\(\int \frac {e^{e^{\frac {2 (4+x \log (2 x+4 x^2+2 x^3))}{\log (2 x+4 x^2+2 x^3)}}+4 e^{\frac {4+x \log (2 x+4 x^2+2 x^3)}{\log (2 x+4 x^2+2 x^3)}} x+4 x^2} ((8 x^2+8 x^3) \log ^2(2 x+4 x^2+2 x^3)+e^{\frac {2 (4+x \log (2 x+4 x^2+2 x^3))}{\log (2 x+4 x^2+2 x^3)}} (-8-24 x+(2 x+2 x^2) \log ^2(2 x+4 x^2+2 x^3))+e^{\frac {4+x \log (2 x+4 x^2+2 x^3)}{\log (2 x+4 x^2+2 x^3)}} (-16 x-48 x^2+(4 x+8 x^2+4 x^3) \log ^2(2 x+4 x^2+2 x^3)))}{(x+x^2) \log ^2(2 x+4 x^2+2 x^3)} \, dx\) [2234]

Optimal result

Integrand size = 299, antiderivative size = 28 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx=e^{\left (e^{x+\frac {4}{\log \left ((2+2 x) \left (x+x^2\right )\right )}}+2 x\right )^2} \] Output:

exp((2*x+exp(4/ln((x^2+x)*(2+2*x))+x))^2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx=e^{\left (e^{x+\frac {4}{\log \left (2 x (1+x)^2\right )}}+2 x\right )^2} \] Input:

Integrate[(E^(E^((2*(4 + x*Log[2*x + 4*x^2 + 2*x^3]))/Log[2*x + 4*x^2 + 2* 
x^3]) + 4*E^((4 + x*Log[2*x + 4*x^2 + 2*x^3])/Log[2*x + 4*x^2 + 2*x^3])*x 
+ 4*x^2)*((8*x^2 + 8*x^3)*Log[2*x + 4*x^2 + 2*x^3]^2 + E^((2*(4 + x*Log[2* 
x + 4*x^2 + 2*x^3]))/Log[2*x + 4*x^2 + 2*x^3])*(-8 - 24*x + (2*x + 2*x^2)* 
Log[2*x + 4*x^2 + 2*x^3]^2) + E^((4 + x*Log[2*x + 4*x^2 + 2*x^3])/Log[2*x 
+ 4*x^2 + 2*x^3])*(-16*x - 48*x^2 + (4*x + 8*x^2 + 4*x^3)*Log[2*x + 4*x^2 
+ 2*x^3]^2)))/((x + x^2)*Log[2*x + 4*x^2 + 2*x^3]^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 7.88 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {2026, 7239, 27, 25, 7257}

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.

\(\displaystyle \int \frac {\exp \left (4 x \exp \left (\frac {x \log \left (2 x^3+4 x^2+2 x\right )+4}{\log \left (2 x^3+4 x^2+2 x\right )}\right )+\exp \left (\frac {2 \left (x \log \left (2 x^3+4 x^2+2 x\right )+4\right )}{\log \left (2 x^3+4 x^2+2 x\right )}\right )+4 x^2\right ) \left (\left (\left (2 x^2+2 x\right ) \log ^2\left (2 x^3+4 x^2+2 x\right )-24 x-8\right ) \exp \left (\frac {2 \left (x \log \left (2 x^3+4 x^2+2 x\right )+4\right )}{\log \left (2 x^3+4 x^2+2 x\right )}\right )+\left (-48 x^2+\left (4 x^3+8 x^2+4 x\right ) \log ^2\left (2 x^3+4 x^2+2 x\right )-16 x\right ) \exp \left (\frac {x \log \left (2 x^3+4 x^2+2 x\right )+4}{\log \left (2 x^3+4 x^2+2 x\right )}\right )+\left (8 x^3+8 x^2\right ) \log ^2\left (2 x^3+4 x^2+2 x\right )\right )}{\left (x^2+x\right ) \log ^2\left (2 x^3+4 x^2+2 x\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {\exp \left (4 x \exp \left (\frac {x \log \left (2 x^3+4 x^2+2 x\right )+4}{\log \left (2 x^3+4 x^2+2 x\right )}\right )+\exp \left (\frac {2 \left (x \log \left (2 x^3+4 x^2+2 x\right )+4\right )}{\log \left (2 x^3+4 x^2+2 x\right )}\right )+4 x^2\right ) \left (\left (\left (2 x^2+2 x\right ) \log ^2\left (2 x^3+4 x^2+2 x\right )-24 x-8\right ) \exp \left (\frac {2 \left (x \log \left (2 x^3+4 x^2+2 x\right )+4\right )}{\log \left (2 x^3+4 x^2+2 x\right )}\right )+\left (-48 x^2+\left (4 x^3+8 x^2+4 x\right ) \log ^2\left (2 x^3+4 x^2+2 x\right )-16 x\right ) \exp \left (\frac {x \log \left (2 x^3+4 x^2+2 x\right )+4}{\log \left (2 x^3+4 x^2+2 x\right )}\right )+\left (8 x^3+8 x^2\right ) \log ^2\left (2 x^3+4 x^2+2 x\right )\right )}{x (x+1) \log ^2\left (2 x^3+4 x^2+2 x\right )}dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 7257

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

Input:

Int[(E^(E^((2*(4 + x*Log[2*x + 4*x^2 + 2*x^3]))/Log[2*x + 4*x^2 + 2*x^3]) 
+ 4*E^((4 + x*Log[2*x + 4*x^2 + 2*x^3])/Log[2*x + 4*x^2 + 2*x^3])*x + 4*x^ 
2)*((8*x^2 + 8*x^3)*Log[2*x + 4*x^2 + 2*x^3]^2 + E^((2*(4 + x*Log[2*x + 4* 
x^2 + 2*x^3]))/Log[2*x + 4*x^2 + 2*x^3])*(-8 - 24*x + (2*x + 2*x^2)*Log[2* 
x + 4*x^2 + 2*x^3]^2) + E^((4 + x*Log[2*x + 4*x^2 + 2*x^3])/Log[2*x + 4*x^ 
2 + 2*x^3])*(-16*x - 48*x^2 + (4*x + 8*x^2 + 4*x^3)*Log[2*x + 4*x^2 + 2*x^ 
3]^2)))/((x + x^2)*Log[2*x + 4*x^2 + 2*x^3]^2),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 2026

Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[u, x])
 

rule 7239

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

rule 7257

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim 
p[q*(F^v/Log[F]), x] /;  !FalseQ[q]] /; FreeQ[F, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(87\) vs. \(2(26)=52\).

Time = 9.85 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.14

method	result	size
risch	\({\mathrm e}^{{\mathrm e}^{\frac {2 x \ln \left (2 x^{3}+4 x^{2}+2 x \right )+8}{\ln \left (2 x^{3}+4 x^{2}+2 x \right )}}+4 x \,{\mathrm e}^{\frac {x \ln \left (2 x^{3}+4 x^{2}+2 x \right )+4}{\ln \left (2 x^{3}+4 x^{2}+2 x \right )}}+4 x^{2}}\)	\(88\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{\frac {2 x \ln \left (2 x^{3}+4 x^{2}+2 x \right )+8}{\ln \left (2 x^{3}+4 x^{2}+2 x \right )}}+4 x \,{\mathrm e}^{\frac {x \ln \left (2 x^{3}+4 x^{2}+2 x \right )+4}{\ln \left (2 x^{3}+4 x^{2}+2 x \right )}}+4 x^{2}}\)	\(89\)

Input:

int((((2*x^2+2*x)*ln(2*x^3+4*x^2+2*x)^2-24*x-8)*exp((x*ln(2*x^3+4*x^2+2*x) 
+4)/ln(2*x^3+4*x^2+2*x))^2+((4*x^3+8*x^2+4*x)*ln(2*x^3+4*x^2+2*x)^2-48*x^2 
-16*x)*exp((x*ln(2*x^3+4*x^2+2*x)+4)/ln(2*x^3+4*x^2+2*x))+(8*x^3+8*x^2)*ln 
(2*x^3+4*x^2+2*x)^2)*exp(exp((x*ln(2*x^3+4*x^2+2*x)+4)/ln(2*x^3+4*x^2+2*x) 
)^2+4*x*exp((x*ln(2*x^3+4*x^2+2*x)+4)/ln(2*x^3+4*x^2+2*x))+4*x^2)/(x^2+x)/ 
ln(2*x^3+4*x^2+2*x)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (25) = 50\).

Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.11 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx=e^{\left (4 \, x^{2} + 4 \, x e^{\left (\frac {x \log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right ) + 4}{\log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right )}\right )} + e^{\left (\frac {2 \, {\left (x \log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right ) + 4\right )}}{\log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right )}\right )}\right )} \] Input:

integrate((((2*x^2+2*x)*log(2*x^3+4*x^2+2*x)^2-24*x-8)*exp((x*log(2*x^3+4* 
x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))^2+((4*x^3+8*x^2+4*x)*log(2*x^3+4*x^2+2*x 
)^2-48*x^2-16*x)*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))+(8*x 
^3+8*x^2)*log(2*x^3+4*x^2+2*x)^2)*exp(exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2 
*x^3+4*x^2+2*x))^2+4*x*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x) 
)+4*x^2)/(x^2+x)/log(2*x^3+4*x^2+2*x)^2,x, algorithm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (22) = 44\).

Time = 5.68 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.93 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx=e^{4 x^{2} + 4 x e^{\frac {x \log {\left (2 x^{3} + 4 x^{2} + 2 x \right )} + 4}{\log {\left (2 x^{3} + 4 x^{2} + 2 x \right )}}} + e^{\frac {2 \left (x \log {\left (2 x^{3} + 4 x^{2} + 2 x \right )} + 4\right )}{\log {\left (2 x^{3} + 4 x^{2} + 2 x \right )}}}} \] Input:

integrate((((2*x**2+2*x)*ln(2*x**3+4*x**2+2*x)**2-24*x-8)*exp((x*ln(2*x**3 
+4*x**2+2*x)+4)/ln(2*x**3+4*x**2+2*x))**2+((4*x**3+8*x**2+4*x)*ln(2*x**3+4 
*x**2+2*x)**2-48*x**2-16*x)*exp((x*ln(2*x**3+4*x**2+2*x)+4)/ln(2*x**3+4*x* 
*2+2*x))+(8*x**3+8*x**2)*ln(2*x**3+4*x**2+2*x)**2)*exp(exp((x*ln(2*x**3+4* 
x**2+2*x)+4)/ln(2*x**3+4*x**2+2*x))**2+4*x*exp((x*ln(2*x**3+4*x**2+2*x)+4) 
/ln(2*x**3+4*x**2+2*x))+4*x**2)/(x**2+x)/ln(2*x**3+4*x**2+2*x)**2,x)
 

Output:

exp(4*x**2 + 4*x*exp((x*log(2*x**3 + 4*x**2 + 2*x) + 4)/log(2*x**3 + 4*x** 
2 + 2*x)) + exp(2*(x*log(2*x**3 + 4*x**2 + 2*x) + 4)/log(2*x**3 + 4*x**2 + 
 2*x)))
 

Maxima [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx=e^{\left (4 \, x^{2} + 4 \, x e^{\left (x + \frac {4}{\log \left (2\right ) + 2 \, \log \left (x + 1\right ) + \log \left (x\right )}\right )} + e^{\left (2 \, x + \frac {8}{\log \left (2\right ) + 2 \, \log \left (x + 1\right ) + \log \left (x\right )}\right )}\right )} \] Input:

integrate((((2*x^2+2*x)*log(2*x^3+4*x^2+2*x)^2-24*x-8)*exp((x*log(2*x^3+4* 
x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))^2+((4*x^3+8*x^2+4*x)*log(2*x^3+4*x^2+2*x 
)^2-48*x^2-16*x)*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))+(8*x 
^3+8*x^2)*log(2*x^3+4*x^2+2*x)^2)*exp(exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2 
*x^3+4*x^2+2*x))^2+4*x*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x) 
)+4*x^2)/(x^2+x)/log(2*x^3+4*x^2+2*x)^2,x, algorithm="maxima")
 

Output:

e^(4*x^2 + 4*x*e^(x + 4/(log(2) + 2*log(x + 1) + log(x))) + e^(2*x + 8/(lo 
g(2) + 2*log(x + 1) + log(x))))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (25) = 50\).

Time = 10.61 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.00 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx=e^{\left (4 \, x^{2} + 4 \, x e^{\left (x + \frac {4}{\log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right )}\right )} + e^{\left (2 \, x + \frac {8}{\log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right )}\right )}\right )} \] Input:

integrate((((2*x^2+2*x)*log(2*x^3+4*x^2+2*x)^2-24*x-8)*exp((x*log(2*x^3+4* 
x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))^2+((4*x^3+8*x^2+4*x)*log(2*x^3+4*x^2+2*x 
)^2-48*x^2-16*x)*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))+(8*x 
^3+8*x^2)*log(2*x^3+4*x^2+2*x)^2)*exp(exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2 
*x^3+4*x^2+2*x))^2+4*x*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x) 
)+4*x^2)/(x^2+x)/log(2*x^3+4*x^2+2*x)^2,x, algorithm="giac")
 

Output:

e^(4*x^2 + 4*x*e^(x + 4/log(2*x^3 + 4*x^2 + 2*x)) + e^(2*x + 8/log(2*x^3 + 
 4*x^2 + 2*x)))
 

Mupad [B] (verification not implemented)

Time = 2.66 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.11 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx={\mathrm {e}}^{{\mathrm {e}}^{\frac {8}{\ln \left (2\,x^3+4\,x^2+2\,x\right )}}\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{4\,x\,{\mathrm {e}}^{\frac {4}{\ln \left (2\,x^3+4\,x^2+2\,x\right )}}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{4\,x^2} \] Input:

int(-(exp(exp((2*(x*log(2*x + 4*x^2 + 2*x^3) + 4))/log(2*x + 4*x^2 + 2*x^3 
)) + 4*x*exp((x*log(2*x + 4*x^2 + 2*x^3) + 4)/log(2*x + 4*x^2 + 2*x^3)) + 
4*x^2)*(exp((2*(x*log(2*x + 4*x^2 + 2*x^3) + 4))/log(2*x + 4*x^2 + 2*x^3)) 
*(24*x - log(2*x + 4*x^2 + 2*x^3)^2*(2*x + 2*x^2) + 8) - log(2*x + 4*x^2 + 
 2*x^3)^2*(8*x^2 + 8*x^3) + exp((x*log(2*x + 4*x^2 + 2*x^3) + 4)/log(2*x + 
 4*x^2 + 2*x^3))*(16*x - log(2*x + 4*x^2 + 2*x^3)^2*(4*x + 8*x^2 + 4*x^3) 
+ 48*x^2)))/(log(2*x + 4*x^2 + 2*x^3)^2*(x + x^2)),x)
 

Output:

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

Reduce [F]

\[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx=\int \frac {\left (\left (\left (2 x^{2}+2 x \right ) \mathrm {log}\left (2 x^{3}+4 x^{2}+2 x \right )^{2}-24 x -8\right ) \left ({\mathrm e}^{\frac {x \,\mathrm {log}\left (2 x^{3}+4 x^{2}+2 x \right )+4}{\mathrm {log}\left (2 x^{3}+4 x^{2}+2 x \right )}}\right )^{2}+\left (\left (4 x^{3}+8 x^{2}+4 x \right ) \mathrm {log}\left (2 x^{3}+4 x^{2}+2 x \right )^{2}-48 x^{2}-16 x \right ) {\mathrm e}^{\frac {x \,\mathrm {log}\left (2 x^{3}+4 x^{2}+2 x \right )+4}{\mathrm {log}\left (2 x^{3}+4 x^{2}+2 x \right )}}+\left (8 x^{3}+8 x^{2}\right ) \mathrm {log}\left (2 x^{3}+4 x^{2}+2 x \right )^{2}\right ) {\mathrm e}^{\left ({\mathrm e}^{\frac {x \,\mathrm {log}\left (2 x^{3}+4 x^{2}+2 x \right )+4}{\mathrm {log}\left (2 x^{3}+4 x^{2}+2 x \right )}}\right )^{2}+4 x \,{\mathrm e}^{\frac {x \,\mathrm {log}\left (2 x^{3}+4 x^{2}+2 x \right )+4}{\mathrm {log}\left (2 x^{3}+4 x^{2}+2 x \right )}}+4 x^{2}}}{\left (x^{2}+x \right ) \mathrm {log}\left (2 x^{3}+4 x^{2}+2 x \right )^{2}}d x \] Input:

int((((2*x^2+2*x)*log(2*x^3+4*x^2+2*x)^2-24*x-8)*exp((x*log(2*x^3+4*x^2+2* 
x)+4)/log(2*x^3+4*x^2+2*x))^2+((4*x^3+8*x^2+4*x)*log(2*x^3+4*x^2+2*x)^2-48 
*x^2-16*x)*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))+(8*x^3+8*x 
^2)*log(2*x^3+4*x^2+2*x)^2)*exp(exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4 
*x^2+2*x))^2+4*x*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))+4*x^ 
2)/(x^2+x)/log(2*x^3+4*x^2+2*x)^2,x)
 

Output:

int((((2*x^2+2*x)*log(2*x^3+4*x^2+2*x)^2-24*x-8)*exp((x*log(2*x^3+4*x^2+2* 
x)+4)/log(2*x^3+4*x^2+2*x))^2+((4*x^3+8*x^2+4*x)*log(2*x^3+4*x^2+2*x)^2-48 
*x^2-16*x)*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))+(8*x^3+8*x 
^2)*log(2*x^3+4*x^2+2*x)^2)*exp(exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4 
*x^2+2*x))^2+4*x*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))+4*x^ 
2)/(x^2+x)/log(2*x^3+4*x^2+2*x)^2,x)