Integrand size = 116, antiderivative size = 24 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=\log (5)-\log \left (5-\log (2 x) \log \left (e^{(2+x)^2}+x\right )\right ) \]
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Time = 0.81 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6820, 6840, 31} \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (5-\log (2 x) \log \left (x+e^{(x+2)^2}\right )\right ) \]
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Rule 31
Rule 6820
Rule 6840
Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (1+2 e^{(2+x)^2} (2+x)\right ) \log (2 x)+\left (e^{(2+x)^2}+x\right ) \log \left (e^{(2+x)^2}+x\right )}{x \left (e^{(2+x)^2}+x\right ) \left (5-\log (2 x) \log \left (e^{(2+x)^2}+x\right )\right )} \, dx \\ & = \text {Subst}\left (\int \frac {1}{5-x} \, dx,x,\log (2 x) \log \left (e^{(2+x)^2}+x\right )\right ) \\ & = -\log \left (5-\log (2 x) \log \left (e^{(2+x)^2}+x\right )\right ) \\ \end{align*}
Time = 0.55 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (5-\log (2 x) \log \left (e^{4+4 x+x^2}+x\right )\right ) \]
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Time = 1.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(-\ln \left (\ln \left (2 x \right ) \ln \left ({\mathrm e}^{x^{2}+4 x +4}+x \right )-5\right )\) | \(23\) |
risch | \(-\ln \left (\ln \left (2 x \right )\right )-\ln \left (\ln \left (x +{\mathrm e}^{\left (2+x \right )^{2}}\right )-\frac {5}{\ln \left (2 x \right )}\right )\) | \(30\) |
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Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (\frac {\log \left (2 \, x\right ) \log \left (x + e^{\left (x^{2} + 4 \, x + 4\right )}\right ) - 5}{\log \left (2 \, x\right )}\right ) - \log \left (\log \left (2 \, x\right )\right ) \]
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Time = 0.42 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=- \log {\left (\log {\left (x + e^{x^{2} + 4 x + 4} \right )} - \frac {5}{\log {\left (2 x \right )}} \right )} - \log {\left (\log {\left (2 x \right )} \right )} \]
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Time = 0.34 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (\frac {{\left (\log \left (2\right ) + \log \left (x\right )\right )} \log \left (x + e^{\left (x^{2} + 4 \, x + 4\right )}\right ) - 5}{\log \left (2\right ) + \log \left (x\right )}\right ) - \log \left (\log \left (2\right ) + \log \left (x\right )\right ) \]
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Time = 0.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (\log \left (2 \, x\right ) \log \left (x + e^{\left (x^{2} + 4 \, x + 4\right )}\right ) - 5\right ) \]
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Time = 14.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\ln \left (\frac {\ln \left (2\,x\right )\,\ln \left (x+{\mathrm {e}}^{{\left (x+2\right )}^2}\right )-5}{\ln \left (2\,x\right )}\right )-\ln \left (\ln \left (2\,x\right )\right ) \]
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