Integrand size = 83, antiderivative size = 20 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=x \left (-3+e^x-\frac {x^2}{2 x+\log (x)}\right ) \]
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\[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=\int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 \left (-11-4 x+4 e^x (1+x)\right )+x \left (4 e^x (1+x)-3 (4+x)\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{(2 x+\log (x))^2} \, dx \\ & = \int \left (e^x (1+x)-\frac {11 x^2}{(2 x+\log (x))^2}-\frac {4 x^3}{(2 x+\log (x))^2}-\frac {3 x (4+x) \log (x)}{(2 x+\log (x))^2}-\frac {3 \log ^2(x)}{(2 x+\log (x))^2}\right ) \, dx \\ & = -\left (3 \int \frac {x (4+x) \log (x)}{(2 x+\log (x))^2} \, dx\right )-3 \int \frac {\log ^2(x)}{(2 x+\log (x))^2} \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+\int e^x (1+x) \, dx \\ & = e^x (1+x)-3 \int \left (1+\frac {4 x^2}{(2 x+\log (x))^2}-\frac {4 x}{2 x+\log (x)}\right ) \, dx-3 \int \left (-\frac {2 x^2 (4+x)}{(2 x+\log (x))^2}+\frac {x (4+x)}{2 x+\log (x)}\right ) \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-\int e^x \, dx \\ & = -e^x-3 x+e^x (1+x)-3 \int \frac {x (4+x)}{2 x+\log (x)} \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx+6 \int \frac {x^2 (4+x)}{(2 x+\log (x))^2} \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-12 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+12 \int \frac {x}{2 x+\log (x)} \, dx \\ & = -e^x-3 x+e^x (1+x)-3 \int \left (\frac {4 x}{2 x+\log (x)}+\frac {x^2}{2 x+\log (x)}\right ) \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx+6 \int \left (\frac {4 x^2}{(2 x+\log (x))^2}+\frac {x^3}{(2 x+\log (x))^2}\right ) \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-12 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+12 \int \frac {x}{2 x+\log (x)} \, dx \\ & = -e^x-3 x+e^x (1+x)-3 \int \frac {x^2}{2 x+\log (x)} \, dx-4 \int \frac {x^3}{(2 x+\log (x))^2} \, dx+6 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-11 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-12 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+24 \int \frac {x^2}{(2 x+\log (x))^2} \, dx \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-3 x+e^x x-\frac {x^3}{2 x+\log (x)} \]
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Time = 1.91 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10
method | result | size |
risch | \({\mathrm e}^{x} x -3 x -\frac {x^{3}}{2 x +\ln \left (x \right )}\) | \(22\) |
parallelrisch | \(\frac {-x^{3}+2 \,{\mathrm e}^{x} x^{2}+x \,{\mathrm e}^{x} \ln \left (x \right )-6 x^{2}-3 x \ln \left (x \right )}{2 x +\ln \left (x \right )}\) | \(39\) |
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Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-\frac {x^{3} - 2 \, x^{2} e^{x} + 6 \, x^{2} - {\left (x e^{x} - 3 \, x\right )} \log \left (x\right )}{2 \, x + \log \left (x\right )} \]
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Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=- \frac {x^{3}}{2 x + \log {\left (x \right )}} + x e^{x} - 3 x \]
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Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-\frac {x^{3} + 6 \, x^{2} - {\left (2 \, x^{2} + x \log \left (x\right )\right )} e^{x} + 3 \, x \log \left (x\right )}{2 \, x + \log \left (x\right )} \]
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Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-\frac {x^{3} - 2 \, x^{2} e^{x} - x e^{x} \log \left (x\right ) + 6 \, x^{2} + 3 \, x \log \left (x\right )}{2 \, x + \log \left (x\right )} \]
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Time = 14.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 3.25 \[ \int \frac {-11 x^2-4 x^3+e^x \left (4 x^2+4 x^3\right )+\left (-12 x-3 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (-3+e^x (1+x)\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=\frac {3}{16\,\left (x+\frac {1}{2}\right )}-\frac {9\,x}{4}+x\,{\mathrm {e}}^x+\frac {\frac {3\,x^3\,\ln \left (x\right )}{2\,x+1}-\frac {x\,\left (x^2-4\,x^3\right )}{2\,x+1}}{2\,x+\ln \left (x\right )}-\frac {3\,x^2}{2} \]
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