Integrand size = 100, antiderivative size = 37 \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=4+\log \left (\frac {x}{2 \left (-x+x^2 \left (-2+\log \left (\frac {x^2+\frac {1}{3} (x+\log (5))}{x}\right )\right )\right )}\right ) \]
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Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6873, 6816} \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=-\log \left (2 x+x \left (-\log \left (x+\frac {\log (5)}{3 x}+\frac {1}{3}\right )\right )+1\right ) \]
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Rule 6816
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {-2 x-3 x^2-3 \log (5)-\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{\left (x+3 x^2+\log (5)\right ) \left (1+2 x-x \log \left (\frac {1}{3}+x+\frac {\log (5)}{3 x}\right )\right )} \, dx \\ & = -\log \left (1+2 x-x \log \left (\frac {1}{3}+x+\frac {\log (5)}{3 x}\right )\right ) \\ \end{align*}
\[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=\int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx \]
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Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70
method | result | size |
norman | \(-\ln \left (\ln \left (\frac {\ln \left (5\right )+3 x^{2}+x}{3 x}\right ) x -2 x -1\right )\) | \(26\) |
parallelrisch | \(-\ln \left (\ln \left (\frac {\ln \left (5\right )+3 x^{2}+x}{3 x}\right ) x -2 x -1\right )\) | \(26\) |
risch | \(-\ln \left (x \right )-\ln \left (\ln \left (\frac {\ln \left (5\right )+3 x^{2}+x}{3 x}\right )-\frac {1+2 x}{x}\right )\) | \(35\) |
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Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=-\log \left (x\right ) - \log \left (\frac {x \log \left (\frac {3 \, x^{2} + x + \log \left (5\right )}{3 \, x}\right ) - 2 \, x - 1}{x}\right ) \]
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Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=- \log {\left (x \right )} - \log {\left (\log {\left (\frac {x^{2} + \frac {x}{3} + \frac {\log {\left (5 \right )}}{3}}{x} \right )} + \frac {- 2 x - 1}{x} \right )} \]
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Time = 0.32 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.03 \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=-\log \left (x\right ) - \log \left (-\frac {x {\left (\log \left (3\right ) + 2\right )} - x \log \left (3 \, x^{2} + x + \log \left (5\right )\right ) + x \log \left (x\right ) + 1}{x}\right ) \]
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Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=-\log \left (-x \log \left (3 \, x^{2} + x + \log \left (5\right )\right ) + x \log \left (3 \, x\right ) + 2 \, x + 1\right ) \]
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Time = 15.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=-\ln \left (\frac {2\,x-x\,\ln \left (\frac {3\,x^2+x+\ln \left (5\right )}{3\,x}\right )+1}{x}\right )-\ln \left (x\right ) \]
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