Integrand size = 23, antiderivative size = 15 \[ \int \frac {8-8 e^3-10 x+5 x \log (x)}{x \log ^3(x)} \, dx=\frac {-4+x+4 \left (e^3+x\right )}{\log ^2(x)} \]
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Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.40, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6873, 6874, 2395, 2334, 2335, 2339, 30} \[ \int \frac {8-8 e^3-10 x+5 x \log (x)}{x \log ^3(x)} \, dx=\frac {5 x}{\log ^2(x)}-\frac {4 \left (1-e^3\right )}{\log ^2(x)} \]
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Rule 30
Rule 2334
Rule 2335
Rule 2339
Rule 2395
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {8 \left (1-e^3\right )-10 x+5 x \log (x)}{x \log ^3(x)} \, dx \\ & = \int \left (-\frac {2 \left (-4+4 e^3+5 x\right )}{x \log ^3(x)}+\frac {5}{\log ^2(x)}\right ) \, dx \\ & = -\left (2 \int \frac {-4+4 e^3+5 x}{x \log ^3(x)} \, dx\right )+5 \int \frac {1}{\log ^2(x)} \, dx \\ & = -\frac {5 x}{\log (x)}-2 \int \left (\frac {5}{\log ^3(x)}+\frac {4 \left (-1+e^3\right )}{x \log ^3(x)}\right ) \, dx+5 \int \frac {1}{\log (x)} \, dx \\ & = -\frac {5 x}{\log (x)}+5 \text {li}(x)-10 \int \frac {1}{\log ^3(x)} \, dx+\left (8 \left (1-e^3\right )\right ) \int \frac {1}{x \log ^3(x)} \, dx \\ & = \frac {5 x}{\log ^2(x)}-\frac {5 x}{\log (x)}+5 \text {li}(x)-5 \int \frac {1}{\log ^2(x)} \, dx+\left (8 \left (1-e^3\right )\right ) \text {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right ) \\ & = -\frac {4 \left (1-e^3\right )}{\log ^2(x)}+\frac {5 x}{\log ^2(x)}+5 \text {li}(x)-5 \int \frac {1}{\log (x)} \, dx \\ & = -\frac {4 \left (1-e^3\right )}{\log ^2(x)}+\frac {5 x}{\log ^2(x)} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.53 \[ \int \frac {8-8 e^3-10 x+5 x \log (x)}{x \log ^3(x)} \, dx=-\frac {4}{\log ^2(x)}+\frac {4 e^3}{\log ^2(x)}+\frac {5 x}{\log ^2(x)} \]
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Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\frac {4 \,{\mathrm e}^{3}+5 x -4}{\ln \left (x \right )^{2}}\) | \(15\) |
risch | \(\frac {4 \,{\mathrm e}^{3}+5 x -4}{\ln \left (x \right )^{2}}\) | \(15\) |
parallelrisch | \(\frac {4 \,{\mathrm e}^{3}+5 x -4}{\ln \left (x \right )^{2}}\) | \(15\) |
default | \(\frac {4 \,{\mathrm e}^{3}}{\ln \left (x \right )^{2}}+\frac {5 x}{\ln \left (x \right )^{2}}-\frac {4}{\ln \left (x \right )^{2}}\) | \(23\) |
parts | \(\frac {4 \,{\mathrm e}^{3}}{\ln \left (x \right )^{2}}+\frac {5 x}{\ln \left (x \right )^{2}}-\frac {4}{\ln \left (x \right )^{2}}\) | \(23\) |
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Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {8-8 e^3-10 x+5 x \log (x)}{x \log ^3(x)} \, dx=\frac {5 \, x + 4 \, e^{3} - 4}{\log \left (x\right )^{2}} \]
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Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {8-8 e^3-10 x+5 x \log (x)}{x \log ^3(x)} \, dx=\frac {5 x - 4 + 4 e^{3}}{\log {\left (x \right )}^{2}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.07 \[ \int \frac {8-8 e^3-10 x+5 x \log (x)}{x \log ^3(x)} \, dx=\frac {4 \, e^{3}}{\log \left (x\right )^{2}} - \frac {4}{\log \left (x\right )^{2}} + 5 \, \Gamma \left (-1, -\log \left (x\right )\right ) + 10 \, \Gamma \left (-2, -\log \left (x\right )\right ) \]
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Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {8-8 e^3-10 x+5 x \log (x)}{x \log ^3(x)} \, dx=\frac {5 \, x + 4 \, e^{3} - 4}{\log \left (x\right )^{2}} \]
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Time = 10.97 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {8-8 e^3-10 x+5 x \log (x)}{x \log ^3(x)} \, dx=\frac {5\,x+4\,{\mathrm {e}}^3-4}{{\ln \left (x\right )}^2} \]
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