Integrand size = 109, antiderivative size = 30 \[ \int \frac {3 x+e^{-5+x} \left (-e^{1-x} x+x^3\right ) \log (x)+\left (e^{-5+x} \left (-e x^2-x^3+e^{1-x} (e+x)\right )+e^{-5+x} \left (-2 x^3-x^4+e \left (-2 x^2-x^3\right )\right ) \log (x)\right ) \log (e+x)}{\left (e x+x^2\right ) \log ^2(e+x)} \, dx=\frac {-3+e^{-5+x} \left (e^{1-x}-x^2\right ) \log (x)}{\log (e+x)} \]
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\[ \int \frac {3 x+e^{-5+x} \left (-e^{1-x} x+x^3\right ) \log (x)+\left (e^{-5+x} \left (-e x^2-x^3+e^{1-x} (e+x)\right )+e^{-5+x} \left (-2 x^3-x^4+e \left (-2 x^2-x^3\right )\right ) \log (x)\right ) \log (e+x)}{\left (e x+x^2\right ) \log ^2(e+x)} \, dx=\int \frac {3 x+e^{-5+x} \left (-e^{1-x} x+x^3\right ) \log (x)+\left (e^{-5+x} \left (-e x^2-x^3+e^{1-x} (e+x)\right )+e^{-5+x} \left (-2 x^3-x^4+e \left (-2 x^2-x^3\right )\right ) \log (x)\right ) \log (e+x)}{\left (e x+x^2\right ) \log ^2(e+x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {3 x+e^{-5+x} \left (-e^{1-x} x+x^3\right ) \log (x)+\left (e^{-5+x} \left (-e x^2-x^3+e^{1-x} (e+x)\right )+e^{-5+x} \left (-2 x^3-x^4+e \left (-2 x^2-x^3\right )\right ) \log (x)\right ) \log (e+x)}{x (e+x) \log ^2(e+x)} \, dx \\ & = \int \frac {3 e^5 x+(e+x) \left (e-e^x x^2\right ) \log (e+x)-x \log (x) \left (e-e^x x^2+e^x x (2+x) (e+x) \log (e+x)\right )}{e^5 x (e+x) \log ^2(e+x)} \, dx \\ & = \frac {\int \frac {3 e^5 x+(e+x) \left (e-e^x x^2\right ) \log (e+x)-x \log (x) \left (e-e^x x^2+e^x x (2+x) (e+x) \log (e+x)\right )}{x (e+x) \log ^2(e+x)} \, dx}{e^5} \\ & = \frac {\int \left (\frac {e \left (3 e^4 x-x \log (x)+e \log (e+x)+x \log (e+x)\right )}{x (e+x) \log ^2(e+x)}+\frac {e^x x \left (x \log (x)-e \log (e+x)-x \log (e+x)-2 e \log (x) \log (e+x)-2 \left (1+\frac {e}{2}\right ) x \log (x) \log (e+x)-x^2 \log (x) \log (e+x)\right )}{(e+x) \log ^2(e+x)}\right ) \, dx}{e^5} \\ & = \frac {\int \frac {e^x x \left (x \log (x)-e \log (e+x)-x \log (e+x)-2 e \log (x) \log (e+x)-2 \left (1+\frac {e}{2}\right ) x \log (x) \log (e+x)-x^2 \log (x) \log (e+x)\right )}{(e+x) \log ^2(e+x)} \, dx}{e^5}+\frac {\int \frac {3 e^4 x-x \log (x)+e \log (e+x)+x \log (e+x)}{x (e+x) \log ^2(e+x)} \, dx}{e^4} \\ & = \frac {\int \frac {e^x x (-((e+x) \log (e+x))-\log (x) (-x+(2+x) (e+x) \log (e+x)))}{(e+x) \log ^2(e+x)} \, dx}{e^5}+\frac {\int \left (\frac {3 e^4-\log (x)}{(e+x) \log ^2(e+x)}+\frac {1}{x \log (e+x)}\right ) \, dx}{e^4} \\ & = \frac {\int \left (\frac {e^x x^2 \log (x)}{(e+x) \log ^2(e+x)}-\frac {e^x x (1+2 \log (x)+x \log (x))}{\log (e+x)}\right ) \, dx}{e^5}+\frac {\int \frac {3 e^4-\log (x)}{(e+x) \log ^2(e+x)} \, dx}{e^4}+\frac {\int \frac {1}{x \log (e+x)} \, dx}{e^4} \\ & = \frac {\int \frac {e^x x^2 \log (x)}{(e+x) \log ^2(e+x)} \, dx}{e^5}-\frac {\int \frac {e^x x (1+2 \log (x)+x \log (x))}{\log (e+x)} \, dx}{e^5}+\frac {\int \frac {1}{x \log (e+x)} \, dx}{e^4}+\frac {\text {Subst}\left (\int \frac {3 e^4-\log (-e+x)}{x \log ^2(x)} \, dx,x,e+x\right )}{e^4} \\ & = \frac {\int \left (-\frac {e^{1+x} \log (x)}{\log ^2(e+x)}+\frac {e^x x \log (x)}{\log ^2(e+x)}+\frac {e^{2+x} \log (x)}{(e+x) \log ^2(e+x)}\right ) \, dx}{e^5}-\frac {\int \left (\frac {e^x x}{\log (e+x)}+\frac {2 e^x x \log (x)}{\log (e+x)}+\frac {e^x x^2 \log (x)}{\log (e+x)}\right ) \, dx}{e^5}+\frac {\int \frac {1}{x \log (e+x)} \, dx}{e^4}+\frac {\text {Subst}\left (\int \frac {3 e^4-\log (-e+x)}{x \log ^2(x)} \, dx,x,e+x\right )}{e^4} \\ & = -\frac {\int \frac {e^{1+x} \log (x)}{\log ^2(e+x)} \, dx}{e^5}+\frac {\int \frac {e^x x \log (x)}{\log ^2(e+x)} \, dx}{e^5}+\frac {\int \frac {e^{2+x} \log (x)}{(e+x) \log ^2(e+x)} \, dx}{e^5}-\frac {\int \frac {e^x x}{\log (e+x)} \, dx}{e^5}-\frac {\int \frac {e^x x^2 \log (x)}{\log (e+x)} \, dx}{e^5}-\frac {2 \int \frac {e^x x \log (x)}{\log (e+x)} \, dx}{e^5}+\frac {\int \frac {1}{x \log (e+x)} \, dx}{e^4}+\frac {\text {Subst}\left (\int \frac {3 e^4-\log (-e+x)}{x \log ^2(x)} \, dx,x,e+x\right )}{e^4} \\ & = -\frac {\int \left (-\frac {e^{1+x}}{\log (e+x)}+\frac {e^x (e+x)}{\log (e+x)}\right ) \, dx}{e^5}-\frac {\int \frac {e^{1+x} \log (x)}{\log ^2(e+x)} \, dx}{e^5}+\frac {\int \frac {e^x x \log (x)}{\log ^2(e+x)} \, dx}{e^5}+\frac {\int \frac {e^{2+x} \log (x)}{(e+x) \log ^2(e+x)} \, dx}{e^5}-\frac {\int \frac {e^x x^2 \log (x)}{\log (e+x)} \, dx}{e^5}-\frac {2 \int \frac {e^x x \log (x)}{\log (e+x)} \, dx}{e^5}+\frac {\int \frac {1}{x \log (e+x)} \, dx}{e^4}+\frac {\text {Subst}\left (\int \frac {3 e^4-\log (-e+x)}{x \log ^2(x)} \, dx,x,e+x\right )}{e^4} \\ & = -\frac {\int \frac {e^{1+x} \log (x)}{\log ^2(e+x)} \, dx}{e^5}+\frac {\int \frac {e^x x \log (x)}{\log ^2(e+x)} \, dx}{e^5}+\frac {\int \frac {e^{2+x} \log (x)}{(e+x) \log ^2(e+x)} \, dx}{e^5}+\frac {\int \frac {e^{1+x}}{\log (e+x)} \, dx}{e^5}-\frac {\int \frac {e^x (e+x)}{\log (e+x)} \, dx}{e^5}-\frac {\int \frac {e^x x^2 \log (x)}{\log (e+x)} \, dx}{e^5}-\frac {2 \int \frac {e^x x \log (x)}{\log (e+x)} \, dx}{e^5}+\frac {\int \frac {1}{x \log (e+x)} \, dx}{e^4}+\frac {\text {Subst}\left (\int \frac {3 e^4-\log (-e+x)}{x \log ^2(x)} \, dx,x,e+x\right )}{e^4} \\ \end{align*}
Time = 1.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {3 x+e^{-5+x} \left (-e^{1-x} x+x^3\right ) \log (x)+\left (e^{-5+x} \left (-e x^2-x^3+e^{1-x} (e+x)\right )+e^{-5+x} \left (-2 x^3-x^4+e \left (-2 x^2-x^3\right )\right ) \log (x)\right ) \log (e+x)}{\left (e x+x^2\right ) \log ^2(e+x)} \, dx=-\frac {3 e^5-e \log (x)+e^x x^2 \log (x)}{e^5 \log (e+x)} \]
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Time = 54.78 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90
method | result | size |
risch | \(\frac {{\mathrm e}^{-4} \ln \left (x \right )-x^{2} {\mathrm e}^{-5+x} \ln \left (x \right )-3}{\ln \left (x +{\mathrm e}\right )}\) | \(27\) |
parallelrisch | \(\frac {-3-x^{2} {\mathrm e}^{-5+x} \ln \left (x \right )+{\mathrm e}^{1-x} \ln \left (x \right ) {\mathrm e}^{-5+x}}{\ln \left (x +{\mathrm e}\right )}\) | \(35\) |
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Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {3 x+e^{-5+x} \left (-e^{1-x} x+x^3\right ) \log (x)+\left (e^{-5+x} \left (-e x^2-x^3+e^{1-x} (e+x)\right )+e^{-5+x} \left (-2 x^3-x^4+e \left (-2 x^2-x^3\right )\right ) \log (x)\right ) \log (e+x)}{\left (e x+x^2\right ) \log ^2(e+x)} \, dx=-\frac {{\left ({\left (x^{2} e^{\left (x - 1\right )} - 1\right )} \log \left (x\right ) + 3 \, e^{4}\right )} e^{\left (-4\right )}}{\log \left (x + e\right )} \]
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Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {3 x+e^{-5+x} \left (-e^{1-x} x+x^3\right ) \log (x)+\left (e^{-5+x} \left (-e x^2-x^3+e^{1-x} (e+x)\right )+e^{-5+x} \left (-2 x^3-x^4+e \left (-2 x^2-x^3\right )\right ) \log (x)\right ) \log (e+x)}{\left (e x+x^2\right ) \log ^2(e+x)} \, dx=- \frac {x^{2} e^{x - 1} \log {\left (x \right )}}{e^{4} \log {\left (x + e \right )}} + \frac {\log {\left (x \right )} - 3 e^{4}}{e^{4} \log {\left (x + e \right )}} \]
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Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {3 x+e^{-5+x} \left (-e^{1-x} x+x^3\right ) \log (x)+\left (e^{-5+x} \left (-e x^2-x^3+e^{1-x} (e+x)\right )+e^{-5+x} \left (-2 x^3-x^4+e \left (-2 x^2-x^3\right )\right ) \log (x)\right ) \log (e+x)}{\left (e x+x^2\right ) \log ^2(e+x)} \, dx=-\frac {{\left (x^{2} e^{x} \log \left (x\right ) - e \log \left (x\right )\right )} e^{\left (-5\right )}}{\log \left (x + e\right )} - \frac {3}{\log \left (x + e\right )} \]
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Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {3 x+e^{-5+x} \left (-e^{1-x} x+x^3\right ) \log (x)+\left (e^{-5+x} \left (-e x^2-x^3+e^{1-x} (e+x)\right )+e^{-5+x} \left (-2 x^3-x^4+e \left (-2 x^2-x^3\right )\right ) \log (x)\right ) \log (e+x)}{\left (e x+x^2\right ) \log ^2(e+x)} \, dx=-\frac {{\left (x^{2} e^{x} \log \left (x\right ) - e \log \left (x\right ) + 3 \, e^{5}\right )} e^{\left (-5\right )}}{\log \left (x + e\right )} \]
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Timed out. \[ \int \frac {3 x+e^{-5+x} \left (-e^{1-x} x+x^3\right ) \log (x)+\left (e^{-5+x} \left (-e x^2-x^3+e^{1-x} (e+x)\right )+e^{-5+x} \left (-2 x^3-x^4+e \left (-2 x^2-x^3\right )\right ) \log (x)\right ) \log (e+x)}{\left (e x+x^2\right ) \log ^2(e+x)} \, dx=\int -\frac {\ln \left (x+\mathrm {e}\right )\,\left ({\mathrm {e}}^{x-5}\,\left (x^2\,\mathrm {e}-{\mathrm {e}}^{1-x}\,\left (x+\mathrm {e}\right )+x^3\right )+{\mathrm {e}}^{x-5}\,\ln \left (x\right )\,\left (\mathrm {e}\,\left (x^3+2\,x^2\right )+2\,x^3+x^4\right )\right )-3\,x+{\mathrm {e}}^{x-5}\,\ln \left (x\right )\,\left (x\,{\mathrm {e}}^{1-x}-x^3\right )}{{\ln \left (x+\mathrm {e}\right )}^2\,\left (x^2+\mathrm {e}\,x\right )} \,d x \]
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