Integrand size = 70, antiderivative size = 28 \[ \int \frac {9-6 x+x^2+18 x^3+16 x^4+3 x^5}{9 x-6 x^2-3 x^3+6 x^4+5 x^5+x^6+\left (9 x+6 x^2+x^3\right ) \log (5 x)} \, dx=\log \left (-1+x+\frac {x-x^2 (1+x)^2}{3+x}-\log (5 x)\right ) \]
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\[ \int \frac {9-6 x+x^2+18 x^3+16 x^4+3 x^5}{9 x-6 x^2-3 x^3+6 x^4+5 x^5+x^6+\left (9 x+6 x^2+x^3\right ) \log (5 x)} \, dx=\int \frac {9-6 x+x^2+18 x^3+16 x^4+3 x^5}{9 x-6 x^2-3 x^3+6 x^4+5 x^5+x^6+\left (9 x+6 x^2+x^3\right ) \log (5 x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {9-6 x+x^2+18 x^3+16 x^4+3 x^5}{x (3+x) \left (3-3 x+2 x^3+x^4+(3+x) \log (5 x)\right )} \, dx \\ & = \int \left (\frac {10}{3-3 x+2 x^3+x^4+3 \log (5 x)+x \log (5 x)}+\frac {3}{x \left (3-3 x+2 x^3+x^4+3 \log (5 x)+x \log (5 x)\right )}-\frac {3 x}{3-3 x+2 x^3+x^4+3 \log (5 x)+x \log (5 x)}+\frac {7 x^2}{3-3 x+2 x^3+x^4+3 \log (5 x)+x \log (5 x)}+\frac {3 x^3}{3-3 x+2 x^3+x^4+3 \log (5 x)+x \log (5 x)}-\frac {39}{(3+x) \left (3-3 x+2 x^3+x^4+3 \log (5 x)+x \log (5 x)\right )}\right ) \, dx \\ & = 3 \int \frac {1}{x \left (3-3 x+2 x^3+x^4+3 \log (5 x)+x \log (5 x)\right )} \, dx-3 \int \frac {x}{3-3 x+2 x^3+x^4+3 \log (5 x)+x \log (5 x)} \, dx+3 \int \frac {x^3}{3-3 x+2 x^3+x^4+3 \log (5 x)+x \log (5 x)} \, dx+7 \int \frac {x^2}{3-3 x+2 x^3+x^4+3 \log (5 x)+x \log (5 x)} \, dx+10 \int \frac {1}{3-3 x+2 x^3+x^4+3 \log (5 x)+x \log (5 x)} \, dx-39 \int \frac {1}{(3+x) \left (3-3 x+2 x^3+x^4+3 \log (5 x)+x \log (5 x)\right )} \, dx \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {9-6 x+x^2+18 x^3+16 x^4+3 x^5}{9 x-6 x^2-3 x^3+6 x^4+5 x^5+x^6+\left (9 x+6 x^2+x^3\right ) \log (5 x)} \, dx=-\log (3+x)+\log \left (3-3 x+2 x^3+x^4+3 \log (5 x)+x \log (5 x)\right ) \]
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Time = 1.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93
method | result | size |
risch | \(\ln \left (\ln \left (5 x \right )+\frac {x^{4}+2 x^{3}-3 x +3}{3+x}\right )\) | \(26\) |
norman | \(-\ln \left (3+x \right )+\ln \left (x^{4}+2 x^{3}+x \ln \left (5 x \right )-3 x +3 \ln \left (5 x \right )+3\right )\) | \(34\) |
parallelrisch | \(-\ln \left (3+x \right )+\ln \left (x^{4}+2 x^{3}+x \ln \left (5 x \right )-3 x +3 \ln \left (5 x \right )+3\right )\) | \(34\) |
derivativedivides | \(-\ln \left (5 x +15\right )+\ln \left (625 x^{4}+1250 x^{3}+625 x \ln \left (5 x \right )+1875 \ln \left (5 x \right )-1875 x +1875\right )\) | \(39\) |
default | \(-\ln \left (5 x +15\right )+\ln \left (625 x^{4}+1250 x^{3}+625 x \ln \left (5 x \right )+1875 \ln \left (5 x \right )-1875 x +1875\right )\) | \(39\) |
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {9-6 x+x^2+18 x^3+16 x^4+3 x^5}{9 x-6 x^2-3 x^3+6 x^4+5 x^5+x^6+\left (9 x+6 x^2+x^3\right ) \log (5 x)} \, dx=\log \left (\frac {x^{4} + 2 \, x^{3} + {\left (x + 3\right )} \log \left (5 \, x\right ) - 3 \, x + 3}{x + 3}\right ) \]
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Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {9-6 x+x^2+18 x^3+16 x^4+3 x^5}{9 x-6 x^2-3 x^3+6 x^4+5 x^5+x^6+\left (9 x+6 x^2+x^3\right ) \log (5 x)} \, dx=\log {\left (\log {\left (5 x \right )} + \frac {x^{4} + 2 x^{3} - 3 x + 3}{x + 3} \right )} \]
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Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {9-6 x+x^2+18 x^3+16 x^4+3 x^5}{9 x-6 x^2-3 x^3+6 x^4+5 x^5+x^6+\left (9 x+6 x^2+x^3\right ) \log (5 x)} \, dx=\log \left (\frac {x^{4} + 2 \, x^{3} + x {\left (\log \left (5\right ) - 3\right )} + {\left (x + 3\right )} \log \left (x\right ) + 3 \, \log \left (5\right ) + 3}{x + 3}\right ) \]
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Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {9-6 x+x^2+18 x^3+16 x^4+3 x^5}{9 x-6 x^2-3 x^3+6 x^4+5 x^5+x^6+\left (9 x+6 x^2+x^3\right ) \log (5 x)} \, dx=\log \left (x^{4} + 2 \, x^{3} + x \log \left (5 \, x\right ) - 3 \, x + 3 \, \log \left (5 \, x\right ) + 3\right ) - \log \left (x + 3\right ) \]
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Time = 13.55 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {9-6 x+x^2+18 x^3+16 x^4+3 x^5}{9 x-6 x^2-3 x^3+6 x^4+5 x^5+x^6+\left (9 x+6 x^2+x^3\right ) \log (5 x)} \, dx=\ln \left (3\,x+\ln \left (5\,x\right )+\frac {39}{x+3}-x^2+x^3-12\right ) \]
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