Integrand size = 170, antiderivative size = 28 \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\frac {x+\log (x)}{5 \left (4+e^5 \left (-5+2 \left (-2+e^x\right )+x+\log (2)\right )\right )} \]
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\[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )+x \left (80+5 e^{10} \log ^2(2)\right )} \, dx \\ & = \int \frac {4 (1+x)+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (-9+x (-8+\log (2))+\log (2))-e^5 \left (1+2 e^x\right ) x \log (x)}{5 x \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx \\ & = \frac {1}{5} \int \frac {4 (1+x)+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (-9+x (-8+\log (2))+\log (2))-e^5 \left (1+2 e^x\right ) x \log (x)}{x \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx \\ & = \frac {1}{5} \int \left (\frac {\left (4-10 e^5+e^5 x+e^5 \log (2)\right ) (x+\log (x))}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}+\frac {1+x-x^2-x \log (x)}{x \left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )}\right ) \, dx \\ & = \frac {1}{5} \int \frac {\left (4-10 e^5+e^5 x+e^5 \log (2)\right ) (x+\log (x))}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx+\frac {1}{5} \int \frac {1+x-x^2-x \log (x)}{x \left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )} \, dx \\ & = \frac {1}{5} \int \frac {1+x-x^2-x \log (x)}{x \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )} \, dx+\frac {1}{5} \int \left (\frac {10 e^5 \left (1-\frac {4+e^5 \log (2)}{10 e^5}\right ) (-x-\log (x))}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}+\frac {e^5 x (x+\log (x))}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}\right ) \, dx \\ & = \frac {1}{5} \int \left (\frac {x}{-2 e^{5+x}-e^5 x-4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )}+\frac {1}{2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )}+\frac {1}{x \left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )}+\frac {\log (x)}{-2 e^{5+x}-e^5 x-4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )}\right ) \, dx+\frac {1}{5} e^5 \int \frac {x (x+\log (x))}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx+\frac {1}{5} \left (-4+e^5 (10-\log (2))\right ) \int \frac {-x-\log (x)}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx \\ & = \frac {1}{5} \int \frac {x}{-2 e^{5+x}-e^5 x-4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )} \, dx+\frac {1}{5} \int \frac {1}{2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )} \, dx+\frac {1}{5} \int \frac {1}{x \left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )} \, dx+\frac {1}{5} \int \frac {\log (x)}{-2 e^{5+x}-e^5 x-4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )} \, dx+\frac {1}{5} e^5 \int \frac {x (x+\log (x))}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx+\frac {1}{5} \left (-4+e^5 (10-\log (2))\right ) \int \frac {-x-\log (x)}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx \\ & = \frac {1}{5} \int \frac {x}{-4-2 e^{5+x}-e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \int \frac {1}{4+2 e^{5+x}+e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \int \frac {1}{x \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )} \, dx-\frac {1}{5} \int \frac {\int -\frac {1}{4+2 e^{5+x}+e^5 (-9+x+\log (2))} \, dx}{x} \, dx+\frac {1}{5} e^5 \int \left (\frac {x^2}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}+\frac {x \log (x)}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}\right ) \, dx+\frac {1}{5} \left (-4+e^5 (10-\log (2))\right ) \int \left (-\frac {x}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}-\frac {\log (x)}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}\right ) \, dx+\frac {1}{5} \log (x) \int \frac {1}{-4-2 e^{5+x}-e^5 (-9+x+\log (2))} \, dx \\ & = \frac {1}{5} \int \frac {x}{-4-2 e^{5+x}-e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \int \frac {1}{4+2 e^{5+x}+e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \int \frac {1}{x \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )} \, dx+\frac {1}{5} \int \frac {\int \frac {1}{4+2 e^{5+x}+e^5 (-9+x+\log (2))} \, dx}{x} \, dx+\frac {1}{5} e^5 \int \frac {x^2}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx+\frac {1}{5} e^5 \int \frac {x \log (x)}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx+\frac {1}{5} \left (4-e^5 (10-\log (2))\right ) \int \frac {x}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx+\frac {1}{5} \left (4-e^5 (10-\log (2))\right ) \int \frac {\log (x)}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx+\frac {1}{5} \log (x) \int \frac {1}{-4-2 e^{5+x}-e^5 (-9+x+\log (2))} \, dx \\ & = \frac {1}{5} \int \frac {x}{-4-2 e^{5+x}-e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \int \frac {1}{4+2 e^{5+x}+e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \int \frac {1}{x \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )} \, dx+\frac {1}{5} \int \frac {\int \frac {1}{4+2 e^{5+x}+e^5 (-9+x+\log (2))} \, dx}{x} \, dx+\frac {1}{5} e^5 \int \frac {x^2}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx-\frac {1}{5} e^5 \int \frac {\int \frac {x}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx}{x} \, dx+\frac {1}{5} \left (4-e^5 (10-\log (2))\right ) \int \frac {x}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx+\frac {1}{5} \left (-4+e^5 (10-\log (2))\right ) \int \frac {\int \frac {1}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx}{x} \, dx+\frac {1}{5} \log (x) \int \frac {1}{-4-2 e^{5+x}-e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \left (e^5 \log (x)\right ) \int \frac {x}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx+\frac {1}{5} \left (\left (4-e^5 (10-\log (2))\right ) \log (x)\right ) \int \frac {1}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx \\ \end{align*}
Time = 1.70 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\frac {x+\log (x)}{5 \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )} \]
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Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50
method | result | size |
parallelrisch | \(\frac {\left (2 x \,{\mathrm e}^{5}+2 \,{\mathrm e}^{5} \ln \left (x \right )\right ) {\mathrm e}^{-5}}{20 \,{\mathrm e}^{5} {\mathrm e}^{x}+10 \,{\mathrm e}^{5} \ln \left (2\right )+10 x \,{\mathrm e}^{5}-90 \,{\mathrm e}^{5}+40}\) | \(42\) |
risch | \(\frac {\ln \left (x \right )}{10 \,{\mathrm e}^{5+x}+5 \,{\mathrm e}^{5} \ln \left (2\right )+5 x \,{\mathrm e}^{5}-45 \,{\mathrm e}^{5}+20}+\frac {x}{10 \,{\mathrm e}^{5+x}+5 \,{\mathrm e}^{5} \ln \left (2\right )+5 x \,{\mathrm e}^{5}-45 \,{\mathrm e}^{5}+20}\) | \(55\) |
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Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\frac {x + \log \left (x\right )}{5 \, {\left ({\left (x - 9\right )} e^{5} + e^{5} \log \left (2\right ) + 2 \, e^{\left (x + 5\right )} + 4\right )}} \]
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Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\frac {x + \log {\left (x \right )}}{5 x e^{5} + 10 e^{5} e^{x} - 45 e^{5} + 20 + 5 e^{5} \log {\left (2 \right )}} \]
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Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\frac {x + \log \left (x\right )}{5 \, {\left (x e^{5} + {\left (\log \left (2\right ) - 9\right )} e^{5} + 2 \, e^{\left (x + 5\right )} + 4\right )}} \]
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Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\frac {x + \log \left (x\right )}{5 \, {\left (x e^{5} + e^{5} \log \left (2\right ) - 9 \, e^{5} + 2 \, e^{\left (x + 5\right )} + 4\right )}} \]
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Timed out. \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\int \frac {4\,x+{\mathrm {e}}^{x+5}\,\left (-2\,x^2+2\,x+2\right )-\ln \left (x\right )\,\left (2\,x\,{\mathrm {e}}^{x+5}+x\,{\mathrm {e}}^5\right )-{\mathrm {e}}^5\,\left (8\,x+9\right )+{\mathrm {e}}^5\,\ln \left (2\right )\,\left (x+1\right )+4}{80\,x-{\mathrm {e}}^5\,\left (360\,x-40\,x^2\right )+{\mathrm {e}}^{10}\,\left (5\,x^3-90\,x^2+405\,x\right )+20\,x\,{\mathrm {e}}^{2\,x+10}-\ln \left (2\right )\,\left ({\mathrm {e}}^{10}\,\left (90\,x-10\,x^2\right )-40\,x\,{\mathrm {e}}^5\right )+{\mathrm {e}}^x\,\left (80\,x\,{\mathrm {e}}^5-{\mathrm {e}}^{10}\,\left (180\,x-20\,x^2\right )+20\,x\,{\mathrm {e}}^{10}\,\ln \left (2\right )\right )+5\,x\,{\mathrm {e}}^{10}\,{\ln \left (2\right )}^2} \,d x \]
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