Integrand size = 95, antiderivative size = 26 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\frac {\log (x)}{1+\frac {8}{1-e^x+2 \log \left (\frac {3 x}{2}\right )}} \]
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\[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {9-10 e^x+e^{2 x}-8 \left (-2+e^x x\right ) \log (x)-4 \left (-5+e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{x \left (9-e^x+2 \log \left (\frac {3 x}{2}\right )\right )^2} \, dx \\ & = \int \left (\frac {1}{x}-\frac {8 (-1+x \log (x))}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )}-\frac {8 \log (x) \left (-2+9 x+2 x \log \left (\frac {3 x}{2}\right )\right )}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2}\right ) \, dx \\ & = \log (x)-8 \int \frac {-1+x \log (x)}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )} \, dx-8 \int \frac {\log (x) \left (-2+9 x+2 x \log \left (\frac {3 x}{2}\right )\right )}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2} \, dx \\ & = \log (x)-8 \int \left (\frac {1}{x \left (-e^x+9 \left (1+\frac {2}{9} \log \left (\frac {3}{2}\right )\right )+2 \log (x)\right )}+\frac {\log (x)}{-9+e^x-2 \log \left (\frac {3 x}{2}\right )}\right ) \, dx-8 \int \left (\frac {9 \log (x)}{\left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2}-\frac {2 \log (x)}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2}+\frac {2 \log (x) \log \left (\frac {3 x}{2}\right )}{\left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2}\right ) \, dx \\ & = \log (x)-8 \int \frac {1}{x \left (-e^x+9 \left (1+\frac {2}{9} \log \left (\frac {3}{2}\right )\right )+2 \log (x)\right )} \, dx-8 \int \frac {\log (x)}{-9+e^x-2 \log \left (\frac {3 x}{2}\right )} \, dx+16 \int \frac {\log (x)}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2} \, dx-16 \int \frac {\log (x) \log \left (\frac {3 x}{2}\right )}{\left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2} \, dx-72 \int \frac {\log (x)}{\left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2} \, dx \\ & = \log (x)-8 \int \frac {1}{x \left (-e^x+9 \left (1+\frac {2}{9} \log \left (\frac {3}{2}\right )\right )+2 \log (x)\right )} \, dx+16 \int \frac {\log (x)}{x \left (-9+e^x-2 \log \left (\frac {3 x}{2}\right )\right )^2} \, dx-16 \text {Subst}\left (\int \frac {\log (2 x)}{-9+e^{2 x}-2 \log (3 x)} \, dx,x,\frac {x}{2}\right )-32 \text {Subst}\left (\int \frac {\log (2 x) \log (3 x)}{\left (-9+e^{2 x}-2 \log (3 x)\right )^2} \, dx,x,\frac {x}{2}\right )-144 \text {Subst}\left (\int \frac {\log (2 x)}{\left (-9+e^{2 x}-2 \log (3 x)\right )^2} \, dx,x,\frac {x}{2}\right ) \\ \end{align*}
Time = 0.87 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\log (x)+\frac {4 \left (-9+e^x-\log \left (\frac {9}{4}\right )\right )}{-9+e^x-2 \log \left (\frac {3 x}{2}\right )} \]
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Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31
method | result | size |
parallelrisch | \(\frac {8 \,{\mathrm e}^{x} \ln \left (x \right )-8 \ln \left (x \right )-16 \ln \left (x \right ) \ln \left (\frac {3 x}{2}\right )}{-72+8 \,{\mathrm e}^{x}-16 \ln \left (\frac {3 x}{2}\right )}\) | \(34\) |
risch | \(\ln \left (x \right )+\frac {-36+8 \ln \left (2\right )-8 \ln \left (3\right )+4 \,{\mathrm e}^{x}}{-9+2 \ln \left (2\right )-2 \ln \left (3\right )+{\mathrm e}^{x}-2 \ln \left (x \right )}\) | \(36\) |
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Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\frac {{\left (e^{x} - 2 \, \log \left (\frac {3}{2}\right ) - 9\right )} \log \left (x\right ) - 2 \, \log \left (x\right )^{2} + 4 \, e^{x} - 8 \, \log \left (\frac {3}{2}\right ) - 36}{e^{x} - 2 \, \log \left (\frac {3}{2}\right ) - 2 \, \log \left (x\right ) - 9} \]
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Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\log {\left (x \right )} + \frac {8 \log {\left (x \right )}}{e^{x} - 2 \log {\left (x \right )} - 9 - 2 \log {\left (3 \right )} + 2 \log {\left (2 \right )}} \]
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Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\frac {8 \, \log \left (x\right )}{e^{x} - 2 \, \log \left (3\right ) + 2 \, \log \left (2\right ) - 2 \, \log \left (x\right ) - 9} + \log \left (x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).
Time = 0.31 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.96 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\frac {e^{x} \log \left (\frac {1}{2} \, x\right ) - 2 \, \log \left (3\right ) \log \left (\frac {1}{2} \, x\right ) - 2 \, \log \left (\frac {1}{2} \, x\right )^{2} + 8 \, \log \left (2\right ) - \log \left (\frac {1}{2} \, x\right )}{e^{x} - 2 \, \log \left (3\right ) - 2 \, \log \left (\frac {1}{2} \, x\right ) - 9} \]
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Time = 12.79 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\frac {\ln \left (x\right )\,\left (2\,\ln \left (\frac {3\,x}{2}\right )-{\mathrm {e}}^x+1\right )}{2\,\ln \left (\frac {3\,x}{2}\right )-{\mathrm {e}}^x+9} \]
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