Integrand size = 83, antiderivative size = 29 \[ \int \frac {2-2 x-2 e^{2-x} x+e^{4 x+x^2} \left (7 x-e^{2-x} x+4 x^2\right )+\left (-x-e^{2-x} x\right ) \log (x)}{2 x+e^{4 x+x^2} x+x \log (x)} \, dx=e^{2-x}-x+\log \left (\frac {1}{9} \left (2+e^{x (4+x)}+\log (x)\right )^2\right ) \]
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\[ \int \frac {2-2 x-2 e^{2-x} x+e^{4 x+x^2} \left (7 x-e^{2-x} x+4 x^2\right )+\left (-x-e^{2-x} x\right ) \log (x)}{2 x+e^{4 x+x^2} x+x \log (x)} \, dx=\int \frac {2-2 x-2 e^{2-x} x+e^{4 x+x^2} \left (7 x-e^{2-x} x+4 x^2\right )+\left (-x-e^{2-x} x\right ) \log (x)}{2 x+e^{4 x+x^2} x+x \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (e^{-x} \left (-e^2+7 e^x+4 e^x x\right )-\frac {2 \left (-1+8 x+4 x^2+4 x \log (x)+2 x^2 \log (x)\right )}{x \left (2+e^{x (4+x)}+\log (x)\right )}\right ) \, dx \\ & = -\left (2 \int \frac {-1+8 x+4 x^2+4 x \log (x)+2 x^2 \log (x)}{x \left (2+e^{x (4+x)}+\log (x)\right )} \, dx\right )+\int e^{-x} \left (-e^2+7 e^x+4 e^x x\right ) \, dx \\ & = -\left (2 \int \left (\frac {8}{2+e^{x (4+x)}+\log (x)}-\frac {1}{x \left (2+e^{x (4+x)}+\log (x)\right )}+\frac {4 x}{2+e^{x (4+x)}+\log (x)}+\frac {4 \log (x)}{2+e^{x (4+x)}+\log (x)}+\frac {2 x \log (x)}{2+e^{x (4+x)}+\log (x)}\right ) \, dx\right )+\int \left (7-e^{2-x}+4 x\right ) \, dx \\ & = 7 x+2 x^2+2 \int \frac {1}{x \left (2+e^{x (4+x)}+\log (x)\right )} \, dx-4 \int \frac {x \log (x)}{2+e^{x (4+x)}+\log (x)} \, dx-8 \int \frac {x}{2+e^{x (4+x)}+\log (x)} \, dx-8 \int \frac {\log (x)}{2+e^{x (4+x)}+\log (x)} \, dx-16 \int \frac {1}{2+e^{x (4+x)}+\log (x)} \, dx-\int e^{2-x} \, dx \\ & = e^{2-x}+7 x+2 x^2+2 \int \frac {1}{x \left (2+e^{x (4+x)}+\log (x)\right )} \, dx-4 \int \frac {x \log (x)}{2+e^{x (4+x)}+\log (x)} \, dx-8 \int \frac {x}{2+e^{x (4+x)}+\log (x)} \, dx-8 \int \frac {\log (x)}{2+e^{x (4+x)}+\log (x)} \, dx-16 \int \frac {1}{2+e^{x (4+x)}+\log (x)} \, dx \\ \end{align*}
Time = 5.83 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {2-2 x-2 e^{2-x} x+e^{4 x+x^2} \left (7 x-e^{2-x} x+4 x^2\right )+\left (-x-e^{2-x} x\right ) \log (x)}{2 x+e^{4 x+x^2} x+x \log (x)} \, dx=e^{2-x}-x+2 \log \left (2+e^{x (4+x)}+\log (x)\right ) \]
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Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83
method | result | size |
risch | \(-x +2 \ln \left (2+{\mathrm e}^{\left (4+x \right ) x}+\ln \left (x \right )\right )+{\mathrm e}^{2-x}\) | \(24\) |
default | \(-x +2 \ln \left (\ln \left (x \right )+{\mathrm e}^{x^{2}+4 x}+2\right )+{\mathrm e}^{2-x}\) | \(26\) |
parallelrisch | \(-x +2 \ln \left (\ln \left (x \right )+{\mathrm e}^{x^{2}+4 x}+2\right )+{\mathrm e}^{2-x}\) | \(26\) |
parts | \(-x +2 \ln \left (\ln \left (x \right )+{\mathrm e}^{x^{2}+4 x}+2\right )+{\mathrm e}^{2-x}\) | \(26\) |
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Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {2-2 x-2 e^{2-x} x+e^{4 x+x^2} \left (7 x-e^{2-x} x+4 x^2\right )+\left (-x-e^{2-x} x\right ) \log (x)}{2 x+e^{4 x+x^2} x+x \log (x)} \, dx=-x + e^{\left (-x + 2\right )} + 2 \, \log \left (e^{\left (x^{2} + 4 \, x\right )} + \log \left (x\right ) + 2\right ) \]
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Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {2-2 x-2 e^{2-x} x+e^{4 x+x^2} \left (7 x-e^{2-x} x+4 x^2\right )+\left (-x-e^{2-x} x\right ) \log (x)}{2 x+e^{4 x+x^2} x+x \log (x)} \, dx=- x + e^{2 - x} + 2 \log {\left (e^{x^{2} + 4 x} + \log {\left (x \right )} + 2 \right )} \]
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Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {2-2 x-2 e^{2-x} x+e^{4 x+x^2} \left (7 x-e^{2-x} x+4 x^2\right )+\left (-x-e^{2-x} x\right ) \log (x)}{2 x+e^{4 x+x^2} x+x \log (x)} \, dx={\left (7 \, x e^{x} + e^{2}\right )} e^{\left (-x\right )} + 2 \, \log \left ({\left (e^{\left (x^{2} + 4 \, x\right )} + \log \left (x\right ) + 2\right )} e^{\left (-4 \, x\right )}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (25) = 50\).
Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.97 \[ \int \frac {2-2 x-2 e^{2-x} x+e^{4 x+x^2} \left (7 x-e^{2-x} x+4 x^2\right )+\left (-x-e^{2-x} x\right ) \log (x)}{2 x+e^{4 x+x^2} x+x \log (x)} \, dx=-{\left (x e^{\left (x^{2} + 4 \, x\right )} - 2 \, e^{\left (x^{2} + 4 \, x\right )} \log \left (e^{\left (x^{2} + 4 \, x\right )} + \log \left (x\right ) + 2\right ) - e^{\left (x^{2} + 3 \, x + 2\right )}\right )} e^{\left (-x^{2} - 4 \, x\right )} \]
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Time = 14.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {2-2 x-2 e^{2-x} x+e^{4 x+x^2} \left (7 x-e^{2-x} x+4 x^2\right )+\left (-x-e^{2-x} x\right ) \log (x)}{2 x+e^{4 x+x^2} x+x \log (x)} \, dx=2\,\ln \left (\ln \left (x\right )+{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{x^2}+2\right )-x+{\mathrm {e}}^{2-x} \]
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