Integrand size = 17, antiderivative size = 16 \[ \int \frac {-40+32 x}{-5 x+2 x^2} \, dx=4 \log \left ((-x+x (-4+2 x))^2\right ) \]
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Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {642} \[ \int \frac {-40+32 x}{-5 x+2 x^2} \, dx=8 \log \left (5 x-2 x^2\right ) \]
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Rule 642
Rubi steps \begin{align*} \text {integral}& = 8 \log \left (5 x-2 x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69 \[ \int \frac {-40+32 x}{-5 x+2 x^2} \, dx=8 (\log (5-2 x)+\log (x)) \]
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Time = 0.39 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69
method | result | size |
default | \(8 \ln \left (x \left (-5+2 x \right )\right )\) | \(11\) |
parallelrisch | \(8 \ln \left (x \right )+8 \ln \left (x -\frac {5}{2}\right )\) | \(12\) |
risch | \(8 \ln \left (2 x^{2}-5 x \right )\) | \(13\) |
norman | \(8 \ln \left (x \right )+8 \ln \left (-5+2 x \right )\) | \(14\) |
meijerg | \(8 \ln \left (x \right )+8 \ln \left (2\right )-8 \ln \left (5\right )+8 i \pi +8 \ln \left (-\frac {2 x}{5}+1\right )\) | \(26\) |
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none
Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {-40+32 x}{-5 x+2 x^2} \, dx=8 \, \log \left (2 \, x^{2} - 5 \, x\right ) \]
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Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {-40+32 x}{-5 x+2 x^2} \, dx=8 \log {\left (2 x^{2} - 5 x \right )} \]
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none
Time = 0.18 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {-40+32 x}{-5 x+2 x^2} \, dx=8 \, \log \left (2 \, x^{2} - 5 \, x\right ) \]
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none
Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {-40+32 x}{-5 x+2 x^2} \, dx=8 \, \log \left ({\left | 2 \, x^{2} - 5 \, x \right |}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {-40+32 x}{-5 x+2 x^2} \, dx=8\,\ln \left (x\,\left (2\,x-5\right )\right ) \]
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