Integrand size = 33, antiderivative size = 21 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x-3 \log \left (1+\frac {1}{4} x \left (2+x+\frac {\log (2)}{20}\right )\right ) \]
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Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6, 1671, 642} \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x-3 \log \left (20 x^2+x (40+\log (2))+80\right ) \]
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Rule 6
Rule 642
Rule 1671
Rubi steps \begin{align*} \text {integral}& = \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+20 x^2+x (40+\log (2))} \, dx \\ & = \int \left (1-\frac {3 (40+40 x+\log (2))}{80+20 x^2+x (40+\log (2))}\right ) \, dx \\ & = x-3 \int \frac {40+40 x+\log (2)}{80+20 x^2+x (40+\log (2))} \, dx \\ & = x-3 \log \left (80+20 x^2+x (40+\log (2))\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x-3 \log \left (80+40 x+20 x^2+x \log (2)\right ) \]
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Time = 2.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90
method | result | size |
risch | \(x -3 \ln \left (80+20 x^{2}+\left (\ln \left (2\right )+40\right ) x \right )\) | \(19\) |
parallelrisch | \(x -3 \ln \left (\frac {x \ln \left (2\right )}{20}+x^{2}+2 x +4\right )\) | \(19\) |
default | \(x -3 \ln \left (x \ln \left (2\right )+20 x^{2}+40 x +80\right )\) | \(20\) |
norman | \(x -3 \ln \left (x \ln \left (2\right )+20 x^{2}+40 x +80\right )\) | \(20\) |
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Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x - 3 \, \log \left (20 \, x^{2} + x \log \left (2\right ) + 40 \, x + 80\right ) \]
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Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x - 3 \log {\left (20 x^{2} + x \left (\log {\left (2 \right )} + 40\right ) + 80 \right )} \]
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Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x - 3 \, \log \left (20 \, x^{2} + x {\left (\log \left (2\right ) + 40\right )} + 80\right ) \]
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Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x - 3 \, \log \left (20 \, x^{2} + x \log \left (2\right ) + 40 \, x + 80\right ) \]
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Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x-3\,\ln \left (\frac {x\,\left (\ln \left (2\right )+40\right )}{20}+x^2+4\right ) \]
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