Integrand size = 73, antiderivative size = 30 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\frac {\left (-10+\frac {x}{2}+\log (x)\right ) \left (e^{e^{4 x^2}}-x+3 \log (x)\right )}{x} \]
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\[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{x^2} \, dx \\ & = \frac {1}{2} \int \left (8 e^{e^{4 x^2}+4 x^2} (-20+x+2 \log (x))-\frac {60-22 e^{e^{4 x^2}}-x+x^2-72 \log (x)+2 e^{e^{4 x^2}} \log (x)+6 \log ^2(x)}{x^2}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {60-22 e^{e^{4 x^2}}-x+x^2-72 \log (x)+2 e^{e^{4 x^2}} \log (x)+6 \log ^2(x)}{x^2} \, dx\right )+4 \int e^{e^{4 x^2}+4 x^2} (-20+x+2 \log (x)) \, dx \\ & = -\left (\frac {1}{2} \int \left (\frac {2 e^{e^{4 x^2}} (-11+\log (x))}{x^2}+\frac {60-x+x^2-72 \log (x)+6 \log ^2(x)}{x^2}\right ) \, dx\right )+4 \int \left (-20 e^{e^{4 x^2}+4 x^2}+e^{e^{4 x^2}+4 x^2} x+2 e^{e^{4 x^2}+4 x^2} \log (x)\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {60-x+x^2-72 \log (x)+6 \log ^2(x)}{x^2} \, dx\right )+4 \int e^{e^{4 x^2}+4 x^2} x \, dx+8 \int e^{e^{4 x^2}+4 x^2} \log (x) \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx-\int \frac {e^{e^{4 x^2}} (-11+\log (x))}{x^2} \, dx \\ & = -\left (\frac {1}{2} \int \left (\frac {60-x+x^2}{x^2}-\frac {72 \log (x)}{x^2}+\frac {6 \log ^2(x)}{x^2}\right ) \, dx\right )+2 \text {Subst}\left (\int e^{e^{4 x}+4 x} \, dx,x,x^2\right )-8 \int \frac {\int e^{e^{4 x^2}+4 x^2} \, dx}{x} \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx+(8 \log (x)) \int e^{e^{4 x^2}+4 x^2} \, dx-\int \left (-\frac {11 e^{e^{4 x^2}}}{x^2}+\frac {e^{e^{4 x^2}} \log (x)}{x^2}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {60-x+x^2}{x^2} \, dx\right )+\frac {1}{2} \text {Subst}\left (\int e^x \, dx,x,e^{4 x^2}\right )-3 \int \frac {\log ^2(x)}{x^2} \, dx-8 \int \frac {\int e^{e^{4 x^2}+4 x^2} \, dx}{x} \, dx+11 \int \frac {e^{e^{4 x^2}}}{x^2} \, dx+36 \int \frac {\log (x)}{x^2} \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx+(8 \log (x)) \int e^{e^{4 x^2}+4 x^2} \, dx-\int \frac {e^{e^{4 x^2}} \log (x)}{x^2} \, dx \\ & = \frac {1}{2} e^{e^{4 x^2}}-\frac {36}{x}-\frac {36 \log (x)}{x}+\frac {3 \log ^2(x)}{x}-\frac {1}{2} \int \left (1+\frac {60}{x^2}-\frac {1}{x}\right ) \, dx-6 \int \frac {\log (x)}{x^2} \, dx-8 \int \frac {\int e^{e^{4 x^2}+4 x^2} \, dx}{x} \, dx+11 \int \frac {e^{e^{4 x^2}}}{x^2} \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx-\log (x) \int \frac {e^{e^{4 x^2}}}{x^2} \, dx+(8 \log (x)) \int e^{e^{4 x^2}+4 x^2} \, dx+\int \frac {\int \frac {e^{e^{4 x^2}}}{x^2} \, dx}{x} \, dx \\ & = \frac {1}{2} e^{e^{4 x^2}}-\frac {x}{2}+\frac {\log (x)}{2}-\frac {30 \log (x)}{x}+\frac {3 \log ^2(x)}{x}-8 \int \frac {\int e^{e^{4 x^2}+4 x^2} \, dx}{x} \, dx+11 \int \frac {e^{e^{4 x^2}}}{x^2} \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx-\log (x) \int \frac {e^{e^{4 x^2}}}{x^2} \, dx+(8 \log (x)) \int e^{e^{4 x^2}+4 x^2} \, dx+\int \frac {\int \frac {e^{e^{4 x^2}}}{x^2} \, dx}{x} \, dx \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\frac {e^{e^{4 x^2}} (-20+x)-x^2+\left (-60+2 e^{e^{4 x^2}}+x\right ) \log (x)+6 \log ^2(x)}{2 x} \]
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Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.47
method | result | size |
risch | \(\frac {3 \ln \left (x \right )^{2}}{x}-\frac {30 \ln \left (x \right )}{x}+\frac {\ln \left (x \right )}{2}-\frac {x}{2}+\frac {\left (x -20+2 \ln \left (x \right )\right ) {\mathrm e}^{{\mathrm e}^{4 x^{2}}}}{2 x}\) | \(44\) |
parallelrisch | \(\frac {6 \ln \left (x \right )^{2}+2 \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{4 x^{2}}}+x \ln \left (x \right )+x \,{\mathrm e}^{{\mathrm e}^{4 x^{2}}}-x^{2}-60 \ln \left (x \right )-20 \,{\mathrm e}^{{\mathrm e}^{4 x^{2}}}}{2 x}\) | \(55\) |
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Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=-\frac {x^{2} - {\left (x + 2 \, \log \left (x\right ) - 20\right )} e^{\left (e^{\left (4 \, x^{2}\right )}\right )} - {\left (x - 60\right )} \log \left (x\right ) - 6 \, \log \left (x\right )^{2}}{2 \, x} \]
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Time = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=- \frac {x}{2} + \frac {\log {\left (x \right )}}{2} + \frac {\left (x + 2 \log {\left (x \right )} - 20\right ) e^{e^{4 x^{2}}}}{2 x} + \frac {3 \log {\left (x \right )}^{2}}{x} - \frac {30 \log {\left (x \right )}}{x} \]
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Time = 0.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=-\frac {1}{2} \, x + \frac {{\left (x + 2 \, \log \left (x\right ) - 20\right )} e^{\left (e^{\left (4 \, x^{2}\right )}\right )} + 6 \, \log \left (x\right )^{2} - 60 \, \log \left (x\right ) - 60}{2 \, x} + \frac {30}{x} + \frac {1}{2} \, \log \left (x\right ) \]
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\[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\int { -\frac {x^{2} - 2 \, {\left (4 \, {\left (x^{3} - 20 \, x^{2}\right )} e^{\left (4 \, x^{2}\right )} + {\left (8 \, x^{2} e^{\left (4 \, x^{2}\right )} - 1\right )} \log \left (x\right ) + 11\right )} e^{\left (e^{\left (4 \, x^{2}\right )}\right )} + 6 \, \log \left (x\right )^{2} - x - 72 \, \log \left (x\right ) + 60}{2 \, x^{2}} \,d x } \]
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Time = 12.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\frac {\ln \left (x\right )}{2}-\frac {x}{2}-\frac {30\,\ln \left (x\right )}{x}+\frac {3\,{\ln \left (x\right )}^2}{x}+\frac {{\mathrm {e}}^{{\mathrm {e}}^{4\,x^2}}\,\left (\frac {x}{2}+\ln \left (x\right )-10\right )}{x} \]
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