Integrand size = 88, antiderivative size = 27 \[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=1+\frac {1}{4} (-1+x)+\log \left (\frac {-4+x+\log (2)-\log (4-x)}{x}\right ) \]
\[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=\int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx \]
Integrate[(64 - 28*x + 8*x^2 - x^3 + (-16 + 8*x - x^2)*Log[2] + (16 - 8*x + x^2)*Log[4 - x])/(-64*x + 32*x^2 - 4*x^3 + (16*x - 4*x^2)*Log[2] + (-16* x + 4*x^2)*Log[4 - x]),x]
Integrate[(64 - 28*x + 8*x^2 - x^3 + (-16 + 8*x - x^2)*Log[2] + (16 - 8*x + x^2)*Log[4 - x])/(-64*x + 32*x^2 - 4*x^3 + (16*x - 4*x^2)*Log[2] + (-16* x + 4*x^2)*Log[4 - x]), x]
Time = 1.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {7292, 27, 25, 7292, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-x^3+8 x^2+\left (x^2-8 x+16\right ) \log (4-x)+\left (-x^2+8 x-16\right ) \log (2)-28 x+64}{-4 x^3+32 x^2+\left (4 x^2-16 x\right ) \log (4-x)+\left (16 x-4 x^2\right ) \log (2)-64 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-x^3+8 x^2+\left (x^2-8 x+16\right ) \log (4-x)+\left (-x^2+8 x-16\right ) \log (2)-28 x+64}{4 (4-x) x \left (x-\log (4-x)-4 \left (1-\frac {\log (2)}{4}\right )\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {-x^3+8 x^2-28 x+\left (x^2-8 x+16\right ) \log (4-x)-\left (x^2-8 x+16\right ) \log (2)+64}{(4-x) x (-x+\log (4-x)-\log (2)+4)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {-x^3+8 x^2-28 x+\left (x^2-8 x+16\right ) \log (4-x)-\left (x^2-8 x+16\right ) \log (2)+64}{(4-x) x (-x+\log (4-x)-\log (2)+4)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {1}{4} \int \frac {x^3-8 x^2+28 x-\left (x^2-8 x+16\right ) \log (4-x)+\left (x^2-8 x+16\right ) \log (2)-64}{(4-x) x \left (x-\log (4-x)-4 \left (1-\frac {\log (2)}{4}\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{4} \int \left (\frac {4-x}{x}-\frac {4 (x-5)}{(x-4) \left (x-\log \left (2-\frac {x}{2}\right )-4\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (x-4 \log (x)+4 \log \left (-x+\log \left (2-\frac {x}{2}\right )+4\right )\right )\) |
Int[(64 - 28*x + 8*x^2 - x^3 + (-16 + 8*x - x^2)*Log[2] + (16 - 8*x + x^2) *Log[4 - x])/(-64*x + 32*x^2 - 4*x^3 + (16*x - 4*x^2)*Log[2] + (-16*x + 4* x^2)*Log[4 - x]),x]
3.22.96.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 | \(\frac {x}{4}-1-\ln \left (-x \right )+\ln \left (x -\ln \left (-x +4\right )-4+\ln \left (2\right )\right )\) | \(26\) |
int(((x^2-8*x+16)*ln(-x+4)+(-x^2+8*x-16)*ln(2)-x^3+8*x^2-28*x+64)/((4*x^2- 16*x)*ln(-x+4)+(-4*x^2+16*x)*ln(2)-4*x^3+32*x^2-64*x),x,method=_RETURNVERB OSE)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=\frac {1}{4} \, x - \log \left (x\right ) + \log \left (-x - \log \left (2\right ) + \log \left (-x + 4\right ) + 4\right ) \]
integrate(((x^2-8*x+16)*log(-x+4)+(-x^2+8*x-16)*log(2)-x^3+8*x^2-28*x+64)/ ((4*x^2-16*x)*log(-x+4)+(-4*x^2+16*x)*log(2)-4*x^3+32*x^2-64*x),x, algorit hm=\
Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=\frac {x}{4} - \log {\left (x \right )} + \log {\left (- x + \log {\left (4 - x \right )} - \log {\left (2 \right )} + 4 \right )} \]
integrate(((x**2-8*x+16)*ln(-x+4)+(-x**2+8*x-16)*ln(2)-x**3+8*x**2-28*x+64 )/((4*x**2-16*x)*ln(-x+4)+(-4*x**2+16*x)*ln(2)-4*x**3+32*x**2-64*x),x)
Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=\frac {1}{4} \, x - \log \left (x\right ) + \log \left (-x - \log \left (2\right ) + \log \left (-x + 4\right ) + 4\right ) \]
integrate(((x^2-8*x+16)*log(-x+4)+(-x^2+8*x-16)*log(2)-x^3+8*x^2-28*x+64)/ ((4*x^2-16*x)*log(-x+4)+(-4*x^2+16*x)*log(2)-4*x^3+32*x^2-64*x),x, algorit hm=\
Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=\frac {1}{4} \, x - \log \left (-x\right ) + \log \left (x + \log \left (2\right ) - \log \left (-x + 4\right ) - 4\right ) - 1 \]
integrate(((x^2-8*x+16)*log(-x+4)+(-x^2+8*x-16)*log(2)-x^3+8*x^2-28*x+64)/ ((4*x^2-16*x)*log(-x+4)+(-4*x^2+16*x)*log(2)-4*x^3+32*x^2-64*x),x, algorit hm=\
Timed out. \[ \int \frac {64-28 x+8 x^2-x^3+\left (-16+8 x-x^2\right ) \log (2)+\left (16-8 x+x^2\right ) \log (4-x)}{-64 x+32 x^2-4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (-16 x+4 x^2\right ) \log (4-x)} \, dx=\int \frac {28\,x-\ln \left (4-x\right )\,\left (x^2-8\,x+16\right )-8\,x^2+x^3+\ln \left (2\right )\,\left (x^2-8\,x+16\right )-64}{64\,x-\ln \left (2\right )\,\left (16\,x-4\,x^2\right )+\ln \left (4-x\right )\,\left (16\,x-4\,x^2\right )-32\,x^2+4\,x^3} \,d x \]
int((28*x - log(4 - x)*(x^2 - 8*x + 16) - 8*x^2 + x^3 + log(2)*(x^2 - 8*x + 16) - 64)/(64*x - log(2)*(16*x - 4*x^2) + log(4 - x)*(16*x - 4*x^2) - 32 *x^2 + 4*x^3),x)