Integrand size = 26, antiderivative size = 22 \[ \int \frac {x^2 \log (8+e)-\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x^2} \, dx=\log (8+e) \left (1+x+\frac {-1+x+\log \left (\frac {\log (4)}{x}\right )}{x}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {14, 2341} \[ \int \frac {x^2 \log (8+e)-\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x^2} \, dx=x \log (8+e)+\frac {\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x}-\frac {\log (8+e)}{x} \]
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Rule 14
Rule 2341
Rubi steps \begin{align*} \text {integral}& = \int \left (\log (8+e)-\frac {\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x^2}\right ) \, dx \\ & = x \log (8+e)-\log (8+e) \int \frac {\log \left (\frac {\log (4)}{x}\right )}{x^2} \, dx \\ & = -\frac {\log (8+e)}{x}+x \log (8+e)+\frac {\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {x^2 \log (8+e)-\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x^2} \, dx=-\frac {\log (8+e)}{x}+x \log (8+e)+\frac {\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x} \]
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Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50
method | result | size |
risch | \(\frac {\ln \left ({\mathrm e}+8\right ) \ln \left (\frac {2 \ln \left (2\right )}{x}\right )}{x}+\frac {\ln \left ({\mathrm e}+8\right ) \left (x^{2}-1\right )}{x}\) | \(33\) |
derivativedivides | \(\frac {\ln \left ({\mathrm e}+8\right ) \ln \left (\frac {2 \ln \left (2\right )}{x}\right )}{x}-\frac {\ln \left ({\mathrm e}+8\right )}{x}+x \ln \left ({\mathrm e}+8\right )\) | \(36\) |
default | \(\frac {\ln \left ({\mathrm e}+8\right ) \ln \left (\frac {2 \ln \left (2\right )}{x}\right )}{x}-\frac {\ln \left ({\mathrm e}+8\right )}{x}+x \ln \left ({\mathrm e}+8\right )\) | \(36\) |
norman | \(\frac {x^{2} \ln \left ({\mathrm e}+8\right )+\ln \left ({\mathrm e}+8\right ) \ln \left (\frac {2 \ln \left (2\right )}{x}\right )-\ln \left ({\mathrm e}+8\right )}{x}\) | \(36\) |
parallelrisch | \(\frac {x^{2} \ln \left ({\mathrm e}+8\right )+\ln \left ({\mathrm e}+8\right ) \ln \left (\frac {2 \ln \left (2\right )}{x}\right )-\ln \left ({\mathrm e}+8\right )}{x}\) | \(36\) |
parts | \(\frac {\ln \left ({\mathrm e}+8\right ) \ln \left (\frac {2 \ln \left (2\right )}{x}\right )}{x}-\frac {\ln \left ({\mathrm e}+8\right )}{x}+x \ln \left ({\mathrm e}+8\right )\) | \(36\) |
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {x^2 \log (8+e)-\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x^2} \, dx=\frac {{\left (x^{2} - 1\right )} \log \left (e + 8\right ) + \log \left (\frac {2 \, \log \left (2\right )}{x}\right ) \log \left (e + 8\right )}{x} \]
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Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {x^2 \log (8+e)-\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x^2} \, dx=x \log {\left (e + 8 \right )} + \frac {\log {\left (\frac {2 \log {\left (2 \right )}}{x} \right )} \log {\left (e + 8 \right )}}{x} - \frac {\log {\left (e + 8 \right )}}{x} \]
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Time = 0.21 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.82 \[ \int \frac {x^2 \log (8+e)-\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x^2} \, dx=x \log \left (e + 8\right ) + \frac {{\left (\frac {\log \left (2\right ) \log \left (\frac {2 \, \log \left (2\right )}{x}\right )}{x} - \frac {\log \left (2\right )}{x}\right )} \log \left (e + 8\right )}{\log \left (2\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (24) = 48\).
Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.36 \[ \int \frac {x^2 \log (8+e)-\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x^2} \, dx=\frac {{\left (\log \left (2\right )^{2} \log \left (e + 8\right ) + \frac {\log \left (2\right )^{2} \log \left (\frac {2 \, \log \left (2\right )}{x}\right ) \log \left (e + 8\right )}{x^{2}} - \frac {\log \left (2\right )^{2} \log \left (e + 8\right )}{x^{2}}\right )} x}{\log \left (2\right )^{2}} \]
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Time = 7.55 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \log (8+e)-\log (8+e) \log \left (\frac {\log (4)}{x}\right )}{x^2} \, dx=\frac {\ln \left (\mathrm {e}+8\right )\,\left (\ln \left (\frac {2\,\ln \left (2\right )}{x}\right )+x^2-1\right )}{x} \]
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