Integrand size = 92, antiderivative size = 31 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^x}{7 x \left (\log (x)+\frac {3 \left (4-e^3+2 \log (x)\right )}{x}\right )} \]
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\[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{63 e^6 x+\left (1008-504 e^3\right ) x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx \\ & = \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{\left (1008-504 e^3+63 e^6\right ) x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx \\ & = \int \frac {e^x \left (-6+\left (11-3 e^3\right ) x+x (5+x) \log (x)\right )}{7 x \left (3 \left (-4+e^3\right )-(6+x) \log (x)\right )^2} \, dx \\ & = \frac {1}{7} \int \frac {e^x \left (-6+\left (11-3 e^3\right ) x+x (5+x) \log (x)\right )}{x \left (3 \left (-4+e^3\right )-(6+x) \log (x)\right )^2} \, dx \\ & = \frac {1}{7} \int \left (\frac {e^x \left (-36-3 e^3 x-x^2\right )}{x (6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2}+\frac {e^x (5+x)}{(6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )}\right ) \, dx \\ & = \frac {1}{7} \int \frac {e^x \left (-36-3 e^3 x-x^2\right )}{x (6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx+\frac {1}{7} \int \frac {e^x (5+x)}{(6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )} \, dx \\ & = \frac {1}{7} \int \left (-\frac {e^x}{\left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2}-\frac {6 e^x}{x \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2}+\frac {3 e^x \left (4-e^3\right )}{(6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2}\right ) \, dx+\frac {1}{7} \int \left (\frac {e^x}{12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)}+\frac {e^x}{(-6-x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )}\right ) \, dx \\ & = -\left (\frac {1}{7} \int \frac {e^x}{\left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx\right )+\frac {1}{7} \int \frac {e^x}{12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)} \, dx+\frac {1}{7} \int \frac {e^x}{(-6-x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )} \, dx-\frac {6}{7} \int \frac {e^x}{x \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx+\frac {1}{7} \left (3 \left (4-e^3\right )\right ) \int \frac {e^x}{(6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx \\ & = -\left (\frac {1}{7} \int \frac {e^x}{\left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx\right )+\frac {1}{7} \int \frac {e^x}{(-6-x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )} \, dx+\frac {1}{7} \int \frac {e^x}{-3 \left (-4+e^3\right )+(6+x) \log (x)} \, dx-\frac {6}{7} \int \frac {e^x}{x \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx+\frac {1}{7} \left (3 \left (4-e^3\right )\right ) \int \frac {e^x}{(6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^x}{7 \left (12-3 e^3+6 \log (x)+x \log (x)\right )} \]
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Time = 0.50 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71
method | result | size |
risch | \(-\frac {{\mathrm e}^{x}}{7 \left (-x \ln \left (x \right )+3 \,{\mathrm e}^{3}-6 \ln \left (x \right )-12\right )}\) | \(22\) |
parallelrisch | \(-\frac {{\mathrm e}^{x}}{7 \left (-x \ln \left (x \right )+3 \,{\mathrm e}^{3}-6 \ln \left (x \right )-12\right )}\) | \(22\) |
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Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.58 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 \, {\left ({\left (x + 6\right )} \log \left (x\right ) - 3 \, e^{3} + 12\right )}} \]
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Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 x \log {\left (x \right )} + 42 \log {\left (x \right )} - 21 e^{3} + 84} \]
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Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.58 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 \, {\left ({\left (x + 6\right )} \log \left (x\right ) - 3 \, e^{3} + 12\right )}} \]
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Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 \, {\left (x \log \left (x\right ) - 3 \, e^{3} + 6 \, \log \left (x\right ) + 12\right )}} \]
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Timed out. \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=-\int \frac {{\mathrm {e}}^x\,\left (3\,x\,{\mathrm {e}}^3-11\,x+6\right )-{\mathrm {e}}^x\,\ln \left (x\right )\,\left (x^2+5\,x\right )}{\left (7\,x^3+84\,x^2+252\,x\right )\,{\ln \left (x\right )}^2+\left (1008\,x-{\mathrm {e}}^3\,\left (42\,x^2+252\,x\right )+168\,x^2\right )\,\ln \left (x\right )+1008\,x-504\,x\,{\mathrm {e}}^3+63\,x\,{\mathrm {e}}^6} \,d x \]
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