\(\int \frac {3-4 x+(2 x-2 x^2) \log (\frac {e^2}{-x^3+x^4})+(1-x) \log (\frac {e^2}{-x^3+x^4}) \log (e^{-2 x} \log (\frac {e^2}{-x^3+x^4}))}{(-1+x) \log (\frac {e^2}{-x^3+x^4}) \log ^2(e^{-2 x} \log (\frac {e^2}{-x^3+x^4}))} \, dx\) [1810]

Optimal result

Integrand size = 125, antiderivative size = 31 \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=1-\frac {x}{\log \left (e^{-2 x} \log \left (\frac {e^2}{x^2 \left (-x+x^2\right )}\right )\right )} \] Output:

1-x/ln(ln(1/x^2/(x^2-x)*exp(1)^2)/exp(2*x))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=-\frac {x}{\log \left (e^{-2 x} \left (2+\log \left (\frac {1}{(-1+x) x^3}\right )\right )\right )} \] Input:

Integrate[(3 - 4*x + (2*x - 2*x^2)*Log[E^2/(-x^3 + x^4)] + (1 - x)*Log[E^2 
/(-x^3 + x^4)]*Log[Log[E^2/(-x^3 + x^4)]/E^(2*x)])/((-1 + x)*Log[E^2/(-x^3 
 + x^4)]*Log[Log[E^2/(-x^3 + x^4)]/E^(2*x)]^2),x]
 

Output:

-(x/Log[(2 + Log[1/((-1 + x)*x^3)])/E^(2*x)])
 

Rubi [F]

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.

\(\displaystyle \int \frac {(1-x) \log \left (\frac {e^2}{x^4-x^3}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{x^4-x^3}\right )\right )+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{x^4-x^3}\right )-4 x+3}{(x-1) \log \left (\frac {e^2}{x^4-x^3}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{x^4-x^3}\right )\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-(1-x) \log \left (\frac {e^2}{x^4-x^3}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{x^4-x^3}\right )\right )-\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{x^4-x^3}\right )+4 x-3}{(1-x) \log \left (\frac {e^2}{(x-1) x^3}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{(x-1) x^3}\right )\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 x \log \left (\frac {1}{(x-1) x^3}\right )-4 x^2-2 x^2 \log \left (\frac {1}{(x-1) x^3}\right )+3}{(x-1) \left (\log \left (\frac {1}{(x-1) x^3}\right )+2\right ) \log ^2\left (e^{-2 x} \left (\log \left (\frac {1}{(x-1) x^3}\right )+2\right )\right )}-\frac {1}{\log \left (e^{-2 x} \left (\log \left (\frac {1}{(x-1) x^3}\right )+2\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -4 \int \frac {1}{\left (\log \left (\frac {1}{(x-1) x^3}\right )+2\right ) \log ^2\left (e^{-2 x} \left (\log \left (\frac {1}{(x-1) x^3}\right )+2\right )\right )}dx-\int \frac {1}{(x-1) \left (\log \left (\frac {1}{(x-1) x^3}\right )+2\right ) \log ^2\left (e^{-2 x} \left (\log \left (\frac {1}{(x-1) x^3}\right )+2\right )\right )}dx-4 \int \frac {x}{\left (\log \left (\frac {1}{(x-1) x^3}\right )+2\right ) \log ^2\left (e^{-2 x} \left (\log \left (\frac {1}{(x-1) x^3}\right )+2\right )\right )}dx-2 \int \frac {x \log \left (\frac {1}{(x-1) x^3}\right )}{\left (\log \left (\frac {1}{(x-1) x^3}\right )+2\right ) \log ^2\left (e^{-2 x} \left (\log \left (\frac {1}{(x-1) x^3}\right )+2\right )\right )}dx-\int \frac {1}{\log \left (e^{-2 x} \left (\log \left (\frac {1}{(x-1) x^3}\right )+2\right )\right )}dx\)

Input:

Int[(3 - 4*x + (2*x - 2*x^2)*Log[E^2/(-x^3 + x^4)] + (1 - x)*Log[E^2/(-x^3 
 + x^4)]*Log[Log[E^2/(-x^3 + x^4)]/E^(2*x)])/((-1 + x)*Log[E^2/(-x^3 + x^4 
)]*Log[Log[E^2/(-x^3 + x^4)]/E^(2*x)]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.52 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90

method	result	size
parallelrisch	\(-\frac {x}{\ln \left (\ln \left (\frac {{\mathrm e}^{2}}{x^{3} \left (-1+x \right )}\right ) {\mathrm e}^{-2 x}\right )}\)	\(28\)
risch	\(\text {Expression too large to display}\)	\(3088\)

Input:

int(((1-x)*ln(exp(1)^2/(x^4-x^3))*ln(ln(exp(1)^2/(x^4-x^3))/exp(2*x))+(-2* 
x^2+2*x)*ln(exp(1)^2/(x^4-x^3))+3-4*x)/(-1+x)/ln(exp(1)^2/(x^4-x^3))/ln(ln 
(exp(1)^2/(x^4-x^3))/exp(2*x))^2,x,method=_RETURNVERBOSE)
 

Output:

-x/ln(ln(exp(1)^2/x^3/(-1+x))/exp(2*x))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=-\frac {x}{\log \left (e^{\left (-2 \, x\right )} \log \left (\frac {e^{2}}{x^{4} - x^{3}}\right )\right )} \] Input:

integrate(((1-x)*log(exp(1)^2/(x^4-x^3))*log(log(exp(1)^2/(x^4-x^3))/exp(2 
*x))+(-2*x^2+2*x)*log(exp(1)^2/(x^4-x^3))+3-4*x)/(-1+x)/log(exp(1)^2/(x^4- 
x^3))/log(log(exp(1)^2/(x^4-x^3))/exp(2*x))^2,x, algorithm="fricas")
 

Output:

-x/log(e^(-2*x)*log(e^2/(x^4 - x^3)))
 

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=- \frac {x}{\log {\left (e^{- 2 x} \log {\left (\frac {e^{2}}{x^{4} - x^{3}} \right )} \right )}} \] Input:

integrate(((1-x)*ln(exp(1)**2/(x**4-x**3))*ln(ln(exp(1)**2/(x**4-x**3))/ex 
p(2*x))+(-2*x**2+2*x)*ln(exp(1)**2/(x**4-x**3))+3-4*x)/(-1+x)/ln(exp(1)**2 
/(x**4-x**3))/ln(ln(exp(1)**2/(x**4-x**3))/exp(2*x))**2,x)
 

Output:

-x/log(exp(-2*x)*log(exp(2)/(x**4 - x**3)))
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=\frac {x}{2 \, x - \log \left (-\log \left (x - 1\right ) - 3 \, \log \left (x\right ) + 2\right )} \] Input:

integrate(((1-x)*log(exp(1)^2/(x^4-x^3))*log(log(exp(1)^2/(x^4-x^3))/exp(2 
*x))+(-2*x^2+2*x)*log(exp(1)^2/(x^4-x^3))+3-4*x)/(-1+x)/log(exp(1)^2/(x^4- 
x^3))/log(log(exp(1)^2/(x^4-x^3))/exp(2*x))^2,x, algorithm="maxima")
 

Output:

x/(2*x - log(-log(x - 1) - 3*log(x) + 2))
 

Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=\frac {x}{2 \, x - \log \left (-\log \left (x^{4} - x^{3}\right ) + 2\right )} \] Input:

integrate(((1-x)*log(exp(1)^2/(x^4-x^3))*log(log(exp(1)^2/(x^4-x^3))/exp(2 
*x))+(-2*x^2+2*x)*log(exp(1)^2/(x^4-x^3))+3-4*x)/(-1+x)/log(exp(1)^2/(x^4- 
x^3))/log(log(exp(1)^2/(x^4-x^3))/exp(2*x))^2,x, algorithm="giac")
 

Output:

x/(2*x - log(-log(x^4 - x^3) + 2))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=-\int \frac {4\,x-\ln \left (-\frac {{\mathrm {e}}^2}{x^3-x^4}\right )\,\left (2\,x-2\,x^2\right )+\ln \left ({\mathrm {e}}^{-2\,x}\,\ln \left (-\frac {{\mathrm {e}}^2}{x^3-x^4}\right )\right )\,\ln \left (-\frac {{\mathrm {e}}^2}{x^3-x^4}\right )\,\left (x-1\right )-3}{{\ln \left ({\mathrm {e}}^{-2\,x}\,\ln \left (-\frac {{\mathrm {e}}^2}{x^3-x^4}\right )\right )}^2\,\ln \left (-\frac {{\mathrm {e}}^2}{x^3-x^4}\right )\,\left (x-1\right )} \,d x \] Input:

int(-(4*x - log(-exp(2)/(x^3 - x^4))*(2*x - 2*x^2) + log(exp(-2*x)*log(-ex 
p(2)/(x^3 - x^4)))*log(-exp(2)/(x^3 - x^4))*(x - 1) - 3)/(log(exp(-2*x)*lo 
g(-exp(2)/(x^3 - x^4)))^2*log(-exp(2)/(x^3 - x^4))*(x - 1)),x)
 

Output:

-int((4*x - log(-exp(2)/(x^3 - x^4))*(2*x - 2*x^2) + log(exp(-2*x)*log(-ex 
p(2)/(x^3 - x^4)))*log(-exp(2)/(x^3 - x^4))*(x - 1) - 3)/(log(exp(-2*x)*lo 
g(-exp(2)/(x^3 - x^4)))^2*log(-exp(2)/(x^3 - x^4))*(x - 1)), x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {3-4 x+\left (2 x-2 x^2\right ) \log \left (\frac {e^2}{-x^3+x^4}\right )+(1-x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log \left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )}{(-1+x) \log \left (\frac {e^2}{-x^3+x^4}\right ) \log ^2\left (e^{-2 x} \log \left (\frac {e^2}{-x^3+x^4}\right )\right )} \, dx=-\frac {x}{\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {e^{2}}{x^{4}-x^{3}}\right )}{e^{2 x}}\right )} \] Input:

int(((1-x)*log(exp(1)^2/(x^4-x^3))*log(log(exp(1)^2/(x^4-x^3))/exp(2*x))+( 
-2*x^2+2*x)*log(exp(1)^2/(x^4-x^3))+3-4*x)/(-1+x)/log(exp(1)^2/(x^4-x^3))/ 
log(log(exp(1)^2/(x^4-x^3))/exp(2*x))^2,x)
 

Output:

( - x)/log(log(e**2/(x**4 - x**3))/e**(2*x))