Integrand size = 127, antiderivative size = 19 \[ \int \frac {-2 x^2+\left (-4 x-2 x^2+2 x^3\right ) \log (2-x)+\left (\left (-8 x^2+4 x^3\right ) \log (2-x)+\left (-8 x+4 x^2\right ) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )\right ) \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}{\left (-2 x+x^2\right ) \log (2-x)+(-2+x) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )} \, dx=2 x^2 \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right ) \] Output:
2*ln(ln(x/ln(2-x))+x)*x^2
Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x^2+\left (-4 x-2 x^2+2 x^3\right ) \log (2-x)+\left (\left (-8 x^2+4 x^3\right ) \log (2-x)+\left (-8 x+4 x^2\right ) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )\right ) \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}{\left (-2 x+x^2\right ) \log (2-x)+(-2+x) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )} \, dx=2 x^2 \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right ) \] Input:
Integrate[(-2*x^2 + (-4*x - 2*x^2 + 2*x^3)*Log[2 - x] + ((-8*x^2 + 4*x^3)* Log[2 - x] + (-8*x + 4*x^2)*Log[2 - x]*Log[x/Log[2 - x]])*Log[x + Log[x/Lo g[2 - x]]])/((-2*x + x^2)*Log[2 - x] + (-2 + x)*Log[2 - x]*Log[x/Log[2 - x ]]),x]
Output:
2*x^2*Log[x + Log[x/Log[2 - x]]]
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.
\(\displaystyle \int \frac {-2 x^2+\left (2 x^3-2 x^2-4 x\right ) \log (2-x)+\left (\left (4 x^2-8 x\right ) \log \left (\frac {x}{\log (2-x)}\right ) \log (2-x)+\left (4 x^3-8 x^2\right ) \log (2-x)\right ) \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}{\left (x^2-2 x\right ) \log (2-x)+(x-2) \log \left (\frac {x}{\log (2-x)}\right ) \log (2-x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 x^2-\left (2 x^3-2 x^2-4 x\right ) \log (2-x)-\left (\left (4 x^2-8 x\right ) \log \left (\frac {x}{\log (2-x)}\right ) \log (2-x)+\left (4 x^3-8 x^2\right ) \log (2-x)\right ) \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}{(2-x) \log (2-x) \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 x \left (x-(x-2) \log (2-x) \left (x+2 \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right ) \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )+1\right )\right )}{(2-x) \log (2-x) \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {x \left (x+(2-x) \log (2-x) \left (x+2 \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right ) \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )+1\right )\right )}{(2-x) \log (2-x) \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (\frac {x \left (\log (2-x) x^2-\log (2-x) x-x-2 \log (2-x)\right )}{(x-2) \log (2-x) \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}+2 x \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\int \frac {x^2}{x+\log \left (\frac {x}{\log (2-x)}\right )}dx+\int \frac {x}{x+\log \left (\frac {x}{\log (2-x)}\right )}dx-2 \int \frac {1}{\log (2-x) \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}dx-4 \int \frac {1}{(x-2) \log (2-x) \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}dx-\int \frac {x}{\log (2-x) \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}dx+2 \int x \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )dx\right )\) |
Input:
Int[(-2*x^2 + (-4*x - 2*x^2 + 2*x^3)*Log[2 - x] + ((-8*x^2 + 4*x^3)*Log[2 - x] + (-8*x + 4*x^2)*Log[2 - x]*Log[x/Log[2 - x]])*Log[x + Log[x/Log[2 - x]]])/((-2*x + x^2)*Log[2 - x] + (-2 + x)*Log[2 - x]*Log[x/Log[2 - x]]),x]
Output:
$Aborted
Time = 10.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05
method | result | size |
parallelrisch | \(2 \ln \left (\ln \left (\frac {x}{\ln \left (2-x \right )}\right )+x \right ) x^{2}\) | \(20\) |
risch | \(2 x^{2} \ln \left (\ln \left (x \right )-\ln \left (\ln \left (2-x \right )\right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i x}{\ln \left (2-x \right )}\right ) \left (-\operatorname {csgn}\left (\frac {i x}{\ln \left (2-x \right )}\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (\frac {i x}{\ln \left (2-x \right )}\right )+\operatorname {csgn}\left (\frac {i}{\ln \left (2-x \right )}\right )\right )}{2}+x \right )\) | \(86\) |
Input:
int((((4*x^2-8*x)*ln(2-x)*ln(x/ln(2-x))+(4*x^3-8*x^2)*ln(2-x))*ln(ln(x/ln( 2-x))+x)+(2*x^3-2*x^2-4*x)*ln(2-x)-2*x^2)/((-2+x)*ln(2-x)*ln(x/ln(2-x))+(x ^2-2*x)*ln(2-x)),x,method=_RETURNVERBOSE)
Output:
2*ln(ln(x/ln(2-x))+x)*x^2
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x^2+\left (-4 x-2 x^2+2 x^3\right ) \log (2-x)+\left (\left (-8 x^2+4 x^3\right ) \log (2-x)+\left (-8 x+4 x^2\right ) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )\right ) \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}{\left (-2 x+x^2\right ) \log (2-x)+(-2+x) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )} \, dx=2 \, x^{2} \log \left (x + \log \left (\frac {x}{\log \left (-x + 2\right )}\right )\right ) \] Input:
integrate((((4*x^2-8*x)*log(2-x)*log(x/log(2-x))+(4*x^3-8*x^2)*log(2-x))*l og(log(x/log(2-x))+x)+(2*x^3-2*x^2-4*x)*log(2-x)-2*x^2)/((-2+x)*log(2-x)*l og(x/log(2-x))+(x^2-2*x)*log(2-x)),x, algorithm="fricas")
Output:
2*x^2*log(x + log(x/log(-x + 2)))
Time = 1.79 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-2 x^2+\left (-4 x-2 x^2+2 x^3\right ) \log (2-x)+\left (\left (-8 x^2+4 x^3\right ) \log (2-x)+\left (-8 x+4 x^2\right ) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )\right ) \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}{\left (-2 x+x^2\right ) \log (2-x)+(-2+x) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )} \, dx=2 x^{2} \log {\left (x + \log {\left (\frac {x}{\log {\left (2 - x \right )}} \right )} \right )} \] Input:
integrate((((4*x**2-8*x)*ln(2-x)*ln(x/ln(2-x))+(4*x**3-8*x**2)*ln(2-x))*ln (ln(x/ln(2-x))+x)+(2*x**3-2*x**2-4*x)*ln(2-x)-2*x**2)/((-2+x)*ln(2-x)*ln(x /ln(2-x))+(x**2-2*x)*ln(2-x)),x)
Output:
2*x**2*log(x + log(x/log(2 - x)))
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x^2+\left (-4 x-2 x^2+2 x^3\right ) \log (2-x)+\left (\left (-8 x^2+4 x^3\right ) \log (2-x)+\left (-8 x+4 x^2\right ) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )\right ) \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}{\left (-2 x+x^2\right ) \log (2-x)+(-2+x) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )} \, dx=2 \, x^{2} \log \left (x + \log \left (x\right ) - \log \left (\log \left (-x + 2\right )\right )\right ) \] Input:
integrate((((4*x^2-8*x)*log(2-x)*log(x/log(2-x))+(4*x^3-8*x^2)*log(2-x))*l og(log(x/log(2-x))+x)+(2*x^3-2*x^2-4*x)*log(2-x)-2*x^2)/((-2+x)*log(2-x)*l og(x/log(2-x))+(x^2-2*x)*log(2-x)),x, algorithm="maxima")
Output:
2*x^2*log(x + log(x) - log(log(-x + 2)))
Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x^2+\left (-4 x-2 x^2+2 x^3\right ) \log (2-x)+\left (\left (-8 x^2+4 x^3\right ) \log (2-x)+\left (-8 x+4 x^2\right ) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )\right ) \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}{\left (-2 x+x^2\right ) \log (2-x)+(-2+x) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )} \, dx=2 \, x^{2} \log \left (x + \log \left (x\right ) - \log \left (\log \left (-x + 2\right )\right )\right ) \] Input:
integrate((((4*x^2-8*x)*log(2-x)*log(x/log(2-x))+(4*x^3-8*x^2)*log(2-x))*l og(log(x/log(2-x))+x)+(2*x^3-2*x^2-4*x)*log(2-x)-2*x^2)/((-2+x)*log(2-x)*l og(x/log(2-x))+(x^2-2*x)*log(2-x)),x, algorithm="giac")
Output:
2*x^2*log(x + log(x) - log(log(-x + 2)))
Time = 0.81 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x^2+\left (-4 x-2 x^2+2 x^3\right ) \log (2-x)+\left (\left (-8 x^2+4 x^3\right ) \log (2-x)+\left (-8 x+4 x^2\right ) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )\right ) \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}{\left (-2 x+x^2\right ) \log (2-x)+(-2+x) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )} \, dx=2\,x^2\,\ln \left (x+\ln \left (\frac {x}{\ln \left (2-x\right )}\right )\right ) \] Input:
int((log(2 - x)*(4*x + 2*x^2 - 2*x^3) + log(x + log(x/log(2 - x)))*(log(2 - x)*(8*x^2 - 4*x^3) + log(x/log(2 - x))*log(2 - x)*(8*x - 4*x^2)) + 2*x^2 )/(log(2 - x)*(2*x - x^2) - log(x/log(2 - x))*log(2 - x)*(x - 2)),x)
Output:
2*x^2*log(x + log(x/log(2 - x)))
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x^2+\left (-4 x-2 x^2+2 x^3\right ) \log (2-x)+\left (\left (-8 x^2+4 x^3\right ) \log (2-x)+\left (-8 x+4 x^2\right ) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )\right ) \log \left (x+\log \left (\frac {x}{\log (2-x)}\right )\right )}{\left (-2 x+x^2\right ) \log (2-x)+(-2+x) \log (2-x) \log \left (\frac {x}{\log (2-x)}\right )} \, dx=2 \,\mathrm {log}\left (\mathrm {log}\left (\frac {x}{\mathrm {log}\left (-x +2\right )}\right )+x \right ) x^{2} \] Input:
int((((4*x^2-8*x)*log(2-x)*log(x/log(2-x))+(4*x^3-8*x^2)*log(2-x))*log(log (x/log(2-x))+x)+(2*x^3-2*x^2-4*x)*log(2-x)-2*x^2)/((-2+x)*log(2-x)*log(x/l og(2-x))+(x^2-2*x)*log(2-x)),x)
Output:
2*log(log(x/log( - x + 2)) + x)*x**2