Integrand size = 141, antiderivative size = 25 \[ \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\log \left (2 x+\log ^2\left (4 \left (-3-e+e^5-e^x\right )\right )+\log (x)\right ) \]
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Time = 0.38 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6, 6820, 6816} \[ \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\log \left (2 x+\log ^2\left (-4 \left (e^x+3+e-e^5\right )\right )+\log (x)\right ) \]
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Rule 6
Rule 6816
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{-2 e^5 x^2+2 e^x x^2+(6+2 e) x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx \\ & = \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{2 e^x x^2+\left (6+2 e-2 e^5\right ) x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx \\ & = \int \frac {\left (3+e-e^5+e^x\right ) (1+2 x)+2 e^x x \log \left (-4 \left (3+e-e^5+e^x\right )\right )}{\left (e^x+3 \left (1-\frac {1}{3} e \left (-1+e^4\right )\right )\right ) x \left (2 x+\log ^2\left (-4 \left (3+e-e^5+e^x\right )\right )+\log (x)\right )} \, dx \\ & = \log \left (2 x+\log ^2\left (-4 \left (3+e-e^5+e^x\right )\right )+\log (x)\right ) \\ \end{align*}
Time = 1.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\log \left (2 x+\log ^2\left (-4 \left (3+e-e^5+e^x\right )\right )+\log (x)\right ) \]
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Time = 5.40 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\ln \left (\ln \left (x \right )+\ln \left (-4 \,{\mathrm e}^{x}+4 \,{\mathrm e}^{5}-4 \,{\mathrm e}-12\right )^{2}+2 x \right )\) | \(25\) |
parallelrisch | \(\ln \left (\frac {\ln \left (-4 \,{\mathrm e}^{x}+4 \,{\mathrm e}^{5}-4 \,{\mathrm e}-12\right )^{2}}{2}+x +\frac {\ln \left (x \right )}{2}\right )\) | \(27\) |
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Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\log \left (\log \left (4 \, e^{5} - 4 \, e - 4 \, e^{x} - 12\right )^{2} + 2 \, x + \log \left (x\right )\right ) \]
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Time = 1.56 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\log {\left (2 x + \log {\left (x \right )} + \log {\left (- 4 e^{x} - 12 - 4 e + 4 e^{5} \right )}^{2} \right )} \]
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Result contains complex when optimal does not.
Time = 0.36 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.28 \[ \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\log \left (-\pi ^{2} + 4 i \, \pi \log \left (2\right ) + 4 \, \log \left (2\right )^{2} - 2 \, {\left (-i \, \pi - 2 \, \log \left (2\right )\right )} \log \left (-e^{5} + e + e^{x} + 3\right ) + \log \left (-e^{5} + e + e^{x} + 3\right )^{2} + 2 \, x + \log \left (x\right )\right ) \]
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Time = 0.46 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\log \left (\log \left (4 \, e^{5} - 4 \, e - 4 \, e^{x} - 12\right )^{2} + 2 \, x + \log \left (x\right )\right ) \]
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Time = 16.84 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\ln \left ({\ln \left (4\,{\mathrm {e}}^5-4\,\mathrm {e}-4\,{\mathrm {e}}^x-12\right )}^2+2\,x+\ln \left (x\right )\right ) \]
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