Integrand size = 47, antiderivative size = 25 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=\log \left (-5-2 x+\frac {1}{2} e^{3+x-3 (1+x)} x-\log (x)\right ) \]
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\[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=\int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1+2 x}{x (5+2 x+\log (x))}+\frac {-4+10 x+4 x^2-\log (x)+2 x \log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}\right ) \, dx \\ & = \int \frac {1+2 x}{x (5+2 x+\log (x))} \, dx+\int \frac {-4+10 x+4 x^2-\log (x)+2 x \log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx \\ & = \log (5+2 x+\log (x))+\int \left (-\frac {4}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}+\frac {10 x}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}+\frac {4 x^2}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}-\frac {\log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}+\frac {2 x \log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}\right ) \, dx \\ & = \log (5+2 x+\log (x))+2 \int \frac {x \log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx-4 \int \frac {1}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx+4 \int \frac {x^2}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx+10 \int \frac {x}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx-\int \frac {\log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx \\ \end{align*}
Time = 0.87 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=\log \left (10+\left (4-e^{-2 x}\right ) x+2 \log (x)\right ) \]
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Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64
method | result | size |
risch | \(\ln \left (\ln \left (x \right )+2 x +5-\frac {{\mathrm e}^{-2 x} x}{2}\right )\) | \(16\) |
parallelrisch | \(\ln \left ({\mathrm e}^{x} {\mathrm e}^{-3 x} x -2 \ln \left (x \right )-4 x -10\right )\) | \(19\) |
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Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=\log \left (-x e^{\left (-2 \, x\right )} + 4 \, x + 2 \, \log \left (x\right ) + 10\right ) \]
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Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=\log {\left (x \right )} + \log {\left (e^{- 2 x} + \frac {- 4 x - 2 \log {\left (x \right )} - 10}{x} \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (17) = 34\).
Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=-2 \, x + \log \left (2 \, x + \log \left (x\right ) + 5\right ) + \log \left (\frac {2 \, {\left (2 \, x + \log \left (x\right ) + 5\right )} e^{\left (2 \, x\right )} - x}{2 \, {\left (2 \, x + \log \left (x\right ) + 5\right )}}\right ) \]
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Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=\log \left (x e^{\left (-2 \, x\right )} - 4 \, x - 2 \, \log \left (x\right ) - 10\right ) \]
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Time = 12.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.60 \[ \int \frac {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx=\ln \left (2\,x+\ln \left (x\right )-\frac {x\,{\mathrm {e}}^{-2\,x}}{2}+5\right ) \]
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