Integrand size = 222, antiderivative size = 27 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=x+\frac {1}{-2 x+\log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \]
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\[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=\int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2 x} \left (-1+2 x+4 x^3\right )-64 \left (1+x^2\right ) (1+\log (5))^2-4 x \left (e^{2 x} x-16 (1+\log (5))^2\right ) \log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )+\left (e^{2 x} x-16 (1+\log (5))^2\right ) \log ^2\left (-x+16 e^{-2 x} (1+\log (5))^2\right )}{\left (e^{2 x} x-16 (1+\log (5))^2\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx \\ & = \int \left (\frac {16 (-1-2 x) (1+\log (5))^2}{x \left (e^{2 x} x-16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}+\frac {-1+2 x+4 x^3-4 x^2 \log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )+x \log ^2\left (-x+16 e^{-2 x} (1+\log (5))^2\right )}{x \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}\right ) \, dx \\ & = \left (16 (1+\log (5))^2\right ) \int \frac {-1-2 x}{x \left (e^{2 x} x-16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\int \frac {-1+2 x+4 x^3-4 x^2 \log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )+x \log ^2\left (-x+16 e^{-2 x} (1+\log (5))^2\right )}{x \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx \\ & = \left (16 (1+\log (5))^2\right ) \int \left (\frac {2}{\left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}+\frac {1}{x \left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}\right ) \, dx+\int \left (1+\frac {-1+2 x}{x \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}\right ) \, dx \\ & = x+\left (16 (1+\log (5))^2\right ) \int \frac {1}{x \left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\left (32 (1+\log (5))^2\right ) \int \frac {1}{\left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\int \frac {-1+2 x}{x \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx \\ & = x+\left (16 (1+\log (5))^2\right ) \int \frac {1}{x \left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\left (32 (1+\log (5))^2\right ) \int \frac {1}{\left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\int \left (\frac {2}{\left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}-\frac {1}{x \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2}\right ) \, dx \\ & = x+2 \int \frac {1}{\left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\left (16 (1+\log (5))^2\right ) \int \frac {1}{x \left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx+\left (32 (1+\log (5))^2\right ) \int \frac {1}{\left (-e^{2 x} x+16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx-\int \frac {1}{x \left (2 x-\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )\right )^2} \, dx \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=x+\frac {1}{-2 x+\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )} \]
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Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15
method | result | size |
risch | \(x -\frac {1}{2 x -\ln \left (\left (4 \ln \left (5\right )+4\right )^{2} {\mathrm e}^{-2 x}-x \right )}\) | \(31\) |
parallelrisch | \(\frac {-1+2 x^{2}-\ln \left (16 \left (\ln \left (5\right )+1\right )^{2} {\mathrm e}^{-2 x}-x \right ) x}{-\ln \left (16 \left (\ln \left (5\right )+1\right )^{2} {\mathrm e}^{-2 x}-x \right )+2 x}\) | \(58\) |
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Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=\frac {2 \, x^{2} - x \log \left (-x + e^{\left (-2 \, x + 2 \, \log \left (4 \, \log \left (5\right ) + 4\right )\right )}\right ) - 1}{2 \, x - \log \left (-x + e^{\left (-2 \, x + 2 \, \log \left (4 \, \log \left (5\right ) + 4\right )\right )}\right )} \]
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Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=x + \frac {1}{- 2 x + \log {\left (- x + \left (4 + 4 \log {\left (5 \right )}\right )^{2} e^{- 2 x} \right )}} \]
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Time = 0.63 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.19 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=\frac {4 \, x^{2} - x \log \left (-x e^{\left (2 \, x\right )} + 16 \, \log \left (5\right )^{2} + 32 \, \log \left (5\right ) + 16\right ) - 1}{4 \, x - \log \left (-x e^{\left (2 \, x\right )} + 16 \, \log \left (5\right )^{2} + 32 \, \log \left (5\right ) + 16\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (31) = 62\).
Time = 1.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.85 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=\frac {2 \, x^{2} - x \log \left (16 \, e^{\left (-2 \, x\right )} \log \left (5\right )^{2} + 32 \, e^{\left (-2 \, x\right )} \log \left (5\right ) - x + 16 \, e^{\left (-2 \, x\right )}\right ) - 1}{2 \, x - \log \left (16 \, e^{\left (-2 \, x\right )} \log \left (5\right )^{2} + 32 \, e^{\left (-2 \, x\right )} \log \left (5\right ) - x + 16 \, e^{\left (-2 \, x\right )}\right )} \]
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Time = 11.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=x-\frac {1}{2\,x-\ln \left (16\,{\mathrm {e}}^{-2\,x}-x+32\,{\mathrm {e}}^{-2\,x}\,\ln \left (5\right )+16\,{\mathrm {e}}^{-2\,x}\,{\ln \left (5\right )}^2\right )} \]
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