Integrand size = 53, antiderivative size = 23 \[ \int \frac {-e^{e^{4-x}}+e^{4+e^{4-x}-x} x \log (x) \log \left (\left (-20+5 e^2+5 \log (5)\right ) \log (x)\right )}{x \log (x)} \, dx=-e^{e^{4-x}} \log \left (5 \left (-4+e^2+\log (5)\right ) \log (x)\right ) \]
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Time = 0.71 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17, number of steps used = 4, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {6873, 6874, 2320, 2225, 2635} \[ \int \frac {-e^{e^{4-x}}+e^{4+e^{4-x}-x} x \log (x) \log \left (\left (-20+5 e^2+5 \log (5)\right ) \log (x)\right )}{x \log (x)} \, dx=-e^{e^{4-x}} \log \left (-5 \left (4-e^2-\log (5)\right ) \log (x)\right ) \]
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Rule 2225
Rule 2320
Rule 2635
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{e^{4-x}-x} \left (-e^x+e^4 x \log (x) \log \left (5 \left (-4+e^2+\log (5)\right ) \log (x)\right )\right )}{x \log (x)} \, dx \\ & = \int \left (-\frac {e^{e^{4-x}}}{x \log (x)}+e^{4+e^{4-x}-x} \log \left (5 \left (-4+e^2+\log (5)\right ) \log (x)\right )\right ) \, dx \\ & = -\int \frac {e^{e^{4-x}}}{x \log (x)} \, dx+\int e^{4+e^{4-x}-x} \log \left (5 \left (-4+e^2+\log (5)\right ) \log (x)\right ) \, dx \\ & = -e^{e^{4-x}} \log \left (-5 \left (4-e^2-\log (5)\right ) \log (x)\right ) \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-e^{e^{4-x}}+e^{4+e^{4-x}-x} x \log (x) \log \left (\left (-20+5 e^2+5 \log (5)\right ) \log (x)\right )}{x \log (x)} \, dx=-e^{e^{4-x}} \log \left (5 \left (-4+e^2+\log (5)\right ) \log (x)\right ) \]
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Time = 0.48 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91
method | result | size |
parallelrisch | \(-{\mathrm e}^{{\mathrm e}^{-x +4}} \ln \left (5 \ln \left (x \right ) \left ({\mathrm e}^{2}+\ln \left (5\right )-4\right )\right )\) | \(21\) |
risch | \(-{\mathrm e}^{{\mathrm e}^{-x +4}} \ln \left (\left (5 \ln \left (5\right )+5 \,{\mathrm e}^{2}-20\right ) \ln \left (x \right )\right )\) | \(24\) |
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Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-e^{e^{4-x}}+e^{4+e^{4-x}-x} x \log (x) \log \left (\left (-20+5 e^2+5 \log (5)\right ) \log (x)\right )}{x \log (x)} \, dx=-e^{\left (e^{\left (-x + 4\right )}\right )} \log \left (5 \, {\left (e^{2} + \log \left (5\right ) - 4\right )} \log \left (x\right )\right ) \]
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Timed out. \[ \int \frac {-e^{e^{4-x}}+e^{4+e^{4-x}-x} x \log (x) \log \left (\left (-20+5 e^2+5 \log (5)\right ) \log (x)\right )}{x \log (x)} \, dx=\text {Timed out} \]
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Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-e^{e^{4-x}}+e^{4+e^{4-x}-x} x \log (x) \log \left (\left (-20+5 e^2+5 \log (5)\right ) \log (x)\right )}{x \log (x)} \, dx=-{\left (\log \left (5\right ) + \log \left (e^{2} + \log \left (5\right ) - 4\right ) + \log \left (\log \left (x\right )\right )\right )} e^{\left (e^{\left (-x + 4\right )}\right )} \]
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Time = 0.32 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74 \[ \int \frac {-e^{e^{4-x}}+e^{4+e^{4-x}-x} x \log (x) \log \left (\left (-20+5 e^2+5 \log (5)\right ) \log (x)\right )}{x \log (x)} \, dx=-e^{\left (e^{\left (-x + 4\right )}\right )} \log \left (5\right ) - e^{\left (e^{\left (-x + 4\right )}\right )} \log \left (e^{2} + \log \left (5\right ) - 4\right ) - e^{\left (e^{\left (-x + 4\right )}\right )} \log \left (\log \left (x\right )\right ) \]
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Time = 10.39 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-e^{e^{4-x}}+e^{4+e^{4-x}-x} x \log (x) \log \left (\left (-20+5 e^2+5 \log (5)\right ) \log (x)\right )}{x \log (x)} \, dx=-{\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^4}\,\left (\ln \left (\ln \left (x\right )\right )+\ln \left (5\,{\mathrm {e}}^2+5\,\ln \left (5\right )-20\right )\right ) \]
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