\(\int \frac {e^{3+e^{2 x^2}-2 e^{x^2} \log (\frac {5}{3+e^x})+\log ^2(\frac {5}{3+e^x})} (3 e^x+e^{2 x}+2 e^{2 x+x^2}+e^{2 x^2} (12 e^x x+4 e^{2 x} x)+(-2 e^{2 x}+e^{x^2} (-12 e^x x-4 e^{2 x} x)) \log (\frac {5}{3+e^x}))}{12+4 e^x} \, dx\) [100]

Optimal result

Integrand size = 135, antiderivative size = 29 \[ \int \frac {e^{3+e^{2 x^2}-2 e^{x^2} \log \left (\frac {5}{3+e^x}\right )+\log ^2\left (\frac {5}{3+e^x}\right )} \left (3 e^x+e^{2 x}+2 e^{2 x+x^2}+e^{2 x^2} \left (12 e^x x+4 e^{2 x} x\right )+\left (-2 e^{2 x}+e^{x^2} \left (-12 e^x x-4 e^{2 x} x\right )\right ) \log \left (\frac {5}{3+e^x}\right )\right )}{12+4 e^x} \, dx=\frac {1}{4} e^{3+x+\left (-e^{x^2}+\log \left (\frac {5}{3+e^x}\right )\right )^2} \] Output:

1/4*exp(x)*exp(3+(ln(5/(3+exp(x)))-exp(x^2))^2)
 

Mathematica [F]

\[ \int \frac {e^{3+e^{2 x^2}-2 e^{x^2} \log \left (\frac {5}{3+e^x}\right )+\log ^2\left (\frac {5}{3+e^x}\right )} \left (3 e^x+e^{2 x}+2 e^{2 x+x^2}+e^{2 x^2} \left (12 e^x x+4 e^{2 x} x\right )+\left (-2 e^{2 x}+e^{x^2} \left (-12 e^x x-4 e^{2 x} x\right )\right ) \log \left (\frac {5}{3+e^x}\right )\right )}{12+4 e^x} \, dx=\int \frac {e^{3+e^{2 x^2}-2 e^{x^2} \log \left (\frac {5}{3+e^x}\right )+\log ^2\left (\frac {5}{3+e^x}\right )} \left (3 e^x+e^{2 x}+2 e^{2 x+x^2}+e^{2 x^2} \left (12 e^x x+4 e^{2 x} x\right )+\left (-2 e^{2 x}+e^{x^2} \left (-12 e^x x-4 e^{2 x} x\right )\right ) \log \left (\frac {5}{3+e^x}\right )\right )}{12+4 e^x} \, dx \] Input:

Integrate[(E^(3 + E^(2*x^2) - 2*E^x^2*Log[5/(3 + E^x)] + Log[5/(3 + E^x)]^ 
2)*(3*E^x + E^(2*x) + 2*E^(2*x + x^2) + E^(2*x^2)*(12*E^x*x + 4*E^(2*x)*x) 
 + (-2*E^(2*x) + E^x^2*(-12*E^x*x - 4*E^(2*x)*x))*Log[5/(3 + E^x)]))/(12 + 
 4*E^x),x]
 

Output:

Integrate[(E^(3 + E^(2*x^2) - 2*E^x^2*Log[5/(3 + E^x)] + Log[5/(3 + E^x)]^ 
2)*(3*E^x + E^(2*x) + 2*E^(2*x + x^2) + E^(2*x^2)*(12*E^x*x + 4*E^(2*x)*x) 
 + (-2*E^(2*x) + E^x^2*(-12*E^x*x - 4*E^(2*x)*x))*Log[5/(3 + E^x)]))/(12 + 
 4*E^x), x]
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(193\) vs. \(2(29)=58\).

Time = 3.92 (sec) , antiderivative size = 193, normalized size of antiderivative = 6.66, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.007, Rules used = {2726}

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.

\(\displaystyle \int \frac {\left (e^{2 x^2} \left (12 e^x x+4 e^{2 x} x\right )+2 e^{x^2+2 x}+\left (e^{x^2} \left (-12 e^x x-4 e^{2 x} x\right )-2 e^{2 x}\right ) \log \left (\frac {5}{e^x+3}\right )+3 e^x+e^{2 x}\right ) \exp \left (e^{2 x^2}-2 e^{x^2} \log \left (\frac {5}{e^x+3}\right )+\log ^2\left (\frac {5}{e^x+3}\right )+3\right )}{4 e^x+12} \, dx\)

\(\Big \downarrow \) 2726

\(\displaystyle \frac {5^{-2 e^{x^2}} \left (\frac {1}{e^x+3}\right )^{1-2 e^{x^2}} e^{e^{2 x^2}+\log ^2\left (\frac {5}{e^x+3}\right )+3} \left (2 e^{2 x^2} \left (3 e^x x+e^{2 x} x\right )+e^{x^2+2 x}-\left (2 e^{x^2} \left (3 e^x x+e^{2 x} x\right )+e^{2 x}\right ) \log \left (\frac {5}{e^x+3}\right )\right )}{4 \left (2 e^{2 x^2} x+\frac {e^{x^2+x}}{e^x+3}-2 e^{x^2} x \log \left (\frac {5}{e^x+3}\right )-\frac {e^x \log \left (\frac {5}{e^x+3}\right )}{e^x+3}\right )}\)

Input:

Int[(E^(3 + E^(2*x^2) - 2*E^x^2*Log[5/(3 + E^x)] + Log[5/(3 + E^x)]^2)*(3* 
E^x + E^(2*x) + 2*E^(2*x + x^2) + E^(2*x^2)*(12*E^x*x + 4*E^(2*x)*x) + (-2 
*E^(2*x) + E^x^2*(-12*E^x*x - 4*E^(2*x)*x))*Log[5/(3 + E^x)]))/(12 + 4*E^x 
),x]
 

Output:

(E^(3 + E^(2*x^2) + Log[5/(3 + E^x)]^2)*((3 + E^x)^(-1))^(1 - 2*E^x^2)*(E^ 
(2*x + x^2) + 2*E^(2*x^2)*(3*E^x*x + E^(2*x)*x) - (E^(2*x) + 2*E^x^2*(3*E^ 
x*x + E^(2*x)*x))*Log[5/(3 + E^x)]))/(4*5^(2*E^x^2)*(E^(x + x^2)/(3 + E^x) 
 + 2*E^(2*x^2)*x - (E^x*Log[5/(3 + E^x)])/(3 + E^x) - 2*E^x^2*x*Log[5/(3 + 
 E^x)]))
 

Defintions of rubi rules used

rule 2726

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, 
 x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
 

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{\ln \left (\frac {5}{3+{\mathrm e}^{x}}\right )^{2}-2 \,{\mathrm e}^{x^{2}} \ln \left (\frac {5}{3+{\mathrm e}^{x}}\right )+{\mathrm e}^{2 x^{2}}+3} {\mathrm e}^{x}}{4}\)	\(40\)
risch	\(\frac {\left (3+{\mathrm e}^{x}\right )^{-2 \ln \left (5\right )} \left (3+{\mathrm e}^{x}\right )^{2 \,{\mathrm e}^{x^{2}}} \left (\frac {1}{25}\right )^{{\mathrm e}^{x^{2}}} {\mathrm e}^{x +3+\ln \left (5\right )^{2}+\ln \left (3+{\mathrm e}^{x}\right )^{2}+{\mathrm e}^{2 x^{2}}}}{4}\)	\(50\)

Input:

int((((-4*x*exp(x)^2-12*exp(x)*x)*exp(x^2)-2*exp(x)^2)*ln(5/(3+exp(x)))+(4 
*x*exp(x)^2+12*exp(x)*x)*exp(x^2)^2+2*exp(x)^2*exp(x^2)+exp(x)^2+3*exp(x)) 
*exp(ln(5/(3+exp(x)))^2-2*exp(x^2)*ln(5/(3+exp(x)))+exp(x^2)^2+3)/(4*exp(x 
)+12),x,method=_RETURNVERBOSE)
 

Output:

1/4*exp(ln(5/(3+exp(x)))^2-2*exp(x^2)*ln(5/(3+exp(x)))+exp(x^2)^2+3)*exp(x 
)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (24) = 48\).

Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.14 \[ \int \frac {e^{3+e^{2 x^2}-2 e^{x^2} \log \left (\frac {5}{3+e^x}\right )+\log ^2\left (\frac {5}{3+e^x}\right )} \left (3 e^x+e^{2 x}+2 e^{2 x+x^2}+e^{2 x^2} \left (12 e^x x+4 e^{2 x} x\right )+\left (-2 e^{2 x}+e^{x^2} \left (-12 e^x x-4 e^{2 x} x\right )\right ) \log \left (\frac {5}{3+e^x}\right )\right )}{12+4 e^x} \, dx=\frac {1}{4} \, e^{\left ({\left (e^{\left (4 \, x\right )} \log \left (\frac {5}{e^{x} + 3}\right )^{2} - 2 \, e^{\left (x^{2} + 4 \, x\right )} \log \left (\frac {5}{e^{x} + 3}\right ) + e^{\left (2 \, x^{2} + 4 \, x\right )} + 3 \, e^{\left (4 \, x\right )}\right )} e^{\left (-4 \, x\right )} + x\right )} \] Input:

integrate((((-4*x*exp(x)^2-12*exp(x)*x)*exp(x^2)-2*exp(x)^2)*log(5/(3+exp( 
x)))+(4*x*exp(x)^2+12*exp(x)*x)*exp(x^2)^2+2*exp(x)^2*exp(x^2)+exp(x)^2+3* 
exp(x))*exp(log(5/(3+exp(x)))^2-2*exp(x^2)*log(5/(3+exp(x)))+exp(x^2)^2+3) 
/(4*exp(x)+12),x, algorithm="fricas")
 

Output:

1/4*e^((e^(4*x)*log(5/(e^x + 3))^2 - 2*e^(x^2 + 4*x)*log(5/(e^x + 3)) + e^ 
(2*x^2 + 4*x) + 3*e^(4*x))*e^(-4*x) + x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{3+e^{2 x^2}-2 e^{x^2} \log \left (\frac {5}{3+e^x}\right )+\log ^2\left (\frac {5}{3+e^x}\right )} \left (3 e^x+e^{2 x}+2 e^{2 x+x^2}+e^{2 x^2} \left (12 e^x x+4 e^{2 x} x\right )+\left (-2 e^{2 x}+e^{x^2} \left (-12 e^x x-4 e^{2 x} x\right )\right ) \log \left (\frac {5}{3+e^x}\right )\right )}{12+4 e^x} \, dx=\text {Timed out} \] Input:

integrate((((-4*x*exp(x)**2-12*exp(x)*x)*exp(x**2)-2*exp(x)**2)*ln(5/(3+ex 
p(x)))+(4*x*exp(x)**2+12*exp(x)*x)*exp(x**2)**2+2*exp(x)**2*exp(x**2)+exp( 
x)**2+3*exp(x))*exp(ln(5/(3+exp(x)))**2-2*exp(x**2)*ln(5/(3+exp(x)))+exp(x 
**2)**2+3)/(4*exp(x)+12),x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (24) = 48\).

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {e^{3+e^{2 x^2}-2 e^{x^2} \log \left (\frac {5}{3+e^x}\right )+\log ^2\left (\frac {5}{3+e^x}\right )} \left (3 e^x+e^{2 x}+2 e^{2 x+x^2}+e^{2 x^2} \left (12 e^x x+4 e^{2 x} x\right )+\left (-2 e^{2 x}+e^{x^2} \left (-12 e^x x-4 e^{2 x} x\right )\right ) \log \left (\frac {5}{3+e^x}\right )\right )}{12+4 e^x} \, dx=\frac {1}{4} \, e^{\left (-2 \, e^{\left (x^{2}\right )} \log \left (5\right ) + \log \left (5\right )^{2} + 2 \, e^{\left (x^{2}\right )} \log \left (e^{x} + 3\right ) - 2 \, \log \left (5\right ) \log \left (e^{x} + 3\right ) + \log \left (e^{x} + 3\right )^{2} + x + e^{\left (2 \, x^{2}\right )} + 3\right )} \] Input:

integrate((((-4*x*exp(x)^2-12*exp(x)*x)*exp(x^2)-2*exp(x)^2)*log(5/(3+exp( 
x)))+(4*x*exp(x)^2+12*exp(x)*x)*exp(x^2)^2+2*exp(x)^2*exp(x^2)+exp(x)^2+3* 
exp(x))*exp(log(5/(3+exp(x)))^2-2*exp(x^2)*log(5/(3+exp(x)))+exp(x^2)^2+3) 
/(4*exp(x)+12),x, algorithm="maxima")
 

Output:

1/4*e^(-2*e^(x^2)*log(5) + log(5)^2 + 2*e^(x^2)*log(e^x + 3) - 2*log(5)*lo 
g(e^x + 3) + log(e^x + 3)^2 + x + e^(2*x^2) + 3)
 

Giac [F]

\[ \int \frac {e^{3+e^{2 x^2}-2 e^{x^2} \log \left (\frac {5}{3+e^x}\right )+\log ^2\left (\frac {5}{3+e^x}\right )} \left (3 e^x+e^{2 x}+2 e^{2 x+x^2}+e^{2 x^2} \left (12 e^x x+4 e^{2 x} x\right )+\left (-2 e^{2 x}+e^{x^2} \left (-12 e^x x-4 e^{2 x} x\right )\right ) \log \left (\frac {5}{3+e^x}\right )\right )}{12+4 e^x} \, dx=\int { \frac {{\left (4 \, {\left (x e^{\left (2 \, x\right )} + 3 \, x e^{x}\right )} e^{\left (2 \, x^{2}\right )} - 2 \, {\left (2 \, {\left (x e^{\left (2 \, x\right )} + 3 \, x e^{x}\right )} e^{\left (x^{2}\right )} + e^{\left (2 \, x\right )}\right )} \log \left (\frac {5}{e^{x} + 3}\right ) + 2 \, e^{\left (x^{2} + 2 \, x\right )} + e^{\left (2 \, x\right )} + 3 \, e^{x}\right )} e^{\left (-2 \, e^{\left (x^{2}\right )} \log \left (\frac {5}{e^{x} + 3}\right ) + \log \left (\frac {5}{e^{x} + 3}\right )^{2} + e^{\left (2 \, x^{2}\right )} + 3\right )}}{4 \, {\left (e^{x} + 3\right )}} \,d x } \] Input:

integrate((((-4*x*exp(x)^2-12*exp(x)*x)*exp(x^2)-2*exp(x)^2)*log(5/(3+exp( 
x)))+(4*x*exp(x)^2+12*exp(x)*x)*exp(x^2)^2+2*exp(x)^2*exp(x^2)+exp(x)^2+3* 
exp(x))*exp(log(5/(3+exp(x)))^2-2*exp(x^2)*log(5/(3+exp(x)))+exp(x^2)^2+3) 
/(4*exp(x)+12),x, algorithm="giac")
 

Output:

integrate(1/4*(4*(x*e^(2*x) + 3*x*e^x)*e^(2*x^2) - 2*(2*(x*e^(2*x) + 3*x*e 
^x)*e^(x^2) + e^(2*x))*log(5/(e^x + 3)) + 2*e^(x^2 + 2*x) + e^(2*x) + 3*e^ 
x)*e^(-2*e^(x^2)*log(5/(e^x + 3)) + log(5/(e^x + 3))^2 + e^(2*x^2) + 3)/(e 
^x + 3), x)
 

Mupad [B] (verification not implemented)

Time = 2.35 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66 \[ \int \frac {e^{3+e^{2 x^2}-2 e^{x^2} \log \left (\frac {5}{3+e^x}\right )+\log ^2\left (\frac {5}{3+e^x}\right )} \left (3 e^x+e^{2 x}+2 e^{2 x+x^2}+e^{2 x^2} \left (12 e^x x+4 e^{2 x} x\right )+\left (-2 e^{2 x}+e^{x^2} \left (-12 e^x x-4 e^{2 x} x\right )\right ) \log \left (\frac {5}{3+e^x}\right )\right )}{12+4 e^x} \, dx=\frac {{\left (\frac {1}{25}\right )}^{{\mathrm {e}}^{x^2}}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x^2}}\,{\mathrm {e}}^{{\ln \left (5\right )}^2}\,{\mathrm {e}}^3\,{\mathrm {e}}^{{\ln \left ({\mathrm {e}}^x+3\right )}^2}\,{\mathrm {e}}^x\,{\left ({\mathrm {e}}^x+3\right )}^{2\,{\mathrm {e}}^{x^2}-2\,\ln \left (5\right )}}{4} \] Input:

int((exp(exp(2*x^2) - 2*exp(x^2)*log(5/(exp(x) + 3)) + log(5/(exp(x) + 3)) 
^2 + 3)*(exp(2*x) + 3*exp(x) + exp(2*x^2)*(4*x*exp(2*x) + 12*x*exp(x)) - l 
og(5/(exp(x) + 3))*(2*exp(2*x) + exp(x^2)*(4*x*exp(2*x) + 12*x*exp(x))) + 
2*exp(2*x)*exp(x^2)))/(4*exp(x) + 12),x)
 

Output:

((1/25)^exp(x^2)*exp(exp(2*x^2))*exp(log(5)^2)*exp(3)*exp(log(exp(x) + 3)^ 
2)*exp(x)*(exp(x) + 3)^(2*exp(x^2) - 2*log(5)))/4
 

Reduce [F]

\[ \int \frac {e^{3+e^{2 x^2}-2 e^{x^2} \log \left (\frac {5}{3+e^x}\right )+\log ^2\left (\frac {5}{3+e^x}\right )} \left (3 e^x+e^{2 x}+2 e^{2 x+x^2}+e^{2 x^2} \left (12 e^x x+4 e^{2 x} x\right )+\left (-2 e^{2 x}+e^{x^2} \left (-12 e^x x-4 e^{2 x} x\right )\right ) \log \left (\frac {5}{3+e^x}\right )\right )}{12+4 e^x} \, dx=\int \frac {\left (\left (\left (-4 x \left ({\mathrm e}^{x}\right )^{2}-12 \,{\mathrm e}^{x} x \right ) {\mathrm e}^{x^{2}}-2 \left ({\mathrm e}^{x}\right )^{2}\right ) \mathrm {log}\left (\frac {5}{3+{\mathrm e}^{x}}\right )+\left (4 x \left ({\mathrm e}^{x}\right )^{2}+12 \,{\mathrm e}^{x} x \right ) \left ({\mathrm e}^{x^{2}}\right )^{2}+2 \left ({\mathrm e}^{x}\right )^{2} {\mathrm e}^{x^{2}}+\left ({\mathrm e}^{x}\right )^{2}+3 \,{\mathrm e}^{x}\right ) {\mathrm e}^{\mathrm {log}\left (\frac {5}{3+{\mathrm e}^{x}}\right )^{2}-2 \,{\mathrm e}^{x^{2}} \mathrm {log}\left (\frac {5}{3+{\mathrm e}^{x}}\right )+\left ({\mathrm e}^{x^{2}}\right )^{2}+3}}{4 \,{\mathrm e}^{x}+12}d x \] Input:

int((((-4*x*exp(x)^2-12*exp(x)*x)*exp(x^2)-2*exp(x)^2)*log(5/(3+exp(x)))+( 
4*x*exp(x)^2+12*exp(x)*x)*exp(x^2)^2+2*exp(x)^2*exp(x^2)+exp(x)^2+3*exp(x) 
)*exp(log(5/(3+exp(x)))^2-2*exp(x^2)*log(5/(3+exp(x)))+exp(x^2)^2+3)/(4*ex 
p(x)+12),x)
 

Output:

int((((-4*x*exp(x)^2-12*exp(x)*x)*exp(x^2)-2*exp(x)^2)*log(5/(3+exp(x)))+( 
4*x*exp(x)^2+12*exp(x)*x)*exp(x^2)^2+2*exp(x)^2*exp(x^2)+exp(x)^2+3*exp(x) 
)*exp(log(5/(3+exp(x)))^2-2*exp(x^2)*log(5/(3+exp(x)))+exp(x^2)^2+3)/(4*ex 
p(x)+12),x)