\(\int \frac {(-e^{2 x}+3 e^x x-2 x^2) \log (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2})+(e^{2 x}-3 e^x x+2 x^3) \log (x) \log (\log (x))+(e^{2 x}-3 e^x x+2 x^2) \log (x) \log (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}) \log (\log (x)) \log (\frac {\log (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2})}{\log (\log (x))})}{(e^{2 x}-3 e^x x+2 x^2) \log (x) \log (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}) \log (\log (x))} \, dx\) [890]

Optimal result

Integrand size = 245, antiderivative size = 26 \[ \int \frac {\left (-e^{2 x}+3 e^x x-2 x^2\right ) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )+\left (e^{2 x}-3 e^x x+2 x^3\right ) \log (x) \log (\log (x))+\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x)) \log \left (\frac {\log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )}{\log (\log (x))}\right )}{\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x))} \, dx=x \log \left (\frac {\log \left (x-\frac {x^3}{\left (-e^x+x\right )^2}\right )}{\log (\log (x))}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {\left (-e^{2 x}+3 e^x x-2 x^2\right ) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )+\left (e^{2 x}-3 e^x x+2 x^3\right ) \log (x) \log (\log (x))+\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x)) \log \left (\frac {\log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )}{\log (\log (x))}\right )}{\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x))} \, dx=x \log \left (\frac {\log \left (\frac {e^x \left (e^x-2 x\right ) x}{\left (e^x-x\right )^2}\right )}{\log (\log (x))}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 3.76 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {7239, 2009}

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.

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

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 2009

\(\displaystyle x \log \left (\frac {\log \left (\frac {e^x \left (e^x-2 x\right ) x}{\left (e^x-x\right )^2}\right )}{\log (\log (x))}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

Maple [F]

\[\int \frac {\left ({\mathrm e}^{2 x}-3 \,{\mathrm e}^{x} x +2 x^{2}\right ) \ln \left (x \right ) \ln \left (\frac {x \,{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x^{2}}{{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x +x^{2}}\right ) \ln \left (\ln \left (x \right )\right ) \ln \left (\frac {\ln \left (\frac {x \,{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x^{2}}{{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x +x^{2}}\right )}{\ln \left (\ln \left (x \right )\right )}\right )+\left ({\mathrm e}^{2 x}-3 \,{\mathrm e}^{x} x +2 x^{3}\right ) \ln \left (x \right ) \ln \left (\ln \left (x \right )\right )+\left (-{\mathrm e}^{2 x}+3 \,{\mathrm e}^{x} x -2 x^{2}\right ) \ln \left (\frac {x \,{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x^{2}}{{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x +x^{2}}\right )}{\left ({\mathrm e}^{2 x}-3 \,{\mathrm e}^{x} x +2 x^{2}\right ) \ln \left (x \right ) \ln \left (\frac {x \,{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x^{2}}{{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x +x^{2}}\right ) \ln \left (\ln \left (x \right )\right )}d x\]

Input:

int(((exp(x)^2-3*exp(x)*x+2*x^2)*ln(x)*ln((x*exp(x)^2-2*exp(x)*x^2)/(exp(x 
)^2-2*exp(x)*x+x^2))*ln(ln(x))*ln(ln((x*exp(x)^2-2*exp(x)*x^2)/(exp(x)^2-2 
*exp(x)*x+x^2))/ln(ln(x)))+(exp(x)^2-3*exp(x)*x+2*x^3)*ln(x)*ln(ln(x))+(-e 
xp(x)^2+3*exp(x)*x-2*x^2)*ln((x*exp(x)^2-2*exp(x)*x^2)/(exp(x)^2-2*exp(x)* 
x+x^2)))/(exp(x)^2-3*exp(x)*x+2*x^2)/ln(x)/ln((x*exp(x)^2-2*exp(x)*x^2)/(e 
xp(x)^2-2*exp(x)*x+x^2))/ln(ln(x)),x)
 

Output:

int(((exp(x)^2-3*exp(x)*x+2*x^2)*ln(x)*ln((x*exp(x)^2-2*exp(x)*x^2)/(exp(x 
)^2-2*exp(x)*x+x^2))*ln(ln(x))*ln(ln((x*exp(x)^2-2*exp(x)*x^2)/(exp(x)^2-2 
*exp(x)*x+x^2))/ln(ln(x)))+(exp(x)^2-3*exp(x)*x+2*x^3)*ln(x)*ln(ln(x))+(-e 
xp(x)^2+3*exp(x)*x-2*x^2)*ln((x*exp(x)^2-2*exp(x)*x^2)/(exp(x)^2-2*exp(x)* 
x+x^2)))/(exp(x)^2-3*exp(x)*x+2*x^2)/ln(x)/ln((x*exp(x)^2-2*exp(x)*x^2)/(e 
xp(x)^2-2*exp(x)*x+x^2))/ln(ln(x)),x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {\left (-e^{2 x}+3 e^x x-2 x^2\right ) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )+\left (e^{2 x}-3 e^x x+2 x^3\right ) \log (x) \log (\log (x))+\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x)) \log \left (\frac {\log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )}{\log (\log (x))}\right )}{\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x))} \, dx=x \log \left (\frac {\log \left (-\frac {2 \, x^{2} e^{x} - x e^{\left (2 \, x\right )}}{x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}}\right )}{\log \left (\log \left (x\right )\right )}\right ) \] Input:

integrate(((exp(x)^2-3*exp(x)*x+2*x^2)*log(x)*log((x*exp(x)^2-2*exp(x)*x^2 
)/(exp(x)^2-2*exp(x)*x+x^2))*log(log(x))*log(log((x*exp(x)^2-2*exp(x)*x^2) 
/(exp(x)^2-2*exp(x)*x+x^2))/log(log(x)))+(exp(x)^2-3*exp(x)*x+2*x^3)*log(x 
)*log(log(x))+(-exp(x)^2+3*exp(x)*x-2*x^2)*log((x*exp(x)^2-2*exp(x)*x^2)/( 
exp(x)^2-2*exp(x)*x+x^2)))/(exp(x)^2-3*exp(x)*x+2*x^2)/log(x)/log((x*exp(x 
)^2-2*exp(x)*x^2)/(exp(x)^2-2*exp(x)*x+x^2))/log(log(x)),x, algorithm="fri 
cas")
 

Output:

x*log(log(-(2*x^2*e^x - x*e^(2*x))/(x^2 - 2*x*e^x + e^(2*x)))/log(log(x)))
 

Sympy [F(-1)]

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

integrate(((exp(x)**2-3*exp(x)*x+2*x**2)*ln(x)*ln((x*exp(x)**2-2*exp(x)*x* 
*2)/(exp(x)**2-2*exp(x)*x+x**2))*ln(ln(x))*ln(ln((x*exp(x)**2-2*exp(x)*x** 
2)/(exp(x)**2-2*exp(x)*x+x**2))/ln(ln(x)))+(exp(x)**2-3*exp(x)*x+2*x**3)*l 
n(x)*ln(ln(x))+(-exp(x)**2+3*exp(x)*x-2*x**2)*ln((x*exp(x)**2-2*exp(x)*x** 
2)/(exp(x)**2-2*exp(x)*x+x**2)))/(exp(x)**2-3*exp(x)*x+2*x**2)/ln(x)/ln((x 
*exp(x)**2-2*exp(x)*x**2)/(exp(x)**2-2*exp(x)*x+x**2))/ln(ln(x)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {\left (-e^{2 x}+3 e^x x-2 x^2\right ) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )+\left (e^{2 x}-3 e^x x+2 x^3\right ) \log (x) \log (\log (x))+\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x)) \log \left (\frac {\log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )}{\log (\log (x))}\right )}{\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x))} \, dx=x \log \left (x + \log \left (x\right ) - 2 \, \log \left (-x + e^{x}\right ) + \log \left (-2 \, x + e^{x}\right )\right ) - x \log \left (\log \left (\log \left (x\right )\right )\right ) \] Input:

integrate(((exp(x)^2-3*exp(x)*x+2*x^2)*log(x)*log((x*exp(x)^2-2*exp(x)*x^2 
)/(exp(x)^2-2*exp(x)*x+x^2))*log(log(x))*log(log((x*exp(x)^2-2*exp(x)*x^2) 
/(exp(x)^2-2*exp(x)*x+x^2))/log(log(x)))+(exp(x)^2-3*exp(x)*x+2*x^3)*log(x 
)*log(log(x))+(-exp(x)^2+3*exp(x)*x-2*x^2)*log((x*exp(x)^2-2*exp(x)*x^2)/( 
exp(x)^2-2*exp(x)*x+x^2)))/(exp(x)^2-3*exp(x)*x+2*x^2)/log(x)/log((x*exp(x 
)^2-2*exp(x)*x^2)/(exp(x)^2-2*exp(x)*x+x^2))/log(log(x)),x, algorithm="max 
ima")
 

Output:

x*log(x + log(x) - 2*log(-x + e^x) + log(-2*x + e^x)) - x*log(log(log(x)))
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (-e^{2 x}+3 e^x x-2 x^2\right ) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )+\left (e^{2 x}-3 e^x x+2 x^3\right ) \log (x) \log (\log (x))+\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x)) \log \left (\frac {\log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )}{\log (\log (x))}\right )}{\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x))} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((exp(x)^2-3*exp(x)*x+2*x^2)*log(x)*log((x*exp(x)^2-2*exp(x)*x^2 
)/(exp(x)^2-2*exp(x)*x+x^2))*log(log(x))*log(log((x*exp(x)^2-2*exp(x)*x^2) 
/(exp(x)^2-2*exp(x)*x+x^2))/log(log(x)))+(exp(x)^2-3*exp(x)*x+2*x^3)*log(x 
)*log(log(x))+(-exp(x)^2+3*exp(x)*x-2*x^2)*log((x*exp(x)^2-2*exp(x)*x^2)/( 
exp(x)^2-2*exp(x)*x+x^2)))/(exp(x)^2-3*exp(x)*x+2*x^2)/log(x)/log((x*exp(x 
)^2-2*exp(x)*x^2)/(exp(x)^2-2*exp(x)*x+x^2))/log(log(x)),x, algorithm="gia 
c")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Not invertible Error: Bad Argument 
Value
 

Mupad [B] (verification not implemented)

Time = 3.77 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {\left (-e^{2 x}+3 e^x x-2 x^2\right ) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )+\left (e^{2 x}-3 e^x x+2 x^3\right ) \log (x) \log (\log (x))+\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x)) \log \left (\frac {\log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )}{\log (\log (x))}\right )}{\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x))} \, dx=x\,\ln \left (\frac {\ln \left (\frac {x\,{\mathrm {e}}^{2\,x}-2\,x^2\,{\mathrm {e}}^x}{{\mathrm {e}}^{2\,x}-2\,x\,{\mathrm {e}}^x+x^2}\right )}{\ln \left (\ln \left (x\right )\right )}\right ) \] Input:

int((log(log(x))*log(x)*(exp(2*x) - 3*x*exp(x) + 2*x^3) - log((x*exp(2*x) 
- 2*x^2*exp(x))/(exp(2*x) - 2*x*exp(x) + x^2))*(exp(2*x) - 3*x*exp(x) + 2* 
x^2) + log(log(x))*log(log((x*exp(2*x) - 2*x^2*exp(x))/(exp(2*x) - 2*x*exp 
(x) + x^2))/log(log(x)))*log((x*exp(2*x) - 2*x^2*exp(x))/(exp(2*x) - 2*x*e 
xp(x) + x^2))*log(x)*(exp(2*x) - 3*x*exp(x) + 2*x^2))/(log(log(x))*log((x* 
exp(2*x) - 2*x^2*exp(x))/(exp(2*x) - 2*x*exp(x) + x^2))*log(x)*(exp(2*x) - 
 3*x*exp(x) + 2*x^2)),x)
 

Output:

x*log(log((x*exp(2*x) - 2*x^2*exp(x))/(exp(2*x) - 2*x*exp(x) + x^2))/log(l 
og(x)))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {\left (-e^{2 x}+3 e^x x-2 x^2\right ) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )+\left (e^{2 x}-3 e^x x+2 x^3\right ) \log (x) \log (\log (x))+\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x)) \log \left (\frac {\log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right )}{\log (\log (x))}\right )}{\left (e^{2 x}-3 e^x x+2 x^2\right ) \log (x) \log \left (\frac {e^{2 x} x-2 e^x x^2}{e^{2 x}-2 e^x x+x^2}\right ) \log (\log (x))} \, dx=\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {e^{2 x} x -2 e^{x} x^{2}}{e^{2 x}-2 e^{x} x +x^{2}}\right )}{\mathrm {log}\left (\mathrm {log}\left (x \right )\right )}\right ) x \] Input:

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

Output:

log(log((e**(2*x)*x - 2*e**x*x**2)/(e**(2*x) - 2*e**x*x + x**2))/log(log(x 
)))*x