Integrand size = 30, antiderivative size = 20 \[ \int \frac {-20-8 x^2}{5 x-x^2-2 x^3+x^2 \log (4)} \, dx=\log \left (\left (2+2 x-\frac {5+x+x \log (4)}{x}\right )^4\right ) \]
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Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6, 1608, 1642, 642} \[ \int \frac {-20-8 x^2}{5 x-x^2-2 x^3+x^2 \log (4)} \, dx=4 \log \left (-2 x^2-x (1-\log (4))+5\right )-4 \log (x) \]
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Rule 6
Rule 642
Rule 1608
Rule 1642
Rubi steps \begin{align*} \text {integral}& = \int \frac {-20-8 x^2}{5 x-2 x^3+x^2 (-1+\log (4))} \, dx \\ & = \int \frac {-20-8 x^2}{x \left (5-2 x^2+x (-1+\log (4))\right )} \, dx \\ & = \int \left (-\frac {4}{x}+\frac {4 (-1-4 x+\log (4))}{5-2 x^2-x (1-\log (4))}\right ) \, dx \\ & = -4 \log (x)+4 \int \frac {-1-4 x+\log (4)}{5-2 x^2+x (-1+\log (4))} \, dx \\ & = -4 \log (x)+4 \log \left (5-2 x^2-x (1-\log (4))\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-20-8 x^2}{5 x-x^2-2 x^3+x^2 \log (4)} \, dx=4 \left (-\log (x)+\log \left (5-x-2 x^2+x \log (4)\right )\right ) \]
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Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10
method | result | size |
default | \(4 \ln \left (-2 x \ln \left (2\right )+2 x^{2}+x -5\right )-4 \ln \left (x \right )\) | \(22\) |
parallelrisch | \(-4 \ln \left (x \right )+4 \ln \left (-x \ln \left (2\right )+x^{2}+\frac {x}{2}-\frac {5}{2}\right )\) | \(22\) |
norman | \(-4 \ln \left (x \right )+4 \ln \left (2 x \ln \left (2\right )-2 x^{2}-x +5\right )\) | \(24\) |
risch | \(-4 \ln \left (-x \right )+4 \ln \left (-5+2 x^{2}+\left (1-2 \ln \left (2\right )\right ) x \right )\) | \(26\) |
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-20-8 x^2}{5 x-x^2-2 x^3+x^2 \log (4)} \, dx=4 \, \log \left (2 \, x^{2} - 2 \, x \log \left (2\right ) + x - 5\right ) - 4 \, \log \left (x\right ) \]
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Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-20-8 x^2}{5 x-x^2-2 x^3+x^2 \log (4)} \, dx=- 4 \log {\left (x \right )} + 4 \log {\left (x^{2} + x \left (\frac {1}{2} - \log {\left (2 \right )}\right ) - \frac {5}{2} \right )} \]
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Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-20-8 x^2}{5 x-x^2-2 x^3+x^2 \log (4)} \, dx=4 \, \log \left (2 \, x^{2} - x {\left (2 \, \log \left (2\right ) - 1\right )} - 5\right ) - 4 \, \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-20-8 x^2}{5 x-x^2-2 x^3+x^2 \log (4)} \, dx=4 \, \log \left ({\left | 2 \, x^{2} - 2 \, x \log \left (2\right ) + x - 5 \right |}\right ) - 4 \, \log \left ({\left | x \right |}\right ) \]
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Time = 8.98 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-20-8 x^2}{5 x-x^2-2 x^3+x^2 \log (4)} \, dx=4\,\ln \left (\frac {x}{2}-x\,\ln \left (2\right )+x^2-\frac {5}{2}\right )-4\,\ln \left (x\right ) \]
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