Integrand size = 86, antiderivative size = 31 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=e^5+\frac {1}{e^5 x^2}-\log ^2\left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right ) \]
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 6820, 14, 6874, 6816, 6818} \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=\frac {1}{e^5 x^2}-\log ^2\left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right ) \]
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Rule 12
Rule 14
Rule 6816
Rule 6818
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{x^3 \log \left (e^{-x/4} x\right )} \, dx}{2 e^5} \\ & = \frac {\int \frac {-4-\frac {e^5 x^2 \left (-4+x+4 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{\log \left (e^{-x/4} x\right )}}{x^3} \, dx}{2 e^5} \\ & = \frac {\int \left (-\frac {4}{x^3}-\frac {e^5 \left (-4+x+4 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{x \log \left (e^{-x/4} x\right )}\right ) \, dx}{2 e^5} \\ & = \frac {1}{e^5 x^2}-\frac {1}{2} \int \frac {\left (-4+x+4 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{x \log \left (e^{-x/4} x\right )} \, dx \\ & = \frac {1}{e^5 x^2}-\log ^2\left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right ) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=\frac {1}{e^5 x^2}-\log ^2\left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right ) \]
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\[\int \frac {\left (\left (-4 x^{2} {\mathrm e}^{5} \ln \left (x \,{\mathrm e}^{-\frac {x}{4}}\right )+\left (-x^{3}+4 x^{2}\right ) {\mathrm e}^{5}\right ) \ln \left (\frac {3 x}{\ln \left (x \,{\mathrm e}^{-\frac {x}{4}}\right )}\right )-4 \ln \left (x \,{\mathrm e}^{-\frac {x}{4}}\right )\right ) {\mathrm e}^{-5}}{2 x^{3} \ln \left (x \,{\mathrm e}^{-\frac {x}{4}}\right )}d x\]
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Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=-\frac {{\left (x^{2} e^{5} \log \left (\frac {3 \, x}{\log \left (x e^{\left (-\frac {1}{4} \, x\right )}\right )}\right )^{2} - 1\right )} e^{\left (-5\right )}}{x^{2}} \]
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Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=- \log {\left (\frac {3 x}{\log {\left (x e^{- \frac {x}{4}} \right )}} \right )}^{2} + \frac {1}{x^{2} e^{5}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (26) = 52\).
Time = 0.36 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.29 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=-{\left (2 \, {\left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )} e^{5} \log \left (x\right ) + e^{5} \log \left (x\right )^{2} + e^{5} \log \left (-x + 4 \, \log \left (x\right )\right )^{2} - 2 \, {\left ({\left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )} e^{5} + e^{5} \log \left (x\right )\right )} \log \left (-x + 4 \, \log \left (x\right )\right ) - \frac {1}{x^{2}}\right )} e^{\left (-5\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (26) = 52\).
Time = 0.32 (sec) , antiderivative size = 456, normalized size of antiderivative = 14.71 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=-\frac {{\left (2 \, \pi ^{2} x^{2} e^{5} \mathrm {sgn}\left (-2 \, \pi + 2 \, \pi \mathrm {sgn}\left (x\right )\right ) \mathrm {sgn}\left (x - 4 \, \log \left ({\left | x \right |}\right )\right ) \mathrm {sgn}\left (x\right ) - 10 \, \pi ^{2} x^{2} e^{5} \mathrm {sgn}\left (-2 \, \pi + 2 \, \pi \mathrm {sgn}\left (x\right )\right ) \mathrm {sgn}\left (x - 4 \, \log \left ({\left | x \right |}\right )\right ) + 4 \, \pi x^{2} \arctan \left (-\frac {2 \, {\left (\pi - \pi \mathrm {sgn}\left (x\right )\right )}}{x - 4 \, \log \left ({\left | x \right |}\right )}\right ) e^{5} \mathrm {sgn}\left (-2 \, \pi + 2 \, \pi \mathrm {sgn}\left (x\right )\right ) \mathrm {sgn}\left (x - 4 \, \log \left ({\left | x \right |}\right )\right ) - 2 \, \pi ^{2} x^{2} e^{5} \mathrm {sgn}\left (-2 \, \pi + 2 \, \pi \mathrm {sgn}\left (x\right )\right ) \mathrm {sgn}\left (x\right ) - 6 \, \pi ^{2} x^{2} e^{5} \mathrm {sgn}\left (-2 \, \pi + 2 \, \pi \mathrm {sgn}\left (x\right )\right ) - 4 \, \pi x^{2} \arctan \left (-\frac {2 \, {\left (\pi - \pi \mathrm {sgn}\left (x\right )\right )}}{x - 4 \, \log \left ({\left | x \right |}\right )}\right ) e^{5} \mathrm {sgn}\left (-2 \, \pi + 2 \, \pi \mathrm {sgn}\left (x\right )\right ) + 2 \, \pi ^{2} x^{2} e^{5} \mathrm {sgn}\left (x - 4 \, \log \left ({\left | x \right |}\right )\right ) + 10 \, \pi ^{2} x^{2} e^{5} \mathrm {sgn}\left (x\right ) - 4 \, \pi x^{2} \arctan \left (-\frac {2 \, {\left (\pi - \pi \mathrm {sgn}\left (x\right )\right )}}{x - 4 \, \log \left ({\left | x \right |}\right )}\right ) e^{5} \mathrm {sgn}\left (x\right ) - 12 \, \pi ^{2} x^{2} e^{5} + 20 \, \pi x^{2} \arctan \left (-\frac {2 \, {\left (\pi - \pi \mathrm {sgn}\left (x\right )\right )}}{x - 4 \, \log \left ({\left | x \right |}\right )}\right ) e^{5} - 4 \, x^{2} \arctan \left (-\frac {2 \, {\left (\pi - \pi \mathrm {sgn}\left (x\right )\right )}}{x - 4 \, \log \left ({\left | x \right |}\right )}\right )^{2} e^{5} - 4 \, x^{2} e^{5} \log \left (12\right ) \log \left (-8 \, \pi ^{2} \mathrm {sgn}\left (x\right ) + 8 \, \pi ^{2} + x^{2} - 8 \, x \log \left ({\left | x \right |}\right ) + 16 \, \log \left ({\left | x \right |}\right )^{2}\right ) + x^{2} e^{5} \log \left (-8 \, \pi ^{2} \mathrm {sgn}\left (x\right ) + 8 \, \pi ^{2} + x^{2} - 8 \, x \log \left ({\left | x \right |}\right ) + 16 \, \log \left ({\left | x \right |}\right )^{2}\right )^{2} + 8 \, x^{2} e^{5} \log \left (12\right ) \log \left ({\left | x \right |}\right ) - 4 \, x^{2} e^{5} \log \left (-8 \, \pi ^{2} \mathrm {sgn}\left (x\right ) + 8 \, \pi ^{2} + x^{2} - 8 \, x \log \left ({\left | x \right |}\right ) + 16 \, \log \left ({\left | x \right |}\right )^{2}\right ) \log \left ({\left | x \right |}\right ) + 4 \, x^{2} e^{5} \log \left ({\left | x \right |}\right )^{2} - 4\right )} e^{\left (-5\right )}}{4 \, x^{2}} \]
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Time = 13.67 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=\frac {{\mathrm {e}}^{-5}}{x^2}-{\ln \left (-\frac {12\,x}{x-4\,\ln \left (x\right )}\right )}^2 \]
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