Integrand size = 82, antiderivative size = 22 \[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=-1+\frac {-1+5 \left (e^4+x+\log (x)\right )^x}{2 x} \]
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\[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 x^2 \left (e^4+x+\log (x)\right )} \, dx \\ & = \frac {1}{2} \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{x^2 \left (e^4+x+\log (x)\right )} \, dx \\ & = \frac {1}{2} \int \left (\frac {1}{x^2}+\frac {5 \left (e^4+x+\log (x)\right )^{-1+x} \left (-e^4+x^2-\log (x)+e^4 x \log \left (e^4+x+\log (x)\right )+x^2 \log \left (e^4+x+\log (x)\right )+x \log (x) \log \left (e^4+x+\log (x)\right )\right )}{x^2}\right ) \, dx \\ & = -\frac {1}{2 x}+\frac {5}{2} \int \frac {\left (e^4+x+\log (x)\right )^{-1+x} \left (-e^4+x^2-\log (x)+e^4 x \log \left (e^4+x+\log (x)\right )+x^2 \log \left (e^4+x+\log (x)\right )+x \log (x) \log \left (e^4+x+\log (x)\right )\right )}{x^2} \, dx \\ & = -\frac {1}{2 x}+\frac {5}{2} \int \left (\frac {\left (-e^4+x^2-\log (x)\right ) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2}+\frac {\left (e^4+x+\log (x)\right )^x \log \left (e^4+x+\log (x)\right )}{x}\right ) \, dx \\ & = -\frac {1}{2 x}+\frac {5}{2} \int \frac {\left (-e^4+x^2-\log (x)\right ) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2} \, dx+\frac {5}{2} \int \frac {\left (e^4+x+\log (x)\right )^x \log \left (e^4+x+\log (x)\right )}{x} \, dx \\ & = -\frac {1}{2 x}+\frac {5}{2} \int \left (\frac {\left (-e^4+x^2\right ) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2}-\frac {\log (x) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2}\right ) \, dx+\frac {5}{2} \int \frac {\left (e^4+x+\log (x)\right )^x \log \left (e^4+x+\log (x)\right )}{x} \, dx \\ & = -\frac {1}{2 x}+\frac {5}{2} \int \frac {\left (-e^4+x^2\right ) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2} \, dx-\frac {5}{2} \int \frac {\log (x) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2} \, dx+\frac {5}{2} \int \frac {\left (e^4+x+\log (x)\right )^x \log \left (e^4+x+\log (x)\right )}{x} \, dx \\ & = -\frac {1}{2 x}-\frac {5}{2} \int \frac {\log (x) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2} \, dx+\frac {5}{2} \int \left (\left (e^4+x+\log (x)\right )^{-1+x}-\frac {e^4 \left (e^4+x+\log (x)\right )^{-1+x}}{x^2}\right ) \, dx+\frac {5}{2} \int \frac {\left (e^4+x+\log (x)\right )^x \log \left (e^4+x+\log (x)\right )}{x} \, dx \\ & = -\frac {1}{2 x}+\frac {5}{2} \int \left (e^4+x+\log (x)\right )^{-1+x} \, dx-\frac {5}{2} \int \frac {\log (x) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2} \, dx+\frac {5}{2} \int \frac {\left (e^4+x+\log (x)\right )^x \log \left (e^4+x+\log (x)\right )}{x} \, dx-\frac {1}{2} \left (5 e^4\right ) \int \frac {\left (e^4+x+\log (x)\right )^{-1+x}}{x^2} \, dx \\ \end{align*}
Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\frac {-1+5 \left (e^4+x+\log (x)\right )^x}{2 x} \]
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Time = 2.95 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {1}{2 x}+\frac {5 \left (\ln \left (x \right )+x +{\mathrm e}^{4}\right )^{x}}{2 x}\) | \(20\) |
parallelrisch | \(\frac {5 \,{\mathrm e}^{x \ln \left (\ln \left (x \right )+x +{\mathrm e}^{4}\right )}-1}{2 x}\) | \(20\) |
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Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\frac {5 \, {\left (x + e^{4} + \log \left (x\right )\right )}^{x} - 1}{2 \, x} \]
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Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\frac {5 e^{x \log {\left (x + \log {\left (x \right )} + e^{4} \right )}}}{2 x} - \frac {1}{2 x} \]
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Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\frac {5 \, {\left (x + e^{4} + \log \left (x\right )\right )}^{x} - 1}{2 \, x} \]
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\[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\int { \frac {5 \, {\left (x^{2} + {\left (x^{2} + x e^{4} + x \log \left (x\right )\right )} \log \left (x + e^{4} + \log \left (x\right )\right ) - e^{4} - \log \left (x\right )\right )} {\left (x + e^{4} + \log \left (x\right )\right )}^{x} + x + e^{4} + \log \left (x\right )}{2 \, {\left (x^{3} + x^{2} e^{4} + x^{2} \log \left (x\right )\right )}} \,d x } \]
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Time = 13.93 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\frac {5\,{\left (x+{\mathrm {e}}^4+\ln \left (x\right )\right )}^x-1}{2\,x} \]
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