Integrand size = 129, antiderivative size = 32 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-x+\frac {-e^{e^x}+\log \left (x^2\right )}{\log \left (\frac {(2+x) \left (x+x^2\right )}{x}\right )} \]
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\[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=\int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{x \left (2+3 x+x^2\right ) \log ^2\left (2+3 x+x^2\right )} \, dx \\ & = \int \left (-1+\frac {(3+2 x) \left (e^{e^x}-\log \left (x^2\right )\right )}{\left (2+3 x+x^2\right ) \log ^2\left (2+3 x+x^2\right )}+\frac {2-e^{e^x+x} x}{x \log \left (2+3 x+x^2\right )}\right ) \, dx \\ & = -x+\int \frac {(3+2 x) \left (e^{e^x}-\log \left (x^2\right )\right )}{\left (2+3 x+x^2\right ) \log ^2\left (2+3 x+x^2\right )} \, dx+\int \frac {2-e^{e^x+x} x}{x \log \left (2+3 x+x^2\right )} \, dx \\ & = -x+\int \left (\frac {e^{e^x} (3+2 x)}{(1+x) (2+x) \log ^2\left (2+3 x+x^2\right )}-\frac {(3+2 x) \log \left (x^2\right )}{(1+x) (2+x) \log ^2\left (2+3 x+x^2\right )}\right ) \, dx+\int \left (-\frac {e^{e^x+x}}{\log \left (2+3 x+x^2\right )}+\frac {2}{x \log \left (2+3 x+x^2\right )}\right ) \, dx \\ & = -x+2 \int \frac {1}{x \log \left (2+3 x+x^2\right )} \, dx+\int \frac {e^{e^x} (3+2 x)}{(1+x) (2+x) \log ^2\left (2+3 x+x^2\right )} \, dx-\int \frac {(3+2 x) \log \left (x^2\right )}{(1+x) (2+x) \log ^2\left (2+3 x+x^2\right )} \, dx-\int \frac {e^{e^x+x}}{\log \left (2+3 x+x^2\right )} \, dx \\ & = -x+2 \int \frac {1}{x \log \left (2+3 x+x^2\right )} \, dx+\int \left (\frac {e^{e^x}}{(1+x) \log ^2\left (2+3 x+x^2\right )}+\frac {e^{e^x}}{(2+x) \log ^2\left (2+3 x+x^2\right )}\right ) \, dx-\int \left (\frac {\log \left (x^2\right )}{(1+x) \log ^2\left (2+3 x+x^2\right )}+\frac {\log \left (x^2\right )}{(2+x) \log ^2\left (2+3 x+x^2\right )}\right ) \, dx-\int \frac {e^{e^x+x}}{\log \left (2+3 x+x^2\right )} \, dx \\ & = -x+2 \int \frac {1}{x \log \left (2+3 x+x^2\right )} \, dx+\int \frac {e^{e^x}}{(1+x) \log ^2\left (2+3 x+x^2\right )} \, dx+\int \frac {e^{e^x}}{(2+x) \log ^2\left (2+3 x+x^2\right )} \, dx-\int \frac {\log \left (x^2\right )}{(1+x) \log ^2\left (2+3 x+x^2\right )} \, dx-\int \frac {\log \left (x^2\right )}{(2+x) \log ^2\left (2+3 x+x^2\right )} \, dx-\int \frac {e^{e^x+x}}{\log \left (2+3 x+x^2\right )} \, dx \\ \end{align*}
Time = 0.43 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-x-\frac {e^{e^x}}{\log \left (2+3 x+x^2\right )}+\frac {\log \left (x^2\right )}{\log \left (2+3 x+x^2\right )} \]
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Time = 59.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53
method | result | size |
parallelrisch | \(\frac {-6 x \ln \left (x^{2}+3 x +2\right )+6 \ln \left (x^{2}\right )-6 \,{\mathrm e}^{{\mathrm e}^{x}}+13 \ln \left (x^{2}+3 x +2\right )}{6 \ln \left (x^{2}+3 x +2\right )}\) | \(49\) |
risch | \(-x +\frac {4 \ln \left (x \right )-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2 \ln \left (x^{2}+3 x +2\right )}-\frac {{\mathrm e}^{{\mathrm e}^{x}}}{\ln \left (x^{2}+3 x +2\right )}\) | \(88\) |
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Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-\frac {x \log \left (x^{2} + 3 \, x + 2\right ) + e^{\left (e^{x}\right )} - \log \left (x^{2}\right )}{\log \left (x^{2} + 3 \, x + 2\right )} \]
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Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=- x - \frac {e^{e^{x}}}{\log {\left (x^{2} + 3 x + 2 \right )}} + \frac {\log {\left (x^{2} \right )}}{\log {\left (x^{2} + 3 x + 2 \right )}} \]
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Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-\frac {x \log \left (x + 2\right ) + x \log \left (x + 1\right ) + e^{\left (e^{x}\right )} - 2 \, \log \left (x\right )}{\log \left (x + 2\right ) + \log \left (x + 1\right )} \]
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Time = 0.37 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-\frac {x \log \left (x^{2} + 3 \, x + 2\right ) + e^{\left (e^{x}\right )} - \log \left (x^{2}\right )}{\log \left (x^{2} + 3 \, x + 2\right )} \]
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Timed out. \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=\int \frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (3\,x+2\,x^2-{\mathrm {e}}^x\,\ln \left (x^2+3\,x+2\right )\,\left (x^3+3\,x^2+2\,x\right )\right )-\ln \left (x^2\right )\,\left (2\,x^2+3\,x\right )+\ln \left (x^2+3\,x+2\right )\,\left (2\,x^2+6\,x+4\right )-{\ln \left (x^2+3\,x+2\right )}^2\,\left (x^3+3\,x^2+2\,x\right )}{{\ln \left (x^2+3\,x+2\right )}^2\,\left (x^3+3\,x^2+2\,x\right )} \,d x \]
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