\(\int \frac {-2 e^{e^2}+e^{\frac {1}{2} e^{-e^2} (2 e^{e^2} x+3 x^2+x^3)} (2 e^{e^2}+6 x+3 x^2)}{16 e^{e^2+\frac {1}{2} e^{-e^2} (2 e^{e^2} x+3 x^2+x^3)}+e^{e^2} (-16-16 x)+(2 e^{e^2+\frac {1}{2} e^{-e^2} (2 e^{e^2} x+3 x^2+x^3)}+e^{e^2} (-2-2 x)) \log (1-e^{\frac {1}{2} e^{-e^2} (2 e^{e^2} x+3 x^2+x^3)}+x)} \, dx\) [2746]

Optimal result

Integrand size = 191, antiderivative size = 30 \[ \int \frac {-2 e^{e^2}+e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )} \left (2 e^{e^2}+6 x+3 x^2\right )}{16 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-16-16 x)+\left (2 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-2-2 x)\right ) \log \left (1-e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+x\right )} \, dx=\log \left (8+\log \left (1-e^{x+\frac {1}{2} e^{-e^2} x^2 (3+x)}+x\right )\right ) \] Output:

ln(ln(x+1-exp(1/2*x^2/exp(exp(2))*(3+x)+x))+8)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-2 e^{e^2}+e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )} \left (2 e^{e^2}+6 x+3 x^2\right )}{16 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-16-16 x)+\left (2 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-2-2 x)\right ) \log \left (1-e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+x\right )} \, dx=\log \left (8+\log \left (1-e^{x+\frac {1}{2} e^{-e^2} x^2 (3+x)}+x\right )\right ) \] Input:

Integrate[(-2*E^E^2 + E^((2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2))*(2*E^E^2 + 6 
*x + 3*x^2))/(16*E^(E^2 + (2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2)) + E^E^2*(-1 
6 - 16*x) + (2*E^(E^2 + (2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2)) + E^E^2*(-2 - 
 2*x))*Log[1 - E^((2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2)) + x]),x]
 

Output:

Log[8 + Log[1 - E^(x + (x^2*(3 + x))/(2*E^E^2)) + x]]
 

Rubi [A] (verified)

Time = 1.39 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {7292, 27, 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.

\(\displaystyle \int \frac {e^{\frac {1}{2} e^{-e^2} \left (x^3+3 x^2+2 e^{e^2} x\right )} \left (3 x^2+6 x+2 e^{e^2}\right )-2 e^{e^2}}{16 e^{\frac {1}{2} e^{-e^2} \left (x^3+3 x^2+2 e^{e^2} x\right )+e^2}+\left (2 e^{\frac {1}{2} e^{-e^2} \left (x^3+3 x^2+2 e^{e^2} x\right )+e^2}+e^{e^2} (-2 x-2)\right ) \log \left (-e^{\frac {1}{2} e^{-e^2} \left (x^3+3 x^2+2 e^{e^2} x\right )}+x+1\right )+e^{e^2} (-16 x-16)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{-e^2} \left (2 e^{e^2}-e^{\frac {1}{2} e^{-e^2} \left (x^3+3 x^2+2 e^{e^2} x\right )} \left (3 x^2+6 x+2 e^{e^2}\right )\right )}{2 \left (-e^{\frac {1}{2} e^{-e^2} (x+3) x^2+x}+x+1\right ) \left (\log \left (-e^{\frac {1}{2} e^{-e^2} (x+3) x^2+x}+x+1\right )+8\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} e^{-e^2} \int \frac {2 e^{e^2}-e^{\frac {1}{2} e^{-e^2} \left (x^3+3 x^2+2 e^{e^2} x\right )} \left (3 x^2+6 x+2 e^{e^2}\right )}{\left (x-e^{\frac {1}{2} e^{-e^2} (x+3) x^2+x}+1\right ) \left (\log \left (x-e^{\frac {1}{2} e^{-e^2} (x+3) x^2+x}+1\right )+8\right )}dx\)

\(\Big \downarrow \) 7235

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

Input:

Int[(-2*E^E^2 + E^((2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2))*(2*E^E^2 + 6*x + 3 
*x^2))/(16*E^(E^2 + (2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2)) + E^E^2*(-16 - 16 
*x) + (2*E^(E^2 + (2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2)) + E^E^2*(-2 - 2*x)) 
*Log[1 - E^((2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2)) + x]),x]
 

Output:

Log[8 + Log[1 - E^(x + (x^2*(3 + x))/(2*E^E^2)) + x]]
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 7235

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

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03

method	result	size
risch	\(\ln \left (\ln \left (-{\mathrm e}^{\frac {x \left (x^{2}+2 \,{\mathrm e}^{{\mathrm e}^{2}}+3 x \right ) {\mathrm e}^{-{\mathrm e}^{2}}}{2}}+x +1\right )+8\right )\)	\(31\)
parallelrisch	\(\ln \left (\ln \left (-{\mathrm e}^{\frac {x \left (x^{2}+2 \,{\mathrm e}^{{\mathrm e}^{2}}+3 x \right ) {\mathrm e}^{-{\mathrm e}^{2}}}{2}}+x +1\right )+8\right )\)	\(31\)
norman	\(\ln \left (\ln \left (-{\mathrm e}^{\frac {\left (2 x \,{\mathrm e}^{{\mathrm e}^{2}}+x^{3}+3 x^{2}\right ) {\mathrm e}^{-{\mathrm e}^{2}}}{2}}+x +1\right )+8\right )\)	\(33\)

Input:

int(((2*exp(exp(2))+3*x^2+6*x)*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp 
(2)))-2*exp(exp(2)))/((2*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/e 
xp(exp(2)))+(-2-2*x)*exp(exp(2)))*ln(-exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/ 
exp(exp(2)))+x+1)+16*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(e 
xp(2)))+(-16*x-16)*exp(exp(2))),x,method=_RETURNVERBOSE)
 

Output:

ln(ln(-exp(1/2*x*(x^2+2*exp(exp(2))+3*x)*exp(-exp(2)))+x+1)+8)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {-2 e^{e^2}+e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )} \left (2 e^{e^2}+6 x+3 x^2\right )}{16 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-16-16 x)+\left (2 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-2-2 x)\right ) \log \left (1-e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+x\right )} \, dx=\log \left (\log \left ({\left ({\left (x + 1\right )} e^{\left (e^{2}\right )} - e^{\left (\frac {1}{2} \, {\left (x^{3} + 3 \, x^{2} + 2 \, {\left (x + e^{2}\right )} e^{\left (e^{2}\right )}\right )} e^{\left (-e^{2}\right )}\right )}\right )} e^{\left (-e^{2}\right )}\right ) + 8\right ) \] Input:

integrate(((2*exp(exp(2))+3*x^2+6*x)*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/e 
xp(exp(2)))-2*exp(exp(2)))/((2*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3* 
x^2)/exp(exp(2)))+(-2-2*x)*exp(exp(2)))*log(-exp(1/2*(2*x*exp(exp(2))+x^3+ 
3*x^2)/exp(exp(2)))+x+1)+16*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2 
)/exp(exp(2)))+(-16*x-16)*exp(exp(2))),x, algorithm="fricas")
 

Output:

log(log(((x + 1)*e^(e^2) - e^(1/2*(x^3 + 3*x^2 + 2*(x + e^2)*e^(e^2))*e^(- 
e^2)))*e^(-e^2)) + 8)
 

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-2 e^{e^2}+e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )} \left (2 e^{e^2}+6 x+3 x^2\right )}{16 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-16-16 x)+\left (2 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-2-2 x)\right ) \log \left (1-e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+x\right )} \, dx=\log {\left (\log {\left (x - e^{\frac {\frac {x^{3}}{2} + \frac {3 x^{2}}{2} + x e^{e^{2}}}{e^{e^{2}}}} + 1 \right )} + 8 \right )} \] Input:

integrate(((2*exp(exp(2))+3*x**2+6*x)*exp(1/2*(2*x*exp(exp(2))+x**3+3*x**2 
)/exp(exp(2)))-2*exp(exp(2)))/((2*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x** 
3+3*x**2)/exp(exp(2)))+(-2-2*x)*exp(exp(2)))*ln(-exp(1/2*(2*x*exp(exp(2))+ 
x**3+3*x**2)/exp(exp(2)))+x+1)+16*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x** 
3+3*x**2)/exp(exp(2)))+(-16*x-16)*exp(exp(2))),x)
 

Output:

log(log(x - exp((x**3/2 + 3*x**2/2 + x*exp(exp(2)))*exp(-exp(2))) + 1) + 8 
)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-2 e^{e^2}+e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )} \left (2 e^{e^2}+6 x+3 x^2\right )}{16 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-16-16 x)+\left (2 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-2-2 x)\right ) \log \left (1-e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+x\right )} \, dx=\log \left (\log \left (x - e^{\left (\frac {1}{2} \, x^{3} e^{\left (-e^{2}\right )} + \frac {3}{2} \, x^{2} e^{\left (-e^{2}\right )} + x\right )} + 1\right ) + 8\right ) \] Input:

integrate(((2*exp(exp(2))+3*x^2+6*x)*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/e 
xp(exp(2)))-2*exp(exp(2)))/((2*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3* 
x^2)/exp(exp(2)))+(-2-2*x)*exp(exp(2)))*log(-exp(1/2*(2*x*exp(exp(2))+x^3+ 
3*x^2)/exp(exp(2)))+x+1)+16*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2 
)/exp(exp(2)))+(-16*x-16)*exp(exp(2))),x, algorithm="maxima")
 

Output:

log(log(x - e^(1/2*x^3*e^(-e^2) + 3/2*x^2*e^(-e^2) + x) + 1) + 8)
 

Giac [F]

\[ \int \frac {-2 e^{e^2}+e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )} \left (2 e^{e^2}+6 x+3 x^2\right )}{16 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-16-16 x)+\left (2 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-2-2 x)\right ) \log \left (1-e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+x\right )} \, dx=\int { -\frac {{\left (3 \, x^{2} + 6 \, x + 2 \, e^{\left (e^{2}\right )}\right )} e^{\left (\frac {1}{2} \, {\left (x^{3} + 3 \, x^{2} + 2 \, x e^{\left (e^{2}\right )}\right )} e^{\left (-e^{2}\right )}\right )} - 2 \, e^{\left (e^{2}\right )}}{2 \, {\left (8 \, {\left (x + 1\right )} e^{\left (e^{2}\right )} + {\left ({\left (x + 1\right )} e^{\left (e^{2}\right )} - e^{\left (\frac {1}{2} \, {\left (x^{3} + 3 \, x^{2} + 2 \, x e^{\left (e^{2}\right )}\right )} e^{\left (-e^{2}\right )} + e^{2}\right )}\right )} \log \left (x - e^{\left (\frac {1}{2} \, {\left (x^{3} + 3 \, x^{2} + 2 \, x e^{\left (e^{2}\right )}\right )} e^{\left (-e^{2}\right )}\right )} + 1\right ) - 8 \, e^{\left (\frac {1}{2} \, {\left (x^{3} + 3 \, x^{2} + 2 \, x e^{\left (e^{2}\right )}\right )} e^{\left (-e^{2}\right )} + e^{2}\right )}\right )}} \,d x } \] Input:

integrate(((2*exp(exp(2))+3*x^2+6*x)*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/e 
xp(exp(2)))-2*exp(exp(2)))/((2*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3* 
x^2)/exp(exp(2)))+(-2-2*x)*exp(exp(2)))*log(-exp(1/2*(2*x*exp(exp(2))+x^3+ 
3*x^2)/exp(exp(2)))+x+1)+16*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2 
)/exp(exp(2)))+(-16*x-16)*exp(exp(2))),x, algorithm="giac")
 

Output:

integrate(-1/2*((3*x^2 + 6*x + 2*e^(e^2))*e^(1/2*(x^3 + 3*x^2 + 2*x*e^(e^2 
))*e^(-e^2)) - 2*e^(e^2))/(8*(x + 1)*e^(e^2) + ((x + 1)*e^(e^2) - e^(1/2*( 
x^3 + 3*x^2 + 2*x*e^(e^2))*e^(-e^2) + e^2))*log(x - e^(1/2*(x^3 + 3*x^2 + 
2*x*e^(e^2))*e^(-e^2)) + 1) - 8*e^(1/2*(x^3 + 3*x^2 + 2*x*e^(e^2))*e^(-e^2 
) + e^2)), x)
 

Mupad [B] (verification not implemented)

Time = 2.88 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-2 e^{e^2}+e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )} \left (2 e^{e^2}+6 x+3 x^2\right )}{16 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-16-16 x)+\left (2 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-2-2 x)\right ) \log \left (1-e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+x\right )} \, dx=\ln \left (\ln \left (x-{\mathrm {e}}^{\frac {{\mathrm {e}}^{-{\mathrm {e}}^2}\,x^3}{2}+\frac {3\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,x^2}{2}+x}+1\right )+8\right ) \] Input:

int((2*exp(exp(2)) - exp(exp(-exp(2))*(x*exp(exp(2)) + (3*x^2)/2 + x^3/2)) 
*(6*x + 2*exp(exp(2)) + 3*x^2))/(exp(exp(2))*(16*x + 16) - 16*exp(exp(-exp 
(2))*(x*exp(exp(2)) + (3*x^2)/2 + x^3/2))*exp(exp(2)) + log(x - exp(exp(-e 
xp(2))*(x*exp(exp(2)) + (3*x^2)/2 + x^3/2)) + 1)*(exp(exp(2))*(2*x + 2) - 
2*exp(exp(-exp(2))*(x*exp(exp(2)) + (3*x^2)/2 + x^3/2))*exp(exp(2)))),x)
 

Output:

log(log(x - exp(x + (3*x^2*exp(-exp(2)))/2 + (x^3*exp(-exp(2)))/2) + 1) + 
8)
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {-2 e^{e^2}+e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )} \left (2 e^{e^2}+6 x+3 x^2\right )}{16 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-16-16 x)+\left (2 e^{e^2+\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+e^{e^2} (-2-2 x)\right ) \log \left (1-e^{\frac {1}{2} e^{-e^2} \left (2 e^{e^2} x+3 x^2+x^3\right )}+x\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (-e^{\frac {2 e^{e^{2}} x +x^{3}+3 x^{2}}{2 e^{e^{2}}}}+x +1\right )+8\right ) \] Input:

int(((2*exp(exp(2))+3*x^2+6*x)*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp 
(2)))-2*exp(exp(2)))/((2*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/e 
xp(exp(2)))+(-2-2*x)*exp(exp(2)))*log(-exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2) 
/exp(exp(2)))+x+1)+16*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp( 
exp(2)))+(-16*x-16)*exp(exp(2))),x)
 

Output:

log(log( - e**((2*e**(e**2)*x + x**3 + 3*x**2)/(2*e**(e**2))) + x + 1) + 8 
)