Integrand size = 111, antiderivative size = 27 \[ \int \frac {(-150-50 x) \log (3)}{9000+5400 x+900 x^2+\left (600+360 x+60 x^2\right ) \log ^2(3)+\left (10+6 x+x^2\right ) \log ^4(3)+\left (\left (-600-360 x-60 x^2\right ) \log (3)+\left (-20-12 x-2 x^2\right ) \log ^3(3)\right ) \log \left (10+6 x+x^2\right )+\left (10+6 x+x^2\right ) \log ^2(3) \log ^2\left (10+6 x+x^2\right )} \, dx=\frac {5}{-6+\frac {1}{5} \log (3) (-\log (3)+\log (5+(1+x) (5+x)))} \]
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Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {12, 6820, 6818} \[ \int \frac {(-150-50 x) \log (3)}{9000+5400 x+900 x^2+\left (600+360 x+60 x^2\right ) \log ^2(3)+\left (10+6 x+x^2\right ) \log ^4(3)+\left (\left (-600-360 x-60 x^2\right ) \log (3)+\left (-20-12 x-2 x^2\right ) \log ^3(3)\right ) \log \left (10+6 x+x^2\right )+\left (10+6 x+x^2\right ) \log ^2(3) \log ^2\left (10+6 x+x^2\right )} \, dx=-\frac {50 \log (3)}{\log (9) \left (-\log (3) \log \left (x^2+6 x+10\right )+30+\log ^2(3)\right )} \]
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Rule 12
Rule 6818
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \log (3) \int \frac {-150-50 x}{9000+5400 x+900 x^2+\left (600+360 x+60 x^2\right ) \log ^2(3)+\left (10+6 x+x^2\right ) \log ^4(3)+\left (\left (-600-360 x-60 x^2\right ) \log (3)+\left (-20-12 x-2 x^2\right ) \log ^3(3)\right ) \log \left (10+6 x+x^2\right )+\left (10+6 x+x^2\right ) \log ^2(3) \log ^2\left (10+6 x+x^2\right )} \, dx \\ & = \log (3) \int \frac {50 (-3-x)}{\left (10+6 x+x^2\right ) \left (30 \left (1+\frac {\log ^2(3)}{30}\right )-\log (3) \log \left (10+6 x+x^2\right )\right )^2} \, dx \\ & = (50 \log (3)) \int \frac {-3-x}{\left (10+6 x+x^2\right ) \left (30 \left (1+\frac {\log ^2(3)}{30}\right )-\log (3) \log \left (10+6 x+x^2\right )\right )^2} \, dx \\ & = -\frac {50 \log (3)}{\log (9) \left (30+\log ^2(3)-\log (3) \log \left (10+6 x+x^2\right )\right )} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {(-150-50 x) \log (3)}{9000+5400 x+900 x^2+\left (600+360 x+60 x^2\right ) \log ^2(3)+\left (10+6 x+x^2\right ) \log ^4(3)+\left (\left (-600-360 x-60 x^2\right ) \log (3)+\left (-20-12 x-2 x^2\right ) \log ^3(3)\right ) \log \left (10+6 x+x^2\right )+\left (10+6 x+x^2\right ) \log ^2(3) \log ^2\left (10+6 x+x^2\right )} \, dx=-\frac {50 \log (3)}{\log (9) \left (30+\log ^2(3)-\log (3) \log \left (10+6 x+x^2\right )\right )} \]
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Time = 0.66 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89
method | result | size |
norman | \(-\frac {25}{\ln \left (3\right )^{2}-\ln \left (3\right ) \ln \left (x^{2}+6 x +10\right )+30}\) | \(24\) |
risch | \(-\frac {25}{\ln \left (3\right )^{2}-\ln \left (3\right ) \ln \left (x^{2}+6 x +10\right )+30}\) | \(24\) |
parallelrisch | \(-\frac {25}{\ln \left (3\right )^{2}-\ln \left (3\right ) \ln \left (x^{2}+6 x +10\right )+30}\) | \(24\) |
default | \(\frac {25}{\ln \left (3\right ) \ln \left (x^{2}+6 x +10\right )-\ln \left (3\right )^{2}-30}\) | \(25\) |
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Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {(-150-50 x) \log (3)}{9000+5400 x+900 x^2+\left (600+360 x+60 x^2\right ) \log ^2(3)+\left (10+6 x+x^2\right ) \log ^4(3)+\left (\left (-600-360 x-60 x^2\right ) \log (3)+\left (-20-12 x-2 x^2\right ) \log ^3(3)\right ) \log \left (10+6 x+x^2\right )+\left (10+6 x+x^2\right ) \log ^2(3) \log ^2\left (10+6 x+x^2\right )} \, dx=-\frac {25}{\log \left (3\right )^{2} - \log \left (3\right ) \log \left (x^{2} + 6 \, x + 10\right ) + 30} \]
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Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {(-150-50 x) \log (3)}{9000+5400 x+900 x^2+\left (600+360 x+60 x^2\right ) \log ^2(3)+\left (10+6 x+x^2\right ) \log ^4(3)+\left (\left (-600-360 x-60 x^2\right ) \log (3)+\left (-20-12 x-2 x^2\right ) \log ^3(3)\right ) \log \left (10+6 x+x^2\right )+\left (10+6 x+x^2\right ) \log ^2(3) \log ^2\left (10+6 x+x^2\right )} \, dx=\frac {25}{\log {\left (3 \right )} \log {\left (x^{2} + 6 x + 10 \right )} - 30 - \log {\left (3 \right )}^{2}} \]
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Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {(-150-50 x) \log (3)}{9000+5400 x+900 x^2+\left (600+360 x+60 x^2\right ) \log ^2(3)+\left (10+6 x+x^2\right ) \log ^4(3)+\left (\left (-600-360 x-60 x^2\right ) \log (3)+\left (-20-12 x-2 x^2\right ) \log ^3(3)\right ) \log \left (10+6 x+x^2\right )+\left (10+6 x+x^2\right ) \log ^2(3) \log ^2\left (10+6 x+x^2\right )} \, dx=-\frac {25 \, \log \left (3\right )}{\log \left (3\right )^{3} - \log \left (3\right )^{2} \log \left (x^{2} + 6 \, x + 10\right ) + 30 \, \log \left (3\right )} \]
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Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {(-150-50 x) \log (3)}{9000+5400 x+900 x^2+\left (600+360 x+60 x^2\right ) \log ^2(3)+\left (10+6 x+x^2\right ) \log ^4(3)+\left (\left (-600-360 x-60 x^2\right ) \log (3)+\left (-20-12 x-2 x^2\right ) \log ^3(3)\right ) \log \left (10+6 x+x^2\right )+\left (10+6 x+x^2\right ) \log ^2(3) \log ^2\left (10+6 x+x^2\right )} \, dx=-\frac {25 \, \log \left (3\right )}{\log \left (3\right )^{3} - \log \left (3\right )^{2} \log \left (x^{2} + 6 \, x + 10\right ) + 30 \, \log \left (3\right )} \]
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Time = 14.66 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {(-150-50 x) \log (3)}{9000+5400 x+900 x^2+\left (600+360 x+60 x^2\right ) \log ^2(3)+\left (10+6 x+x^2\right ) \log ^4(3)+\left (\left (-600-360 x-60 x^2\right ) \log (3)+\left (-20-12 x-2 x^2\right ) \log ^3(3)\right ) \log \left (10+6 x+x^2\right )+\left (10+6 x+x^2\right ) \log ^2(3) \log ^2\left (10+6 x+x^2\right )} \, dx=\frac {25}{\ln \left (3\right )\,\left (\ln \left (x^2+6\,x+10\right )-\frac {{\ln \left (3\right )}^2+30}{\ln \left (3\right )}\right )} \]
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