Integrand size = 80, antiderivative size = 24 \[ \int \frac {2 e^6 (17+8 x-\log (x))}{\left (18 x+4 x^2-x \log (x)\right ) \left (54 x+12 x^2-3 x \log (x)+\frac {2 e^6 \left (-18 x-4 x^2+x \log (x)\right )}{18 x+4 x^2-x \log (x)}\right )} \, dx=\log \left (3-\frac {e^6}{x \left (9+2 x-\frac {\log (x)}{2}\right )}\right ) \]
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\[ \int \frac {2 e^6 (17+8 x-\log (x))}{\left (18 x+4 x^2-x \log (x)\right ) \left (54 x+12 x^2-3 x \log (x)+\frac {2 e^6 \left (-18 x-4 x^2+x \log (x)\right )}{18 x+4 x^2-x \log (x)}\right )} \, dx=\int \frac {2 e^6 (17+8 x-\log (x))}{\left (18 x+4 x^2-x \log (x)\right ) \left (54 x+12 x^2-3 x \log (x)+\frac {2 e^6 \left (-18 x-4 x^2+x \log (x)\right )}{18 x+4 x^2-x \log (x)}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \left (2 e^6\right ) \int \frac {17+8 x-\log (x)}{\left (18 x+4 x^2-x \log (x)\right ) \left (54 x+12 x^2-3 x \log (x)+\frac {2 e^6 \left (-18 x-4 x^2+x \log (x)\right )}{18 x+4 x^2-x \log (x)}\right )} \, dx \\ & = \left (2 e^6\right ) \int \frac {-17-8 x+\log (x)}{x (18+4 x-\log (x)) \left (2 e^6-6 x (9+2 x)+3 x \log (x)\right )} \, dx \\ & = \left (2 e^6\right ) \int \left (\frac {1-4 x}{2 e^6 x (18+4 x-\log (x))}+\frac {-2 e^6+3 x-12 x^2}{2 e^6 x \left (2 e^6-54 x-12 x^2+3 x \log (x)\right )}\right ) \, dx \\ & = \int \frac {1-4 x}{x (18+4 x-\log (x))} \, dx+\int \frac {-2 e^6+3 x-12 x^2}{x \left (2 e^6-54 x-12 x^2+3 x \log (x)\right )} \, dx \\ & = -\log (18+4 x-\log (x))+\int \left (\frac {2 e^6}{x \left (-2 e^6+54 x+12 x^2-3 x \log (x)\right )}+\frac {12 x}{-2 e^6+54 x+12 x^2-3 x \log (x)}+\frac {3}{2 e^6-54 x-12 x^2+3 x \log (x)}\right ) \, dx \\ & = -\log (18+4 x-\log (x))+3 \int \frac {1}{2 e^6-54 x-12 x^2+3 x \log (x)} \, dx+12 \int \frac {x}{-2 e^6+54 x+12 x^2-3 x \log (x)} \, dx+\left (2 e^6\right ) \int \frac {1}{x \left (-2 e^6+54 x+12 x^2-3 x \log (x)\right )} \, dx \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(59\) vs. \(2(24)=48\).
Time = 0.48 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.46 \[ \int \frac {2 e^6 (17+8 x-\log (x))}{\left (18 x+4 x^2-x \log (x)\right ) \left (54 x+12 x^2-3 x \log (x)+\frac {2 e^6 \left (-18 x-4 x^2+x \log (x)\right )}{18 x+4 x^2-x \log (x)}\right )} \, dx=2 e^6 \left (-\frac {\log (x)}{2 e^6}-\frac {\log (18+4 x-\log (x))}{2 e^6}+\frac {\log \left (2 e^6-54 x-12 x^2+3 x \log (x)\right )}{2 e^6}\right ) \]
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Time = 1.33 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83
method | result | size |
parallelrisch | \(\ln \left ({\mathrm e}^{-\ln \left (-\frac {x \left (\ln \left (x \right )-4 x -18\right )}{2}\right )+6}-3\right )\) | \(20\) |
default | \(2 \,{\mathrm e}^{6} \left (-\frac {{\mathrm e}^{-6} \ln \left (x \right )}{2}-\frac {{\mathrm e}^{-6} \ln \left (\ln \left (x \right )-4 x -18\right )}{2}+\frac {{\mathrm e}^{-6} \ln \left (3 x \ln \left (x \right )-12 x^{2}+2 \,{\mathrm e}^{6}-54 x \right )}{2}\right )\) | \(53\) |
risch | \(-\ln \left (\ln \left (x \right )-4 x -18\right )-\ln \left (2\right )+\ln \left (-\frac {\ln \left (x \right )}{4}+x +\frac {9}{2}\right )+\frac {i \pi \,\operatorname {csgn}\left (i x \left (\frac {\ln \left (x \right )}{4}-x -\frac {9}{2}\right )\right ) \left (\operatorname {csgn}\left (i x \left (\frac {\ln \left (x \right )}{4}-x -\frac {9}{2}\right )\right )+\operatorname {csgn}\left (i x \right )\right ) \left (\operatorname {csgn}\left (i x \left (\frac {\ln \left (x \right )}{4}-x -\frac {9}{2}\right )\right )-\operatorname {csgn}\left (i \left (\frac {\ln \left (x \right )}{4}-x -\frac {9}{2}\right )\right )\right )}{2}-6+\ln \left (\frac {2 \,{\mathrm e}^{6} {\mathrm e}^{-\frac {i \pi \operatorname {csgn}\left (i x \left (\frac {\ln \left (x \right )}{4}-x -\frac {9}{2}\right )\right )^{3}}{2}} {\mathrm e}^{-\frac {i \pi \operatorname {csgn}\left (i x \left (\frac {\ln \left (x \right )}{4}-x -\frac {9}{2}\right )\right )^{2} \operatorname {csgn}\left (i x \right )}{2}} {\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (i x \left (\frac {\ln \left (x \right )}{4}-x -\frac {9}{2}\right )\right )^{2} \operatorname {csgn}\left (i \left (\frac {\ln \left (x \right )}{4}-x -\frac {9}{2}\right )\right )}{2}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i x \left (\frac {\ln \left (x \right )}{4}-x -\frac {9}{2}\right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (\frac {\ln \left (x \right )}{4}-x -\frac {9}{2}\right )\right )}{2}}}{x \left (-\frac {\ln \left (x \right )}{4}+x +\frac {9}{2}\right )}-3\right )\) | \(230\) |
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Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {2 e^6 (17+8 x-\log (x))}{\left (18 x+4 x^2-x \log (x)\right ) \left (54 x+12 x^2-3 x \log (x)+\frac {2 e^6 \left (-18 x-4 x^2+x \log (x)\right )}{18 x+4 x^2-x \log (x)}\right )} \, dx=-\log \left (-4 \, x + \log \left (x\right ) - 18\right ) + \log \left (-\frac {12 \, x^{2} - 3 \, x \log \left (x\right ) + 54 \, x - 2 \, e^{6}}{x}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (17) = 34\).
Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {2 e^6 (17+8 x-\log (x))}{\left (18 x+4 x^2-x \log (x)\right ) \left (54 x+12 x^2-3 x \log (x)+\frac {2 e^6 \left (-18 x-4 x^2+x \log (x)\right )}{18 x+4 x^2-x \log (x)}\right )} \, dx=- \log {\left (\log {\left (x \right )} + \frac {- 12 x^{2} - 54 x}{3 x} \right )} + \log {\left (\log {\left (x \right )} + \frac {- 12 x^{2} - 54 x + 2 e^{6}}{3 x} \right )} \]
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Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {2 e^6 (17+8 x-\log (x))}{\left (18 x+4 x^2-x \log (x)\right ) \left (54 x+12 x^2-3 x \log (x)+\frac {2 e^6 \left (-18 x-4 x^2+x \log (x)\right )}{18 x+4 x^2-x \log (x)}\right )} \, dx=-{\left (e^{\left (-6\right )} \log \left (-4 \, x + \log \left (x\right ) - 18\right ) - e^{\left (-6\right )} \log \left (-\frac {12 \, x^{2} - 3 \, x \log \left (x\right ) + 54 \, x - 2 \, e^{6}}{3 \, x}\right )\right )} e^{6} \]
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Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {2 e^6 (17+8 x-\log (x))}{\left (18 x+4 x^2-x \log (x)\right ) \left (54 x+12 x^2-3 x \log (x)+\frac {2 e^6 \left (-18 x-4 x^2+x \log (x)\right )}{18 x+4 x^2-x \log (x)}\right )} \, dx=\log \left (-12 \, x^{2} + 3 \, x \log \left (x\right ) - 54 \, x + 2 \, e^{6}\right ) - \log \left (4 \, x - \log \left (x\right ) + 18\right ) - \log \left (x\right ) \]
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Time = 14.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {2 e^6 (17+8 x-\log (x))}{\left (18 x+4 x^2-x \log (x)\right ) \left (54 x+12 x^2-3 x \log (x)+\frac {2 e^6 \left (-18 x-4 x^2+x \log (x)\right )}{18 x+4 x^2-x \log (x)}\right )} \, dx=\ln \left (54\,x-2\,{\mathrm {e}}^6-3\,x\,\ln \left (x\right )+12\,x^2\right )-\ln \left (x\right )-\ln \left (4\,x-\ln \left (x\right )+18\right ) \]
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