Integrand size = 38, antiderivative size = 20 \[ \int \frac {2+8 x+8 x^2+\left (-2-8 x-8 x^2\right ) \log (x)+4 x \log ^2(x)}{x \log ^3(x)} \, dx=\log (2)-\left (1-\frac {1+2 x}{\log (x)}\right )^2 \]
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Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80, number of steps used = 23, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {6874, 2395, 2334, 2335, 2339, 30, 2343, 2346, 2209} \[ \int \frac {2+8 x+8 x^2+\left (-2-8 x-8 x^2\right ) \log (x)+4 x \log ^2(x)}{x \log ^3(x)} \, dx=-\frac {4 x^2}{\log ^2(x)}-\frac {4 x}{\log ^2(x)}-\frac {1}{\log ^2(x)}+\frac {4 x}{\log (x)}+\frac {2}{\log (x)} \]
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Rule 30
Rule 2209
Rule 2334
Rule 2335
Rule 2339
Rule 2343
Rule 2346
Rule 2395
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2 (1+2 x)^2}{x \log ^3(x)}-\frac {2 (1+2 x)^2}{x \log ^2(x)}+\frac {4}{\log (x)}\right ) \, dx \\ & = 2 \int \frac {(1+2 x)^2}{x \log ^3(x)} \, dx-2 \int \frac {(1+2 x)^2}{x \log ^2(x)} \, dx+4 \int \frac {1}{\log (x)} \, dx \\ & = 4 \operatorname {LogIntegral}(x)+2 \int \left (\frac {4}{\log ^3(x)}+\frac {1}{x \log ^3(x)}+\frac {4 x}{\log ^3(x)}\right ) \, dx-2 \int \left (\frac {4}{\log ^2(x)}+\frac {1}{x \log ^2(x)}+\frac {4 x}{\log ^2(x)}\right ) \, dx \\ & = 4 \operatorname {LogIntegral}(x)+2 \int \frac {1}{x \log ^3(x)} \, dx-2 \int \frac {1}{x \log ^2(x)} \, dx+8 \int \frac {1}{\log ^3(x)} \, dx+8 \int \frac {x}{\log ^3(x)} \, dx-8 \int \frac {1}{\log ^2(x)} \, dx-8 \int \frac {x}{\log ^2(x)} \, dx \\ & = -\frac {4 x}{\log ^2(x)}-\frac {4 x^2}{\log ^2(x)}+\frac {8 x}{\log (x)}+\frac {8 x^2}{\log (x)}+4 \operatorname {LogIntegral}(x)+2 \text {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right )-2 \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )+4 \int \frac {1}{\log ^2(x)} \, dx+8 \int \frac {x}{\log ^2(x)} \, dx-8 \int \frac {1}{\log (x)} \, dx-16 \int \frac {x}{\log (x)} \, dx \\ & = -\frac {1}{\log ^2(x)}-\frac {4 x}{\log ^2(x)}-\frac {4 x^2}{\log ^2(x)}+\frac {2}{\log (x)}+\frac {4 x}{\log (x)}-4 \operatorname {LogIntegral}(x)+4 \int \frac {1}{\log (x)} \, dx+16 \int \frac {x}{\log (x)} \, dx-16 \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = -16 \operatorname {ExpIntegralEi}(2 \log (x))-\frac {1}{\log ^2(x)}-\frac {4 x}{\log ^2(x)}-\frac {4 x^2}{\log ^2(x)}+\frac {2}{\log (x)}+\frac {4 x}{\log (x)}+16 \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = -\frac {1}{\log ^2(x)}-\frac {4 x}{\log ^2(x)}-\frac {4 x^2}{\log ^2(x)}+\frac {2}{\log (x)}+\frac {4 x}{\log (x)} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80 \[ \int \frac {2+8 x+8 x^2+\left (-2-8 x-8 x^2\right ) \log (x)+4 x \log ^2(x)}{x \log ^3(x)} \, dx=-\frac {1}{\log ^2(x)}-\frac {4 x}{\log ^2(x)}-\frac {4 x^2}{\log ^2(x)}+\frac {2}{\log (x)}+\frac {4 x}{\log (x)} \]
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Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25
method | result | size |
norman | \(\frac {-1-4 x -4 x^{2}+4 x \ln \left (x \right )+2 \ln \left (x \right )}{\ln \left (x \right )^{2}}\) | \(25\) |
parallelrisch | \(\frac {-1-4 x -4 x^{2}+4 x \ln \left (x \right )+2 \ln \left (x \right )}{\ln \left (x \right )^{2}}\) | \(25\) |
risch | \(-\frac {4 x^{2}-4 x \ln \left (x \right )+4 x -2 \ln \left (x \right )+1}{\ln \left (x \right )^{2}}\) | \(26\) |
default | \(\frac {4 x}{\ln \left (x \right )}-\frac {4 x^{2}}{\ln \left (x \right )^{2}}+\frac {2}{\ln \left (x \right )}-\frac {4 x}{\ln \left (x \right )^{2}}-\frac {1}{\ln \left (x \right )^{2}}\) | \(37\) |
parts | \(\frac {4 x}{\ln \left (x \right )}-\frac {4 x^{2}}{\ln \left (x \right )^{2}}+\frac {2}{\ln \left (x \right )}-\frac {4 x}{\ln \left (x \right )^{2}}-\frac {1}{\ln \left (x \right )^{2}}\) | \(37\) |
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none
Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {2+8 x+8 x^2+\left (-2-8 x-8 x^2\right ) \log (x)+4 x \log ^2(x)}{x \log ^3(x)} \, dx=-\frac {4 \, x^{2} - 2 \, {\left (2 \, x + 1\right )} \log \left (x\right ) + 4 \, x + 1}{\log \left (x\right )^{2}} \]
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Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {2+8 x+8 x^2+\left (-2-8 x-8 x^2\right ) \log (x)+4 x \log ^2(x)}{x \log ^3(x)} \, dx=\frac {- 4 x^{2} - 4 x + \left (4 x + 2\right ) \log {\left (x \right )} - 1}{\log {\left (x \right )}^{2}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.22 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.50 \[ \int \frac {2+8 x+8 x^2+\left (-2-8 x-8 x^2\right ) \log (x)+4 x \log ^2(x)}{x \log ^3(x)} \, dx=\frac {2}{\log \left (x\right )} - \frac {1}{\log \left (x\right )^{2}} + 4 \, {\rm Ei}\left (\log \left (x\right )\right ) - 8 \, \Gamma \left (-1, -\log \left (x\right )\right ) - 16 \, \Gamma \left (-1, -2 \, \log \left (x\right )\right ) - 8 \, \Gamma \left (-2, -\log \left (x\right )\right ) - 32 \, \Gamma \left (-2, -2 \, \log \left (x\right )\right ) \]
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Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {2+8 x+8 x^2+\left (-2-8 x-8 x^2\right ) \log (x)+4 x \log ^2(x)}{x \log ^3(x)} \, dx=-\frac {4 \, x^{2} - 4 \, x \log \left (x\right ) + 4 \, x - 2 \, \log \left (x\right ) + 1}{\log \left (x\right )^{2}} \]
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Time = 7.43 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {2+8 x+8 x^2+\left (-2-8 x-8 x^2\right ) \log (x)+4 x \log ^2(x)}{x \log ^3(x)} \, dx=\frac {4\,x+2}{\ln \left (x\right )}-\frac {{\left (2\,x+1\right )}^2}{{\ln \left (x\right )}^2} \]
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