Integrand size = 56, antiderivative size = 20 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {1}{5} \log \left (\log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )\right ) \]
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {1}{5} \log \left (\log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )\right ) \]
Integrate[(-2 + 4*x + (8 - 4*x)*Log[-2 + x])/((10 - 25*x + 10*x^2)*Log[-2 + x]*Log[Log[-2 + x]^2/(1 - 4*x + 4*x^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.
\(\displaystyle \int \frac {4 x+(8-4 x) \log (x-2)-2}{\left (10 x^2-25 x+10\right ) \log (x-2) \log \left (\frac {\log ^2(x-2)}{4 x^2-4 x+1}\right )} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {4 (-2 x+2 x \log (x-2)-4 \log (x-2)+1)}{15 (2 x-1) \log (x-2) \log \left (\frac {\log ^2(x-2)}{(2 x-1)^2}\right )}-\frac {2 (-2 x+2 x \log (x-2)-4 \log (x-2)+1)}{15 (x-2) \log (x-2) \log \left (\frac {\log ^2(x-2)}{(2 x-1)^2}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{5} \int \frac {1}{(x-2) \log (x-2) \log \left (\frac {\log ^2(x-2)}{(2 x-1)^2}\right )}dx-\frac {4}{5} \int \frac {1}{(2 x-1) \log \left (\frac {\log ^2(x-2)}{(2 x-1)^2}\right )}dx\) |
Int[(-2 + 4*x + (8 - 4*x)*Log[-2 + x])/((10 - 25*x + 10*x^2)*Log[-2 + x]*L og[Log[-2 + x]^2/(1 - 4*x + 4*x^2)]),x]
3.18.10.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.82 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20
method | result | size |
norman | \(\frac {\ln \left (\ln \left (\frac {\ln \left (-2+x \right )^{2}}{4 x^{2}-4 x +1}\right )\right )}{5}\) | \(24\) |
parallelrisch | \(\frac {\ln \left (\ln \left (\frac {\ln \left (-2+x \right )^{2}}{4 x^{2}-4 x +1}\right )\right )}{5}\) | \(24\) |
default | \(\frac {\ln \left (\ln \left (\frac {\ln \left (-2+x \right )^{2}}{4 \left (-2+x \right )^{2}-15+12 x}\right )\right )}{5}\) | \(26\) |
risch | \(\frac {\ln \left (\ln \left (\ln \left (-2+x \right )\right )-\frac {i \left (\pi \,\operatorname {csgn}\left (\frac {i}{\left (x -\frac {1}{2}\right )^{2}}\right ) \operatorname {csgn}\left (i \ln \left (-2+x \right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (-2+x \right )^{2}}{\left (x -\frac {1}{2}\right )^{2}}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{\left (x -\frac {1}{2}\right )^{2}}\right ) \operatorname {csgn}\left (\frac {i \ln \left (-2+x \right )^{2}}{\left (x -\frac {1}{2}\right )^{2}}\right )^{2}-\pi \operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )\right )^{2} \operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )^{2}\right )+2 \pi \,\operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )\right ) \operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )^{2}\right )^{2}-\pi \operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )^{2}\right )^{3}+\pi \operatorname {csgn}\left (i \ln \left (-2+x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (-2+x \right )^{2}\right )-2 \pi \,\operatorname {csgn}\left (i \ln \left (-2+x \right )\right ) \operatorname {csgn}\left (i \ln \left (-2+x \right )^{2}\right )^{2}+\pi \operatorname {csgn}\left (i \ln \left (-2+x \right )^{2}\right )^{3}-\pi \,\operatorname {csgn}\left (i \ln \left (-2+x \right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (-2+x \right )^{2}}{\left (x -\frac {1}{2}\right )^{2}}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i \ln \left (-2+x \right )^{2}}{\left (x -\frac {1}{2}\right )^{2}}\right )^{3}-4 i \ln \left (2\right )-4 i \ln \left (x -\frac {1}{2}\right )\right )}{4}\right )}{5}\) | \(255\) |
int(((-4*x+8)*ln(-2+x)+4*x-2)/(10*x^2-25*x+10)/ln(-2+x)/ln(ln(-2+x)^2/(4*x ^2-4*x+1)),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {1}{5} \, \log \left (\log \left (\frac {\log \left (x - 2\right )^{2}}{4 \, x^{2} - 4 \, x + 1}\right )\right ) \]
integrate(((-4*x+8)*log(-2+x)+4*x-2)/(10*x^2-25*x+10)/log(-2+x)/log(log(-2 +x)^2/(4*x^2-4*x+1)),x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {\log {\left (\log {\left (\frac {\log {\left (x - 2 \right )}^{2}}{4 x^{2} - 4 x + 1} \right )} \right )}}{5} \]
integrate(((-4*x+8)*ln(-2+x)+4*x-2)/(10*x**2-25*x+10)/ln(-2+x)/ln(ln(-2+x) **2/(4*x**2-4*x+1)),x)
Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {1}{5} \, \log \left (\log \left (2 \, x - 1\right ) - \log \left (\log \left (x - 2\right )\right )\right ) \]
integrate(((-4*x+8)*log(-2+x)+4*x-2)/(10*x^2-25*x+10)/log(-2+x)/log(log(-2 +x)^2/(4*x^2-4*x+1)),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {1}{5} \, \log \left (\log \left (4 \, x^{2} - 4 \, x + 1\right ) - \log \left (\log \left (x - 2\right )^{2}\right )\right ) \]
integrate(((-4*x+8)*log(-2+x)+4*x-2)/(10*x^2-25*x+10)/log(-2+x)/log(log(-2 +x)^2/(4*x^2-4*x+1)),x, algorithm=\
Time = 9.68 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {\ln \left (\ln \left (\frac {{\ln \left (x-2\right )}^2}{4\,x^2-4\,x+1}\right )\right )}{5} \]