Integrand size = 171, antiderivative size = 24 \[ \int \frac {e^{2 x} (4+16 x)+e^{2 x} \left (x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )+\left (e^{2 x} \left (-4-16 x-16 x^2\right ) \log \left (3 x+6 x^2\right )+e^{2 x} \left (4+17 x+20 x^2+4 x^3\right ) \log ^2\left (3 x+6 x^2\right )\right ) \log \left (\frac {-4+(4+x) \log \left (3 x+6 x^2\right )}{\log \left (3 x+6 x^2\right )}\right )}{(-4-8 x) \log \left (3 x+6 x^2\right )+\left (4+9 x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )} \, dx=e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \]
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\[ \int \frac {e^{2 x} (4+16 x)+e^{2 x} \left (x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )+\left (e^{2 x} \left (-4-16 x-16 x^2\right ) \log \left (3 x+6 x^2\right )+e^{2 x} \left (4+17 x+20 x^2+4 x^3\right ) \log ^2\left (3 x+6 x^2\right )\right ) \log \left (\frac {-4+(4+x) \log \left (3 x+6 x^2\right )}{\log \left (3 x+6 x^2\right )}\right )}{(-4-8 x) \log \left (3 x+6 x^2\right )+\left (4+9 x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )} \, dx=\int \frac {e^{2 x} (4+16 x)+e^{2 x} \left (x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )+\left (e^{2 x} \left (-4-16 x-16 x^2\right ) \log \left (3 x+6 x^2\right )+e^{2 x} \left (4+17 x+20 x^2+4 x^3\right ) \log ^2\left (3 x+6 x^2\right )\right ) \log \left (\frac {-4+(4+x) \log \left (3 x+6 x^2\right )}{\log \left (3 x+6 x^2\right )}\right )}{(-4-8 x) \log \left (3 x+6 x^2\right )+\left (4+9 x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2 x} \left (-4-16 x-x (1+2 x) \log ^2(3 x (1+2 x))-(1+2 x)^2 \log (3 x (1+2 x)) (-4+(4+x) \log (3 x (1+2 x))) \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right )\right )}{(1+2 x) \log (3 x (1+2 x)) (4-(4+x) \log (3 x (1+2 x)))} \, dx \\ & = \int \left (\frac {e^{2 x} \left (4+16 x+x \log ^2(3 x (1+2 x))+2 x^2 \log ^2(3 x (1+2 x))\right )}{(1+2 x) \log (3 x (1+2 x)) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))}+e^{2 x} (1+2 x) \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right )\right ) \, dx \\ & = \int \frac {e^{2 x} \left (4+16 x+x \log ^2(3 x (1+2 x))+2 x^2 \log ^2(3 x (1+2 x))\right )}{(1+2 x) \log (3 x (1+2 x)) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx+\int e^{2 x} (1+2 x) \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx \\ & = \int \frac {e^{2 x} \left (-4-16 x-x (1+2 x) \log ^2(3 x (1+2 x))\right )}{(1+2 x) \log (3 x (1+2 x)) (4-(4+x) \log (3 x (1+2 x)))} \, dx+\int \left (e^{2 x} \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right )+2 e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right )\right ) \, dx \\ & = 2 \int e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx+\int \left (\frac {e^{2 x} x}{4+x}+\frac {e^{2 x} (-1-4 x)}{(1+2 x) \log (3 x (1+2 x))}+\frac {4 e^{2 x} x}{(4+x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))}+\frac {e^{2 x} (4+x) (1+4 x)}{(1+2 x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))}\right ) \, dx+\int e^{2 x} \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx \\ & = 2 \int e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx+4 \int \frac {e^{2 x} x}{(4+x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx+\int \frac {e^{2 x} x}{4+x} \, dx+\int \frac {e^{2 x} (-1-4 x)}{(1+2 x) \log (3 x (1+2 x))} \, dx+\int \frac {e^{2 x} (4+x) (1+4 x)}{(1+2 x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx+\int e^{2 x} \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx \\ & = 2 \int e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx+4 \int \left (\frac {e^{2 x}}{-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x))}-\frac {4 e^{2 x}}{(4+x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))}\right ) \, dx+\int \left (e^{2 x}-\frac {4 e^{2 x}}{4+x}\right ) \, dx+\int \left (-\frac {2 e^{2 x}}{\log (3 x (1+2 x))}+\frac {e^{2 x}}{(1+2 x) \log (3 x (1+2 x))}\right ) \, dx+\int \left (\frac {15 e^{2 x}}{2 (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))}+\frac {2 e^{2 x} x}{-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x))}-\frac {7 e^{2 x}}{2 (1+2 x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))}\right ) \, dx+\int e^{2 x} \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx \\ & = -\left (2 \int \frac {e^{2 x}}{\log (3 x (1+2 x))} \, dx\right )+2 \int \frac {e^{2 x} x}{-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x))} \, dx+2 \int e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx-\frac {7}{2} \int \frac {e^{2 x}}{(1+2 x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx-4 \int \frac {e^{2 x}}{4+x} \, dx+4 \int \frac {e^{2 x}}{-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x))} \, dx+\frac {15}{2} \int \frac {e^{2 x}}{-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x))} \, dx-16 \int \frac {e^{2 x}}{(4+x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx+\int e^{2 x} \, dx+\int \frac {e^{2 x}}{(1+2 x) \log (3 x (1+2 x))} \, dx+\int e^{2 x} \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx \\ & = \frac {e^{2 x}}{2}-\frac {4 \text {Ei}(2 (4+x))}{e^8}-2 \int \frac {e^{2 x}}{\log (3 x (1+2 x))} \, dx+2 \int \frac {e^{2 x} x}{-4+(4+x) \log (3 x (1+2 x))} \, dx+2 \int e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx-\frac {7}{2} \int \frac {e^{2 x}}{(1+2 x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx+4 \int \frac {e^{2 x}}{-4+(4+x) \log (3 x (1+2 x))} \, dx+\frac {15}{2} \int \frac {e^{2 x}}{-4+(4+x) \log (3 x (1+2 x))} \, dx-16 \int \frac {e^{2 x}}{(4+x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx+\int \frac {e^{2 x}}{(1+2 x) \log (3 x (1+2 x))} \, dx+\int e^{2 x} \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 x} (4+16 x)+e^{2 x} \left (x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )+\left (e^{2 x} \left (-4-16 x-16 x^2\right ) \log \left (3 x+6 x^2\right )+e^{2 x} \left (4+17 x+20 x^2+4 x^3\right ) \log ^2\left (3 x+6 x^2\right )\right ) \log \left (\frac {-4+(4+x) \log \left (3 x+6 x^2\right )}{\log \left (3 x+6 x^2\right )}\right )}{(-4-8 x) \log \left (3 x+6 x^2\right )+\left (4+9 x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )} \, dx=e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \]
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Time = 94.42 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54
method | result | size |
parallelrisch | \(\ln \left (\frac {\left (4+x \right ) \ln \left (6 x^{2}+3 x \right )-4}{\ln \left (6 x^{2}+3 x \right )}\right ) {\mathrm e}^{2 x} x\) | \(37\) |
risch | \(\text {Expression too large to display}\) | \(2063\) |
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Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{2 x} (4+16 x)+e^{2 x} \left (x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )+\left (e^{2 x} \left (-4-16 x-16 x^2\right ) \log \left (3 x+6 x^2\right )+e^{2 x} \left (4+17 x+20 x^2+4 x^3\right ) \log ^2\left (3 x+6 x^2\right )\right ) \log \left (\frac {-4+(4+x) \log \left (3 x+6 x^2\right )}{\log \left (3 x+6 x^2\right )}\right )}{(-4-8 x) \log \left (3 x+6 x^2\right )+\left (4+9 x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )} \, dx=x e^{\left (2 \, x\right )} \log \left (\frac {{\left (x + 4\right )} \log \left (6 \, x^{2} + 3 \, x\right ) - 4}{\log \left (6 \, x^{2} + 3 \, x\right )}\right ) \]
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Timed out. \[ \int \frac {e^{2 x} (4+16 x)+e^{2 x} \left (x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )+\left (e^{2 x} \left (-4-16 x-16 x^2\right ) \log \left (3 x+6 x^2\right )+e^{2 x} \left (4+17 x+20 x^2+4 x^3\right ) \log ^2\left (3 x+6 x^2\right )\right ) \log \left (\frac {-4+(4+x) \log \left (3 x+6 x^2\right )}{\log \left (3 x+6 x^2\right )}\right )}{(-4-8 x) \log \left (3 x+6 x^2\right )+\left (4+9 x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (23) = 46\).
Time = 0.34 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.42 \[ \int \frac {e^{2 x} (4+16 x)+e^{2 x} \left (x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )+\left (e^{2 x} \left (-4-16 x-16 x^2\right ) \log \left (3 x+6 x^2\right )+e^{2 x} \left (4+17 x+20 x^2+4 x^3\right ) \log ^2\left (3 x+6 x^2\right )\right ) \log \left (\frac {-4+(4+x) \log \left (3 x+6 x^2\right )}{\log \left (3 x+6 x^2\right )}\right )}{(-4-8 x) \log \left (3 x+6 x^2\right )+\left (4+9 x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )} \, dx=x e^{\left (2 \, x\right )} \log \left (x {\left (\log \left (3\right ) + \log \left (2 \, x + 1\right )\right )} + {\left (x + 4\right )} \log \left (x\right ) + 4 \, \log \left (3\right ) + 4 \, \log \left (2 \, x + 1\right ) - 4\right ) - x e^{\left (2 \, x\right )} \log \left (\log \left (3\right ) + \log \left (2 \, x + 1\right ) + \log \left (x\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (23) = 46\).
Time = 0.91 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \[ \int \frac {e^{2 x} (4+16 x)+e^{2 x} \left (x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )+\left (e^{2 x} \left (-4-16 x-16 x^2\right ) \log \left (3 x+6 x^2\right )+e^{2 x} \left (4+17 x+20 x^2+4 x^3\right ) \log ^2\left (3 x+6 x^2\right )\right ) \log \left (\frac {-4+(4+x) \log \left (3 x+6 x^2\right )}{\log \left (3 x+6 x^2\right )}\right )}{(-4-8 x) \log \left (3 x+6 x^2\right )+\left (4+9 x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )} \, dx=x e^{\left (2 \, x\right )} \log \left (x \log \left (6 \, x^{2} + 3 \, x\right ) + 4 \, \log \left (6 \, x^{2} + 3 \, x\right ) - 4\right ) - x e^{\left (2 \, x\right )} \log \left (\log \left (6 \, x^{2} + 3 \, x\right )\right ) \]
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Time = 12.68 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{2 x} (4+16 x)+e^{2 x} \left (x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )+\left (e^{2 x} \left (-4-16 x-16 x^2\right ) \log \left (3 x+6 x^2\right )+e^{2 x} \left (4+17 x+20 x^2+4 x^3\right ) \log ^2\left (3 x+6 x^2\right )\right ) \log \left (\frac {-4+(4+x) \log \left (3 x+6 x^2\right )}{\log \left (3 x+6 x^2\right )}\right )}{(-4-8 x) \log \left (3 x+6 x^2\right )+\left (4+9 x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )} \, dx=x\,{\mathrm {e}}^{2\,x}\,\ln \left (\frac {\ln \left (6\,x^2+3\,x\right )\,\left (x+4\right )-4}{\ln \left (6\,x^2+3\,x\right )}\right ) \]
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