Integrand size = 314, antiderivative size = 31 \[ \int \frac {-80 e^2-64 x+e^x \left (8 x^3+4 x^4+e^2 \left (4 x^2+4 x^3\right )\right )+\left (144 e^2+136 x+e^x \left (-6 x^3-4 x^4+e^2 \left (-4 x^2-4 x^3\right )\right )\right ) \log \left (e^2+x\right )+\left (-100 e^2-100 x+e^x \left (x^3+x^4+e^2 \left (x^2+x^3\right )\right )\right ) \log ^2\left (e^2+x\right )+\left (32 e^2+32 x\right ) \log ^3\left (e^2+x\right )+\left (-4 e^2-4 x\right ) \log ^4\left (e^2+x\right )}{400 e^2+400 x+e^x \left (40 e^2 x^2+40 x^3\right )+e^{2 x} \left (e^2 x^4+x^5\right )+\left (-640 e^2-640 x+e^x \left (-32 e^2 x^2-32 x^3\right )\right ) \log \left (e^2+x\right )+\left (416 e^2+416 x+e^x \left (8 e^2 x^2+8 x^3\right )\right ) \log ^2\left (e^2+x\right )+\left (-128 e^2-128 x\right ) \log ^3\left (e^2+x\right )+\left (16 e^2+16 x\right ) \log ^4\left (e^2+x\right )} \, dx=\frac {x}{-4+\frac {\left (-e^x-\frac {4}{x^2}\right ) x^2}{\left (-2+\log \left (e^2+x\right )\right )^2}} \]
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\[ \int \frac {-80 e^2-64 x+e^x \left (8 x^3+4 x^4+e^2 \left (4 x^2+4 x^3\right )\right )+\left (144 e^2+136 x+e^x \left (-6 x^3-4 x^4+e^2 \left (-4 x^2-4 x^3\right )\right )\right ) \log \left (e^2+x\right )+\left (-100 e^2-100 x+e^x \left (x^3+x^4+e^2 \left (x^2+x^3\right )\right )\right ) \log ^2\left (e^2+x\right )+\left (32 e^2+32 x\right ) \log ^3\left (e^2+x\right )+\left (-4 e^2-4 x\right ) \log ^4\left (e^2+x\right )}{400 e^2+400 x+e^x \left (40 e^2 x^2+40 x^3\right )+e^{2 x} \left (e^2 x^4+x^5\right )+\left (-640 e^2-640 x+e^x \left (-32 e^2 x^2-32 x^3\right )\right ) \log \left (e^2+x\right )+\left (416 e^2+416 x+e^x \left (8 e^2 x^2+8 x^3\right )\right ) \log ^2\left (e^2+x\right )+\left (-128 e^2-128 x\right ) \log ^3\left (e^2+x\right )+\left (16 e^2+16 x\right ) \log ^4\left (e^2+x\right )} \, dx=\int \frac {-80 e^2-64 x+e^x \left (8 x^3+4 x^4+e^2 \left (4 x^2+4 x^3\right )\right )+\left (144 e^2+136 x+e^x \left (-6 x^3-4 x^4+e^2 \left (-4 x^2-4 x^3\right )\right )\right ) \log \left (e^2+x\right )+\left (-100 e^2-100 x+e^x \left (x^3+x^4+e^2 \left (x^2+x^3\right )\right )\right ) \log ^2\left (e^2+x\right )+\left (32 e^2+32 x\right ) \log ^3\left (e^2+x\right )+\left (-4 e^2-4 x\right ) \log ^4\left (e^2+x\right )}{400 e^2+400 x+e^x \left (40 e^2 x^2+40 x^3\right )+e^{2 x} \left (e^2 x^4+x^5\right )+\left (-640 e^2-640 x+e^x \left (-32 e^2 x^2-32 x^3\right )\right ) \log \left (e^2+x\right )+\left (416 e^2+416 x+e^x \left (8 e^2 x^2+8 x^3\right )\right ) \log ^2\left (e^2+x\right )+\left (-128 e^2-128 x\right ) \log ^3\left (e^2+x\right )+\left (16 e^2+16 x\right ) \log ^4\left (e^2+x\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (2-\log \left (e^2+x\right )\right ) \left (-40 e^2-32 x+2 e^{2+x} x^2 (1+x)+2 e^x x^3 (2+x)-\left (e^2+x\right ) \left (-52+e^x x^2 (1+x)\right ) \log \left (e^2+x\right )-24 \left (e^2+x\right ) \log ^2\left (e^2+x\right )+4 \left (e^2+x\right ) \log ^3\left (e^2+x\right )\right )}{\left (e^2+x\right ) \left (20+e^x x^2-16 \log \left (e^2+x\right )+4 \log ^2\left (e^2+x\right )\right )^2} \, dx \\ & = \int \left (\frac {\left (2-\log \left (e^2+x\right )\right ) \left (2 e^2+4 \left (1+\frac {e^2}{2}\right ) x+2 x^2-e^2 \log \left (e^2+x\right )-\left (1+e^2\right ) x \log \left (e^2+x\right )-x^2 \log \left (e^2+x\right )\right )}{\left (e^2+x\right ) \left (20+e^x x^2-16 \log \left (e^2+x\right )+4 \log ^2\left (e^2+x\right )\right )}+\frac {4 \left (2-\log \left (e^2+x\right )\right )^2 \left (-10 e^2-14 \left (1+\frac {5 e^2}{14}\right ) x-5 x^2+8 e^2 \log \left (e^2+x\right )+10 \left (1+\frac {2 e^2}{5}\right ) x \log \left (e^2+x\right )+4 x^2 \log \left (e^2+x\right )-2 e^2 \log ^2\left (e^2+x\right )-2 \left (1+\frac {e^2}{2}\right ) x \log ^2\left (e^2+x\right )-x^2 \log ^2\left (e^2+x\right )\right )}{\left (e^2+x\right ) \left (20+e^x x^2-16 \log \left (e^2+x\right )+4 \log ^2\left (e^2+x\right )\right )^2}\right ) \, dx \\ & = 4 \int \frac {\left (2-\log \left (e^2+x\right )\right )^2 \left (-10 e^2-14 \left (1+\frac {5 e^2}{14}\right ) x-5 x^2+8 e^2 \log \left (e^2+x\right )+10 \left (1+\frac {2 e^2}{5}\right ) x \log \left (e^2+x\right )+4 x^2 \log \left (e^2+x\right )-2 e^2 \log ^2\left (e^2+x\right )-2 \left (1+\frac {e^2}{2}\right ) x \log ^2\left (e^2+x\right )-x^2 \log ^2\left (e^2+x\right )\right )}{\left (e^2+x\right ) \left (20+e^x x^2-16 \log \left (e^2+x\right )+4 \log ^2\left (e^2+x\right )\right )^2} \, dx+\int \frac {\left (2-\log \left (e^2+x\right )\right ) \left (2 e^2+4 \left (1+\frac {e^2}{2}\right ) x+2 x^2-e^2 \log \left (e^2+x\right )-\left (1+e^2\right ) x \log \left (e^2+x\right )-x^2 \log \left (e^2+x\right )\right )}{\left (e^2+x\right ) \left (20+e^x x^2-16 \log \left (e^2+x\right )+4 \log ^2\left (e^2+x\right )\right )} \, dx \\ & = 4 \int \frac {\left (2-\log \left (e^2+x\right )\right )^2 \left (-5 e^2 (2+x)-x (14+5 x)+2 \left (2 e^2 (2+x)+x (5+2 x)\right ) \log \left (e^2+x\right )-(2+x) \left (e^2+x\right ) \log ^2\left (e^2+x\right )\right )}{\left (e^2+x\right ) \left (20+e^x x^2-16 \log \left (e^2+x\right )+4 \log ^2\left (e^2+x\right )\right )^2} \, dx+\int \frac {\left (2-\log \left (e^2+x\right )\right ) \left (2 \left (e^2 (1+x)+x (2+x)\right )-(1+x) \left (e^2+x\right ) \log \left (e^2+x\right )\right )}{\left (e^2+x\right ) \left (20+e^x x^2-16 \log \left (e^2+x\right )+4 \log ^2\left (e^2+x\right )\right )} \, dx \\ & = \text {Too large to display} \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {-80 e^2-64 x+e^x \left (8 x^3+4 x^4+e^2 \left (4 x^2+4 x^3\right )\right )+\left (144 e^2+136 x+e^x \left (-6 x^3-4 x^4+e^2 \left (-4 x^2-4 x^3\right )\right )\right ) \log \left (e^2+x\right )+\left (-100 e^2-100 x+e^x \left (x^3+x^4+e^2 \left (x^2+x^3\right )\right )\right ) \log ^2\left (e^2+x\right )+\left (32 e^2+32 x\right ) \log ^3\left (e^2+x\right )+\left (-4 e^2-4 x\right ) \log ^4\left (e^2+x\right )}{400 e^2+400 x+e^x \left (40 e^2 x^2+40 x^3\right )+e^{2 x} \left (e^2 x^4+x^5\right )+\left (-640 e^2-640 x+e^x \left (-32 e^2 x^2-32 x^3\right )\right ) \log \left (e^2+x\right )+\left (416 e^2+416 x+e^x \left (8 e^2 x^2+8 x^3\right )\right ) \log ^2\left (e^2+x\right )+\left (-128 e^2-128 x\right ) \log ^3\left (e^2+x\right )+\left (16 e^2+16 x\right ) \log ^4\left (e^2+x\right )} \, dx=-\frac {x \left (-2+\log \left (e^2+x\right )\right )^2}{20+e^x x^2-16 \log \left (e^2+x\right )+4 \log ^2\left (e^2+x\right )} \]
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Time = 1.78 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35
method | result | size |
risch | \(-\frac {x}{4}+\frac {x \left ({\mathrm e}^{x} x^{2}+4\right )}{4 \,{\mathrm e}^{x} x^{2}+16 \ln \left (x +{\mathrm e}^{2}\right )^{2}-64 \ln \left (x +{\mathrm e}^{2}\right )+80}\) | \(42\) |
parallelrisch | \(\frac {2 x^{2} {\mathrm e}^{2} {\mathrm e}^{x}+8 \ln \left (x +{\mathrm e}^{2}\right )^{2} {\mathrm e}^{2}-4 \ln \left (x +{\mathrm e}^{2}\right )^{2} x -32 \ln \left (x +{\mathrm e}^{2}\right ) {\mathrm e}^{2}+16 \ln \left (x +{\mathrm e}^{2}\right ) x +40 \,{\mathrm e}^{2}-16 x}{4 \,{\mathrm e}^{x} x^{2}+16 \ln \left (x +{\mathrm e}^{2}\right )^{2}-64 \ln \left (x +{\mathrm e}^{2}\right )+80}\) | \(84\) |
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Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {-80 e^2-64 x+e^x \left (8 x^3+4 x^4+e^2 \left (4 x^2+4 x^3\right )\right )+\left (144 e^2+136 x+e^x \left (-6 x^3-4 x^4+e^2 \left (-4 x^2-4 x^3\right )\right )\right ) \log \left (e^2+x\right )+\left (-100 e^2-100 x+e^x \left (x^3+x^4+e^2 \left (x^2+x^3\right )\right )\right ) \log ^2\left (e^2+x\right )+\left (32 e^2+32 x\right ) \log ^3\left (e^2+x\right )+\left (-4 e^2-4 x\right ) \log ^4\left (e^2+x\right )}{400 e^2+400 x+e^x \left (40 e^2 x^2+40 x^3\right )+e^{2 x} \left (e^2 x^4+x^5\right )+\left (-640 e^2-640 x+e^x \left (-32 e^2 x^2-32 x^3\right )\right ) \log \left (e^2+x\right )+\left (416 e^2+416 x+e^x \left (8 e^2 x^2+8 x^3\right )\right ) \log ^2\left (e^2+x\right )+\left (-128 e^2-128 x\right ) \log ^3\left (e^2+x\right )+\left (16 e^2+16 x\right ) \log ^4\left (e^2+x\right )} \, dx=-\frac {x \log \left (x + e^{2}\right )^{2} - 4 \, x \log \left (x + e^{2}\right ) + 4 \, x}{x^{2} e^{x} + 4 \, \log \left (x + e^{2}\right )^{2} - 16 \, \log \left (x + e^{2}\right ) + 20} \]
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Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {-80 e^2-64 x+e^x \left (8 x^3+4 x^4+e^2 \left (4 x^2+4 x^3\right )\right )+\left (144 e^2+136 x+e^x \left (-6 x^3-4 x^4+e^2 \left (-4 x^2-4 x^3\right )\right )\right ) \log \left (e^2+x\right )+\left (-100 e^2-100 x+e^x \left (x^3+x^4+e^2 \left (x^2+x^3\right )\right )\right ) \log ^2\left (e^2+x\right )+\left (32 e^2+32 x\right ) \log ^3\left (e^2+x\right )+\left (-4 e^2-4 x\right ) \log ^4\left (e^2+x\right )}{400 e^2+400 x+e^x \left (40 e^2 x^2+40 x^3\right )+e^{2 x} \left (e^2 x^4+x^5\right )+\left (-640 e^2-640 x+e^x \left (-32 e^2 x^2-32 x^3\right )\right ) \log \left (e^2+x\right )+\left (416 e^2+416 x+e^x \left (8 e^2 x^2+8 x^3\right )\right ) \log ^2\left (e^2+x\right )+\left (-128 e^2-128 x\right ) \log ^3\left (e^2+x\right )+\left (16 e^2+16 x\right ) \log ^4\left (e^2+x\right )} \, dx=\frac {- x \log {\left (x + e^{2} \right )}^{2} + 4 x \log {\left (x + e^{2} \right )} - 4 x}{x^{2} e^{x} + 4 \log {\left (x + e^{2} \right )}^{2} - 16 \log {\left (x + e^{2} \right )} + 20} \]
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Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {-80 e^2-64 x+e^x \left (8 x^3+4 x^4+e^2 \left (4 x^2+4 x^3\right )\right )+\left (144 e^2+136 x+e^x \left (-6 x^3-4 x^4+e^2 \left (-4 x^2-4 x^3\right )\right )\right ) \log \left (e^2+x\right )+\left (-100 e^2-100 x+e^x \left (x^3+x^4+e^2 \left (x^2+x^3\right )\right )\right ) \log ^2\left (e^2+x\right )+\left (32 e^2+32 x\right ) \log ^3\left (e^2+x\right )+\left (-4 e^2-4 x\right ) \log ^4\left (e^2+x\right )}{400 e^2+400 x+e^x \left (40 e^2 x^2+40 x^3\right )+e^{2 x} \left (e^2 x^4+x^5\right )+\left (-640 e^2-640 x+e^x \left (-32 e^2 x^2-32 x^3\right )\right ) \log \left (e^2+x\right )+\left (416 e^2+416 x+e^x \left (8 e^2 x^2+8 x^3\right )\right ) \log ^2\left (e^2+x\right )+\left (-128 e^2-128 x\right ) \log ^3\left (e^2+x\right )+\left (16 e^2+16 x\right ) \log ^4\left (e^2+x\right )} \, dx=-\frac {x \log \left (x + e^{2}\right )^{2} - 4 \, x \log \left (x + e^{2}\right ) + 4 \, x}{x^{2} e^{x} + 4 \, \log \left (x + e^{2}\right )^{2} - 16 \, \log \left (x + e^{2}\right ) + 20} \]
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Exception generated. \[ \int \frac {-80 e^2-64 x+e^x \left (8 x^3+4 x^4+e^2 \left (4 x^2+4 x^3\right )\right )+\left (144 e^2+136 x+e^x \left (-6 x^3-4 x^4+e^2 \left (-4 x^2-4 x^3\right )\right )\right ) \log \left (e^2+x\right )+\left (-100 e^2-100 x+e^x \left (x^3+x^4+e^2 \left (x^2+x^3\right )\right )\right ) \log ^2\left (e^2+x\right )+\left (32 e^2+32 x\right ) \log ^3\left (e^2+x\right )+\left (-4 e^2-4 x\right ) \log ^4\left (e^2+x\right )}{400 e^2+400 x+e^x \left (40 e^2 x^2+40 x^3\right )+e^{2 x} \left (e^2 x^4+x^5\right )+\left (-640 e^2-640 x+e^x \left (-32 e^2 x^2-32 x^3\right )\right ) \log \left (e^2+x\right )+\left (416 e^2+416 x+e^x \left (8 e^2 x^2+8 x^3\right )\right ) \log ^2\left (e^2+x\right )+\left (-128 e^2-128 x\right ) \log ^3\left (e^2+x\right )+\left (16 e^2+16 x\right ) \log ^4\left (e^2+x\right )} \, dx=\text {Exception raised: TypeError} \]
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Time = 13.45 (sec) , antiderivative size = 454, normalized size of antiderivative = 14.65 \[ \int \frac {-80 e^2-64 x+e^x \left (8 x^3+4 x^4+e^2 \left (4 x^2+4 x^3\right )\right )+\left (144 e^2+136 x+e^x \left (-6 x^3-4 x^4+e^2 \left (-4 x^2-4 x^3\right )\right )\right ) \log \left (e^2+x\right )+\left (-100 e^2-100 x+e^x \left (x^3+x^4+e^2 \left (x^2+x^3\right )\right )\right ) \log ^2\left (e^2+x\right )+\left (32 e^2+32 x\right ) \log ^3\left (e^2+x\right )+\left (-4 e^2-4 x\right ) \log ^4\left (e^2+x\right )}{400 e^2+400 x+e^x \left (40 e^2 x^2+40 x^3\right )+e^{2 x} \left (e^2 x^4+x^5\right )+\left (-640 e^2-640 x+e^x \left (-32 e^2 x^2-32 x^3\right )\right ) \log \left (e^2+x\right )+\left (416 e^2+416 x+e^x \left (8 e^2 x^2+8 x^3\right )\right ) \log ^2\left (e^2+x\right )+\left (-128 e^2-128 x\right ) \log ^3\left (e^2+x\right )+\left (16 e^2+16 x\right ) \log ^4\left (e^2+x\right )} \, dx=\frac {32\,x^4\,{\mathrm {e}}^x+64\,x\,{\mathrm {e}}^2+8\,x^6\,{\mathrm {e}}^{2\,x}+4\,x^7\,{\mathrm {e}}^{2\,x}+x^8\,{\mathrm {e}}^{2\,x}+x^8\,{\mathrm {e}}^{3\,x}+x^9\,{\mathrm {e}}^{3\,x}+\frac {x^{10}\,{\mathrm {e}}^{3\,x}}{4}+32\,x^3\,{\mathrm {e}}^{x+2}+16\,x^5\,{\mathrm {e}}^{2\,x+2}+12\,x^4\,{\mathrm {e}}^{2\,x+4}+12\,x^6\,{\mathrm {e}}^{2\,x+2}+4\,x^3\,{\mathrm {e}}^{2\,x+6}+12\,x^5\,{\mathrm {e}}^{2\,x+4}+3\,x^7\,{\mathrm {e}}^{2\,x+2}+4\,x^4\,{\mathrm {e}}^{2\,x+6}+3\,x^6\,{\mathrm {e}}^{2\,x+4}+3\,x^7\,{\mathrm {e}}^{3\,x+2}+x^5\,{\mathrm {e}}^{2\,x+6}+3\,x^6\,{\mathrm {e}}^{3\,x+4}+3\,x^8\,{\mathrm {e}}^{3\,x+2}+x^5\,{\mathrm {e}}^{3\,x+6}+3\,x^7\,{\mathrm {e}}^{3\,x+4}+\frac {3\,x^9\,{\mathrm {e}}^{3\,x+2}}{4}+x^6\,{\mathrm {e}}^{3\,x+6}+\frac {3\,x^8\,{\mathrm {e}}^{3\,x+4}}{4}+\frac {x^7\,{\mathrm {e}}^{3\,x+6}}{4}+64\,x^2}{\left (4\,{\ln \left (x+{\mathrm {e}}^2\right )}^2-16\,\ln \left (x+{\mathrm {e}}^2\right )+x^2\,{\mathrm {e}}^x+20\right )\,\left (64\,x+64\,{\mathrm {e}}^2+16\,x^3\,{\mathrm {e}}^x+4\,x^5\,{\mathrm {e}}^{2\,x}+4\,x^6\,{\mathrm {e}}^{2\,x}+x^7\,{\mathrm {e}}^{2\,x}+16\,x^2\,{\mathrm {e}}^{x+2}+12\,x^4\,{\mathrm {e}}^{2\,x+2}+12\,x^3\,{\mathrm {e}}^{2\,x+4}+12\,x^5\,{\mathrm {e}}^{2\,x+2}+4\,x^2\,{\mathrm {e}}^{2\,x+6}+12\,x^4\,{\mathrm {e}}^{2\,x+4}+3\,x^6\,{\mathrm {e}}^{2\,x+2}+4\,x^3\,{\mathrm {e}}^{2\,x+6}+3\,x^5\,{\mathrm {e}}^{2\,x+4}+x^4\,{\mathrm {e}}^{2\,x+6}\right )}-\frac {x}{4} \]
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