\(\int e^{1-e^x+e^{1-e^x+x+x \log (2 e^{2 x} x^2)}+x+x \log (2 e^{2 x} x^2)} (3-e^x+2 x+\log (2 e^{2 x} x^2)) \, dx\) [1744]

Optimal result

Integrand size = 68, antiderivative size = 25 \[ \int e^{1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+x+x \log \left (2 e^{2 x} x^2\right )} \left (3-e^x+2 x+\log \left (2 e^{2 x} x^2\right )\right ) \, dx=e^{e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}} \] Output:

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

Mathematica [A] (verified)

Time = 5.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int e^{1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+x+x \log \left (2 e^{2 x} x^2\right )} \left (3-e^x+2 x+\log \left (2 e^{2 x} x^2\right )\right ) \, dx=e^{2^x e^{1-e^x+x} \left (e^{2 x} x^2\right )^x} \] Input:

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

Output:

E^(2^x*E^(1 - E^x + x)*(E^(2*x)*x^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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\({\mathrm e}^{{\mathrm e}^{x \ln \left (2 \,{\mathrm e}^{2 x} x^{2}\right )+1-{\mathrm e}^{x}+x}}\)	\(22\)
default	\({\mathrm e}^{{\mathrm e}^{x \ln \left (2 \,{\mathrm e}^{2 x} x^{2}\right )+1-{\mathrm e}^{x}+x}}\)	\(22\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{x \ln \left (2 \,{\mathrm e}^{2 x} x^{2}\right )+1-{\mathrm e}^{x}+x}}\)	\(22\)
risch	\({\mathrm e}^{2^{x} x^{2 x} \left ({\mathrm e}^{x}\right )^{2 x} {\mathrm e}^{1+\frac {i x \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )^{2} \pi }{2}-\frac {i x \operatorname {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )^{3} \pi }{2}+i x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} \pi +\frac {i x \operatorname {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )^{2} \operatorname {csgn}\left (i x^{2}\right ) \pi }{2}-\frac {i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}-\frac {i x \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3} \pi }{2}-\frac {i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}-\frac {i x \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i x^{2}\right ) \pi }{2}-\frac {i x \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2} \pi }{2}+i x \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \pi -{\mathrm e}^{x}+x}}\)	\(234\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int e^{1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+x+x \log \left (2 e^{2 x} x^2\right )} \left (3-e^x+2 x+\log \left (2 e^{2 x} x^2\right )\right ) \, dx=e^{\left (e^{\left (x \log \left (2 \, x^{2} e^{\left (2 \, x\right )}\right ) + x - e^{x} + 1\right )}\right )} \] Input:

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

Output:

e^(e^(x*log(2*x^2*e^(2*x)) + x - e^x + 1))
 

Sympy [F(-1)]

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

integrate((ln(2*exp(2*x)*x**2)-exp(x)+2*x+3)*exp(x*ln(2*exp(2*x)*x**2)+1-e 
xp(x)+x)*exp(exp(x*ln(2*exp(2*x)*x**2)+1-exp(x)+x)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int e^{1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+x+x \log \left (2 e^{2 x} x^2\right )} \left (3-e^x+2 x+\log \left (2 e^{2 x} x^2\right )\right ) \, dx=e^{\left (e^{\left (2 \, x^{2} + x \log \left (2\right ) + 2 \, x \log \left (x\right ) + x - e^{x} + 1\right )}\right )} \] Input:

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

Output:

e^(e^(2*x^2 + x*log(2) + 2*x*log(x) + x - e^x + 1))
 

Giac [F]

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

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

Output:

integrate((2*x - e^x + log(2*x^2*e^(2*x)) + 3)*e^(x*log(2*x^2*e^(2*x)) + x 
 + e^(x*log(2*x^2*e^(2*x)) + x - e^x + 1) - e^x + 1), x)
 

Mupad [B] (verification not implemented)

Time = 3.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int e^{1-e^x+e^{1-e^x+x+x \log \left (2 e^{2 x} x^2\right )}+x+x \log \left (2 e^{2 x} x^2\right )} \left (3-e^x+2 x+\log \left (2 e^{2 x} x^2\right )\right ) \, dx={\mathrm {e}}^{\mathrm {e}\,{\mathrm {e}}^{2\,x^2}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\mathrm {e}}^x\,{\left (2\,x^2\right )}^x} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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