Integrand size = 299, antiderivative size = 28 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx=e^{\left (e^{x+\frac {4}{\log \left ((2+2 x) \left (x+x^2\right )\right )}}+2 x\right )^2} \]
Time = 0.36 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx=e^{\left (e^{x+\frac {4}{\log \left (2 x (1+x)^2\right )}}+2 x\right )^2} \]
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]
Time = 7.67 (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 |
\(\displaystyle e^{\left (2 x+e^{x+\frac {4}{\log \left (2 x (x+1)^2\right )}}\right )^2}\) |
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]
3.23.34.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(87\) vs. \(2(26)=52\).
Time = 5.33 (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 | \({\mathrm e}^{{\mathrm e}^{\frac {2 x \ln \left (2 x^{3}+4 x^{2}+2 x \right )+8}{\ln \left (2 x^{3}+4 x^{2}+2 x \right )}}+4 x \,{\mathrm e}^{\frac {x \ln \left (2 x^{3}+4 x^{2}+2 x \right )+4}{\ln \left (2 x^{3}+4 x^{2}+2 x \right )}}+4 x^{2}}\) | \(89\) |
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)
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)
Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.11 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx=e^{\left (4 \, x^{2} + 4 \, x e^{\left (\frac {x \log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right ) + 4}{\log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right )}\right )} + e^{\left (\frac {2 \, {\left (x \log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right ) + 4\right )}}{\log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right )}\right )}\right )} \]
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=\
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)))
Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (22) = 44\).
Time = 5.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.93 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx=e^{4 x^{2} + 4 x e^{\frac {x \log {\left (2 x^{3} + 4 x^{2} + 2 x \right )} + 4}{\log {\left (2 x^{3} + 4 x^{2} + 2 x \right )}}} + e^{\frac {2 \left (x \log {\left (2 x^{3} + 4 x^{2} + 2 x \right )} + 4\right )}{\log {\left (2 x^{3} + 4 x^{2} + 2 x \right )}}}} \]
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)
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)))
Time = 0.60 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx=e^{\left (4 \, x^{2} + 4 \, x e^{\left (x + \frac {4}{\log \left (2\right ) + 2 \, \log \left (x + 1\right ) + \log \left (x\right )}\right )} + e^{\left (2 \, x + \frac {8}{\log \left (2\right ) + 2 \, \log \left (x + 1\right ) + \log \left (x\right )}\right )}\right )} \]
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=\
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))))
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (25) = 50\).
Time = 16.98 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.00 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx=e^{\left (4 \, x^{2} + 4 \, x e^{\left (x + \frac {4}{\log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right )}\right )} + e^{\left (2 \, x + \frac {8}{\log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right )}\right )}\right )} \]
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=\
Time = 11.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.11 \[ \int \frac {e^{e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}}+4 e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} x+4 x^2} \left (\left (8 x^2+8 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )+e^{\frac {2 \left (4+x \log \left (2 x+4 x^2+2 x^3\right )\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-8-24 x+\left (2 x+2 x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )+e^{\frac {4+x \log \left (2 x+4 x^2+2 x^3\right )}{\log \left (2 x+4 x^2+2 x^3\right )}} \left (-16 x-48 x^2+\left (4 x+8 x^2+4 x^3\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )\right )\right )}{\left (x+x^2\right ) \log ^2\left (2 x+4 x^2+2 x^3\right )} \, dx={\mathrm {e}}^{{\mathrm {e}}^{\frac {8}{\ln \left (2\,x^3+4\,x^2+2\,x\right )}}\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{4\,x\,{\mathrm {e}}^{\frac {4}{\ln \left (2\,x^3+4\,x^2+2\,x\right )}}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{4\,x^2} \]
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)