Integrand size = 60, antiderivative size = 29 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=-1-x+\frac {x}{\log (2)}+\frac {-4+x}{5 x \log \left (\frac {8}{x^2}\right )} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.23 (sec) , antiderivative size = 106, normalized size of antiderivative = 3.66, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 6874, 2395, 2343, 2347, 2209, 2339, 30} \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=\frac {\operatorname {ExpIntegralEi}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right )}{5 \sqrt {2} \sqrt {\frac {1}{x^2}} x}-\frac {\log (16) \operatorname {ExpIntegralEi}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right )}{20 \sqrt {2} \sqrt {\frac {1}{x^2}} x \log (2)}+\frac {1}{5 \log \left (\frac {8}{x^2}\right )}-\frac {4}{5 x \log \left (\frac {8}{x^2}\right )}+\frac {x (1-\log (2))}{\log (2)} \]
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Rule 12
Rule 30
Rule 2209
Rule 2339
Rule 2343
Rule 2347
Rule 2395
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{x^2 \log ^2\left (\frac {8}{x^2}\right )} \, dx}{5 \log (2)} \\ & = \frac {\int \left (-5 (-1+\log (2))+\frac {2 (-4+x) \log (2)}{x^2 \log ^2\left (\frac {8}{x^2}\right )}+\frac {\log (16)}{x^2 \log \left (\frac {8}{x^2}\right )}\right ) \, dx}{5 \log (2)} \\ & = \frac {x (1-\log (2))}{\log (2)}+\frac {2}{5} \int \frac {-4+x}{x^2 \log ^2\left (\frac {8}{x^2}\right )} \, dx+\frac {\log (16) \int \frac {1}{x^2 \log \left (\frac {8}{x^2}\right )} \, dx}{5 \log (2)} \\ & = \frac {x (1-\log (2))}{\log (2)}+\frac {2}{5} \int \left (-\frac {4}{x^2 \log ^2\left (\frac {8}{x^2}\right )}+\frac {1}{x \log ^2\left (\frac {8}{x^2}\right )}\right ) \, dx-\frac {\log (16) \text {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (\frac {8}{x^2}\right )\right )}{20 \sqrt {2} \sqrt {\frac {1}{x^2}} x \log (2)} \\ & = \frac {x (1-\log (2))}{\log (2)}-\frac {\text {Ei}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right ) \log (16)}{20 \sqrt {2} \sqrt {\frac {1}{x^2}} x \log (2)}+\frac {2}{5} \int \frac {1}{x \log ^2\left (\frac {8}{x^2}\right )} \, dx-\frac {8}{5} \int \frac {1}{x^2 \log ^2\left (\frac {8}{x^2}\right )} \, dx \\ & = \frac {x (1-\log (2))}{\log (2)}-\frac {\text {Ei}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right ) \log (16)}{20 \sqrt {2} \sqrt {\frac {1}{x^2}} x \log (2)}-\frac {4}{5 x \log \left (\frac {8}{x^2}\right )}-\frac {1}{5} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {8}{x^2}\right )\right )-\frac {4}{5} \int \frac {1}{x^2 \log \left (\frac {8}{x^2}\right )} \, dx \\ & = \frac {x (1-\log (2))}{\log (2)}-\frac {\text {Ei}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right ) \log (16)}{20 \sqrt {2} \sqrt {\frac {1}{x^2}} x \log (2)}+\frac {1}{5 \log \left (\frac {8}{x^2}\right )}-\frac {4}{5 x \log \left (\frac {8}{x^2}\right )}+\frac {\text {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (\frac {8}{x^2}\right )\right )}{5 \sqrt {2} \sqrt {\frac {1}{x^2}} x} \\ & = \frac {\text {Ei}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right )}{5 \sqrt {2} \sqrt {\frac {1}{x^2}} x}+\frac {x (1-\log (2))}{\log (2)}-\frac {\text {Ei}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right ) \log (16)}{20 \sqrt {2} \sqrt {\frac {1}{x^2}} x \log (2)}+\frac {1}{5 \log \left (\frac {8}{x^2}\right )}-\frac {4}{5 x \log \left (\frac {8}{x^2}\right )} \\ \end{align*}
Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=\frac {-5 x (-1+\log (2))+\frac {(-4+x) \log (2)}{x \log \left (\frac {8}{x^2}\right )}}{5 \log (2)} \]
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Time = 1.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93
method | result | size |
risch | \(-x +\frac {x}{\ln \left (2\right )}+\frac {x -4}{5 x \ln \left (\frac {8}{x^{2}}\right )}\) | \(27\) |
norman | \(\frac {-\frac {4}{5}+\frac {x}{5}-\frac {\left (\ln \left (2\right )-1\right ) x^{2} \ln \left (\frac {8}{x^{2}}\right )}{\ln \left (2\right )}}{\ln \left (\frac {8}{x^{2}}\right ) x}\) | \(37\) |
parallelrisch | \(-\frac {10 \ln \left (2\right ) x^{2} \ln \left (\frac {8}{x^{2}}\right )-10 \ln \left (\frac {8}{x^{2}}\right ) x^{2}-2 x \ln \left (2\right )+8 \ln \left (2\right )}{10 \ln \left (2\right ) x \ln \left (\frac {8}{x^{2}}\right )}\) | \(52\) |
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Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=\frac {{\left (x - 4\right )} \log \left (2\right ) - 5 \, {\left (x^{2} \log \left (2\right ) - x^{2}\right )} \log \left (\frac {8}{x^{2}}\right )}{5 \, x \log \left (2\right ) \log \left (\frac {8}{x^{2}}\right )} \]
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Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=\frac {x \left (1 - \log {\left (2 \right )}\right )}{\log {\left (2 \right )}} + \frac {x - 4}{5 x \log {\left (\frac {8}{x^{2}} \right )}} \]
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Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=-\frac {5 \, x \log \left (2\right ) - 5 \, x - \frac {x \log \left (2\right ) - 4 \, \log \left (2\right )}{3 \, x \log \left (2\right ) - 2 \, x \log \left (x\right )}}{5 \, \log \left (2\right )} \]
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Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=-\frac {5 \, x {\left (\log \left (2\right ) - 1\right )} - \frac {x \log \left (2\right ) - 4 \, \log \left (2\right )}{3 \, x \log \left (2\right ) - x \log \left (x^{2}\right )}}{5 \, \log \left (2\right )} \]
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Time = 14.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{5 x^2 \log (2) \log ^2\left (\frac {8}{x^2}\right )} \, dx=-\frac {x\,\left (\ln \left (32\right )-5\right )}{5\,\ln \left (2\right )}-\frac {\frac {x\,\ln \left (16\right )}{5}-\frac {x^2\,\ln \left (2\right )}{5}}{x^2\,\ln \left (2\right )\,\ln \left (\frac {8}{x^2}\right )} \]
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