Integrand size = 110, antiderivative size = 25 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=\frac {25 e^4 \log (x)}{\left (1-\frac {e^x}{2}\right ) \left (3+x^2\right )} \]
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\[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=\int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {50 e^4 \left (-\left (\left (-2+e^x\right ) \left (3+x^2\right )\right )+x \left (-4 x+e^x \left (3+2 x+x^2\right )\right ) \log (x)\right )}{\left (2-e^x\right )^2 x \left (3+x^2\right )^2} \, dx \\ & = \left (50 e^4\right ) \int \frac {-\left (\left (-2+e^x\right ) \left (3+x^2\right )\right )+x \left (-4 x+e^x \left (3+2 x+x^2\right )\right ) \log (x)}{\left (2-e^x\right )^2 x \left (3+x^2\right )^2} \, dx \\ & = \left (50 e^4\right ) \int \left (\frac {2 \log (x)}{\left (-2+e^x\right )^2 \left (3+x^2\right )}+\frac {-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)}{\left (-2+e^x\right ) x \left (3+x^2\right )^2}\right ) \, dx \\ & = \left (50 e^4\right ) \int \frac {-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)}{\left (-2+e^x\right ) x \left (3+x^2\right )^2} \, dx+\left (100 e^4\right ) \int \frac {\log (x)}{\left (-2+e^x\right )^2 \left (3+x^2\right )} \, dx \\ & = \left (50 e^4\right ) \int \left (\frac {-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)}{9 \left (-2+e^x\right ) x}-\frac {x \left (-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)\right )}{3 \left (-2+e^x\right ) \left (3+x^2\right )^2}-\frac {x \left (-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)\right )}{9 \left (-2+e^x\right ) \left (3+x^2\right )}\right ) \, dx-\left (100 e^4\right ) \int \frac {i \left (\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx+\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx\right )}{2 \sqrt {3} x} \, dx+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{\sqrt {3}} \\ & = \frac {1}{9} \left (50 e^4\right ) \int \frac {-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)}{\left (-2+e^x\right ) x} \, dx-\frac {1}{9} \left (50 e^4\right ) \int \frac {x \left (-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)\right )}{\left (-2+e^x\right ) \left (3+x^2\right )} \, dx-\frac {1}{3} \left (50 e^4\right ) \int \frac {x \left (-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)\right )}{\left (-2+e^x\right ) \left (3+x^2\right )^2} \, dx-\frac {\left (50 i e^4\right ) \int \frac {\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx+\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{x} \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{\sqrt {3}} \\ & = \frac {1}{9} \left (50 e^4\right ) \int \left (-\frac {3}{\left (-2+e^x\right ) x}-\frac {x}{-2+e^x}+\frac {3 \log (x)}{-2+e^x}+\frac {2 x \log (x)}{-2+e^x}+\frac {x^2 \log (x)}{-2+e^x}\right ) \, dx-\frac {1}{9} \left (50 e^4\right ) \int \left (-\frac {3 x}{\left (-2+e^x\right ) \left (3+x^2\right )}-\frac {x^3}{\left (-2+e^x\right ) \left (3+x^2\right )}+\frac {3 x^2 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )}+\frac {2 x^3 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )}+\frac {x^4 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )}\right ) \, dx-\frac {1}{3} \left (50 e^4\right ) \int \left (-\frac {3 x}{\left (-2+e^x\right ) \left (3+x^2\right )^2}-\frac {x^3}{\left (-2+e^x\right ) \left (3+x^2\right )^2}+\frac {3 x^2 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )^2}+\frac {2 x^3 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )^2}+\frac {x^4 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )^2}\right ) \, dx-\frac {\left (50 i e^4\right ) \int \left (\frac {\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx}{x}+\frac {\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{x}\right ) \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{\sqrt {3}} \\ & = -\left (\frac {1}{9} \left (50 e^4\right ) \int \frac {x}{-2+e^x} \, dx\right )+\frac {1}{9} \left (50 e^4\right ) \int \frac {x^3}{\left (-2+e^x\right ) \left (3+x^2\right )} \, dx+\frac {1}{9} \left (50 e^4\right ) \int \frac {x^2 \log (x)}{-2+e^x} \, dx-\frac {1}{9} \left (50 e^4\right ) \int \frac {x^4 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )} \, dx+\frac {1}{9} \left (100 e^4\right ) \int \frac {x \log (x)}{-2+e^x} \, dx-\frac {1}{9} \left (100 e^4\right ) \int \frac {x^3 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )} \, dx-\frac {1}{3} \left (50 e^4\right ) \int \frac {1}{\left (-2+e^x\right ) x} \, dx+\frac {1}{3} \left (50 e^4\right ) \int \frac {x^3}{\left (-2+e^x\right ) \left (3+x^2\right )^2} \, dx+\frac {1}{3} \left (50 e^4\right ) \int \frac {x}{\left (-2+e^x\right ) \left (3+x^2\right )} \, dx+\frac {1}{3} \left (50 e^4\right ) \int \frac {\log (x)}{-2+e^x} \, dx-\frac {1}{3} \left (50 e^4\right ) \int \frac {x^4 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )^2} \, dx-\frac {1}{3} \left (50 e^4\right ) \int \frac {x^2 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )} \, dx-\frac {1}{3} \left (100 e^4\right ) \int \frac {x^3 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )^2} \, dx+\left (50 e^4\right ) \int \frac {x}{\left (-2+e^x\right ) \left (3+x^2\right )^2} \, dx-\left (50 e^4\right ) \int \frac {x^2 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )^2} \, dx-\frac {\left (50 i e^4\right ) \int \frac {\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx}{x} \, dx}{\sqrt {3}}-\frac {\left (50 i e^4\right ) \int \frac {\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{x} \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{\sqrt {3}} \\ & = \text {Too large to display} \\ \end{align*}
Time = 0.52 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=-\frac {50 e^4 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )} \]
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Time = 1.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80
method | result | size |
risch | \(-\frac {50 \,{\mathrm e}^{4} \ln \left (x \right )}{\left (x^{2}+3\right ) \left ({\mathrm e}^{x}-2\right )}\) | \(20\) |
parallelrisch | \(-\frac {50 \,{\mathrm e}^{4} \ln \left (x \right )}{\left (x^{2}+3\right ) \left ({\mathrm e}^{x}-2\right )}\) | \(20\) |
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=\frac {50 \, e^{8} \log \left (x\right )}{2 \, {\left (x^{2} + 3\right )} e^{4} - {\left (x^{2} + 3\right )} e^{\left (x + 4\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=- \frac {50 e^{4} \log {\left (x \right )}}{- 2 x^{2} + \left (x^{2} + 3\right ) e^{x} - 6} \]
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=\frac {50 \, e^{4} \log \left (x\right )}{2 \, x^{2} - {\left (x^{2} + 3\right )} e^{x} + 6} \]
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Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=-\frac {50 \, e^{4} \log \left (x\right )}{x^{2} e^{x} - 2 \, x^{2} + 3 \, e^{x} - 6} \]
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Time = 13.48 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=-\frac {50\,{\mathrm {e}}^4\,\ln \left (x\right )}{\left (x^2+3\right )\,\left ({\mathrm {e}}^x-2\right )} \]
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