Integrand size = 100, antiderivative size = 21 \[ \int \frac {e^{16+2 x} x+8 e^{16} x^2+e^{4 e^{-2 x}} \left (-4 e^{2 x}+2 e^{2 x} \log (x)\right )}{e^{16+2 x} x^2+e^{4 e^{-2 x}} \left (4 e^{2 x} x-4 e^{2 x} x \log (x)+e^{2 x} x \log ^2(x)\right )} \, dx=\log \left (e^{16-4 e^{-2 x}} x+(-2+\log (x))^2\right ) \]
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\[ \int \frac {e^{16+2 x} x+8 e^{16} x^2+e^{4 e^{-2 x}} \left (-4 e^{2 x}+2 e^{2 x} \log (x)\right )}{e^{16+2 x} x^2+e^{4 e^{-2 x}} \left (4 e^{2 x} x-4 e^{2 x} x \log (x)+e^{2 x} x \log ^2(x)\right )} \, dx=\int \frac {e^{16+2 x} x+8 e^{16} x^2+e^{4 e^{-2 x}} \left (-4 e^{2 x}+2 e^{2 x} \log (x)\right )}{e^{16+2 x} x^2+e^{4 e^{-2 x}} \left (4 e^{2 x} x-4 e^{2 x} x \log (x)+e^{2 x} x \log ^2(x)\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-2 x} \left (e^{2 (8+x)} x+8 e^{16} x^2+2 e^{4 e^{-2 x}+2 x} (-2+\log (x))\right )}{x \left (e^{16} x+e^{4 e^{-2 x}} (-2+\log (x))^2\right )} \, dx \\ & = \int \left (\frac {8 e^{16-2 x} x}{4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)}+\frac {-4 e^{4 e^{-2 x}}+e^{16} x+2 e^{4 e^{-2 x}} \log (x)}{x \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )}\right ) \, dx \\ & = 8 \int \frac {e^{16-2 x} x}{4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)} \, dx+\int \frac {-4 e^{4 e^{-2 x}}+e^{16} x+2 e^{4 e^{-2 x}} \log (x)}{x \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )} \, dx \\ & = 8 \int \frac {e^{16-2 x} x}{4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)} \, dx+\int \left (\frac {2}{x (-2+\log (x))}+\frac {e^{16} (-4+\log (x))}{(-2+\log (x)) \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )}\right ) \, dx \\ & = 2 \int \frac {1}{x (-2+\log (x))} \, dx+8 \int \frac {e^{16-2 x} x}{4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)} \, dx+e^{16} \int \frac {-4+\log (x)}{(-2+\log (x)) \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )} \, dx \\ & = 2 \text {Subst}\left (\int \frac {1}{x} \, dx,x,-2+\log (x)\right )+8 \int \frac {e^{16-2 x} x}{4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)} \, dx+e^{16} \int \left (-\frac {4}{(-2+\log (x)) \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )}+\frac {\log (x)}{(-2+\log (x)) \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )}\right ) \, dx \\ & = 2 \log (2-\log (x))+8 \int \frac {e^{16-2 x} x}{4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)} \, dx+e^{16} \int \frac {\log (x)}{(-2+\log (x)) \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )} \, dx-\left (4 e^{16}\right ) \int \frac {1}{(-2+\log (x)) \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )} \, dx \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(21)=42\).
Time = 1.52 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.52 \[ \int \frac {e^{16+2 x} x+8 e^{16} x^2+e^{4 e^{-2 x}} \left (-4 e^{2 x}+2 e^{2 x} \log (x)\right )}{e^{16+2 x} x^2+e^{4 e^{-2 x}} \left (4 e^{2 x} x-4 e^{2 x} x \log (x)+e^{2 x} x \log ^2(x)\right )} \, dx=-4 e^{-2 x}+\log \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right ) \]
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Time = 25.66 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.90
method | result | size |
risch | \(2 \ln \left (\ln \left (x \right )-2\right )-4 \,{\mathrm e}^{-2 x}+\ln \left ({\mathrm e}^{4 \,{\mathrm e}^{-2 x}}+\frac {x \,{\mathrm e}^{16}}{\ln \left (x \right )^{2}-4 \ln \left (x \right )+4}\right )\) | \(40\) |
parallelrisch | \({\mathrm e}^{-2 x} \left (-4+\ln \left (\left (\ln \left (x \right )^{2} {\mathrm e}^{4 \,{\mathrm e}^{-2 x}}-4 \ln \left (x \right ) {\mathrm e}^{4 \,{\mathrm e}^{-2 x}}+x \,{\mathrm e}^{16}+4 \,{\mathrm e}^{4 \,{\mathrm e}^{-2 x}}\right ) {\mathrm e}^{-16}\right ) {\mathrm e}^{2 x}\right )\) | \(56\) |
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Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (19) = 38\).
Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.29 \[ \int \frac {e^{16+2 x} x+8 e^{16} x^2+e^{4 e^{-2 x}} \left (-4 e^{2 x}+2 e^{2 x} \log (x)\right )}{e^{16+2 x} x^2+e^{4 e^{-2 x}} \left (4 e^{2 x} x-4 e^{2 x} x \log (x)+e^{2 x} x \log ^2(x)\right )} \, dx={\left (e^{\left (2 \, x + 16\right )} \log \left (\frac {x e^{16} + {\left (\log \left (x\right )^{2} - 4 \, \log \left (x\right ) + 4\right )} e^{\left (4 \, e^{\left (-2 \, x\right )}\right )}}{\log \left (x\right )^{2} - 4 \, \log \left (x\right ) + 4}\right ) + 2 \, e^{\left (2 \, x + 16\right )} \log \left (\log \left (x\right ) - 2\right ) - 4 \, e^{16}\right )} e^{\left (-2 \, x - 16\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 0.49 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {e^{16+2 x} x+8 e^{16} x^2+e^{4 e^{-2 x}} \left (-4 e^{2 x}+2 e^{2 x} \log (x)\right )}{e^{16+2 x} x^2+e^{4 e^{-2 x}} \left (4 e^{2 x} x-4 e^{2 x} x \log (x)+e^{2 x} x \log ^2(x)\right )} \, dx=\log {\left (\frac {x e^{16}}{\log {\left (x \right )}^{2} - 4 \log {\left (x \right )} + 4} + e^{4 e^{- 2 x}} \right )} + 2 \log {\left (\log {\left (x \right )} - 2 \right )} - 4 e^{- 2 x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (19) = 38\).
Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.43 \[ \int \frac {e^{16+2 x} x+8 e^{16} x^2+e^{4 e^{-2 x}} \left (-4 e^{2 x}+2 e^{2 x} \log (x)\right )}{e^{16+2 x} x^2+e^{4 e^{-2 x}} \left (4 e^{2 x} x-4 e^{2 x} x \log (x)+e^{2 x} x \log ^2(x)\right )} \, dx=-4 \, e^{\left (-2 \, x\right )} + \log \left (\frac {x e^{16} + {\left (\log \left (x\right )^{2} - 4 \, \log \left (x\right ) + 4\right )} e^{\left (4 \, e^{\left (-2 \, x\right )}\right )}}{\log \left (x\right )^{2} - 4 \, \log \left (x\right ) + 4}\right ) + 2 \, \log \left (\log \left (x\right ) - 2\right ) \]
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\[ \int \frac {e^{16+2 x} x+8 e^{16} x^2+e^{4 e^{-2 x}} \left (-4 e^{2 x}+2 e^{2 x} \log (x)\right )}{e^{16+2 x} x^2+e^{4 e^{-2 x}} \left (4 e^{2 x} x-4 e^{2 x} x \log (x)+e^{2 x} x \log ^2(x)\right )} \, dx=\int { \frac {8 \, x^{2} e^{16} + x e^{\left (2 \, x + 16\right )} + 2 \, {\left (e^{\left (2 \, x\right )} \log \left (x\right ) - 2 \, e^{\left (2 \, x\right )}\right )} e^{\left (4 \, e^{\left (-2 \, x\right )}\right )}}{x^{2} e^{\left (2 \, x + 16\right )} + {\left (x e^{\left (2 \, x\right )} \log \left (x\right )^{2} - 4 \, x e^{\left (2 \, x\right )} \log \left (x\right ) + 4 \, x e^{\left (2 \, x\right )}\right )} e^{\left (4 \, e^{\left (-2 \, x\right )}\right )}} \,d x } \]
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Timed out. \[ \int \frac {e^{16+2 x} x+8 e^{16} x^2+e^{4 e^{-2 x}} \left (-4 e^{2 x}+2 e^{2 x} \log (x)\right )}{e^{16+2 x} x^2+e^{4 e^{-2 x}} \left (4 e^{2 x} x-4 e^{2 x} x \log (x)+e^{2 x} x \log ^2(x)\right )} \, dx=\int \frac {x\,{\mathrm {e}}^{2\,x+16}+8\,x^2\,{\mathrm {e}}^{16}-{\mathrm {e}}^{4\,{\mathrm {e}}^{-2\,x}}\,\left (4\,{\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^{2\,x}\,\ln \left (x\right )\right )}{{\mathrm {e}}^{4\,{\mathrm {e}}^{-2\,x}}\,\left (x\,{\mathrm {e}}^{2\,x}\,{\ln \left (x\right )}^2-4\,x\,{\mathrm {e}}^{2\,x}\,\ln \left (x\right )+4\,x\,{\mathrm {e}}^{2\,x}\right )+x^2\,{\mathrm {e}}^{2\,x+16}} \,d x \]
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