Integrand size = 109, antiderivative size = 28 \[ \int \frac {e^{\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e^{2 x} \left (-224 e+200 x+224 x^2\right )+e^{2 x} \left (-84 e+80 x+84 x^2\right ) \log \left (\frac {\log (3)}{x}\right )+e^{2 x} \left (-8 e+8 x+8 x^2\right ) \log ^2\left (\frac {\log (3)}{x}\right )\right )}{25+10 \log \left (\frac {\log (3)}{x}\right )+\log ^2\left (\frac {\log (3)}{x}\right )} \, dx=4 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (-e+x^2\right ) \]
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\[ \int \frac {e^{\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e^{2 x} \left (-224 e+200 x+224 x^2\right )+e^{2 x} \left (-84 e+80 x+84 x^2\right ) \log \left (\frac {\log (3)}{x}\right )+e^{2 x} \left (-8 e+8 x+8 x^2\right ) \log ^2\left (\frac {\log (3)}{x}\right )\right )}{25+10 \log \left (\frac {\log (3)}{x}\right )+\log ^2\left (\frac {\log (3)}{x}\right )} \, dx=\int \frac {e^{\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e^{2 x} \left (-224 e+200 x+224 x^2\right )+e^{2 x} \left (-84 e+80 x+84 x^2\right ) \log \left (\frac {\log (3)}{x}\right )+e^{2 x} \left (-8 e+8 x+8 x^2\right ) \log ^2\left (\frac {\log (3)}{x}\right )\right )}{25+10 \log \left (\frac {\log (3)}{x}\right )+\log ^2\left (\frac {\log (3)}{x}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e^{2 x} \left (-224 e+200 x+224 x^2\right )+e^{2 x} \left (-84 e+80 x+84 x^2\right ) \log \left (\frac {\log (3)}{x}\right )+e^{2 x} \left (-8 e+8 x+8 x^2\right ) \log ^2\left (\frac {\log (3)}{x}\right )\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx \\ & = \int \left (-\frac {224 e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {200 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {224 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}-\frac {84 e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \log \left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {80 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x \log \left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {84 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2 \log \left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}-\frac {8 e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \log ^2\left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {8 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x \log ^2\left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {8 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2 \log ^2\left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}\right ) \, dx \\ & = -\left (8 \int \frac {e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \log ^2\left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx\right )+8 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x \log ^2\left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+8 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2 \log ^2\left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+80 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x \log \left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx-84 \int \frac {e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \log \left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+84 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2 \log \left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+200 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx-224 \int \frac {e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+224 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx \\ & = -\left (8 \int \left (e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}+\frac {25 e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}-\frac {10 e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{5+\log \left (\frac {\log (3)}{x}\right )}\right ) \, dx\right )+8 \int \left (e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x+\frac {25 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}-\frac {10 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x}{5+\log \left (\frac {\log (3)}{x}\right )}\right ) \, dx+8 \int \left (e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2+\frac {25 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}-\frac {10 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{5+\log \left (\frac {\log (3)}{x}\right )}\right ) \, dx+80 \int \left (-\frac {5 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x}{5+\log \left (\frac {\log (3)}{x}\right )}\right ) \, dx-84 \int \left (-\frac {5 e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{5+\log \left (\frac {\log (3)}{x}\right )}\right ) \, dx+84 \int \left (-\frac {5 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{5+\log \left (\frac {\log (3)}{x}\right )}\right ) \, dx+200 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx-224 \int \frac {e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+224 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx \\ & = -\left (8 \int e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \, dx\right )+8 \int e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x \, dx+8 \int e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2 \, dx+80 \int \frac {e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{5+\log \left (\frac {\log (3)}{x}\right )} \, dx-80 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{5+\log \left (\frac {\log (3)}{x}\right )} \, dx-84 \int \frac {e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{5+\log \left (\frac {\log (3)}{x}\right )} \, dx+84 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{5+\log \left (\frac {\log (3)}{x}\right )} \, dx-200 \int \frac {e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+2 \left (200 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx\right )+200 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx-224 \int \frac {e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+224 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx-400 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+420 \int \frac {e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx-420 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e^{2 x} \left (-224 e+200 x+224 x^2\right )+e^{2 x} \left (-84 e+80 x+84 x^2\right ) \log \left (\frac {\log (3)}{x}\right )+e^{2 x} \left (-8 e+8 x+8 x^2\right ) \log ^2\left (\frac {\log (3)}{x}\right )\right )}{25+10 \log \left (\frac {\log (3)}{x}\right )+\log ^2\left (\frac {\log (3)}{x}\right )} \, dx=-4 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e-x^2\right ) \]
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Time = 23.41 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29
method | result | size |
risch | \(-4 \left ({\mathrm e}-x^{2}\right ) {\mathrm e}^{\frac {x \left (-2 \ln \left (\ln \left (3\right )\right )+2 \ln \left (x \right )-11\right )}{-\ln \left (\ln \left (3\right )\right )+\ln \left (x \right )-5}}\) | \(36\) |
parallelrisch | \(4 \,{\mathrm e}^{\frac {x}{\ln \left (\frac {\ln \left (3\right )}{x}\right )+5}} x^{2} {\mathrm e}^{2 x}-4 \,{\mathrm e} \,{\mathrm e}^{\frac {x}{\ln \left (\frac {\ln \left (3\right )}{x}\right )+5}} {\mathrm e}^{2 x}\) | \(47\) |
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e^{2 x} \left (-224 e+200 x+224 x^2\right )+e^{2 x} \left (-84 e+80 x+84 x^2\right ) \log \left (\frac {\log (3)}{x}\right )+e^{2 x} \left (-8 e+8 x+8 x^2\right ) \log ^2\left (\frac {\log (3)}{x}\right )\right )}{25+10 \log \left (\frac {\log (3)}{x}\right )+\log ^2\left (\frac {\log (3)}{x}\right )} \, dx=4 \, {\left (x^{2} - e\right )} e^{\left (2 \, x + \frac {x}{\log \left (\frac {\log \left (3\right )}{x}\right ) + 5}\right )} \]
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Timed out. \[ \int \frac {e^{\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e^{2 x} \left (-224 e+200 x+224 x^2\right )+e^{2 x} \left (-84 e+80 x+84 x^2\right ) \log \left (\frac {\log (3)}{x}\right )+e^{2 x} \left (-8 e+8 x+8 x^2\right ) \log ^2\left (\frac {\log (3)}{x}\right )\right )}{25+10 \log \left (\frac {\log (3)}{x}\right )+\log ^2\left (\frac {\log (3)}{x}\right )} \, dx=\text {Timed out} \]
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Time = 0.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e^{2 x} \left (-224 e+200 x+224 x^2\right )+e^{2 x} \left (-84 e+80 x+84 x^2\right ) \log \left (\frac {\log (3)}{x}\right )+e^{2 x} \left (-8 e+8 x+8 x^2\right ) \log ^2\left (\frac {\log (3)}{x}\right )\right )}{25+10 \log \left (\frac {\log (3)}{x}\right )+\log ^2\left (\frac {\log (3)}{x}\right )} \, dx=4 \, {\left (x^{2} - e\right )} e^{\left (2 \, x - \frac {x}{\log \left (x\right ) - \log \left (\log \left (3\right )\right ) - 5}\right )} \]
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\[ \int \frac {e^{\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e^{2 x} \left (-224 e+200 x+224 x^2\right )+e^{2 x} \left (-84 e+80 x+84 x^2\right ) \log \left (\frac {\log (3)}{x}\right )+e^{2 x} \left (-8 e+8 x+8 x^2\right ) \log ^2\left (\frac {\log (3)}{x}\right )\right )}{25+10 \log \left (\frac {\log (3)}{x}\right )+\log ^2\left (\frac {\log (3)}{x}\right )} \, dx=\int { \frac {4 \, {\left (2 \, {\left (x^{2} + x - e\right )} e^{\left (2 \, x\right )} \log \left (\frac {\log \left (3\right )}{x}\right )^{2} + {\left (21 \, x^{2} + 20 \, x - 21 \, e\right )} e^{\left (2 \, x\right )} \log \left (\frac {\log \left (3\right )}{x}\right ) + 2 \, {\left (28 \, x^{2} + 25 \, x - 28 \, e\right )} e^{\left (2 \, x\right )}\right )} e^{\left (\frac {x}{\log \left (\frac {\log \left (3\right )}{x}\right ) + 5}\right )}}{\log \left (\frac {\log \left (3\right )}{x}\right )^{2} + 10 \, \log \left (\frac {\log \left (3\right )}{x}\right ) + 25} \,d x } \]
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Timed out. \[ \int \frac {e^{\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e^{2 x} \left (-224 e+200 x+224 x^2\right )+e^{2 x} \left (-84 e+80 x+84 x^2\right ) \log \left (\frac {\log (3)}{x}\right )+e^{2 x} \left (-8 e+8 x+8 x^2\right ) \log ^2\left (\frac {\log (3)}{x}\right )\right )}{25+10 \log \left (\frac {\log (3)}{x}\right )+\log ^2\left (\frac {\log (3)}{x}\right )} \, dx=\int \frac {{\mathrm {e}}^{\frac {x}{\ln \left (\frac {\ln \left (3\right )}{x}\right )+5}}\,\left ({\mathrm {e}}^{2\,x}\,\left (8\,x^2+8\,x-8\,\mathrm {e}\right )\,{\ln \left (\frac {\ln \left (3\right )}{x}\right )}^2+{\mathrm {e}}^{2\,x}\,\left (84\,x^2+80\,x-84\,\mathrm {e}\right )\,\ln \left (\frac {\ln \left (3\right )}{x}\right )+{\mathrm {e}}^{2\,x}\,\left (224\,x^2+200\,x-224\,\mathrm {e}\right )\right )}{{\ln \left (\frac {\ln \left (3\right )}{x}\right )}^2+10\,\ln \left (\frac {\ln \left (3\right )}{x}\right )+25} \,d x \]
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