Integrand size = 104, antiderivative size = 34 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=e^{x \left (-x+\frac {1-e^{2 e^{2 x}}+4 x (5+\log (4))}{4+x}\right )} \]
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\[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=\int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}\right ) \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}\right ) \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{(4+x)^2} \, dx \\ & = \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{(4+x)^2} \, dx \\ & = \int \left (\frac {4 \exp \left (2 \left (e^{2 x}+x\right )+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x}{-4-x}-\frac {4 \exp \left (2 e^{2 x}+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2}+\frac {4 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2}+\frac {128 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x}{(4+x)^2}+\frac {4 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x^2}{(4+x)^2}-\frac {2 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x^3}{(4+x)^2}+\frac {4 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x (8+x) \log (4)}{(4+x)^2}\right ) \, dx \\ & = -\left (2 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x^3}{(4+x)^2} \, dx\right )+4 \int \frac {\exp \left (2 \left (e^{2 x}+x\right )+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x}{-4-x} \, dx-4 \int \frac {\exp \left (2 e^{2 x}+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+4 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+4 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x^2}{(4+x)^2} \, dx+128 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x}{(4+x)^2} \, dx+(4 \log (4)) \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x (8+x)}{(4+x)^2} \, dx \\ & = -\left (2 \int \left (-8 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )+\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x-\frac {64 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2}+\frac {48 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x}\right ) \, dx\right )-4 \int \frac {\exp \left (2 e^{2 x}+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+4 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+4 \int \left (-\exp \left (2 \left (e^{2 x}+x\right )+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )+\frac {4 \exp \left (2 \left (e^{2 x}+x\right )+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x}\right ) \, dx+4 \int \left (\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )+\frac {16 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2}-\frac {8 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x}\right ) \, dx+128 \int \left (-\frac {4 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2}+\frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x}\right ) \, dx+(4 \log (4)) \int \left (\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )-\frac {16 \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2}\right ) \, dx \\ & = -\left (2 \int \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) x \, dx\right )-4 \int \exp \left (2 \left (e^{2 x}+x\right )+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) \, dx+4 \int \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) \, dx-4 \int \frac {\exp \left (2 e^{2 x}+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+4 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+16 \int \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) \, dx+16 \int \frac {\exp \left (2 \left (e^{2 x}+x\right )+\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x} \, dx-32 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x} \, dx+64 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx-96 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x} \, dx+128 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+128 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{4+x} \, dx-512 \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx+(4 \log (4)) \int \exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right ) \, dx-(64 \log (4)) \int \frac {\exp \left (\frac {x-e^{2 e^{2 x}} x-x^3+16 x^2 \left (1+\frac {\log (2)}{2}\right )}{4+x}\right )}{(4+x)^2} \, dx \\ \end{align*}
Time = 0.62 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=2^{\frac {8 x^2}{4+x}} e^{-\frac {x \left (-1+e^{2 e^{2 x}}-16 x+x^2\right )}{4+x}} \]
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Time = 28.17 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97
method | result | size |
risch | \({\mathrm e}^{\frac {x \left (8 x \ln \left (2\right )-x^{2}-{\mathrm e}^{2 \,{\mathrm e}^{2 x}}+16 x +1\right )}{4+x}}\) | \(33\) |
parallelrisch | \({\mathrm e}^{\frac {x \left (8 x \ln \left (2\right )-x^{2}-{\mathrm e}^{2 \,{\mathrm e}^{2 x}}+16 x +1\right )}{4+x}}\) | \(33\) |
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Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=e^{\left (-\frac {x^{3} - 8 \, x^{2} \log \left (2\right ) - 16 \, x^{2} + x e^{\left (2 \, e^{\left (2 \, x\right )}\right )} - x}{x + 4}\right )} \]
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Time = 0.49 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=e^{\frac {- x^{3} + 8 x^{2} \log {\left (2 \right )} + 16 x^{2} - x e^{2 e^{2 x}} + x}{x + 4}} \]
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Time = 0.45 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.68 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=\frac {1}{4294967296} \, e^{\left (-x^{2} + 8 \, x \log \left (2\right ) + 20 \, x + \frac {4 \, e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{x + 4} + \frac {128 \, \log \left (2\right )}{x + 4} + \frac {316}{x + 4} - e^{\left (2 \, e^{\left (2 \, x\right )}\right )} - 79\right )} \]
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Time = 0.38 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.65 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=e^{\left (-\frac {x^{3}}{x + 4} + \frac {8 \, x^{2} \log \left (2\right )}{x + 4} + \frac {16 \, x^{2}}{x + 4} - \frac {x e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{x + 4} + \frac {x}{x + 4}\right )} \]
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Time = 12.48 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.74 \[ \int \frac {e^{\frac {x-e^{2 e^{2 x}} x+16 x^2-x^3+4 x^2 \log (4)}{4+x}} \left (4+128 x+4 x^2-2 x^3+e^{2 e^{2 x}} \left (-4+e^{2 x} \left (-16 x-4 x^2\right )\right )+\left (32 x+4 x^2\right ) \log (4)\right )}{16+8 x+x^2} \, dx=2^{\frac {8\,x^2}{x+4}}\,{\mathrm {e}}^{\frac {x}{x+4}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}}{x+4}}\,{\mathrm {e}}^{-\frac {x^3}{x+4}}\,{\mathrm {e}}^{\frac {16\,x^2}{x+4}} \]
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