Integrand size = 66, antiderivative size = 28 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {e^x \left (2 x-\left (16+e^5\right ) x\right )}{-1+(5-x) \log (x)} \]
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\[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (14+e^5\right ) \left (6+\left (-5-5 x+x^2\right ) \log (x)\right )}{(1+(-5+x) \log (x))^2} \, dx \\ & = \left (14+e^5\right ) \int \frac {e^x \left (6+\left (-5-5 x+x^2\right ) \log (x)\right )}{(1+(-5+x) \log (x))^2} \, dx \\ & = \left (14+e^5\right ) \int \left (\frac {e^x \left (-25+11 x-x^2\right )}{(-5+x) (1-5 \log (x)+x \log (x))^2}+\frac {e^x \left (-5-5 x+x^2\right )}{(-5+x) (1-5 \log (x)+x \log (x))}\right ) \, dx \\ & = \left (14+e^5\right ) \int \frac {e^x \left (-25+11 x-x^2\right )}{(-5+x) (1-5 \log (x)+x \log (x))^2} \, dx+\left (14+e^5\right ) \int \frac {e^x \left (-5-5 x+x^2\right )}{(-5+x) (1-5 \log (x)+x \log (x))} \, dx \\ & = \left (14+e^5\right ) \int \left (\frac {6 e^x}{(1-5 \log (x)+x \log (x))^2}+\frac {5 e^x}{(-5+x) (1-5 \log (x)+x \log (x))^2}-\frac {e^x x}{(1-5 \log (x)+x \log (x))^2}\right ) \, dx+\left (14+e^5\right ) \int \left (-\frac {5 e^x}{(-5+x) (1-5 \log (x)+x \log (x))}+\frac {e^x x}{1-5 \log (x)+x \log (x)}\right ) \, dx \\ & = \left (-14-e^5\right ) \int \frac {e^x x}{(1-5 \log (x)+x \log (x))^2} \, dx+\left (14+e^5\right ) \int \frac {e^x x}{1-5 \log (x)+x \log (x)} \, dx+\left (5 \left (14+e^5\right )\right ) \int \frac {e^x}{(-5+x) (1-5 \log (x)+x \log (x))^2} \, dx-\left (5 \left (14+e^5\right )\right ) \int \frac {e^x}{(-5+x) (1-5 \log (x)+x \log (x))} \, dx+\left (6 \left (14+e^5\right )\right ) \int \frac {e^x}{(1-5 \log (x)+x \log (x))^2} \, dx \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {e^x \left (14+e^5\right ) x}{1-5 \log (x)+x \log (x)} \]
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75
method | result | size |
norman | \(\frac {\left (14+{\mathrm e}^{5}\right ) x \,{\mathrm e}^{x}}{x \ln \left (x \right )-5 \ln \left (x \right )+1}\) | \(21\) |
risch | \(\frac {\left (14+{\mathrm e}^{5}\right ) x \,{\mathrm e}^{x}}{x \ln \left (x \right )-5 \ln \left (x \right )+1}\) | \(21\) |
parallelrisch | \(\frac {x \,{\mathrm e}^{5} {\mathrm e}^{x}+14 \,{\mathrm e}^{x} x}{x \ln \left (x \right )-5 \ln \left (x \right )+1}\) | \(26\) |
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Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {{\left (x e^{5} + 14 \, x\right )} e^{x}}{{\left (x - 5\right )} \log \left (x\right ) + 1} \]
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Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {\left (14 x + x e^{5}\right ) e^{x}}{x \log {\left (x \right )} - 5 \log {\left (x \right )} + 1} \]
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Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {x {\left (e^{5} + 14\right )} e^{x}}{{\left (x - 5\right )} \log \left (x\right ) + 1} \]
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Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {x e^{\left (x + 5\right )} + 14 \, x e^{x}}{x \log \left (x\right ) - 5 \, \log \left (x\right ) + 1} \]
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Timed out. \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\int \frac {{\mathrm {e}}^x\,\left (6\,{\mathrm {e}}^5+84\right )-{\mathrm {e}}^x\,\ln \left (x\right )\,\left (70\,x+{\mathrm {e}}^5\,\left (-x^2+5\,x+5\right )-14\,x^2+70\right )}{\left (x^2-10\,x+25\right )\,{\ln \left (x\right )}^2+\left (2\,x-10\right )\,\ln \left (x\right )+1} \,d x \]
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