Integrand size = 25, antiderivative size = 23 \[ \int \frac {6-25 x+(4-10 x) \log (-4+10 x)}{-2+5 x} \, dx=1-x-2 \left (-3-e^2+x+x \log (-4+10 x)\right ) \]
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Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6874, 45, 2436, 2332} \[ \int \frac {6-25 x+(4-10 x) \log (-4+10 x)}{-2+5 x} \, dx=-3 x-\frac {4}{5} \log (2-5 x)+\frac {2}{5} (2-5 x) \log (10 x-4) \]
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Rule 45
Rule 2332
Rule 2436
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {6-25 x}{-2+5 x}-2 \log (-4+10 x)\right ) \, dx \\ & = -(2 \int \log (-4+10 x) \, dx)+\int \frac {6-25 x}{-2+5 x} \, dx \\ & = -\left (\frac {1}{5} \text {Subst}(\int \log (x) \, dx,x,-4+10 x)\right )+\int \left (-5-\frac {4}{-2+5 x}\right ) \, dx \\ & = -3 x-\frac {4}{5} \log (2-5 x)+\frac {2}{5} (2-5 x) \log (-4+10 x) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {6-25 x+(4-10 x) \log (-4+10 x)}{-2+5 x} \, dx=-3 x-\frac {4}{5} \log (2-5 x)+\frac {2}{5} (2-5 x) \log (-4+10 x) \]
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Time = 0.12 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61
method | result | size |
norman | \(-3 x -2 x \ln \left (10 x -4\right )\) | \(14\) |
risch | \(-3 x -2 x \ln \left (10 x -4\right )\) | \(14\) |
parallelrisch | \(-3 x -2 x \ln \left (10 x -4\right )\) | \(14\) |
derivativedivides | \(-\frac {\left (10 x -4\right ) \ln \left (10 x -4\right )}{5}-3 x +\frac {6}{5}-\frac {4 \ln \left (10 x -4\right )}{5}\) | \(27\) |
default | \(-\frac {\left (10 x -4\right ) \ln \left (10 x -4\right )}{5}-3 x +\frac {6}{5}-\frac {4 \ln \left (10 x -4\right )}{5}\) | \(27\) |
parts | \(-3 x -\frac {4 \ln \left (5 x -2\right )}{5}-\frac {\left (10 x -4\right ) \ln \left (10 x -4\right )}{5}-\frac {4}{5}\) | \(27\) |
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Time = 0.23 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {6-25 x+(4-10 x) \log (-4+10 x)}{-2+5 x} \, dx=-2 \, x \log \left (10 \, x - 4\right ) - 3 \, x \]
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Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int \frac {6-25 x+(4-10 x) \log (-4+10 x)}{-2+5 x} \, dx=- 2 x \log {\left (10 x - 4 \right )} - 3 x \]
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Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (18) = 36\).
Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {6-25 x+(4-10 x) \log (-4+10 x)}{-2+5 x} \, dx=-\frac {2}{5} \, {\left (5 \, x + 2 \, \log \left (5 \, x - 2\right )\right )} \log \left (10 \, x - 4\right ) + \frac {4}{5} \, \log \left (2\right ) \log \left (5 \, x - 2\right ) + \frac {4}{5} \, \log \left (5 \, x - 2\right )^{2} - 3 \, x \]
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Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {6-25 x+(4-10 x) \log (-4+10 x)}{-2+5 x} \, dx=-2 \, x \log \left (10 \, x - 4\right ) - 3 \, x \]
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Time = 12.13 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {6-25 x+(4-10 x) \log (-4+10 x)}{-2+5 x} \, dx=-x\,\left (2\,\ln \left (10\,x-4\right )+3\right ) \]
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