Integrand size = 111, antiderivative size = 27 \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {3 x}{5+e^{-8+50 x^2}+x-\log (5)-\log \left (x^2\right )} \]
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\[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {3 e^8 \left (-e^{50 x^2} \left (-1+100 x^2\right )-e^8 (-7+\log (5))-e^8 \log \left (x^2\right )\right )}{\left (e^{50 x^2}+e^8 (5+x-\log (5))-e^8 \log \left (x^2\right )\right )^2} \, dx \\ & = \left (3 e^8\right ) \int \frac {-e^{50 x^2} \left (-1+100 x^2\right )-e^8 (-7+\log (5))-e^8 \log \left (x^2\right )}{\left (e^{50 x^2}+e^8 (5+x-\log (5))-e^8 \log \left (x^2\right )\right )^2} \, dx \\ & = \left (3 e^8\right ) \int \left (\frac {1-100 x^2}{e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )}+\frac {e^8 \left (2-x+100 x^3+500 x^2 \left (1-\frac {\log (5)}{5}\right )-100 x^2 \log \left (x^2\right )\right )}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}\right ) \, dx \\ & = \left (3 e^8\right ) \int \frac {1-100 x^2}{e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )} \, dx+\left (3 e^{16}\right ) \int \frac {2-x+100 x^3+500 x^2 \left (1-\frac {\log (5)}{5}\right )-100 x^2 \log \left (x^2\right )}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx \\ & = \left (3 e^8\right ) \int \left (\frac {1}{e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )}+\frac {100 x^2}{-e^{50 x^2}-e^8 x-5 e^8 \left (1-\frac {\log (5)}{5}\right )+e^8 \log \left (x^2\right )}\right ) \, dx+\left (3 e^{16}\right ) \int \left (\frac {2}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}-\frac {x}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}+\frac {100 x^3}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}+\frac {100 x^2 (5-\log (5))}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}-\frac {100 x^2 \log \left (x^2\right )}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}\right ) \, dx \\ & = \left (3 e^8\right ) \int \frac {1}{e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )} \, dx+\left (300 e^8\right ) \int \frac {x^2}{-e^{50 x^2}-e^8 x-5 e^8 \left (1-\frac {\log (5)}{5}\right )+e^8 \log \left (x^2\right )} \, dx-\left (3 e^{16}\right ) \int \frac {x}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx+\left (6 e^{16}\right ) \int \frac {1}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx+\left (300 e^{16}\right ) \int \frac {x^3}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx-\left (300 e^{16}\right ) \int \frac {x^2 \log \left (x^2\right )}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx+\left (300 e^{16} (5-\log (5))\right ) \int \frac {x^2}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx \\ \end{align*}
Time = 0.53 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {3 e^8 x}{e^{50 x^2}+e^8 (5+x-\log (5))-e^8 \log \left (x^2\right )} \]
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Time = 3.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(-\frac {3 x}{-{\mathrm e}^{50 x^{2}-8}+\ln \left (x^{2}\right )+\ln \left (5\right )-x -5}\) | \(29\) |
risch | \(\frac {6 x}{i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 \,{\mathrm e}^{2 \left (5 x -2\right ) \left (2+5 x \right )}-2 \ln \left (5\right )+2 x -4 \ln \left (x \right )+10}\) | \(83\) |
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Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {3 \, x}{x + e^{\left (50 \, x^{2} - 8\right )} - \log \left (5\right ) - \log \left (x^{2}\right ) + 5} \]
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Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {3 x}{x + e^{50 x^{2} - 8} - \log {\left (x^{2} \right )} - \log {\left (5 \right )} + 5} \]
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Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {3 \, x e^{8}}{x e^{8} - {\left (\log \left (5\right ) - 5\right )} e^{8} - 2 \, e^{8} \log \left (x\right ) + e^{\left (50 \, x^{2}\right )}} \]
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Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {3 \, x e^{8}}{x e^{8} - e^{8} \log \left (5\right ) - e^{8} \log \left (x^{2}\right ) + 5 \, e^{8} + e^{\left (50 \, x^{2}\right )}} \]
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Timed out. \[ \int \frac {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\int -\frac {3\,\ln \left (x^2\right )+3\,\ln \left (5\right )+{\mathrm {e}}^{50\,x^2-8}\,\left (300\,x^2-3\right )-21}{10\,x+{\mathrm {e}}^{100\,x^2-16}-\ln \left (x^2\right )\,\left (2\,x-2\,\ln \left (5\right )+2\,{\mathrm {e}}^{50\,x^2-8}+10\right )-\ln \left (5\right )\,\left (2\,x+10\right )+{\mathrm {e}}^{50\,x^2-8}\,\left (2\,x-2\,\ln \left (5\right )+10\right )+{\ln \left (x^2\right )}^2+{\ln \left (5\right )}^2+x^2+25} \,d x \]
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