Integrand size = 66, antiderivative size = 19 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=\frac {x^3 \log (5 x)}{-3+\frac {x}{2}+\log (x)} \]
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\[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=\int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {2 x^2 (-6+x+2 (-10+x) \log (5 x)+\log (x) (2+6 \log (5 x)))}{(6-x-2 \log (x))^2} \, dx \\ & = 2 \int \frac {x^2 (-6+x+2 (-10+x) \log (5 x)+\log (x) (2+6 \log (5 x)))}{(6-x-2 \log (x))^2} \, dx \\ & = 2 \int \left (\frac {x^2}{-6+x+2 \log (x)}+\frac {2 x^2 (-10+x+3 \log (x)) \log (5 x)}{(-6+x+2 \log (x))^2}\right ) \, dx \\ & = 2 \int \frac {x^2}{-6+x+2 \log (x)} \, dx+4 \int \frac {x^2 (-10+x+3 \log (x)) \log (5 x)}{(-6+x+2 \log (x))^2} \, dx \\ & = 2 \int \frac {x^2}{-6+x+2 \log (x)} \, dx+4 \int \left (-\frac {10 x^2 \log (5 x)}{(-6+x+2 \log (x))^2}+\frac {x^3 \log (5 x)}{(-6+x+2 \log (x))^2}+\frac {3 x^2 \log (x) \log (5 x)}{(-6+x+2 \log (x))^2}\right ) \, dx \\ & = 2 \int \frac {x^2}{-6+x+2 \log (x)} \, dx+4 \int \frac {x^3 \log (5 x)}{(-6+x+2 \log (x))^2} \, dx+12 \int \frac {x^2 \log (x) \log (5 x)}{(-6+x+2 \log (x))^2} \, dx-40 \int \frac {x^2 \log (5 x)}{(-6+x+2 \log (x))^2} \, dx \\ \end{align*}
Time = 2.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=\frac {2 x^3 \log (5 x)}{-6+x+2 \log (x)} \]
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Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {2 x^{3} \ln \left (5 x \right )}{-6+2 \ln \left (x \right )+x}\) | \(19\) |
risch | \(x^{3}-\frac {\left (-6-2 \ln \left (5\right )+x \right ) x^{3}}{-6+2 \ln \left (x \right )+x}\) | \(26\) |
default | \(\frac {2 x^{3} \ln \left (x \right )}{-6+2 \ln \left (x \right )+x}+\frac {2 \ln \left (5\right ) x^{3}}{-6+2 \ln \left (x \right )+x}\) | \(34\) |
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Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=\frac {2 \, {\left (x^{3} \log \left (5\right ) + x^{3} \log \left (x\right )\right )}}{x + 2 \, \log \left (x\right ) - 6} \]
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Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=x^{3} + \frac {- x^{4} + 2 x^{3} \log {\left (5 \right )} + 6 x^{3}}{x + 2 \log {\left (x \right )} - 6} \]
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Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=\frac {2 \, {\left (x^{3} \log \left (5\right ) + x^{3} \log \left (x\right )\right )}}{x + 2 \, \log \left (x\right ) - 6} \]
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Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=x^{3} - \frac {x^{4} - 2 \, x^{3} \log \left (5\right ) - 6 \, x^{3}}{x + 2 \, \log \left (x\right ) - 6} \]
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Time = 12.87 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=\frac {2\,x^3\,\left (\ln \left (5\right )+\ln \left (x\right )\right )}{x+2\,\ln \left (x\right )-6} \]
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