Integrand size = 70, antiderivative size = 31 \[ \int \frac {1}{8} \left (36 x-4 e^4 x-4 x \log (4)+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx=\frac {1}{2} x^2 \left (9-e^4-e^{e^{x/4}}-\log (4)\right ) \log (x) \]
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Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26, number of steps used = 12, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {6, 12, 2341, 6874, 2326, 2634} \[ \int \frac {1}{8} \left (36 x-4 e^4 x-4 x \log (4)+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx=\frac {1}{2} x^2 \left (9-e^4-\log (4)\right ) \log (x)-\frac {1}{2} e^{e^{x/4}} x^2 \log (x) \]
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Rule 6
Rule 12
Rule 2326
Rule 2341
Rule 2634
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{8} \left (\left (36-4 e^4\right ) x-4 x \log (4)+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx \\ & = \int \frac {1}{8} \left (x \left (36-4 e^4-4 \log (4)\right )+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx \\ & = \frac {1}{8} \int \left (x \left (36-4 e^4-4 \log (4)\right )+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx \\ & = \frac {1}{4} x^2 \left (9-e^4-\log (4)\right )+\frac {1}{8} \int \left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x) \, dx+\frac {1}{8} \int e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right ) \, dx \\ & = \frac {1}{4} x^2 \left (9-e^4-\log (4)\right )+\frac {1}{8} \int \left (\left (72-8 e^4\right ) x-8 x \log (4)\right ) \log (x) \, dx+\frac {1}{8} \int \left (-4 e^{e^{x/4}} x-e^{e^{x/4}} x \left (8+e^{x/4} x\right ) \log (x)\right ) \, dx \\ & = \frac {1}{4} x^2 \left (9-e^4-\log (4)\right )-\frac {1}{8} \int e^{e^{x/4}} x \left (8+e^{x/4} x\right ) \log (x) \, dx+\frac {1}{8} \int x \left (72-8 e^4-8 \log (4)\right ) \log (x) \, dx-\frac {1}{2} \int e^{e^{x/4}} x \, dx \\ & = \frac {1}{4} x^2 \left (9-e^4-\log (4)\right )-\frac {1}{2} e^{e^{x/4}} x^2 \log (x)+\frac {1}{8} \int 4 e^{e^{x/4}} x \, dx-\frac {1}{2} \int e^{e^{x/4}} x \, dx+\left (9-e^4-\log (4)\right ) \int x \log (x) \, dx \\ & = -\frac {1}{2} e^{e^{x/4}} x^2 \log (x)+\frac {1}{2} x^2 \left (9-e^4-\log (4)\right ) \log (x) \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{8} \left (36 x-4 e^4 x-4 x \log (4)+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx=-\frac {1}{2} x^2 \left (-9+e^4+e^{e^{x/4}}+\log (4)\right ) \log (x) \]
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Time = 1.37 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26
method | result | size |
risch | \(-\frac {\ln \left (x \right ) x^{2} {\mathrm e}^{{\mathrm e}^{\frac {x}{4}}}}{2}-\frac {x^{2} {\mathrm e}^{4} \ln \left (x \right )}{2}-x^{2} \ln \left (2\right ) \ln \left (x \right )+\frac {9 x^{2} \ln \left (x \right )}{2}\) | \(39\) |
parallelrisch | \(-\frac {\ln \left (x \right ) x^{2} {\mathrm e}^{{\mathrm e}^{\frac {x}{4}}}}{2}-\frac {x^{2} {\mathrm e}^{4} \ln \left (x \right )}{2}-x^{2} \ln \left (2\right ) \ln \left (x \right )+\frac {9 x^{2} \ln \left (x \right )}{2}\) | \(39\) |
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Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {1}{8} \left (36 x-4 e^4 x-4 x \log (4)+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx=-\frac {1}{2} \, x^{2} e^{\left (e^{\left (\frac {1}{4} \, x\right )}\right )} \log \left (x\right ) - \frac {1}{2} \, {\left (x^{2} e^{4} + 2 \, x^{2} \log \left (2\right ) - 9 \, x^{2}\right )} \log \left (x\right ) \]
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Time = 2.93 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {1}{8} \left (36 x-4 e^4 x-4 x \log (4)+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx=- \frac {x^{2} e^{e^{\frac {x}{4}}} \log {\left (x \right )}}{2} + \left (- \frac {x^{2} e^{4}}{2} - x^{2} \log {\left (2 \right )} + \frac {9 x^{2}}{2}\right ) \log {\left (x \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (20) = 40\).
Time = 0.21 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.19 \[ \int \frac {1}{8} \left (36 x-4 e^4 x-4 x \log (4)+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx=-\frac {1}{2} \, x^{2} e^{\left (e^{\left (\frac {1}{4} \, x\right )}\right )} \log \left (x\right ) + \frac {1}{4} \, x^{2} {\left (e^{4} + 2 \, \log \left (2\right ) - 9\right )} - \frac {1}{4} \, x^{2} e^{4} - \frac {1}{2} \, x^{2} \log \left (2\right ) + \frac {9}{4} \, x^{2} - \frac {1}{2} \, {\left (x^{2} e^{4} + 2 \, x^{2} \log \left (2\right ) - 9 \, x^{2}\right )} \log \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {1}{8} \left (36 x-4 e^4 x-4 x \log (4)+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx=-\frac {1}{2} \, x^{2} e^{4} \log \left (x\right ) - \frac {1}{2} \, x^{2} e^{\left (e^{\left (\frac {1}{4} \, x\right )}\right )} \log \left (x\right ) - x^{2} \log \left (2\right ) \log \left (x\right ) + \frac {9}{2} \, x^{2} \log \left (x\right ) \]
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Time = 14.55 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {1}{8} \left (36 x-4 e^4 x-4 x \log (4)+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx=-\frac {x^2\,\ln \left (x\right )\,\left (2\,{\mathrm {e}}^4+\ln \left (16\right )+2\,{\mathrm {e}}^{{\mathrm {e}}^{x/4}}-18\right )}{4} \]
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