Integrand size = 88, antiderivative size = 27 \[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=1+\frac {1}{4} (-1+x)+\log \left (\frac {-4+x+\log (2)-\log (4-x)}{x}\right ) \]
[Out]
Time = 0.45 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {6873, 12, 6874, 45, 6816} \[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=\frac {x}{4}-\log (x)+\log \left (-x+\log \left (2-\frac {x}{2}\right )+4\right ) \]
[In]
[Out]
Rule 12
Rule 45
Rule 6816
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{4 (4-x) x \left (x-4 \left (1-\frac {\log (2)}{4}\right )-\log (4-x)\right )} \, dx \\ & = \frac {1}{4} \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{(4-x) x \left (x-4 \left (1-\frac {\log (2)}{4}\right )-\log (4-x)\right )} \, dx \\ & = \frac {1}{4} \int \left (\frac {-4+x}{x}+\frac {4 (-5+x)}{(-4+x) \left (-4+x-\log \left (2-\frac {x}{2}\right )\right )}\right ) \, dx \\ & = \frac {1}{4} \int \frac {-4+x}{x} \, dx+\int \frac {-5+x}{(-4+x) \left (-4+x-\log \left (2-\frac {x}{2}\right )\right )} \, dx \\ & = \log \left (4-x+\log \left (2-\frac {x}{2}\right )\right )+\frac {1}{4} \int \left (1-\frac {4}{x}\right ) \, dx \\ & = \frac {x}{4}-\log (x)+\log \left (4-x+\log \left (2-\frac {x}{2}\right )\right ) \\ \end{align*}
\[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=\int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx \]
[In]
[Out]
Time = 0.61 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85
method | result | size |
norman | \(\frac {x}{4}-\ln \left (x \right )+\ln \left (x -\ln \left (-x +4\right )-4+\ln \left (2\right )\right )\) | \(23\) |
parallelrisch | \(2-\ln \left (x \right )+\ln \left (x -\ln \left (-x +4\right )-4+\ln \left (2\right )\right )+\frac {x}{4}\) | \(24\) |
risch | \(\frac {x}{4}-\ln \left (x \right )+\ln \left (-x +\ln \left (-x +4\right )-\ln \left (2\right )+4\right )\) | \(25\) |
derivativedivides | \(\frac {x}{4}-1-\ln \left (-x \right )+\ln \left (x -\ln \left (-x +4\right )-4+\ln \left (2\right )\right )\) | \(26\) |
default | \(\frac {x}{4}-1-\ln \left (-x \right )+\ln \left (x -\ln \left (-x +4\right )-4+\ln \left (2\right )\right )\) | \(26\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=\frac {1}{4} \, x - \log \left (x\right ) + \log \left (-x - \log \left (2\right ) + \log \left (-x + 4\right ) + 4\right ) \]
[In]
[Out]
Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=\frac {x}{4} - \log {\left (x \right )} + \log {\left (- x + \log {\left (4 - x \right )} - \log {\left (2 \right )} + 4 \right )} \]
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=\frac {1}{4} \, x - \log \left (x\right ) + \log \left (-x - \log \left (2\right ) + \log \left (-x + 4\right ) + 4\right ) \]
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=\frac {1}{4} \, x - \log \left (-x\right ) + \log \left (x + \log \left (2\right ) - \log \left (-x + 4\right ) - 4\right ) - 1 \]
[In]
[Out]
Timed out. \[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=\int \frac {28\,x-\ln \left (4-x\right )\,\left (x^2-8\,x+16\right )-8\,x^2+x^3+\ln \left (2\right )\,\left (x^2-8\,x+16\right )-64}{64\,x-\ln \left (2\right )\,\left (16\,x-4\,x^2\right )+\ln \left (4-x\right )\,\left (16\,x-4\,x^2\right )-32\,x^2+4\,x^3} \,d x \]
[In]
[Out]