\(\int \frac {(\frac {e^x}{2+e^x+x})^{\frac {2 e^{-3+x^2}}{x}} (e^{-3+x^2} (2 x+2 x^2)+e^{-3+x^2} (-4-2 x+8 x^2+4 x^3+e^x (-2+4 x^2)) \log (\frac {e^x}{2+e^x+x}))}{2 x^2+e^x x^2+x^3} \, dx\) [1754]

Optimal result

Integrand size = 109, antiderivative size = 25 \[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx=\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \] Output:

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

Mathematica [A] (verified)

Time = 5.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx=\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \] Input:

Integrate[((E^x/(2 + E^x + x))^((2*E^(-3 + x^2))/x)*(E^(-3 + x^2)*(2*x + 2 
*x^2) + E^(-3 + x^2)*(-4 - 2*x + 8*x^2 + 4*x^3 + E^x*(-2 + 4*x^2))*Log[E^x 
/(2 + E^x + x)]))/(2*x^2 + E^x*x^2 + x^3),x]
 

Output:

(E^x/(2 + E^x + x))^((2*E^(-3 + 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.

\(\displaystyle \int \frac {\left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \left (e^{x^2-3} \left (2 x^2+2 x\right )+e^{x^2-3} \left (4 x^3+8 x^2+e^x \left (4 x^2-2\right )-2 x-4\right ) \log \left (\frac {e^x}{x+e^x+2}\right )\right )}{x^3+e^x x^2+2 x^2} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x+e^x+2}+\frac {2 e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x \left (x+e^x+2\right )}+\frac {8 e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \log \left (\frac {e^x}{x+e^x+2}\right )}{x+e^x+2}+\frac {4 e^{x^2+x-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \log \left (\frac {e^x}{x+e^x+2}\right )}{x+e^x+2}+\frac {4 e^{x^2-3} x \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \log \left (\frac {e^x}{x+e^x+2}\right )}{x+e^x+2}-\frac {2 e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \log \left (\frac {e^x}{x+e^x+2}\right )}{x \left (x+e^x+2\right )}-\frac {4 e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \log \left (\frac {e^x}{x+e^x+2}\right )}{x^2 \left (x+e^x+2\right )}-\frac {2 e^{x^2+x-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}} \log \left (\frac {e^x}{x+e^x+2}\right )}{x^2 \left (x+e^x+2\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 8 \log \left (\frac {e^x}{x+e^x+2}\right ) \int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x+e^x+2}dx+2 \int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x+e^x+2}dx+4 \log \left (\frac {e^x}{x+e^x+2}\right ) \int \frac {e^{x^2+x-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x+e^x+2}dx-4 \log \left (\frac {e^x}{x+e^x+2}\right ) \int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x^2 \left (x+e^x+2\right )}dx-2 \log \left (\frac {e^x}{x+e^x+2}\right ) \int \frac {e^{x^2+x-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x^2 \left (x+e^x+2\right )}dx-2 \log \left (\frac {e^x}{x+e^x+2}\right ) \int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x \left (x+e^x+2\right )}dx+2 \int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x \left (x+e^x+2\right )}dx+4 \log \left (\frac {e^x}{x+e^x+2}\right ) \int \frac {e^{x^2-3} x \left (\frac {e^x}{x+e^x+2}\right )^{\frac {2 e^{x^2-3}}{x}}}{x+e^x+2}dx-4 \int \frac {\int e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}dx}{x+e^x+2}dx-4 \int \frac {x \int e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}dx}{x+e^x+2}dx-8 \int \frac {\int e^{x^2-x-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}dx}{x+e^x+2}dx-8 \int \frac {x \int e^{x^2-x-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}dx}{x+e^x+2}dx+2 \int \frac {\int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}}{x^2}dx}{x+e^x+2}dx+2 \int \frac {x \int \frac {e^{x^2-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}}{x^2}dx}{x+e^x+2}dx+4 \int \frac {\int \frac {e^{x^2-x-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}}{x^2}dx}{x+e^x+2}dx+4 \int \frac {x \int \frac {e^{x^2-x-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}}{x^2}dx}{x+e^x+2}dx+2 \int \frac {\int \frac {e^{x^2-x-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}}{x}dx}{x+e^x+2}dx+2 \int \frac {x \int \frac {e^{x^2-x-3} \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}}{x}dx}{x+e^x+2}dx-4 \int \frac {\int e^{x^2-x-3} x \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}dx}{x+e^x+2}dx-4 \int \frac {x \int e^{x^2-x-3} x \left (\frac {e^x}{x+e^x+2}\right )^{1+\frac {2 e^{x^2-3}}{x}}dx}{x+e^x+2}dx\)

Input:

Int[((E^x/(2 + E^x + x))^((2*E^(-3 + x^2))/x)*(E^(-3 + x^2)*(2*x + 2*x^2) 
+ E^(-3 + x^2)*(-4 - 2*x + 8*x^2 + 4*x^3 + E^x*(-2 + 4*x^2))*Log[E^x/(2 + 
E^x + x)]))/(2*x^2 + E^x*x^2 + x^3),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 58.43 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\({\mathrm e}^{\frac {2 \,{\mathrm e}^{x^{2}-3} \ln \left (\frac {{\mathrm e}^{x}}{{\mathrm e}^{x}+2+x}\right )}{x}}\)	\(25\)
risch	\(\left ({\mathrm e}^{x}+2+x \right )^{-\frac {2 \,{\mathrm e}^{x^{2}-3}}{x}} \left ({\mathrm e}^{x}\right )^{\frac {2 \,{\mathrm e}^{x^{2}-3}}{x}} {\mathrm e}^{-\frac {i {\mathrm e}^{x^{2}-3} \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}+2+x}\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}+2+x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}+2+x}\right )+\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}+2+x}\right )\right )}{x}}\)	\(112\)

Input:

int((((4*x^2-2)*exp(x)+4*x^3+8*x^2-2*x-4)*exp(x^2-3)*ln(exp(x)/(exp(x)+2+x 
))+(2*x^2+2*x)*exp(x^2-3))*exp(exp(x^2-3)*ln(exp(x)/(exp(x)+2+x))/x)^2/(ex 
p(x)*x^2+x^3+2*x^2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx=\left (\frac {e^{x}}{x + e^{x} + 2}\right )^{\frac {2 \, e^{\left (x^{2} - 3\right )}}{x}} \] Input:

integrate((((4*x^2-2)*exp(x)+4*x^3+8*x^2-2*x-4)*exp(x^2-3)*log(exp(x)/(exp 
(x)+2+x))+(2*x^2+2*x)*exp(x^2-3))*exp(exp(x^2-3)*log(exp(x)/(exp(x)+2+x))/ 
x)^2/(exp(x)*x^2+x^3+2*x^2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx=\text {Timed out} \] Input:

integrate((((4*x**2-2)*exp(x)+4*x**3+8*x**2-2*x-4)*exp(x**2-3)*ln(exp(x)/( 
exp(x)+2+x))+(2*x**2+2*x)*exp(x**2-3))*exp(exp(x**2-3)*ln(exp(x)/(exp(x)+2 
+x))/x)**2/(exp(x)*x**2+x**3+2*x**2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx=e^{\left (-\frac {2 \, e^{\left (x^{2} - 3\right )} \log \left (x + e^{x} + 2\right )}{x} + 2 \, e^{\left (x^{2} - 3\right )}\right )} \] Input:

integrate((((4*x^2-2)*exp(x)+4*x^3+8*x^2-2*x-4)*exp(x^2-3)*log(exp(x)/(exp 
(x)+2+x))+(2*x^2+2*x)*exp(x^2-3))*exp(exp(x^2-3)*log(exp(x)/(exp(x)+2+x))/ 
x)^2/(exp(x)*x^2+x^3+2*x^2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx=\int { \frac {2 \, {\left ({\left (2 \, x^{3} + 4 \, x^{2} + {\left (2 \, x^{2} - 1\right )} e^{x} - x - 2\right )} e^{\left (x^{2} - 3\right )} \log \left (\frac {e^{x}}{x + e^{x} + 2}\right ) + {\left (x^{2} + x\right )} e^{\left (x^{2} - 3\right )}\right )} \left (\frac {e^{x}}{x + e^{x} + 2}\right )^{\frac {2 \, e^{\left (x^{2} - 3\right )}}{x}}}{x^{3} + x^{2} e^{x} + 2 \, x^{2}} \,d x } \] Input:

integrate((((4*x^2-2)*exp(x)+4*x^3+8*x^2-2*x-4)*exp(x^2-3)*log(exp(x)/(exp 
(x)+2+x))+(2*x^2+2*x)*exp(x^2-3))*exp(exp(x^2-3)*log(exp(x)/(exp(x)+2+x))/ 
x)^2/(exp(x)*x^2+x^3+2*x^2),x, algorithm="giac")
 

Output:

integrate(2*((2*x^3 + 4*x^2 + (2*x^2 - 1)*e^x - x - 2)*e^(x^2 - 3)*log(e^x 
/(x + e^x + 2)) + (x^2 + x)*e^(x^2 - 3))*(e^x/(x + e^x + 2))^(2*e^(x^2 - 3 
)/x)/(x^3 + x^2*e^x + 2*x^2), x)
 

Mupad [B] (verification not implemented)

Time = 3.97 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx={\mathrm {e}}^{2\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-3}}\,{\left (\frac {1}{x+{\mathrm {e}}^x+2}\right )}^{\frac {2\,{\mathrm {e}}^{x^2-3}}{x}} \] Input:

int((exp((2*exp(x^2 - 3)*log(exp(x)/(x + exp(x) + 2)))/x)*(exp(x^2 - 3)*(2 
*x + 2*x^2) + exp(x^2 - 3)*log(exp(x)/(x + exp(x) + 2))*(exp(x)*(4*x^2 - 2 
) - 2*x + 8*x^2 + 4*x^3 - 4)))/(x^2*exp(x) + 2*x^2 + x^3),x)
 

Output:

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

Reduce [F]

\[ \int \frac {\left (\frac {e^x}{2+e^x+x}\right )^{\frac {2 e^{-3+x^2}}{x}} \left (e^{-3+x^2} \left (2 x+2 x^2\right )+e^{-3+x^2} \left (-4-2 x+8 x^2+4 x^3+e^x \left (-2+4 x^2\right )\right ) \log \left (\frac {e^x}{2+e^x+x}\right )\right )}{2 x^2+e^x x^2+x^3} \, dx=\int \frac {\left (\left (\left (4 x^{2}-2\right ) {\mathrm e}^{x}+4 x^{3}+8 x^{2}-2 x -4\right ) {\mathrm e}^{x^{2}-3} \mathrm {log}\left (\frac {{\mathrm e}^{x}}{{\mathrm e}^{x}+2+x}\right )+\left (2 x^{2}+2 x \right ) {\mathrm e}^{x^{2}-3}\right ) \left ({\mathrm e}^{\frac {{\mathrm e}^{x^{2}-3} \mathrm {log}\left (\frac {{\mathrm e}^{x}}{{\mathrm e}^{x}+2+x}\right )}{x}}\right )^{2}}{{\mathrm e}^{x} x^{2}+x^{3}+2 x^{2}}d x \] Input:

int((((4*x^2-2)*exp(x)+4*x^3+8*x^2-2*x-4)*exp(x^2-3)*log(exp(x)/(exp(x)+2+ 
x))+(2*x^2+2*x)*exp(x^2-3))*exp(exp(x^2-3)*log(exp(x)/(exp(x)+2+x))/x)^2/( 
exp(x)*x^2+x^3+2*x^2),x)
 

Output:

int((((4*x^2-2)*exp(x)+4*x^3+8*x^2-2*x-4)*exp(x^2-3)*log(exp(x)/(exp(x)+2+ 
x))+(2*x^2+2*x)*exp(x^2-3))*exp(exp(x^2-3)*log(exp(x)/(exp(x)+2+x))/x)^2/( 
exp(x)*x^2+x^3+2*x^2),x)