Integrand size = 108, antiderivative size = 22 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 e^x}{(-3+x) \log (x) (-1+x \log (x))} \]
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\[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {20 e^x \left (-3+x+(10-3 x) x \log (x)+x \left (3-5 x+x^2\right ) \log ^2(x)\right )}{(3-x)^2 x \log ^2(x) (1-x \log (x))^2} \, dx \\ & = 20 \int \frac {e^x \left (-3+x+(10-3 x) x \log (x)+x \left (3-5 x+x^2\right ) \log ^2(x)\right )}{(3-x)^2 x \log ^2(x) (1-x \log (x))^2} \, dx \\ & = 20 \int \left (\frac {e^x}{(-3+x) x \log ^2(x)}+\frac {e^x (4-x)}{(-3+x)^2 \log (x)}+\frac {e^x (-1-x)}{(-3+x) (-1+x \log (x))^2}+\frac {e^x (-4+x) x}{(-3+x)^2 (-1+x \log (x))}\right ) \, dx \\ & = 20 \int \frac {e^x}{(-3+x) x \log ^2(x)} \, dx+20 \int \frac {e^x (4-x)}{(-3+x)^2 \log (x)} \, dx+20 \int \frac {e^x (-1-x)}{(-3+x) (-1+x \log (x))^2} \, dx+20 \int \frac {e^x (-4+x) x}{(-3+x)^2 (-1+x \log (x))} \, dx \\ & = 20 \int \left (\frac {e^x}{3 (-3+x) \log ^2(x)}-\frac {e^x}{3 x \log ^2(x)}\right ) \, dx+20 \int \left (\frac {e^x}{(-3+x)^2 \log (x)}-\frac {e^x}{(-3+x) \log (x)}\right ) \, dx+20 \int \left (-\frac {e^x}{(-1+x \log (x))^2}-\frac {4 e^x}{(-3+x) (-1+x \log (x))^2}\right ) \, dx+20 \int \left (\frac {e^x}{-1+x \log (x)}-\frac {3 e^x}{(-3+x)^2 (-1+x \log (x))}+\frac {2 e^x}{(-3+x) (-1+x \log (x))}\right ) \, dx \\ & = \frac {20}{3} \int \frac {e^x}{(-3+x) \log ^2(x)} \, dx-\frac {20}{3} \int \frac {e^x}{x \log ^2(x)} \, dx+20 \int \frac {e^x}{(-3+x)^2 \log (x)} \, dx-20 \int \frac {e^x}{(-3+x) \log (x)} \, dx-20 \int \frac {e^x}{(-1+x \log (x))^2} \, dx+20 \int \frac {e^x}{-1+x \log (x)} \, dx+40 \int \frac {e^x}{(-3+x) (-1+x \log (x))} \, dx-60 \int \frac {e^x}{(-3+x)^2 (-1+x \log (x))} \, dx-80 \int \frac {e^x}{(-3+x) (-1+x \log (x))^2} \, dx \\ \end{align*}
Time = 0.69 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 e^x}{(-3+x) \log (x) (-1+x \log (x))} \]
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Time = 1.54 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {20 \,{\mathrm e}^{x}}{\left (-3+x \right ) \ln \left (x \right ) \left (x \ln \left (x \right )-1\right )}\) | \(22\) |
parallelrisch | \(\frac {20 \,{\mathrm e}^{x}}{\ln \left (x \right ) \left (x^{2} \ln \left (x \right )-3 x \ln \left (x \right )-x +3\right )}\) | \(27\) |
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 \, e^{x}}{{\left (x^{2} - 3 \, x\right )} \log \left (x\right )^{2} - {\left (x - 3\right )} \log \left (x\right )} \]
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Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 e^{x}}{x^{2} \log {\left (x \right )}^{2} - 3 x \log {\left (x \right )}^{2} - x \log {\left (x \right )} + 3 \log {\left (x \right )}} \]
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Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 \, e^{x}}{{\left (x^{2} - 3 \, x\right )} \log \left (x\right )^{2} - {\left (x - 3\right )} \log \left (x\right )} \]
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Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 \, e^{x}}{x^{2} \log \left (x\right )^{2} - 3 \, x \log \left (x\right )^{2} - x \log \left (x\right ) + 3 \, \log \left (x\right )} \]
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Time = 7.40 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=-\frac {20\,{\mathrm {e}}^x}{\left (\ln \left (x\right )-x\,{\ln \left (x\right )}^2\right )\,\left (x-3\right )} \]
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