Integrand size = 199, antiderivative size = 31 \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=2 x+\frac {6}{x+\frac {3 (5-x)}{\left (e^4-\log (2)\right ) \log (x)}} \]
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\[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=\int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {2 \left (9 (-5+x) \left (-e^4-5 x+x^2+\log (2)\right )-3 x \left (-3-10 x+2 x^2\right ) \left (e^4-\log (2)\right ) \log (x)+x \left (-3+x^2\right ) \left (e^4-\log (2)\right )^2 \log ^2(x)\right )}{x \left (3 (-5+x)-x \left (e^4-\log (2)\right ) \log (x)\right )^2} \, dx \\ & = 2 \int \frac {9 (-5+x) \left (-e^4-5 x+x^2+\log (2)\right )-3 x \left (-3-10 x+2 x^2\right ) \left (e^4-\log (2)\right ) \log (x)+x \left (-3+x^2\right ) \left (e^4-\log (2)\right )^2 \log ^2(x)}{x \left (3 (-5+x)-x \left (e^4-\log (2)\right ) \log (x)\right )^2} \, dx \\ & = 2 \int \left (\frac {-3+x^2}{x^2}+\frac {9 (5-x) \left (-15+x \left (e^4-\log (2)\right )\right )}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2}+\frac {9 (10-x)}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )}\right ) \, dx \\ & = 2 \int \frac {-3+x^2}{x^2} \, dx+18 \int \frac {(5-x) \left (-15+x \left (e^4-\log (2)\right )\right )}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2} \, dx+18 \int \frac {10-x}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )} \, dx \\ & = 2 \int \left (1-\frac {3}{x^2}\right ) \, dx+18 \int \left (-\frac {75}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2}+\frac {5 \left (3+e^4-\log (2)\right )}{x \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2}-\frac {e^4 \left (1-\frac {\log (2)}{e^4}\right )}{\left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2}\right ) \, dx+18 \int \left (\frac {1}{x \left (-15+3 x-e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )}+\frac {10}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )}\right ) \, dx \\ & = \frac {6}{x}+2 x+18 \int \frac {1}{x \left (-15+3 x-e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )} \, dx+180 \int \frac {1}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )} \, dx-1350 \int \frac {1}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2} \, dx-\left (18 \left (e^4-\log (2)\right )\right ) \int \frac {1}{\left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2} \, dx+\left (90 \left (3+e^4-\log (2)\right )\right ) \int \frac {1}{x \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2} \, dx \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=2 \left (\frac {3}{x}+x+\frac {9 (-5+x)}{x \left (15-3 x+e^4 x \log (x)-x \log (2) \log (x)\right )}\right ) \]
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Time = 1.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29
method | result | size |
risch | \(\frac {2 x^{2}+6}{x}+\frac {18 x -90}{x \left (x \,{\mathrm e}^{4} \ln \left (x \right )-x \ln \left (2\right ) \ln \left (x \right )-3 x +15\right )}\) | \(40\) |
default | \(\frac {2 \left (3 \,{\mathrm e}^{4}-3 \ln \left (2\right )\right ) \ln \left (x \right )+2 \left ({\mathrm e}^{4}-\ln \left (2\right )\right ) \ln \left (x \right ) x^{2}-6 x^{2}+10 \left ({\mathrm e}^{4}-\ln \left (2\right )\right ) \ln \left (x \right ) x +150}{x \,{\mathrm e}^{4} \ln \left (x \right )-x \ln \left (2\right ) \ln \left (x \right )-3 x +15}\) | \(67\) |
norman | \(\frac {\left (6 \,{\mathrm e}^{4}-6 \ln \left (2\right )\right ) \ln \left (x \right )+\left (2 \,{\mathrm e}^{4}-2 \ln \left (2\right )\right ) x^{2} \ln \left (x \right )+\left (10 \,{\mathrm e}^{4}-10 \ln \left (2\right )\right ) x \ln \left (x \right )-6 x^{2}+150}{x \,{\mathrm e}^{4} \ln \left (x \right )-x \ln \left (2\right ) \ln \left (x \right )-3 x +15}\) | \(69\) |
parallelrisch | \(\frac {6 x^{2} {\mathrm e}^{4} \ln \left (x \right )+450-6 x^{2} \ln \left (2\right ) \ln \left (x \right )+30 x \,{\mathrm e}^{4} \ln \left (x \right )-30 x \ln \left (2\right ) \ln \left (x \right )+18 \,{\mathrm e}^{4} \ln \left (x \right )-18 \ln \left (2\right ) \ln \left (x \right )-18 x^{2}}{3 x \,{\mathrm e}^{4} \ln \left (x \right )-3 x \ln \left (2\right ) \ln \left (x \right )-9 x +45}\) | \(74\) |
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Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=-\frac {2 \, {\left (3 \, x^{2} - {\left ({\left (x^{2} + 3\right )} e^{4} - {\left (x^{2} + 3\right )} \log \left (2\right )\right )} \log \left (x\right ) - 15 \, x\right )}}{{\left (x e^{4} - x \log \left (2\right )\right )} \log \left (x\right ) - 3 \, x + 15} \]
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Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=2 x + \frac {18 x - 90}{- 3 x^{2} + 15 x + \left (- x^{2} \log {\left (2 \right )} + x^{2} e^{4}\right ) \log {\left (x \right )}} + \frac {6}{x} \]
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Time = 0.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=-\frac {2 \, {\left (3 \, x^{2} - {\left (x^{2} {\left (e^{4} - \log \left (2\right )\right )} + 3 \, e^{4} - 3 \, \log \left (2\right )\right )} \log \left (x\right ) - 15 \, x\right )}}{x {\left (e^{4} - \log \left (2\right )\right )} \log \left (x\right ) - 3 \, x + 15} \]
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Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (28) = 56\).
Time = 0.31 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.94 \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=\frac {2 \, {\left (x^{2} e^{4} \log \left (x\right ) - x^{2} \log \left (2\right ) \log \left (x\right ) - 3 \, x^{2} + 3 \, e^{4} \log \left (x\right ) - 3 \, \log \left (2\right ) \log \left (x\right ) + 15 \, x\right )}}{x e^{4} \log \left (x\right ) - x \log \left (2\right ) \log \left (x\right ) - 3 \, x + 15} \]
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Timed out. \[ \int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx=\int \frac {450\,x+\ln \left (2\right )\,\left (18\,x-90\right )-{\ln \left (x\right )}^2\,\left ({\mathrm {e}}^8\,\left (6\,x-2\,x^3\right )+{\ln \left (2\right )}^2\,\left (6\,x-2\,x^3\right )-{\mathrm {e}}^4\,\ln \left (2\right )\,\left (12\,x-4\,x^3\right )\right )+\ln \left (x\right )\,\left ({\mathrm {e}}^4\,\left (-12\,x^3+60\,x^2+18\,x\right )-\ln \left (2\right )\,\left (-12\,x^3+60\,x^2+18\,x\right )\right )-180\,x^2+18\,x^3-{\mathrm {e}}^4\,\left (18\,x-90\right )}{225\,x+{\ln \left (x\right )}^2\,\left (x^3\,{\ln \left (2\right )}^2+x^3\,{\mathrm {e}}^8-2\,x^3\,{\mathrm {e}}^4\,\ln \left (2\right )\right )+\ln \left (x\right )\,\left ({\mathrm {e}}^4\,\left (30\,x^2-6\,x^3\right )-\ln \left (2\right )\,\left (30\,x^2-6\,x^3\right )\right )-90\,x^2+9\,x^3} \,d x \]
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