Integrand size = 357, antiderivative size = 35 \[ \int \frac {e^{15} \left (10-24 x+6 x^2\right )+e^{15} \left (-10+22 x-4 x^2\right ) \log (5)+e^{15} \left (-10+22 x-4 x^2\right ) \log (5-x)}{-5 x+6 x^2-x^3+\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log (5-x)+\left (e^5 \left (-15 x+18 x^2-3 x^3\right )+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log \left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{10} \left (-15 x+18 x^2-3 x^3\right )+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log ^2\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{15} \left (-5 x+6 x^2-x^3\right )+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5)+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5-x)\right ) \log ^3\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )} \, dx=\frac {x^2}{\left (\frac {x}{e^5}+x \log \left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )\right )^2} \]
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Time = 1.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {6820, 12, 6840, 32} \[ \int \frac {e^{15} \left (10-24 x+6 x^2\right )+e^{15} \left (-10+22 x-4 x^2\right ) \log (5)+e^{15} \left (-10+22 x-4 x^2\right ) \log (5-x)}{-5 x+6 x^2-x^3+\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log (5-x)+\left (e^5 \left (-15 x+18 x^2-3 x^3\right )+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log \left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{10} \left (-15 x+18 x^2-3 x^3\right )+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log ^2\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{15} \left (-5 x+6 x^2-x^3\right )+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5)+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5-x)\right ) \log ^3\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )} \, dx=\frac {e^{10}}{\left (e^5 \log \left (\frac {(1-x) x}{1-\log (5 (5-x))}\right )+1\right )^2} \]
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Rule 12
Rule 32
Rule 6820
Rule 6840
Rubi steps \begin{align*} \text {integral}& = \int \frac {2 e^{15} \left (x (12-11 \log (5))+5 (-1+\log (5))+x^2 (-3+\log (25))+\left (5-11 x+2 x^2\right ) \log (5-x)\right )}{x \left (5-6 x+x^2\right ) (1-\log (-5 (-5+x))) \left (1+e^5 \log \left (\frac {(-1+x) x}{-1+\log (-5 (-5+x))}\right )\right )^3} \, dx \\ & = \left (2 e^{15}\right ) \int \frac {x (12-11 \log (5))+5 (-1+\log (5))+x^2 (-3+\log (25))+\left (5-11 x+2 x^2\right ) \log (5-x)}{x \left (5-6 x+x^2\right ) (1-\log (-5 (-5+x))) \left (1+e^5 \log \left (\frac {(-1+x) x}{-1+\log (-5 (-5+x))}\right )\right )^3} \, dx \\ & = -\left (\left (2 e^{10}\right ) \text {Subst}\left (\int \frac {1}{(1+x)^3} \, dx,x,e^5 \log \left (\frac {(-1+x) x}{-1+\log (-5 (-5+x))}\right )\right )\right ) \\ & = \frac {e^{10}}{\left (1+e^5 \log \left (\frac {(1-x) x}{1-\log (5 (5-x))}\right )\right )^2} \\ \end{align*}
\[ \int \frac {e^{15} \left (10-24 x+6 x^2\right )+e^{15} \left (-10+22 x-4 x^2\right ) \log (5)+e^{15} \left (-10+22 x-4 x^2\right ) \log (5-x)}{-5 x+6 x^2-x^3+\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log (5-x)+\left (e^5 \left (-15 x+18 x^2-3 x^3\right )+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log \left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{10} \left (-15 x+18 x^2-3 x^3\right )+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log ^2\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{15} \left (-5 x+6 x^2-x^3\right )+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5)+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5-x)\right ) \log ^3\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )} \, dx=\int \frac {e^{15} \left (10-24 x+6 x^2\right )+e^{15} \left (-10+22 x-4 x^2\right ) \log (5)+e^{15} \left (-10+22 x-4 x^2\right ) \log (5-x)}{-5 x+6 x^2-x^3+\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log (5-x)+\left (e^5 \left (-15 x+18 x^2-3 x^3\right )+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log \left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{10} \left (-15 x+18 x^2-3 x^3\right )+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log ^2\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{15} \left (-5 x+6 x^2-x^3\right )+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5)+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5-x)\right ) \log ^3\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )} \, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. \(104\) vs. \(2(36)=72\).
Time = 85.06 (sec) , antiderivative size = 105, normalized size of antiderivative = 3.00
method | result | size |
parallelrisch | \(-\frac {300 \ln \left (\frac {x \left (-1+x \right )}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{2} {\mathrm e}^{20}+600 \ln \left (\frac {x \left (-1+x \right )}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right ) {\mathrm e}^{15}}{300 \left ({\mathrm e}^{10} \ln \left (\frac {x \left (-1+x \right )}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{2}+2 \,{\mathrm e}^{5} \ln \left (\frac {x \left (-1+x \right )}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )+1\right )}\) | \(105\) |
default | \(-\frac {4 \,{\mathrm e}^{15} {\mathrm e}^{-5}}{\left ({\mathrm e}^{5} \pi \,\operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right ) x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )+{\mathrm e}^{5} \pi \,\operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right ) x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{2}+{\mathrm e}^{5} \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )+{\mathrm e}^{5} \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{2}-{\mathrm e}^{5} \pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{2}-{\mathrm e}^{5} \pi \operatorname {csgn}\left (\frac {i x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{3}+{\mathrm e}^{5} \pi \,\operatorname {csgn}\left (\frac {i x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right ) x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{2}+{\mathrm e}^{5} \pi \operatorname {csgn}\left (\frac {i \left (-1+x \right ) x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{3}+2 i {\mathrm e}^{5} \ln \left (1-x \right )+2 i {\mathrm e}^{5} \ln \left (-x \right )-2 i {\mathrm e}^{5} \ln \left (\ln \left (5-x \right )+\ln \left (5\right )-1\right )+2 i\right )^{2}}\) | \(334\) |
risch | \(-\frac {4 \,{\mathrm e}^{10}}{\left (2 i {\mathrm e}^{5} \ln \left (\ln \left (5-x \right )+\ln \left (5\right )-1\right )+{\mathrm e}^{5} \pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{2}-{\mathrm e}^{5} \pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right ) \operatorname {csgn}\left (i \left (-1+x \right )\right )-{\mathrm e}^{5} \pi \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{3}+{\mathrm e}^{5} \pi \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{2} \operatorname {csgn}\left (i \left (-1+x \right )\right )+{\mathrm e}^{5} \pi \,\operatorname {csgn}\left (\frac {i \left (-1+x \right )}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right ) x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{2}-{\mathrm e}^{5} \pi \,\operatorname {csgn}\left (\frac {i \left (-1+x \right )}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right ) x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right ) \operatorname {csgn}\left (i x \right )-{\mathrm e}^{5} \pi \operatorname {csgn}\left (\frac {i \left (-1+x \right ) x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{3}+{\mathrm e}^{5} \pi \operatorname {csgn}\left (\frac {i \left (-1+x \right ) x}{\ln \left (5-x \right )+\ln \left (5\right )-1}\right )^{2} \operatorname {csgn}\left (i x \right )-2 i {\mathrm e}^{5} \ln \left (x \right )-2 i {\mathrm e}^{5} \ln \left (-1+x \right )-2 i\right )^{2}}\) | \(342\) |
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Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.66 \[ \int \frac {e^{15} \left (10-24 x+6 x^2\right )+e^{15} \left (-10+22 x-4 x^2\right ) \log (5)+e^{15} \left (-10+22 x-4 x^2\right ) \log (5-x)}{-5 x+6 x^2-x^3+\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log (5-x)+\left (e^5 \left (-15 x+18 x^2-3 x^3\right )+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log \left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{10} \left (-15 x+18 x^2-3 x^3\right )+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log ^2\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{15} \left (-5 x+6 x^2-x^3\right )+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5)+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5-x)\right ) \log ^3\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )} \, dx=\frac {e^{10}}{e^{10} \log \left (\frac {x^{2} - x}{\log \left (5\right ) + \log \left (-x + 5\right ) - 1}\right )^{2} + 2 \, e^{5} \log \left (\frac {x^{2} - x}{\log \left (5\right ) + \log \left (-x + 5\right ) - 1}\right ) + 1} \]
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Time = 0.42 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37 \[ \int \frac {e^{15} \left (10-24 x+6 x^2\right )+e^{15} \left (-10+22 x-4 x^2\right ) \log (5)+e^{15} \left (-10+22 x-4 x^2\right ) \log (5-x)}{-5 x+6 x^2-x^3+\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log (5-x)+\left (e^5 \left (-15 x+18 x^2-3 x^3\right )+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log \left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{10} \left (-15 x+18 x^2-3 x^3\right )+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log ^2\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{15} \left (-5 x+6 x^2-x^3\right )+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5)+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5-x)\right ) \log ^3\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )} \, dx=\frac {e^{10}}{e^{10} \log {\left (\frac {x^{2} - x}{\log {\left (5 - x \right )} - 1 + \log {\left (5 \right )}} \right )}^{2} + 2 e^{5} \log {\left (\frac {x^{2} - x}{\log {\left (5 - x \right )} - 1 + \log {\left (5 \right )}} \right )} + 1} \]
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Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (34) = 68\).
Time = 0.53 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.49 \[ \int \frac {e^{15} \left (10-24 x+6 x^2\right )+e^{15} \left (-10+22 x-4 x^2\right ) \log (5)+e^{15} \left (-10+22 x-4 x^2\right ) \log (5-x)}{-5 x+6 x^2-x^3+\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log (5-x)+\left (e^5 \left (-15 x+18 x^2-3 x^3\right )+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log \left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{10} \left (-15 x+18 x^2-3 x^3\right )+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log ^2\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{15} \left (-5 x+6 x^2-x^3\right )+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5)+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5-x)\right ) \log ^3\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )} \, dx=\frac {e^{10}}{e^{10} \log \left (x - 1\right )^{2} + e^{10} \log \left (x\right )^{2} + e^{10} \log \left (\log \left (5\right ) + \log \left (-x + 5\right ) - 1\right )^{2} + 2 \, {\left (e^{10} \log \left (x\right ) + e^{5}\right )} \log \left (x - 1\right ) + 2 \, e^{5} \log \left (x\right ) - 2 \, {\left (e^{10} \log \left (x - 1\right ) + e^{10} \log \left (x\right ) + e^{5}\right )} \log \left (\log \left (5\right ) + \log \left (-x + 5\right ) - 1\right ) + 1} \]
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Leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (34) = 68\).
Time = 10.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.46 \[ \int \frac {e^{15} \left (10-24 x+6 x^2\right )+e^{15} \left (-10+22 x-4 x^2\right ) \log (5)+e^{15} \left (-10+22 x-4 x^2\right ) \log (5-x)}{-5 x+6 x^2-x^3+\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log (5-x)+\left (e^5 \left (-15 x+18 x^2-3 x^3\right )+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log \left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{10} \left (-15 x+18 x^2-3 x^3\right )+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log ^2\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{15} \left (-5 x+6 x^2-x^3\right )+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5)+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5-x)\right ) \log ^3\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )} \, dx=\frac {e^{10}}{e^{10} \log \left (x^{2} - x\right )^{2} - 2 \, e^{10} \log \left (x^{2} - x\right ) \log \left (\log \left (5\right ) + \log \left (-x + 5\right ) - 1\right ) + e^{10} \log \left (\log \left (5\right ) + \log \left (-x + 5\right ) - 1\right )^{2} + 2 \, e^{5} \log \left (x^{2} - x\right ) - 2 \, e^{5} \log \left (\log \left (5\right ) + \log \left (-x + 5\right ) - 1\right ) + 1} \]
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Time = 15.63 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91 \[ \int \frac {e^{15} \left (10-24 x+6 x^2\right )+e^{15} \left (-10+22 x-4 x^2\right ) \log (5)+e^{15} \left (-10+22 x-4 x^2\right ) \log (5-x)}{-5 x+6 x^2-x^3+\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log (5-x)+\left (e^5 \left (-15 x+18 x^2-3 x^3\right )+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^5 \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log \left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{10} \left (-15 x+18 x^2-3 x^3\right )+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5)+e^{10} \left (15 x-18 x^2+3 x^3\right ) \log (5-x)\right ) \log ^2\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )+\left (e^{15} \left (-5 x+6 x^2-x^3\right )+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5)+e^{15} \left (5 x-6 x^2+x^3\right ) \log (5-x)\right ) \log ^3\left (\frac {-x+x^2}{-1+\log (5)+\log (5-x)}\right )} \, dx=\frac {{\mathrm {e}}^{10}}{{\left ({\mathrm {e}}^5\,\ln \left (-\frac {x-x^2}{\ln \left (5\right )+\ln \left (5-x\right )-1}\right )+1\right )}^2} \]
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