Integrand size = 49, antiderivative size = 24 \[ \int \frac {-10+3 x-2 \log (3)-x \log \left (x^2\right )}{-25 x-5 x^2-5 x \log (3)+\left (5 x+x^2+x \log (3)\right ) \log \left (x^2\right )} \, dx=\log \left (\frac {\frac {11}{2}+e^4}{(5+x+\log (3)) \left (-5+\log \left (x^2\right )\right )}\right ) \]
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Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {6, 6873, 6874, 2339, 29} \[ \int \frac {-10+3 x-2 \log (3)-x \log \left (x^2\right )}{-25 x-5 x^2-5 x \log (3)+\left (5 x+x^2+x \log (3)\right ) \log \left (x^2\right )} \, dx=-\log \left (5-\log \left (x^2\right )\right )-\log (x+5+\log (3)) \]
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Rule 6
Rule 29
Rule 2339
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-10+3 x-2 \log (3)-x \log \left (x^2\right )}{-5 x^2+x (-25-5 \log (3))+\left (5 x+x^2+x \log (3)\right ) \log \left (x^2\right )} \, dx \\ & = \int \frac {-3 x+10 \left (1+\frac {\log (3)}{5}\right )+x \log \left (x^2\right )}{x (5+x+\log (3)) \left (5-\log \left (x^2\right )\right )} \, dx \\ & = \int \left (\frac {1}{-5-x-\log (3)}-\frac {2}{x \left (-5+\log \left (x^2\right )\right )}\right ) \, dx \\ & = -\log (5+x+\log (3))-2 \int \frac {1}{x \left (-5+\log \left (x^2\right )\right )} \, dx \\ & = -\log (5+x+\log (3))-\text {Subst}\left (\int \frac {1}{x} \, dx,x,-5+\log \left (x^2\right )\right ) \\ & = -\log (5+x+\log (3))-\log \left (5-\log \left (x^2\right )\right ) \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-10+3 x-2 \log (3)-x \log \left (x^2\right )}{-25 x-5 x^2-5 x \log (3)+\left (5 x+x^2+x \log (3)\right ) \log \left (x^2\right )} \, dx=-\log (5+x+\log (3))-\log \left (5-\log \left (x^2\right )\right ) \]
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Time = 0.52 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79
method | result | size |
norman | \(-\ln \left (\ln \left (x^{2}\right )-5\right )-\ln \left (\ln \left (3\right )+5+x \right )\) | \(19\) |
risch | \(-\ln \left (\ln \left (x^{2}\right )-5\right )-\ln \left (\ln \left (3\right )+5+x \right )\) | \(19\) |
parallelrisch | \(-\ln \left (\ln \left (x^{2}\right )-5\right )-\ln \left (\ln \left (3\right )+5+x \right )\) | \(19\) |
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Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {-10+3 x-2 \log (3)-x \log \left (x^2\right )}{-25 x-5 x^2-5 x \log (3)+\left (5 x+x^2+x \log (3)\right ) \log \left (x^2\right )} \, dx=-\log \left (x + \log \left (3\right ) + 5\right ) - \log \left (\log \left (x^{2}\right ) - 5\right ) \]
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Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-10+3 x-2 \log (3)-x \log \left (x^2\right )}{-25 x-5 x^2-5 x \log (3)+\left (5 x+x^2+x \log (3)\right ) \log \left (x^2\right )} \, dx=- \log {\left (\log {\left (x^{2} \right )} - 5 \right )} - \log {\left (x + \log {\left (3 \right )} + 5 \right )} \]
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Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-10+3 x-2 \log (3)-x \log \left (x^2\right )}{-25 x-5 x^2-5 x \log (3)+\left (5 x+x^2+x \log (3)\right ) \log \left (x^2\right )} \, dx=-\log \left (x + \log \left (3\right ) + 5\right ) - \log \left (\log \left (x\right ) - \frac {5}{2}\right ) \]
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Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {-10+3 x-2 \log (3)-x \log \left (x^2\right )}{-25 x-5 x^2-5 x \log (3)+\left (5 x+x^2+x \log (3)\right ) \log \left (x^2\right )} \, dx=-\log \left (x + \log \left (3\right ) + 5\right ) - \log \left (\log \left (x^{2}\right ) - 5\right ) \]
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Time = 10.81 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {-10+3 x-2 \log (3)-x \log \left (x^2\right )}{-25 x-5 x^2-5 x \log (3)+\left (5 x+x^2+x \log (3)\right ) \log \left (x^2\right )} \, dx=-\ln \left (\ln \left (x^2\right )-5\right )-\ln \left (x+\ln \left (3\right )+5\right ) \]
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