\(\int \frac {-2+4 x+(8-4 x) \log (-2+x)}{(10-25 x+10 x^2) \log (-2+x) \log (\frac {\log ^2(-2+x)}{1-4 x+4 x^2})} \, dx\) [1710]

Optimal result

Integrand size = 56, antiderivative size = 20 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {1}{5} \log \left (\log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )\right ) \]

[Out]

1/5*ln(ln(ln(-2+x)^2/(-1+2*x)^2))

Rubi [F]

\[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx \]

[In]

Int[(-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]

[Out]

(-4*Defer[Int][1/((-1 + 2*x)*Log[Log[-2 + x]^2/(-1 + 2*x)^2]), x])/5 + (2*Defer[Int][1/((-2 + x)*Log[-2 + x]*L
og[Log[-2 + x]^2/(-1 + 2*x)^2]), x])/5

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 (1-2 x-4 \log (-2+x)+2 x \log (-2+x))}{15 (-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {4 (1-2 x-4 \log (-2+x)+2 x \log (-2+x))}{15 (-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx \\ & = -\left (\frac {2}{15} \int \frac {1-2 x-4 \log (-2+x)+2 x \log (-2+x)}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx\right )+\frac {4}{15} \int \frac {1-2 x-4 \log (-2+x)+2 x \log (-2+x)}{(-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx \\ & = -\left (\frac {2}{15} \int \left (-\frac {4}{(-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {2 x}{(-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {1}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}-\frac {2 x}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx\right )+\frac {4}{15} \int \left (-\frac {4}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {2 x}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {1}{(-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}-\frac {2 x}{(-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx \\ & = -\left (\frac {2}{15} \int \frac {1}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx\right )-\frac {4}{15} \int \frac {x}{(-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx+\frac {4}{15} \int \frac {x}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx+\frac {4}{15} \int \frac {1}{(-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx+\frac {8}{15} \int \frac {1}{(-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx+\frac {8}{15} \int \frac {x}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx-\frac {8}{15} \int \frac {x}{(-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx-\frac {16}{15} \int \frac {1}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx \\ & = -\left (\frac {2}{15} \int \frac {1}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx\right )-\frac {4}{15} \int \left (\frac {1}{\log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {2}{(-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx+\frac {4}{15} \int \left (\frac {1}{\log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {2}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx+\frac {4}{15} \int \frac {1}{(-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx+\frac {8}{15} \int \left (\frac {1}{2 \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {1}{2 (-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx-\frac {8}{15} \int \left (\frac {1}{2 \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}+\frac {1}{2 (-1+2 x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )}\right ) \, dx+\frac {8}{15} \int \frac {1}{(-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx-\frac {16}{15} \int \frac {1}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx \\ & = -\left (\frac {2}{15} \int \frac {1}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx\right )+\frac {4}{15} \int \frac {1}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx+\frac {8}{15} \int \frac {1}{(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx-\frac {16}{15} \int \frac {1}{(-1+2 x) \log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {1}{5} \log \left (\log \left (\frac {\log ^2(-2+x)}{(-1+2 x)^2}\right )\right ) \]

[In]

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]

[Out]

Log[Log[Log[-2 + x]^2/(-1 + 2*x)^2]]/5

Maple [A] (verified)

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	\(\frac {\ln \left (\ln \left (\ln \left (-2+x \right )\right )-\frac {i \left (\pi \,\operatorname {csgn}\left (\frac {i}{\left (x -\frac {1}{2}\right )^{2}}\right ) \operatorname {csgn}\left (i \ln \left (-2+x \right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (-2+x \right )^{2}}{\left (x -\frac {1}{2}\right )^{2}}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{\left (x -\frac {1}{2}\right )^{2}}\right ) \operatorname {csgn}\left (\frac {i \ln \left (-2+x \right )^{2}}{\left (x -\frac {1}{2}\right )^{2}}\right )^{2}-\pi \operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )\right )^{2} \operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )^{2}\right )+2 \pi \,\operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )\right ) \operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )^{2}\right )^{2}-\pi \operatorname {csgn}\left (i \left (x -\frac {1}{2}\right )^{2}\right )^{3}+\pi \operatorname {csgn}\left (i \ln \left (-2+x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (-2+x \right )^{2}\right )-2 \pi \,\operatorname {csgn}\left (i \ln \left (-2+x \right )\right ) \operatorname {csgn}\left (i \ln \left (-2+x \right )^{2}\right )^{2}+\pi \operatorname {csgn}\left (i \ln \left (-2+x \right )^{2}\right )^{3}-\pi \,\operatorname {csgn}\left (i \ln \left (-2+x \right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (-2+x \right )^{2}}{\left (x -\frac {1}{2}\right )^{2}}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i \ln \left (-2+x \right )^{2}}{\left (x -\frac {1}{2}\right )^{2}}\right )^{3}-4 i \ln \left (2\right )-4 i \ln \left (x -\frac {1}{2}\right )\right )}{4}\right )}{5}\)	\(255\)

[In]

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)

[Out]

1/5*ln(ln(ln(-2+x)^2/(4*x^2-4*x+1)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {1}{5} \, \log \left (\log \left (\frac {\log \left (x - 2\right )^{2}}{4 \, x^{2} - 4 \, x + 1}\right )\right ) \]

[In]

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="f
ricas")

[Out]

1/5*log(log(log(x - 2)^2/(4*x^2 - 4*x + 1)))

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {\log {\left (\log {\left (\frac {\log {\left (x - 2 \right )}^{2}}{4 x^{2} - 4 x + 1} \right )} \right )}}{5} \]

[In]

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)

[Out]

log(log(log(x - 2)**2/(4*x**2 - 4*x + 1)))/5

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {1}{5} \, \log \left (\log \left (2 \, x - 1\right ) - \log \left (\log \left (x - 2\right )\right )\right ) \]

[In]

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="m
axima")

[Out]

1/5*log(log(2*x - 1) - log(log(x - 2)))

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {1}{5} \, \log \left (\log \left (4 \, x^{2} - 4 \, x + 1\right ) - \log \left (\log \left (x - 2\right )^{2}\right )\right ) \]

[In]

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="g
iac")

[Out]

1/5*log(log(4*x^2 - 4*x + 1) - log(log(x - 2)^2))

Mupad [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-2+4 x+(8-4 x) \log (-2+x)}{\left (10-25 x+10 x^2\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{1-4 x+4 x^2}\right )} \, dx=\frac {\ln \left (\ln \left (\frac {{\ln \left (x-2\right )}^2}{4\,x^2-4\,x+1}\right )\right )}{5} \]

[In]

int(-(log(x - 2)*(4*x - 8) - 4*x + 2)/(log(log(x - 2)^2/(4*x^2 - 4*x + 1))*log(x - 2)*(10*x^2 - 25*x + 10)),x)

[Out]

log(log(log(x - 2)^2/(4*x^2 - 4*x + 1)))/5