Integrand size = 147, antiderivative size = 29 \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=\frac {8 x}{5 (2+x) \left (1-x+\frac {e^x}{\log (4)}-\log (4)\right )} \]
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\[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=\int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {8 \log (4) \left (-e^x \left (-2+2 x+x^2\right )+x^2 \log (4)-2 \log ^2(4) \left (1-\frac {\log (16)}{2 \log ^2(4)}\right )\right )}{5 (2+x)^2 \left (e^x-\log (4) (-1+x+\log (4))\right )^2} \, dx \\ & = \frac {1}{5} (8 \log (4)) \int \frac {-e^x \left (-2+2 x+x^2\right )+x^2 \log (4)-2 \log ^2(4) \left (1-\frac {\log (16)}{2 \log ^2(4)}\right )}{(2+x)^2 \left (e^x-\log (4) (-1+x+\log (4))\right )^2} \, dx \\ & = \frac {1}{5} (8 \log (4)) \int \left (\frac {2-2 x-x^2}{(2+x)^2 \left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )}+\frac {x \left (-x^2 \log (4)-2 \log ^2(4)-x \log ^2(4)+\log (256)\right )}{(2+x)^2 \left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )^2}\right ) \, dx \\ & = \frac {1}{5} (8 \log (4)) \int \frac {2-2 x-x^2}{(2+x)^2 \left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )} \, dx+\frac {1}{5} (8 \log (4)) \int \frac {x \left (-x^2 \log (4)-2 \log ^2(4)-x \log ^2(4)+\log (256)\right )}{(2+x)^2 \left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )^2} \, dx \\ & = \frac {1}{5} (8 \log (4)) \int \frac {x (-x \log (4)+(2-\log (4)) \log (4))}{(2+x) \left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )^2} \, dx+\frac {1}{5} (8 \log (4)) \int \left (\frac {1}{-e^x+x \log (4)-(1-\log (4)) \log (4)}+\frac {2}{(2+x)^2 \left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )}+\frac {2}{(2+x) \left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )}\right ) \, dx \\ & = \frac {1}{5} (8 \log (4)) \int \frac {1}{-e^x+x \log (4)-(1-\log (4)) \log (4)} \, dx+\frac {1}{5} (8 \log (4)) \int \left (-\frac {x \log (4)}{\left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )^2}+\frac {(4-\log (4)) \log (4)}{\left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )^2}+\frac {2 (-4+\log (4)) \log (4)}{(2+x) \left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )^2}\right ) \, dx+\frac {1}{5} (16 \log (4)) \int \frac {1}{(2+x)^2 \left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )} \, dx+\frac {1}{5} (16 \log (4)) \int \frac {1}{(2+x) \left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )} \, dx \\ & = \frac {1}{5} (8 \log (4)) \int \frac {1}{-e^x+\log (4) (-1+x+\log (4))} \, dx+\frac {1}{5} (16 \log (4)) \int \frac {1}{(2+x)^2 \left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )} \, dx+\frac {1}{5} (16 \log (4)) \int \frac {1}{(2+x) \left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )} \, dx-\frac {1}{5} \left (8 \log ^2(4)\right ) \int \frac {x}{\left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )^2} \, dx+\frac {1}{5} \left (8 (4-\log (4)) \log ^2(4)\right ) \int \frac {1}{\left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )^2} \, dx-\frac {1}{5} \left (16 (4-\log (4)) \log ^2(4)\right ) \int \frac {1}{(2+x) \left (e^x-x \log (4)+(1-\log (4)) \log (4)\right )^2} \, dx \\ \end{align*}
Time = 2.47 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.93 \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=-\frac {8 x \left (x^2 \log (4)+2 \log ^2(4)+x \log ^2(4)-\log (256)\right )}{5 (2+x)^2 (-2+x+\log (4)) \left (-e^x+\log (4) (-1+x+\log (4))\right )} \]
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Time = 0.43 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14
method | result | size |
norman | \(-\frac {16 x \ln \left (2\right )}{5 \left (2+x \right ) \left (4 \ln \left (2\right )^{2}+2 x \ln \left (2\right )-2 \ln \left (2\right )-{\mathrm e}^{x}\right )}\) | \(33\) |
risch | \(-\frac {16 x \ln \left (2\right )}{5 \left (2+x \right ) \left (4 \ln \left (2\right )^{2}+2 x \ln \left (2\right )-2 \ln \left (2\right )-{\mathrm e}^{x}\right )}\) | \(33\) |
parallelrisch | \(-\frac {16 x \ln \left (2\right )}{5 \left (4 x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+8 \ln \left (2\right )^{2}+2 x \ln \left (2\right )-{\mathrm e}^{x} x -4 \ln \left (2\right )-2 \,{\mathrm e}^{x}\right )}\) | \(47\) |
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Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=-\frac {16 \, x \log \left (2\right )}{5 \, {\left (4 \, {\left (x + 2\right )} \log \left (2\right )^{2} - {\left (x + 2\right )} e^{x} + 2 \, {\left (x^{2} + x - 2\right )} \log \left (2\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (24) = 48\).
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=\frac {16 x \log {\left (2 \right )}}{- 10 x^{2} \log {\left (2 \right )} - 20 x \log {\left (2 \right )}^{2} - 10 x \log {\left (2 \right )} + \left (5 x + 10\right ) e^{x} - 40 \log {\left (2 \right )}^{2} + 20 \log {\left (2 \right )}} \]
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Time = 0.35 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=-\frac {16 \, x \log \left (2\right )}{5 \, {\left (2 \, x^{2} \log \left (2\right ) + 2 \, {\left (2 \, \log \left (2\right )^{2} + \log \left (2\right )\right )} x - {\left (x + 2\right )} e^{x} + 8 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right )\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=-\frac {16 \, x \log \left (2\right )}{5 \, {\left (2 \, x^{2} \log \left (2\right ) + 4 \, x \log \left (2\right )^{2} - x e^{x} + 2 \, x \log \left (2\right ) + 8 \, \log \left (2\right )^{2} - 2 \, e^{x} - 4 \, \log \left (2\right )\right )}} \]
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Timed out. \[ \int \frac {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx=-\int \frac {128\,{\ln \left (2\right )}^3-4\,{\ln \left (2\right )}^2\,\left (8\,x^2+16\right )+2\,{\mathrm {e}}^x\,\ln \left (2\right )\,\left (8\,x^2+16\,x-16\right )}{4\,{\ln \left (2\right )}^2\,\left (5\,x^4+10\,x^3-15\,x^2-20\,x+20\right )+{\mathrm {e}}^{2\,x}\,\left (5\,x^2+20\,x+20\right )+16\,{\ln \left (2\right )}^4\,\left (5\,x^2+20\,x+20\right )-{\mathrm {e}}^x\,\left (2\,\ln \left (2\right )\,\left (10\,x^3+30\,x^2-40\right )+4\,{\ln \left (2\right )}^2\,\left (10\,x^2+40\,x+40\right )\right )+8\,{\ln \left (2\right )}^3\,\left (10\,x^3+30\,x^2-40\right )} \,d x \]
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