Integrand size = 102, antiderivative size = 25 \[ \int \frac {\left (48600 x+20250 e^{2+2 x} x\right ) \log (\log (x))+\left (48600 x+e^{2+2 x} \left (20250 x-81000 x^2\right )\right ) \log (x) \log ^2(\log (x))}{\left (248832+518400 e^{2+2 x}+432000 e^{4+4 x}+180000 e^{6+6 x}+37500 e^{8+8 x}+3125 e^{10+10 x}\right ) \log (x)} \, dx=\frac {25 x^2 \log ^2(\log (x))}{\left (4+\frac {5}{3} e^{2+2 x}\right )^4} \]
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\[ \int \frac {\left (48600 x+20250 e^{2+2 x} x\right ) \log (\log (x))+\left (48600 x+e^{2+2 x} \left (20250 x-81000 x^2\right )\right ) \log (x) \log ^2(\log (x))}{\left (248832+518400 e^{2+2 x}+432000 e^{4+4 x}+180000 e^{6+6 x}+37500 e^{8+8 x}+3125 e^{10+10 x}\right ) \log (x)} \, dx=\int \frac {\left (48600 x+20250 e^{2+2 x} x\right ) \log (\log (x))+\left (48600 x+e^{2+2 x} \left (20250 x-81000 x^2\right )\right ) \log (x) \log ^2(\log (x))}{\left (248832+518400 e^{2+2 x}+432000 e^{4+4 x}+180000 e^{6+6 x}+37500 e^{8+8 x}+3125 e^{10+10 x}\right ) \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {4050 x \log (\log (x)) \left (12+5 e^{2+2 x}-\left (-12+5 e^{2+2 x} (-1+4 x)\right ) \log (x) \log (\log (x))\right )}{\left (12+5 e^{2+2 x}\right )^5 \log (x)} \, dx \\ & = 4050 \int \frac {x \log (\log (x)) \left (12+5 e^{2+2 x}-\left (-12+5 e^{2+2 x} (-1+4 x)\right ) \log (x) \log (\log (x))\right )}{\left (12+5 e^{2+2 x}\right )^5 \log (x)} \, dx \\ & = 4050 \int \left (\frac {48 x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^5}-\frac {x \log (\log (x)) (-1-\log (x) \log (\log (x))+4 x \log (x) \log (\log (x)))}{\left (12+5 e^{2+2 x}\right )^4 \log (x)}\right ) \, dx \\ & = -\left (4050 \int \frac {x \log (\log (x)) (-1-\log (x) \log (\log (x))+4 x \log (x) \log (\log (x)))}{\left (12+5 e^{2+2 x}\right )^4 \log (x)} \, dx\right )+194400 \int \frac {x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^5} \, dx \\ & = -\left (4050 \int \left (-\frac {x \log (\log (x))}{\left (12+5 e^{2+2 x}\right )^4 \log (x)}-\frac {x \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^4}+\frac {4 x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^4}\right ) \, dx\right )+194400 \int \frac {x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^5} \, dx \\ & = 4050 \int \frac {x \log (\log (x))}{\left (12+5 e^{2+2 x}\right )^4 \log (x)} \, dx+4050 \int \frac {x \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^4} \, dx-16200 \int \frac {x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^4} \, dx+194400 \int \frac {x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^5} \, dx \\ \end{align*}
Time = 0.41 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {\left (48600 x+20250 e^{2+2 x} x\right ) \log (\log (x))+\left (48600 x+e^{2+2 x} \left (20250 x-81000 x^2\right )\right ) \log (x) \log ^2(\log (x))}{\left (248832+518400 e^{2+2 x}+432000 e^{4+4 x}+180000 e^{6+6 x}+37500 e^{8+8 x}+3125 e^{10+10 x}\right ) \log (x)} \, dx=\frac {2025 x^2 \log ^2(\log (x))}{\left (12+5 e^{2+2 x}\right )^4} \]
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Time = 79.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
risch | \(\frac {2025 x^{2} \ln \left (\ln \left (x \right )\right )^{2}}{\left (5 \,{\mathrm e}^{2+2 x}+12\right )^{4}}\) | \(23\) |
parallelrisch | \(\frac {2025 x^{2} \ln \left (\ln \left (x \right )\right )^{2}}{625 \,{\mathrm e}^{8 x +8}+6000 \,{\mathrm e}^{6+6 x}+21600 \,{\mathrm e}^{4+4 x}+34560 \,{\mathrm e}^{2+2 x}+20736}\) | \(47\) |
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Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.25 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \frac {\left (48600 x+20250 e^{2+2 x} x\right ) \log (\log (x))+\left (48600 x+e^{2+2 x} \left (20250 x-81000 x^2\right )\right ) \log (x) \log ^2(\log (x))}{\left (248832+518400 e^{2+2 x}+432000 e^{4+4 x}+180000 e^{6+6 x}+37500 e^{8+8 x}+3125 e^{10+10 x}\right ) \log (x)} \, dx=\frac {2025 \, x^{2} \log \left (\log \left (x\right )\right )^{2}}{625 \, e^{\left (8 \, x + 8\right )} + 6000 \, e^{\left (6 \, x + 6\right )} + 21600 \, e^{\left (4 \, x + 4\right )} + 34560 \, e^{\left (2 \, x + 2\right )} + 20736} \]
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Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (24) = 48\).
Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int \frac {\left (48600 x+20250 e^{2+2 x} x\right ) \log (\log (x))+\left (48600 x+e^{2+2 x} \left (20250 x-81000 x^2\right )\right ) \log (x) \log ^2(\log (x))}{\left (248832+518400 e^{2+2 x}+432000 e^{4+4 x}+180000 e^{6+6 x}+37500 e^{8+8 x}+3125 e^{10+10 x}\right ) \log (x)} \, dx=\frac {81 x^{2} \log {\left (\log {\left (x \right )} \right )}^{2}}{\frac {6912 e^{2 x + 2}}{5} + 864 e^{4 x + 4} + 240 e^{6 x + 6} + 25 e^{8 x + 8} + \frac {20736}{25}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \frac {\left (48600 x+20250 e^{2+2 x} x\right ) \log (\log (x))+\left (48600 x+e^{2+2 x} \left (20250 x-81000 x^2\right )\right ) \log (x) \log ^2(\log (x))}{\left (248832+518400 e^{2+2 x}+432000 e^{4+4 x}+180000 e^{6+6 x}+37500 e^{8+8 x}+3125 e^{10+10 x}\right ) \log (x)} \, dx=\frac {2025 \, x^{2} \log \left (\log \left (x\right )\right )^{2}}{625 \, e^{\left (8 \, x + 8\right )} + 6000 \, e^{\left (6 \, x + 6\right )} + 21600 \, e^{\left (4 \, x + 4\right )} + 34560 \, e^{\left (2 \, x + 2\right )} + 20736} \]
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Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \frac {\left (48600 x+20250 e^{2+2 x} x\right ) \log (\log (x))+\left (48600 x+e^{2+2 x} \left (20250 x-81000 x^2\right )\right ) \log (x) \log ^2(\log (x))}{\left (248832+518400 e^{2+2 x}+432000 e^{4+4 x}+180000 e^{6+6 x}+37500 e^{8+8 x}+3125 e^{10+10 x}\right ) \log (x)} \, dx=\frac {2025 \, x^{2} \log \left (\log \left (x\right )\right )^{2}}{625 \, e^{\left (8 \, x + 8\right )} + 6000 \, e^{\left (6 \, x + 6\right )} + 21600 \, e^{\left (4 \, x + 4\right )} + 34560 \, e^{\left (2 \, x + 2\right )} + 20736} \]
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Time = 11.42 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \frac {\left (48600 x+20250 e^{2+2 x} x\right ) \log (\log (x))+\left (48600 x+e^{2+2 x} \left (20250 x-81000 x^2\right )\right ) \log (x) \log ^2(\log (x))}{\left (248832+518400 e^{2+2 x}+432000 e^{4+4 x}+180000 e^{6+6 x}+37500 e^{8+8 x}+3125 e^{10+10 x}\right ) \log (x)} \, dx=\frac {81\,x^2\,{\ln \left (\ln \left (x\right )\right )}^2}{25\,\left (\frac {6912\,{\mathrm {e}}^{2\,x+2}}{125}+\frac {864\,{\mathrm {e}}^{4\,x+4}}{25}+\frac {48\,{\mathrm {e}}^{6\,x+6}}{5}+{\mathrm {e}}^{8\,x+8}+\frac {20736}{625}\right )} \]
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