Integrand size = 37, antiderivative size = 18 \[ \int \frac {5-2 x+\left (-35+18 x+6 x^2\right ) \log (x)+(5-4 x) \log (x) \log (\log (x))}{5 \log (x)} \, dx=\frac {1}{5} x (-5+2 x) (7+x-\log (\log (x))) \]
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Time = 0.12 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.83, number of steps used = 17, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {12, 6874, 2367, 2335, 2346, 2209, 2600, 2602} \[ \int \frac {5-2 x+\left (-35+18 x+6 x^2\right ) \log (x)+(5-4 x) \log (x) \log (\log (x))}{5 \log (x)} \, dx=\frac {2 x^3}{5}+\frac {9 x^2}{5}-\frac {2}{5} x^2 \log (\log (x))-7 x+x \log (\log (x)) \]
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Rule 12
Rule 2209
Rule 2335
Rule 2346
Rule 2367
Rule 2600
Rule 2602
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {5-2 x+\left (-35+18 x+6 x^2\right ) \log (x)+(5-4 x) \log (x) \log (\log (x))}{\log (x)} \, dx \\ & = \frac {1}{5} \int \left (\frac {5-2 x-35 \log (x)+18 x \log (x)+6 x^2 \log (x)}{\log (x)}-(-5+4 x) \log (\log (x))\right ) \, dx \\ & = \frac {1}{5} \int \frac {5-2 x-35 \log (x)+18 x \log (x)+6 x^2 \log (x)}{\log (x)} \, dx-\frac {1}{5} \int (-5+4 x) \log (\log (x)) \, dx \\ & = \frac {1}{5} \int \left (-35+18 x+6 x^2+\frac {5-2 x}{\log (x)}\right ) \, dx-\frac {1}{5} \int (-5 \log (\log (x))+4 x \log (\log (x))) \, dx \\ & = -7 x+\frac {9 x^2}{5}+\frac {2 x^3}{5}+\frac {1}{5} \int \frac {5-2 x}{\log (x)} \, dx-\frac {4}{5} \int x \log (\log (x)) \, dx+\int \log (\log (x)) \, dx \\ & = -7 x+\frac {9 x^2}{5}+\frac {2 x^3}{5}+x \log (\log (x))-\frac {2}{5} x^2 \log (\log (x))+\frac {1}{5} \int \left (\frac {5}{\log (x)}-\frac {2 x}{\log (x)}\right ) \, dx+\frac {2}{5} \int \frac {x}{\log (x)} \, dx-\int \frac {1}{\log (x)} \, dx \\ & = -7 x+\frac {9 x^2}{5}+\frac {2 x^3}{5}+x \log (\log (x))-\frac {2}{5} x^2 \log (\log (x))-\operatorname {LogIntegral}(x)-\frac {2}{5} \int \frac {x}{\log (x)} \, dx+\frac {2}{5} \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )+\int \frac {1}{\log (x)} \, dx \\ & = -7 x+\frac {9 x^2}{5}+\frac {2 x^3}{5}+\frac {2}{5} \operatorname {ExpIntegralEi}(2 \log (x))+x \log (\log (x))-\frac {2}{5} x^2 \log (\log (x))-\frac {2}{5} \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = -7 x+\frac {9 x^2}{5}+\frac {2 x^3}{5}+x \log (\log (x))-\frac {2}{5} x^2 \log (\log (x)) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.83 \[ \int \frac {5-2 x+\left (-35+18 x+6 x^2\right ) \log (x)+(5-4 x) \log (x) \log (\log (x))}{5 \log (x)} \, dx=-7 x+\frac {9 x^2}{5}+\frac {2 x^3}{5}+x \log (\log (x))-\frac {2}{5} x^2 \log (\log (x)) \]
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Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.56
method | result | size |
default | \(-7 x -\frac {2 x^{2} \ln \left (\ln \left (x \right )\right )}{5}+x \ln \left (\ln \left (x \right )\right )+\frac {9 x^{2}}{5}+\frac {2 x^{3}}{5}\) | \(28\) |
norman | \(-7 x -\frac {2 x^{2} \ln \left (\ln \left (x \right )\right )}{5}+x \ln \left (\ln \left (x \right )\right )+\frac {9 x^{2}}{5}+\frac {2 x^{3}}{5}\) | \(28\) |
parallelrisch | \(-7 x -\frac {2 x^{2} \ln \left (\ln \left (x \right )\right )}{5}+x \ln \left (\ln \left (x \right )\right )+\frac {9 x^{2}}{5}+\frac {2 x^{3}}{5}\) | \(28\) |
parts | \(-7 x -\frac {2 x^{2} \ln \left (\ln \left (x \right )\right )}{5}+x \ln \left (\ln \left (x \right )\right )+\frac {9 x^{2}}{5}+\frac {2 x^{3}}{5}\) | \(28\) |
risch | \(\frac {\left (-2 x^{2}+5 x \right ) \ln \left (\ln \left (x \right )\right )}{5}+\frac {2 x^{3}}{5}+\frac {9 x^{2}}{5}-7 x\) | \(29\) |
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Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.56 \[ \int \frac {5-2 x+\left (-35+18 x+6 x^2\right ) \log (x)+(5-4 x) \log (x) \log (\log (x))}{5 \log (x)} \, dx=\frac {2}{5} \, x^{3} + \frac {9}{5} \, x^{2} - \frac {1}{5} \, {\left (2 \, x^{2} - 5 \, x\right )} \log \left (\log \left (x\right )\right ) - 7 \, x \]
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Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61 \[ \int \frac {5-2 x+\left (-35+18 x+6 x^2\right ) \log (x)+(5-4 x) \log (x) \log (\log (x))}{5 \log (x)} \, dx=\frac {2 x^{3}}{5} + \frac {9 x^{2}}{5} - 7 x + \left (- \frac {2 x^{2}}{5} + x\right ) \log {\left (\log {\left (x \right )} \right )} \]
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Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {5-2 x+\left (-35+18 x+6 x^2\right ) \log (x)+(5-4 x) \log (x) \log (\log (x))}{5 \log (x)} \, dx=\frac {2}{5} \, x^{3} - \frac {2}{5} \, x^{2} \log \left (\log \left (x\right )\right ) + \frac {9}{5} \, x^{2} + x \log \left (\log \left (x\right )\right ) - 7 \, x \]
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Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {5-2 x+\left (-35+18 x+6 x^2\right ) \log (x)+(5-4 x) \log (x) \log (\log (x))}{5 \log (x)} \, dx=\frac {2}{5} \, x^{3} - \frac {2}{5} \, x^{2} \log \left (\log \left (x\right )\right ) + \frac {9}{5} \, x^{2} + x \log \left (\log \left (x\right )\right ) - 7 \, x \]
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Time = 12.61 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {5-2 x+\left (-35+18 x+6 x^2\right ) \log (x)+(5-4 x) \log (x) \log (\log (x))}{5 \log (x)} \, dx=\frac {x\,\left (2\,x-5\right )\,\left (x-\ln \left (\ln \left (x\right )\right )+7\right )}{5} \]
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