right ) \log \left (2 \left (-1+x^3\right )\right )-\left (-1+x^3\right ) \log ^2\left (2 \left (-1+x^3\right )\right )}{\left (1-x^3\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2} \, dx \\ & = \int \left (\frac {25 \left (1+\frac {e^2}{5}\right )+10 x+\left (1+15 e^2\right ) x^2-25 \left (1+\frac {e^2}{5}\right ) x^3-10 x^4-x^5-10 \log \left (2 \left (-1+x^3\right )\right )-2 x \log \left (2 \left (-1+x^3\right )\right )+10 x^3 \log \left (2 \left (-1+x^3\right )\right )+2 x^4 \log \left (2 \left (-1+x^3\right )\right )+\log ^2\left (2 \left (-1+x^3\right )\right )-x^3 \log ^2\left (2 \left (-1+x^3\right )\right )}{3 (1-x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {(2+x) \left (25 \left (1+\frac {e^2}{5}\right )+10 x+\left (1+15 e^2\right ) x^2-25 \left (1+\frac {e^2}{5}\right ) x^3-10 x^4-x^5-10 \log \left (2 \left (-1+x^3\right )\right )-2 x \log \left (2 \left (-1+x^3\right )\right )+10 x^3 \log \left (2 \left (-1+x^3\right )\right )+2 x^4 \log \left (2 \left (-1+x^3\right )\right )+\log ^2\left (2 \left (-1+x^3\right )\right )-x^3 \log ^2\left (2 \left (-1+x^3\right )\right )\right )}{3 \left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}\right ) \, dx \\ & = \frac {1}{3} \int \frac {25 \left (1+\frac {e^2}{5}\right )+10 x+\left (1+15 e^2\right ) x^2-25 \left (1+\frac {e^2}{5}\right ) x^3-10 x^4-x^5-10 \log \left (2 \left (-1+x^3\right )\right )-2 x \log \left (2 \left (-1+x^3\right )\right )+10 x^3 \log \left (2 \left (-1+x^3\right )\right )+2 x^4 \log \left (2 \left (-1+x^3\right )\right )+\log ^2\left (2 \left (-1+x^3\right )\right )-x^3 \log ^2\left (2 \left (-1+x^3\right )\right )}{(1-x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2} \, dx+\frac {1}{3} \int \frac {(2+x) \left (25 \left (1+\frac {e^2}{5}\right )+10 x+\left (1+15 e^2\right ) x^2-25 \left (1+\frac {e^2}{5}\right ) x^3-10 x^4-x^5-10 \log \left (2 \left (-1+x^3\right )\right )-2 x \log \left (2 \left (-1+x^3\right )\right )+10 x^3 \log \left (2 \left (-1+x^3\right )\right )+2 x^4 \log \left (2 \left (-1+x^3\right )\right )+\log ^2\left (2 \left (-1+x^3\right )\right )-x^3 \log ^2\left (2 \left (-1+x^3\right )\right )\right )}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2} \, dx \\ & = \frac {1}{3} \int \frac {-(5+x)^2 \left (-1+x^3\right )-5 e^2 \left (-1-3 x^2+x^3\right )+2 \left (-5-x+5 x^3+x^4\right ) \log \left (2 \left (-1+x^3\right )\right )-\left (-1+x^3\right ) \log ^2\left (2 \left (-1+x^3\right )\right )}{(1-x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2} \, dx+\frac {1}{3} \int \frac {(2+x) \left (-(5+x)^2 \left (-1+x^3\right )-5 e^2 \left (-1-3 x^2+x^3\right )+2 \left (-5-x+5 x^3+x^4\right ) \log \left (2 \left (-1+x^3\right )\right )-\left (-1+x^3\right ) \log ^2\left (2 \left (-1+x^3\right )\right )\right )}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2} \, dx \\ & = \frac {1}{3} \int \left (-\frac {25 \left (1+\frac {e^2}{5}\right )}{(-1+x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}-\frac {10 x}{(-1+x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}-\frac {\left (1+15 e^2\right ) x^2}{(-1+x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {25 \left (1+\frac {e^2}{5}\right ) x^3}{(-1+x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {10 x^4}{(-1+x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {x^5}{(-1+x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {10 \log \left (-2+2 x^3\right )}{(-1+x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {2 x \log \left (-2+2 x^3\right )}{(-1+x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {10 x^3 \log \left (-2+2 x^3\right )}{(1-x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {2 x^4 \log \left (-2+2 x^3\right )}{(1-x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {\log ^2\left (-2+2 x^3\right )}{(1-x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {x^3 \log ^2\left (-2+2 x^3\right )}{(-1+x) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}\right ) \, dx+\frac {1}{3} \int \left (\frac {50 \left (1+\frac {e^2}{5}\right )}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {45 \left (1+\frac {e^2}{9}\right ) x}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {12 \left (1+\frac {5 e^2}{2}\right ) x^2}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}-\frac {49 \left (1-\frac {5 e^2}{49}\right ) x^3}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}-\frac {45 \left (1+\frac {e^2}{9}\right ) x^4}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}-\frac {12 x^5}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}-\frac {x^6}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {20 \log \left (-2+2 x^3\right )}{\left (-1-x-x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {14 x \log \left (-2+2 x^3\right )}{\left (-1-x-x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {2 x^2 \log \left (-2+2 x^3\right )}{\left (-1-x-x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {20 x^3 \log \left (-2+2 x^3\right )}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {14 x^4 \log \left (-2+2 x^3\right )}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {2 x^5 \log \left (-2+2 x^3\right )}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {2 x^3 \log ^2\left (-2+2 x^3\right )}{\left (-1-x-x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {x^4 \log ^2\left (-2+2 x^3\right )}{\left (-1-x-x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {2 \log ^2\left (-2+2 x^3\right )}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}+\frac {x \log ^2\left (-2+2 x^3\right )}{\left (1+x+x^2\right ) \left (5+x-\log \left (-2+2 x^3\right )\right )^2}\right ) \, dx \\ & = \text {Too large to display} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=x+\frac {5 e^2}{-5-x+\log \left (2 \left (-1+x^3\right )\right )} \]
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Time = 4.78 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79
method | result | size |
risch | \(x -\frac {5 \,{\mathrm e}^{2}}{5+x -\ln \left (2 x^{3}-2\right )}\) | \(22\) |
parallelrisch | \(-\frac {\ln \left (2 x^{3}-2\right ) x -x^{2}+5 \,{\mathrm e}^{2}-5 x}{5+x -\ln \left (2 x^{3}-2\right )}\) | \(41\) |
norman | \(\frac {x^{2}-25+5 \ln \left (2 x^{3}-2\right )-\ln \left (2 x^{3}-2\right ) x -5 \,{\mathrm e}^{2}}{5+x -\ln \left (2 x^{3}-2\right )}\) | \(47\) |
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Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=\frac {x^{2} - x \log \left (2 \, x^{3} - 2\right ) + 5 \, x - 5 \, e^{2}}{x - \log \left (2 \, x^{3} - 2\right ) + 5} \]
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Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=x + \frac {5 e^{2}}{- x + \log {\left (2 x^{3} - 2 \right )} - 5} \]
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Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (26) = 52\).
Time = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.04 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=\frac {x^{2} - x {\left (\log \left (2\right ) - 5\right )} - x \log \left (x^{2} + x + 1\right ) - x \log \left (x - 1\right ) - 5 \, e^{2}}{x - \log \left (2\right ) - \log \left (x^{2} + x + 1\right ) - \log \left (x - 1\right ) + 5} \]
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Time = 0.32 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=\frac {x^{2} - x \log \left (2 \, x^{3} - 2\right ) + 5 \, x - 5 \, e^{2}}{x - \log \left (2 \, x^{3} - 2\right ) + 5} \]
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Time = 10.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=x-\frac {5\,{\mathrm {e}}^2}{x-\ln \left (2\,x^3-2\right )+5} \]
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