Integrand size = 96, antiderivative size = 31 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=\frac {\left (-e^x+x\right ) \left (-x+\frac {3 x}{\log (x)}-\frac {16 x}{5+\log (x)}\right )}{x} \]
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Leaf count is larger than twice the leaf count of optimal. \(65\) vs. \(2(31)=62\).
Time = 1.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.10, number of steps used = 39, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6873, 6874, 2334, 2336, 2209, 2407, 2335, 2326} \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=-x-\frac {e^x \left (-x \log ^4(x)-23 x \log ^3(x)-75 x \log ^2(x)+75 x \log (x)\right )}{x \log ^2(x) (\log (x)+5)^2}+\frac {3 x}{\log (x)}-\frac {16 x}{\log (x)+5} \]
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Rule 2209
Rule 2326
Rule 2334
Rule 2335
Rule 2336
Rule 2407
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{x \log ^2(x) (5+\log (x))^2} \, dx \\ & = \int \left (-\frac {62}{(5+\log (x))^2}-\frac {75}{\log ^2(x) (5+\log (x))^2}+\frac {45}{\log (x) (5+\log (x))^2}-\frac {23 \log (x)}{(5+\log (x))^2}-\frac {\log ^2(x)}{(5+\log (x))^2}+\frac {e^x \left (75+30 \log (x)-75 x \log (x)-13 \log ^2(x)+75 x \log ^2(x)+23 x \log ^3(x)+x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2}\right ) \, dx \\ & = -\left (23 \int \frac {\log (x)}{(5+\log (x))^2} \, dx\right )+45 \int \frac {1}{\log (x) (5+\log (x))^2} \, dx-62 \int \frac {1}{(5+\log (x))^2} \, dx-75 \int \frac {1}{\log ^2(x) (5+\log (x))^2} \, dx-\int \frac {\log ^2(x)}{(5+\log (x))^2} \, dx+\int \frac {e^x \left (75+30 \log (x)-75 x \log (x)-13 \log ^2(x)+75 x \log ^2(x)+23 x \log ^3(x)+x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2} \, dx \\ & = \frac {62 x}{5+\log (x)}-\frac {e^x \left (75 x \log (x)-75 x \log ^2(x)-23 x \log ^3(x)-x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2}-23 \int \left (-\frac {5}{(5+\log (x))^2}+\frac {1}{5+\log (x)}\right ) \, dx+45 \int \left (\frac {1}{25 \log (x)}-\frac {1}{5 (5+\log (x))^2}-\frac {1}{25 (5+\log (x))}\right ) \, dx-62 \int \frac {1}{5+\log (x)} \, dx-75 \int \left (\frac {1}{25 \log ^2(x)}-\frac {2}{125 \log (x)}+\frac {1}{25 (5+\log (x))^2}+\frac {2}{125 (5+\log (x))}\right ) \, dx-\int \left (1+\frac {25}{(5+\log (x))^2}-\frac {10}{5+\log (x)}\right ) \, dx \\ & = -x+\frac {62 x}{5+\log (x)}-\frac {e^x \left (75 x \log (x)-75 x \log ^2(x)-23 x \log ^3(x)-x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2}+\frac {6}{5} \int \frac {1}{\log (x)} \, dx-\frac {6}{5} \int \frac {1}{5+\log (x)} \, dx+\frac {9}{5} \int \frac {1}{\log (x)} \, dx-\frac {9}{5} \int \frac {1}{5+\log (x)} \, dx-3 \int \frac {1}{\log ^2(x)} \, dx-3 \int \frac {1}{(5+\log (x))^2} \, dx-9 \int \frac {1}{(5+\log (x))^2} \, dx+10 \int \frac {1}{5+\log (x)} \, dx-23 \int \frac {1}{5+\log (x)} \, dx-25 \int \frac {1}{(5+\log (x))^2} \, dx-62 \text {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )+115 \int \frac {1}{(5+\log (x))^2} \, dx \\ & = -x-\frac {62 \text {Ei}(5+\log (x))}{e^5}+\frac {3 x}{\log (x)}-\frac {16 x}{5+\log (x)}-\frac {e^x \left (75 x \log (x)-75 x \log ^2(x)-23 x \log ^3(x)-x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2}+3 \text {li}(x)-\frac {6}{5} \text {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )-\frac {9}{5} \text {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )-3 \int \frac {1}{\log (x)} \, dx-3 \int \frac {1}{5+\log (x)} \, dx-9 \int \frac {1}{5+\log (x)} \, dx+10 \text {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )-23 \text {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )-25 \int \frac {1}{5+\log (x)} \, dx+115 \int \frac {1}{5+\log (x)} \, dx \\ & = -x-\frac {78 \text {Ei}(5+\log (x))}{e^5}+\frac {3 x}{\log (x)}-\frac {16 x}{5+\log (x)}-\frac {e^x \left (75 x \log (x)-75 x \log ^2(x)-23 x \log ^3(x)-x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2}-3 \text {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )-9 \text {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )-25 \text {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )+115 \text {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right ) \\ & = -x+\frac {3 x}{\log (x)}-\frac {16 x}{5+\log (x)}-\frac {e^x \left (75 x \log (x)-75 x \log ^2(x)-23 x \log ^3(x)-x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=\frac {\left (e^x-x\right ) \left (-15+18 \log (x)+\log ^2(x)\right )}{\log (x) (5+\log (x))} \]
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Time = 0.58 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00
method | result | size |
risch | \({\mathrm e}^{x}-x -\frac {\left (x -{\mathrm e}^{x}\right ) \left (13 \ln \left (x \right )-15\right )}{\left (5+\ln \left (x \right )\right ) \ln \left (x \right )}\) | \(31\) |
norman | \(\frac {{\mathrm e}^{x} \ln \left (x \right )^{2}+15 x -18 x \ln \left (x \right )-x \ln \left (x \right )^{2}+18 \,{\mathrm e}^{x} \ln \left (x \right )-15 \,{\mathrm e}^{x}}{\ln \left (x \right ) \left (5+\ln \left (x \right )\right )}\) | \(45\) |
parallelrisch | \(-\frac {x \ln \left (x \right )^{2}-{\mathrm e}^{x} \ln \left (x \right )^{2}+18 x \ln \left (x \right )-18 \,{\mathrm e}^{x} \ln \left (x \right )-15 x +15 \,{\mathrm e}^{x}}{\ln \left (x \right ) \left (5+\ln \left (x \right )\right )}\) | \(46\) |
default | \(-\frac {x \left (\ln \left (x \right )^{2}+18 \ln \left (x \right )-15\right )}{\left (5+\ln \left (x \right )\right ) \ln \left (x \right )}+\frac {{\mathrm e}^{x} \ln \left (x \right )^{2}+18 \,{\mathrm e}^{x} \ln \left (x \right )-15 \,{\mathrm e}^{x}}{\ln \left (x \right ) \left (5+\ln \left (x \right )\right )}\) | \(54\) |
parts | \(-\frac {x \left (\ln \left (x \right )^{2}+18 \ln \left (x \right )-15\right )}{\left (5+\ln \left (x \right )\right ) \ln \left (x \right )}+\frac {{\mathrm e}^{x} \ln \left (x \right )^{2}+18 \,{\mathrm e}^{x} \ln \left (x \right )-15 \,{\mathrm e}^{x}}{\ln \left (x \right ) \left (5+\ln \left (x \right )\right )}\) | \(54\) |
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Time = 0.35 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=-\frac {{\left (x - e^{x}\right )} \log \left (x\right )^{2} + 18 \, {\left (x - e^{x}\right )} \log \left (x\right ) - 15 \, x + 15 \, e^{x}}{\log \left (x\right )^{2} + 5 \, \log \left (x\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=- x + \frac {- 13 x \log {\left (x \right )} + 15 x}{\log {\left (x \right )}^{2} + 5 \log {\left (x \right )}} + \frac {\left (\log {\left (x \right )}^{2} + 18 \log {\left (x \right )} - 15\right ) e^{x}}{\log {\left (x \right )}^{2} + 5 \log {\left (x \right )}} \]
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Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=-\frac {x \log \left (x\right )^{2} - {\left (\log \left (x\right )^{2} + 18 \, \log \left (x\right ) - 15\right )} e^{x} + 18 \, x \log \left (x\right ) - 15 \, x}{\log \left (x\right )^{2} + 5 \, \log \left (x\right )} \]
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Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=-\frac {x \log \left (x\right )^{2} - e^{x} \log \left (x\right )^{2} + 18 \, x \log \left (x\right ) - 18 \, e^{x} \log \left (x\right ) - 15 \, x + 15 \, e^{x}}{\log \left (x\right )^{2} + 5 \, \log \left (x\right )} \]
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Time = 11.52 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=-\frac {\left (x-{\mathrm {e}}^x\right )\,\left ({\ln \left (x\right )}^2+18\,\ln \left (x\right )-15\right )}{\ln \left (x\right )\,\left (\ln \left (x\right )+5\right )} \]
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