Integrand size = 74, antiderivative size = 21 \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\log \left (\log \left (x+\frac {1}{4} (2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right ) \]
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\[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4 \left (1+\frac {1}{4} (2-\log (5))\right )}{\left (4 x-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )}+\frac {2 (-2+\log (5))}{x \left (4 x-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )}+\frac {(2-\log (5)) \log (x)}{\left (4 x-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )}\right ) \, dx \\ & = (2-\log (5)) \int \frac {\log (x)}{\left (4 x-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )} \, dx+(6-\log (5)) \int \frac {1}{\left (4 x-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )} \, dx+(2 (-2+\log (5))) \int \frac {1}{x \left (4 x-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )} \, dx \\ & = (2-\log (5)) \int \frac {\log (x)}{(4 x-(-2+x) (-2+\log (5)) \log (x)) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )} \, dx+(6-\log (5)) \int \frac {1}{(4 x-(-2+x) (-2+\log (5)) \log (x)) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )} \, dx+(2 (-2+\log (5))) \int \frac {1}{x (4 x-(-2+x) (-2+\log (5)) \log (x)) \log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )} \, dx \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\log \left (\log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )\right ) \]
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Time = 2.62 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\ln \left (\ln \left (\frac {\left (2-x \right ) \left (\ln \left (5\right )-2\right ) \ln \left (x \right )}{4}+x \right )\right )\) | \(18\) |
norman | \(\ln \left (\ln \left (\frac {\left (2-x \right ) \ln \left (5 \,{\mathrm e}^{-2}\right ) \ln \left (x \right )}{4}+x \right )\right )\) | \(21\) |
parallelrisch | \(\ln \left (\ln \left (\frac {\left (2-x \right ) \ln \left (5 \,{\mathrm e}^{-2}\right ) \ln \left (x \right )}{4}+x \right )\right )\) | \(21\) |
default | \(\ln \left (2 \ln \left (2\right )-\ln \left (-x \ln \left (5 \,{\mathrm e}^{-2}\right ) \ln \left (x \right )+2 \ln \left (5 \,{\mathrm e}^{-2}\right ) \ln \left (x \right )+4 x \right )\right )\) | \(37\) |
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Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\log \left (\log \left (-\frac {1}{4} \, {\left ({\left (x - 2\right )} \log \left (5\right ) - 2 \, x + 4\right )} \log \left (x\right ) + x\right )\right ) \]
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Time = 0.45 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\log {\left (\log {\left (x + \left (\frac {1}{2} - \frac {x}{4}\right ) \log {\left (x \right )} \log {\left (\frac {5}{e^{2}} \right )} \right )} \right )} \]
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Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\log \left (-2 \, \log \left (2\right ) + \log \left (-{\left (x {\left (\log \left (5\right ) - 2\right )} - 2 \, \log \left (5\right ) + 4\right )} \log \left (x\right ) + 4 \, x\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (16) = 32\).
Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\log \left (2 \, \log \left (2\right ) - \log \left (-x \log \left (5\right ) \log \left (x\right ) + 2 \, x \log \left (x\right ) + 2 \, \log \left (5\right ) \log \left (x\right ) + 4 \, x - 4 \, \log \left (x\right )\right )\right ) \]
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Timed out. \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=-\int \frac {\ln \left (5\,{\mathrm {e}}^{-2}\right )\,\left (x-2\right )-4\,x+x\,\ln \left (5\,{\mathrm {e}}^{-2}\right )\,\ln \left (x\right )}{\ln \left (x-\frac {\ln \left (5\,{\mathrm {e}}^{-2}\right )\,\ln \left (x\right )\,\left (x-2\right )}{4}\right )\,\left (4\,x^2+\ln \left (5\,{\mathrm {e}}^{-2}\right )\,\ln \left (x\right )\,\left (2\,x-x^2\right )\right )} \,d x \]
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