Integrand size = 173, antiderivative size = 29 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=-3+e-\frac {64 x^4 (2+x)^2}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^2} \]
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\[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=\int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {128 x^3 (2+x) \left (-2 x^2 (1+x)-e^x \left (-2+3 x+3 x^2\right )+\left (e^x+x\right ) (4+3 x) \log \left (\frac {e^x+x}{x}\right )\right )}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx \\ & = 128 \int \frac {x^3 (2+x) \left (-2 x^2 (1+x)-e^x \left (-2+3 x+3 x^2\right )+\left (e^x+x\right ) (4+3 x) \log \left (\frac {e^x+x}{x}\right )\right )}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx \\ & = 128 \int \left (\frac {(-1+x) x^4 (2+x)^2}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3}-\frac {x^3 (2+x) \left (-2+3 x+3 x^2-4 \log \left (\frac {e^x+x}{x}\right )-3 x \log \left (\frac {e^x+x}{x}\right )\right )}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3}\right ) \, dx \\ & = 128 \int \frac {(-1+x) x^4 (2+x)^2}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx-128 \int \frac {x^3 (2+x) \left (-2+3 x+3 x^2-4 \log \left (\frac {e^x+x}{x}\right )-3 x \log \left (\frac {e^x+x}{x}\right )\right )}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx \\ & = 128 \int \left (-\frac {4 x^4}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3}+\frac {3 x^6}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3}+\frac {x^7}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3}\right ) \, dx-128 \int \left (-\frac {x^3 (2+x)^2}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3}+\frac {x^3 \left (8+10 x+3 x^2\right )}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^2}\right ) \, dx \\ & = 128 \int \frac {x^3 (2+x)^2}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx+128 \int \frac {x^7}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx-128 \int \frac {x^3 \left (8+10 x+3 x^2\right )}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^2} \, dx+384 \int \frac {x^6}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx-512 \int \frac {x^4}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx \\ & = 128 \int \left (\frac {4 x^3}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3}+\frac {4 x^4}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3}+\frac {x^5}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3}\right ) \, dx-128 \int \left (\frac {8 x^3}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^2}+\frac {10 x^4}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^2}+\frac {3 x^5}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^2}\right ) \, dx+128 \int \frac {x^7}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx+384 \int \frac {x^6}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx-512 \int \frac {x^4}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx \\ & = 128 \int \frac {x^5}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx+128 \int \frac {x^7}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx+384 \int \frac {x^6}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx-384 \int \frac {x^5}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^2} \, dx+512 \int \frac {x^3}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx+512 \int \frac {x^4}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx-512 \int \frac {x^4}{\left (e^x+x\right ) \left (x-\log \left (\frac {e^x+x}{x}\right )\right )^3} \, dx-1024 \int \frac {x^3}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^2} \, dx-1280 \int \frac {x^4}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^2} \, dx \\ \end{align*}
Time = 0.52 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=-\frac {64 x^4 (2+x)^2}{\left (-x+\log \left (\frac {e^x+x}{x}\right )\right )^2} \]
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Time = 1.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66
method | result | size |
parallelrisch | \(\frac {-128 x^{6}-512 x^{5}-512 x^{4}}{2 x^{2}-4 x \ln \left (\frac {{\mathrm e}^{x}+x}{x}\right )+2 \ln \left (\frac {{\mathrm e}^{x}+x}{x}\right )^{2}}\) | \(48\) |
risch | \(-\frac {256 \left (x^{2}+4 x +4\right ) x^{4}}{{\left (i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x}+x \right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+x \right )}{x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+x \right )}{x}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+x \right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+x \right )}{x}\right )}^{2}+i \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+x \right )}{x}\right )}^{3}+2 x +2 \ln \left (x \right )-2 \ln \left ({\mathrm e}^{x}+x \right )\right )}^{2}}\) | \(127\) |
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Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=-\frac {64 \, {\left (x^{6} + 4 \, x^{5} + 4 \, x^{4}\right )}}{x^{2} - 2 \, x \log \left (\frac {x + e^{x}}{x}\right ) + \log \left (\frac {x + e^{x}}{x}\right )^{2}} \]
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Time = 0.13 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=\frac {- 64 x^{6} - 256 x^{5} - 256 x^{4}}{x^{2} - 2 x \log {\left (\frac {x + e^{x}}{x} \right )} + \log {\left (\frac {x + e^{x}}{x} \right )}^{2}} \]
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Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=-\frac {64 \, {\left (x^{6} + 4 \, x^{5} + 4 \, x^{4}\right )}}{x^{2} - 2 \, {\left (x + \log \left (x\right )\right )} \log \left (x + e^{x}\right ) + \log \left (x + e^{x}\right )^{2} + 2 \, x \log \left (x\right ) + \log \left (x\right )^{2}} \]
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Time = 0.62 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=-\frac {64 \, {\left (x^{6} + 4 \, x^{5} + 4 \, x^{4}\right )}}{x^{2} - 2 \, x \log \left (\frac {x + e^{x}}{x}\right ) + \log \left (\frac {x + e^{x}}{x}\right )^{2}} \]
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Time = 9.26 (sec) , antiderivative size = 540, normalized size of antiderivative = 18.62 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=\frac {\frac {64\,x\,\left (x+{\mathrm {e}}^x\right )\,\left (24\,x^5\,{\mathrm {e}}^x+48\,x^6\,{\mathrm {e}}^x+43\,x^7\,{\mathrm {e}}^x+19\,x^8\,{\mathrm {e}}^x+3\,x^9\,{\mathrm {e}}^x-8\,x^3\,{\mathrm {e}}^{2\,x}+22\,x^4\,{\mathrm {e}}^{2\,x}+47\,x^5\,{\mathrm {e}}^{2\,x}+18\,x^6\,{\mathrm {e}}^{2\,x}+16\,x^7+30\,x^8+12\,x^9\right )}{{\left ({\mathrm {e}}^x+x^2\right )}^3}-\frac {64\,x\,\ln \left (\frac {x+{\mathrm {e}}^x}{x}\right )\,\left (x+{\mathrm {e}}^x\right )\,\left (40\,x^4\,{\mathrm {e}}^x+68\,x^5\,{\mathrm {e}}^x+49\,x^6\,{\mathrm {e}}^x+19\,x^7\,{\mathrm {e}}^x+3\,x^8\,{\mathrm {e}}^x+32\,x^3\,{\mathrm {e}}^{2\,x}+50\,x^4\,{\mathrm {e}}^{2\,x}+18\,x^5\,{\mathrm {e}}^{2\,x}+24\,x^6+40\,x^7+15\,x^8\right )}{{\left ({\mathrm {e}}^x+x^2\right )}^3}}{x-\ln \left (\frac {x+{\mathrm {e}}^x}{x}\right )}+\frac {\frac {64\,x^4\,\left (x+2\right )\,\left (3\,x^2\,{\mathrm {e}}^x-2\,{\mathrm {e}}^x+3\,x\,{\mathrm {e}}^x+2\,x^2+2\,x^3\right )}{{\mathrm {e}}^x+x^2}-\frac {64\,x^4\,\ln \left (\frac {x+{\mathrm {e}}^x}{x}\right )\,\left (x+{\mathrm {e}}^x\right )\,\left (3\,x^2+10\,x+8\right )}{{\mathrm {e}}^x+x^2}}{x^2-2\,x\,\ln \left (\frac {x+{\mathrm {e}}^x}{x}\right )+{\ln \left (\frac {x+{\mathrm {e}}^x}{x}\right )}^2}-2048\,x^4-3200\,x^5-1152\,x^6-\frac {64\,\left (-3\,x^{11}+41\,x^{10}+13\,x^9-188\,x^8-28\,x^7+144\,x^6\right )}{\left ({\mathrm {e}}^x+x^2\right )\,\left (2\,x-x^2\right )}-\frac {64\,\left (-3\,x^{15}+8\,x^{14}+13\,x^{13}-46\,x^{12}+4\,x^{11}+56\,x^{10}-32\,x^9\right )}{\left (2\,x-x^2\right )\,\left ({\mathrm {e}}^{3\,x}+3\,x^4\,{\mathrm {e}}^x+3\,x^2\,{\mathrm {e}}^{2\,x}+x^6\right )}+\frac {64\,\left (-6\,x^{13}+31\,x^{12}+12\,x^{11}-151\,x^{10}+50\,x^9+144\,x^8-80\,x^7\right )}{\left (2\,x-x^2\right )\,\left ({\mathrm {e}}^{2\,x}+2\,x^2\,{\mathrm {e}}^x+x^4\right )} \]
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