Integrand size = 123, antiderivative size = 23 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=10 x^4 \left (x+\frac {x}{-\frac {1}{5}+\log \left (-1+e^{5 x}\right )}\right ) \]
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\[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=\int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {50 x^4 \left (-4+e^{5 x} (4+25 x)-15 \left (-1+e^{5 x}\right ) \log \left (-1+e^{5 x}\right )-25 \left (-1+e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )\right )}{\left (1-e^{5 x}\right ) \left (1-5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx \\ & = 50 \int \frac {x^4 \left (-4+e^{5 x} (4+25 x)-15 \left (-1+e^{5 x}\right ) \log \left (-1+e^{5 x}\right )-25 \left (-1+e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )\right )}{\left (1-e^{5 x}\right ) \left (1-5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx \\ & = 50 \int \left (-\frac {5 x^5}{\left (-1+e^x\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}+\frac {5 \left (4+3 e^x+2 e^{2 x}+e^{3 x}\right ) x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}-\frac {x^4 \left (4+25 x-15 \log \left (-1+e^{5 x}\right )-25 \log ^2\left (-1+e^{5 x}\right )\right )}{\left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}\right ) \, dx \\ & = -\left (50 \int \frac {x^4 \left (4+25 x-15 \log \left (-1+e^{5 x}\right )-25 \log ^2\left (-1+e^{5 x}\right )\right )}{\left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx\right )-250 \int \frac {x^5}{\left (-1+e^x\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx+250 \int \frac {\left (4+3 e^x+2 e^{2 x}+e^{3 x}\right ) x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx \\ & = -\left (50 \int \left (-x^4+\frac {25 x^5}{\left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}-\frac {5 x^4}{-1+5 \log \left (-1+e^{5 x}\right )}\right ) \, dx\right )-250 \int \frac {x^5}{\left (-1+e^x\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx+250 \int \left (\frac {4 x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}+\frac {3 e^x x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}+\frac {2 e^{2 x} x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}+\frac {e^{3 x} x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}\right ) \, dx \\ & = 10 x^5-250 \int \frac {x^5}{\left (-1+e^x\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx+250 \int \frac {e^{3 x} x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx+250 \int \frac {x^4}{-1+5 \log \left (-1+e^{5 x}\right )} \, dx+500 \int \frac {e^{2 x} x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx+750 \int \frac {e^x x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx+1000 \int \frac {x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx-1250 \int \frac {x^5}{\left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx \\ \end{align*}
Time = 0.74 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=-50 \left (-\frac {x^5}{5}-\frac {x^5}{-1+5 \log \left (-1+e^{5 x}\right )}\right ) \]
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Time = 0.71 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09
method | result | size |
risch | \(10 x^{5}+\frac {50 x^{5}}{-1+5 \ln \left ({\mathrm e}^{5 x}-1\right )}\) | \(25\) |
parallelrisch | \(\frac {2500 \ln \left ({\mathrm e}^{5 x}-1\right ) x^{5}+2000 x^{5}}{-50+250 \ln \left ({\mathrm e}^{5 x}-1\right )}\) | \(34\) |
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Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=\frac {10 \, {\left (5 \, x^{5} \log \left (e^{\left (5 \, x\right )} - 1\right ) + 4 \, x^{5}\right )}}{5 \, \log \left (e^{\left (5 \, x\right )} - 1\right ) - 1} \]
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Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=10 x^{5} + \frac {50 x^{5}}{5 \log {\left (e^{5 x} - 1 \right )} - 1} \]
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Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (23) = 46\).
Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 3.04 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=\frac {10 \, {\left (5 \, x^{5} \log \left (e^{\left (4 \, x\right )} + e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x} + 1\right ) + 5 \, x^{5} \log \left (e^{x} - 1\right ) + 4 \, x^{5}\right )}}{5 \, \log \left (e^{\left (4 \, x\right )} + e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x} + 1\right ) + 5 \, \log \left (e^{x} - 1\right ) - 1} \]
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Time = 0.36 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=\frac {10 \, {\left (5 \, x^{5} \log \left (e^{\left (5 \, x\right )} - 1\right ) + 4 \, x^{5}\right )}}{5 \, \log \left (e^{\left (5 \, x\right )} - 1\right ) - 1} \]
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Time = 8.73 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.43 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=\frac {10\,x^4\,{\mathrm {e}}^{-5\,x}\,\left ({\mathrm {e}}^{5\,x}+5\,x\,{\mathrm {e}}^{5\,x}-1\right )-50\,x^4\,\ln \left ({\mathrm {e}}^{5\,x}-1\right )\,{\mathrm {e}}^{-5\,x}\,\left ({\mathrm {e}}^{5\,x}-1\right )}{5\,\ln \left ({\mathrm {e}}^{5\,x}-1\right )-1}-10\,x^4\,{\mathrm {e}}^{-5\,x}+10\,x^4+10\,x^5 \]
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