\(\int \frac {e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x-2 x \log (3)-4 e^{2-x^2} x^2 \log (3)+(1+e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x) \log (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x))}{(x+e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x^2) \log (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x)) \log (x \log (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x)))} \, dx\) [2043]

Optimal result

Integrand size = 255, antiderivative size = 28 \[ \int \frac {e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x-2 x \log (3)-4 e^{2-x^2} x^2 \log (3)+\left (1+e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right ) \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right )}{\left (x+e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x^2\right ) \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right ) \log \left (x \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right )\right )} \, dx=\log \left (\log \left (x \log \left (-3^{2 e^{2-x^2}-2 x}-x\right )\right )\right ) \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 1.95 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {7239, 7235}

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.

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

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\frac {e^{-x^2} \left (e^{x^2} \left (9^x-9^{e^{2-x^2}} \log (9)\right )-4 e^2 9^{e^{2-x^2}} x \log (3)\right )}{\left (9^{e^{2-x^2}}+9^x x\right ) \log \left (-9^{e^{2-x^2}-x}-x\right )}+\frac {1}{x}}{\log \left (x \log \left (-9^{e^{2-x^2}-x}-x\right )\right )}dx\)

\(\Big \downarrow \) 7235

\(\displaystyle \log \left (\log \left (x \log \left (-9^{e^{2-x^2}-x}-x\right )\right )\right )\)

Input:

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

Output:

Log[Log[x*Log[-9^(E^(2 - x^2) - x) - x]]]
 

Defintions of rubi rules used

rule 7235

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L 
og[RemoveContent[y, x]], x] /;  !FalseQ[q]]
 

rule 7239

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

Maple [F]

\[\int \frac {\left (x \,{\mathrm e}^{-2 \ln \left (3\right ) {\mathrm e}^{-x^{2}+2}+2 x \ln \left (3\right )}+1\right ) \ln \left (\left (-x \,{\mathrm e}^{-2 \ln \left (3\right ) {\mathrm e}^{-x^{2}+2}+2 x \ln \left (3\right )}-1\right ) {\mathrm e}^{2 \ln \left (3\right ) {\mathrm e}^{-x^{2}+2}-2 x \ln \left (3\right )}\right )+x \,{\mathrm e}^{-2 \ln \left (3\right ) {\mathrm e}^{-x^{2}+2}+2 x \ln \left (3\right )}-4 x^{2} \ln \left (3\right ) {\mathrm e}^{-x^{2}+2}-2 x \ln \left (3\right )}{\left (x^{2} {\mathrm e}^{-2 \ln \left (3\right ) {\mathrm e}^{-x^{2}+2}+2 x \ln \left (3\right )}+x \right ) \ln \left (\left (-x \,{\mathrm e}^{-2 \ln \left (3\right ) {\mathrm e}^{-x^{2}+2}+2 x \ln \left (3\right )}-1\right ) {\mathrm e}^{2 \ln \left (3\right ) {\mathrm e}^{-x^{2}+2}-2 x \ln \left (3\right )}\right ) \ln \left (x \ln \left (\left (-x \,{\mathrm e}^{-2 \ln \left (3\right ) {\mathrm e}^{-x^{2}+2}+2 x \ln \left (3\right )}-1\right ) {\mathrm e}^{2 \ln \left (3\right ) {\mathrm e}^{-x^{2}+2}-2 x \ln \left (3\right )}\right )\right )}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x-2 x \log (3)-4 e^{2-x^2} x^2 \log (3)+\left (1+e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right ) \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right )}{\left (x+e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x^2\right ) \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right ) \log \left (x \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right )\right )} \, dx=\log \left (\log \left (x \log \left (-x - e^{\left (-2 \, x \log \left (3\right ) + 2 \, e^{\left (-x^{2} + 2\right )} \log \left (3\right )\right )}\right )\right )\right ) \] Input:

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

Output:

log(log(x*log(-x - e^(-2*x*log(3) + 2*e^(-x^2 + 2)*log(3)))))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x-2 x \log (3)-4 e^{2-x^2} x^2 \log (3)+\left (1+e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right ) \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right )}{\left (x+e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x^2\right ) \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right ) \log \left (x \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right )\right )} \, dx=\log \left (\log \left (-2 \, x \log \left (3\right ) + \log \left (-3^{2 \, x} x - 3^{2 \, e^{\left (-x^{2} + 2\right )}}\right )\right ) + \log \left (x\right )\right ) \] Input:

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

Output:

log(log(-2*x*log(3) + log(-3^(2*x)*x - 3^(2*e^(-x^2 + 2)))) + log(x))
 

Giac [F]

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

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

Output:

sage0*x
 

Mupad [B] (verification not implemented)

Time = 3.72 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x-2 x \log (3)-4 e^{2-x^2} x^2 \log (3)+\left (1+e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right ) \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right )}{\left (x+e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x^2\right ) \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right ) \log \left (x \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right )\right )} \, dx=\ln \left (\ln \left (x\,\ln \left (-3^{2\,x}\,x-3^{2\,{\mathrm {e}}^2\,{\mathrm {e}}^{-x^2}}\right )-2\,x^2\,\ln \left (3\right )\right )\right ) \] Input:

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

Output:

log(log(x*log(- 3^(2*x)*x - 3^(2*exp(2)*exp(-x^2))) - 2*x^2*log(3)))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x-2 x \log (3)-4 e^{2-x^2} x^2 \log (3)+\left (1+e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right ) \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right )}{\left (x+e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x^2\right ) \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right ) \log \left (x \log \left (e^{2 e^{2-x^2} \log (3)-2 x \log (3)} \left (-1-e^{-2 e^{2-x^2} \log (3)+2 x \log (3)} x\right )\right )\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\frac {-e^{\frac {2 \,\mathrm {log}\left (3\right ) e^{2}}{e^{x^{2}}}}-3^{2 x} x}{3^{2 x}}\right ) x \right )\right ) \] Input:

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

Output:

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