Integrand size = 103, antiderivative size = 30 \[ \int \frac {\left (-12 x^2+12 x^2 \log (1-x)\right ) \log (5-4 x+(-2+2 x) \log (1-x))+\left (30 x-24 x^2+\left (-12 x+12 x^2\right ) \log (1-x)\right ) \log ^2(5-4 x+(-2+2 x) \log (1-x))}{5-4 x+(-2+2 x) \log (1-x)} \, dx=3 x^2 \log ^2(2-x-(1-x) (-3+2 \log (1-x))) \] Output:
3*x^2*ln(2-x-(1-x)*(2*ln(1-x)-3))^2
Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {\left (-12 x^2+12 x^2 \log (1-x)\right ) \log (5-4 x+(-2+2 x) \log (1-x))+\left (30 x-24 x^2+\left (-12 x+12 x^2\right ) \log (1-x)\right ) \log ^2(5-4 x+(-2+2 x) \log (1-x))}{5-4 x+(-2+2 x) \log (1-x)} \, dx=3 x^2 \log ^2(5-4 x+2 (-1+x) \log (1-x)) \] Input:
Integrate[((-12*x^2 + 12*x^2*Log[1 - x])*Log[5 - 4*x + (-2 + 2*x)*Log[1 - x]] + (30*x - 24*x^2 + (-12*x + 12*x^2)*Log[1 - x])*Log[5 - 4*x + (-2 + 2* x)*Log[1 - x]]^2)/(5 - 4*x + (-2 + 2*x)*Log[1 - x]),x]
Output:
3*x^2*Log[5 - 4*x + 2*(-1 + x)*Log[1 - x]]^2
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.
\(\displaystyle \int \frac {\left (-24 x^2+\left (12 x^2-12 x\right ) \log (1-x)+30 x\right ) \log ^2(-4 x+(2 x-2) \log (1-x)+5)+\left (12 x^2 \log (1-x)-12 x^2\right ) \log (-4 x+(2 x-2) \log (1-x)+5)}{-4 x+(2 x-2) \log (1-x)+5} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {12 x^2 (\log (1-x)-1) \log (-4 x+2 (x-1) \log (1-x)+5)}{-4 x+2 x \log (1-x)-2 \log (1-x)+5}+6 x \log ^2(-4 x+2 (x-1) \log (1-x)+5)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -12 \int \frac {x^2 \log (-4 x+2 (x-1) \log (1-x)+5)}{2 \log (1-x) x-4 x-2 \log (1-x)+5}dx+12 \int \frac {x^2 \log (1-x) \log (-4 x+2 (x-1) \log (1-x)+5)}{2 \log (1-x) x-4 x-2 \log (1-x)+5}dx+6 \int x \log ^2(-4 x+2 (x-1) \log (1-x)+5)dx\) |
Input:
Int[((-12*x^2 + 12*x^2*Log[1 - x])*Log[5 - 4*x + (-2 + 2*x)*Log[1 - x]] + (30*x - 24*x^2 + (-12*x + 12*x^2)*Log[1 - x])*Log[5 - 4*x + (-2 + 2*x)*Log [1 - x]]^2)/(5 - 4*x + (-2 + 2*x)*Log[1 - x]),x]
Output:
$Aborted
Time = 0.38 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87
method | result | size |
risch | \(3 x^{2} \ln \left (\left (-2+2 x \right ) \ln \left (1-x \right )-4 x +5\right )^{2}\) | \(26\) |
parallelrisch | \(3 x^{2} \ln \left (\left (-2+2 x \right ) \ln \left (1-x \right )-4 x +5\right )^{2}\) | \(26\) |
Input:
int((((12*x^2-12*x)*ln(1-x)-24*x^2+30*x)*ln((-2+2*x)*ln(1-x)-4*x+5)^2+(12* x^2*ln(1-x)-12*x^2)*ln((-2+2*x)*ln(1-x)-4*x+5))/((-2+2*x)*ln(1-x)-4*x+5),x ,method=_RETURNVERBOSE)
Output:
3*x^2*ln((-2+2*x)*ln(1-x)-4*x+5)^2
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {\left (-12 x^2+12 x^2 \log (1-x)\right ) \log (5-4 x+(-2+2 x) \log (1-x))+\left (30 x-24 x^2+\left (-12 x+12 x^2\right ) \log (1-x)\right ) \log ^2(5-4 x+(-2+2 x) \log (1-x))}{5-4 x+(-2+2 x) \log (1-x)} \, dx=3 \, x^{2} \log \left (2 \, {\left (x - 1\right )} \log \left (-x + 1\right ) - 4 \, x + 5\right )^{2} \] Input:
integrate((((12*x^2-12*x)*log(1-x)-24*x^2+30*x)*log((2*x-2)*log(1-x)-4*x+5 )^2+(12*x^2*log(1-x)-12*x^2)*log((2*x-2)*log(1-x)-4*x+5))/((2*x-2)*log(1-x )-4*x+5),x, algorithm="fricas")
Output:
3*x^2*log(2*(x - 1)*log(-x + 1) - 4*x + 5)^2
Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {\left (-12 x^2+12 x^2 \log (1-x)\right ) \log (5-4 x+(-2+2 x) \log (1-x))+\left (30 x-24 x^2+\left (-12 x+12 x^2\right ) \log (1-x)\right ) \log ^2(5-4 x+(-2+2 x) \log (1-x))}{5-4 x+(-2+2 x) \log (1-x)} \, dx=3 x^{2} \log {\left (- 4 x + \left (2 x - 2\right ) \log {\left (1 - x \right )} + 5 \right )}^{2} \] Input:
integrate((((12*x**2-12*x)*ln(1-x)-24*x**2+30*x)*ln((2*x-2)*ln(1-x)-4*x+5) **2+(12*x**2*ln(1-x)-12*x**2)*ln((2*x-2)*ln(1-x)-4*x+5))/((2*x-2)*ln(1-x)- 4*x+5),x)
Output:
3*x**2*log(-4*x + (2*x - 2)*log(1 - x) + 5)**2
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-12 x^2+12 x^2 \log (1-x)\right ) \log (5-4 x+(-2+2 x) \log (1-x))+\left (30 x-24 x^2+\left (-12 x+12 x^2\right ) \log (1-x)\right ) \log ^2(5-4 x+(-2+2 x) \log (1-x))}{5-4 x+(-2+2 x) \log (1-x)} \, dx=3 \, x^{2} \log \left (2 \, x {\left (\log \left (-x + 1\right ) - 2\right )} - 2 \, \log \left (-x + 1\right ) + 5\right )^{2} \] Input:
integrate((((12*x^2-12*x)*log(1-x)-24*x^2+30*x)*log((2*x-2)*log(1-x)-4*x+5 )^2+(12*x^2*log(1-x)-12*x^2)*log((2*x-2)*log(1-x)-4*x+5))/((2*x-2)*log(1-x )-4*x+5),x, algorithm="maxima")
Output:
3*x^2*log(2*x*(log(-x + 1) - 2) - 2*log(-x + 1) + 5)^2
Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-12 x^2+12 x^2 \log (1-x)\right ) \log (5-4 x+(-2+2 x) \log (1-x))+\left (30 x-24 x^2+\left (-12 x+12 x^2\right ) \log (1-x)\right ) \log ^2(5-4 x+(-2+2 x) \log (1-x))}{5-4 x+(-2+2 x) \log (1-x)} \, dx=3 \, x^{2} \log \left (2 \, x \log \left (-x + 1\right ) - 4 \, x - 2 \, \log \left (-x + 1\right ) + 5\right )^{2} \] Input:
integrate((((12*x^2-12*x)*log(1-x)-24*x^2+30*x)*log((2*x-2)*log(1-x)-4*x+5 )^2+(12*x^2*log(1-x)-12*x^2)*log((2*x-2)*log(1-x)-4*x+5))/((2*x-2)*log(1-x )-4*x+5),x, algorithm="giac")
Output:
3*x^2*log(2*x*log(-x + 1) - 4*x - 2*log(-x + 1) + 5)^2
Time = 3.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {\left (-12 x^2+12 x^2 \log (1-x)\right ) \log (5-4 x+(-2+2 x) \log (1-x))+\left (30 x-24 x^2+\left (-12 x+12 x^2\right ) \log (1-x)\right ) \log ^2(5-4 x+(-2+2 x) \log (1-x))}{5-4 x+(-2+2 x) \log (1-x)} \, dx=3\,x^2\,{\ln \left (\ln \left (1-x\right )\,\left (2\,x-2\right )-4\,x+5\right )}^2 \] Input:
int(-(log(log(1 - x)*(2*x - 2) - 4*x + 5)^2*(log(1 - x)*(12*x - 12*x^2) - 30*x + 24*x^2) + log(log(1 - x)*(2*x - 2) - 4*x + 5)*(12*x^2 - 12*x^2*log( 1 - x)))/(log(1 - x)*(2*x - 2) - 4*x + 5),x)
Output:
3*x^2*log(log(1 - x)*(2*x - 2) - 4*x + 5)^2
Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-12 x^2+12 x^2 \log (1-x)\right ) \log (5-4 x+(-2+2 x) \log (1-x))+\left (30 x-24 x^2+\left (-12 x+12 x^2\right ) \log (1-x)\right ) \log ^2(5-4 x+(-2+2 x) \log (1-x))}{5-4 x+(-2+2 x) \log (1-x)} \, dx=3 \mathrm {log}\left (2 \,\mathrm {log}\left (1-x \right ) x -2 \,\mathrm {log}\left (1-x \right )-4 x +5\right )^{2} x^{2} \] Input:
int((((12*x^2-12*x)*log(1-x)-24*x^2+30*x)*log((2*x-2)*log(1-x)-4*x+5)^2+(1 2*x^2*log(1-x)-12*x^2)*log((2*x-2)*log(1-x)-4*x+5))/((2*x-2)*log(1-x)-4*x+ 5),x)
Output:
3*log(2*log( - x + 1)*x - 2*log( - x + 1) - 4*x + 5)**2*x**2