\(\int \frac {e^{-\frac {2 (-5 e^{e^{x^2}}-5 x+x^2)}{e^{e^{x^2}}+x}} (2 e^{2 e^{x^2}} \log (x+\log (2))+2 x^2 \log (x+\log (2))+(-2 x^3-2 x^2 \log (2)) \log ^2(x+\log (2))+e^{e^{x^2}} (4 x \log (x+\log (2))+(-4 x^2-4 x \log (2)+e^{x^2} (4 x^4+4 x^3 \log (2))) \log ^2(x+\log (2))))}{x^3+x^2 \log (2)+e^{2 e^{x^2}} (x+\log (2))+e^{e^{x^2}} (2 x^2+2 x \log (2))} \, dx\) [6466]

Optimal result

Integrand size = 180, antiderivative size = 30 \[ \int \frac {e^{-\frac {2 \left (-5 e^{e^{x^2}}-5 x+x^2\right )}{e^{e^{x^2}}+x}} \left (2 e^{2 e^{x^2}} \log (x+\log (2))+2 x^2 \log (x+\log (2))+\left (-2 x^3-2 x^2 \log (2)\right ) \log ^2(x+\log (2))+e^{e^{x^2}} \left (4 x \log (x+\log (2))+\left (-4 x^2-4 x \log (2)+e^{x^2} \left (4 x^4+4 x^3 \log (2)\right )\right ) \log ^2(x+\log (2))\right )\right )}{x^3+x^2 \log (2)+e^{2 e^{x^2}} (x+\log (2))+e^{e^{x^2}} \left (2 x^2+2 x \log (2)\right )} \, dx=-5+e^{10-\frac {2 x^2}{e^{e^{x^2}}+x}} \log ^2(x+\log (2)) \]

[Out]

ln(ln(2)+x)^2/exp(x^2/(exp(exp(x^2))+x)-5)^2-5

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(182\) vs. \(2(30)=60\).

Time = 2.83 (sec) , antiderivative size = 182, normalized size of antiderivative = 6.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6820, 12, 2326} \[ \int \frac {e^{-\frac {2 \left (-5 e^{e^{x^2}}-5 x+x^2\right )}{e^{e^{x^2}}+x}} \left (2 e^{2 e^{x^2}} \log (x+\log (2))+2 x^2 \log (x+\log (2))+\left (-2 x^3-2 x^2 \log (2)\right ) \log ^2(x+\log (2))+e^{e^{x^2}} \left (4 x \log (x+\log (2))+\left (-4 x^2-4 x \log (2)+e^{x^2} \left (4 x^4+4 x^3 \log (2)\right )\right ) \log ^2(x+\log (2))\right )\right )}{x^3+x^2 \log (2)+e^{2 e^{x^2}} (x+\log (2))+e^{e^{x^2}} \left (2 x^2+2 x \log (2)\right )} \, dx=-\frac {e^{\frac {2 \left (5 e^{e^{x^2}}+(5-x) x\right )}{e^{e^{x^2}}+x}} x \left (-e^{x^2+e^{x^2}} x^2 (2 x+\log (4))+e^{e^{x^2}} (2 x+\log (4))+x (x+\log (2))\right ) \log ^2(x+\log (2))}{\left (e^{e^{x^2}}+x\right )^2 \left (\frac {10 e^{x^2+e^{x^2}} x-2 x+5}{e^{e^{x^2}}+x}-\frac {\left (2 e^{x^2+e^{x^2}} x+1\right ) \left (5 e^{e^{x^2}}+(5-x) x\right )}{\left (e^{e^{x^2}}+x\right )^2}\right ) (x+\log (2))} \]

[In]

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

[Out]

-((E^((2*(5*E^E^x^2 + (5 - x)*x))/(E^E^x^2 + x))*x*(x*(x + Log[2]) + E^E^x^2*(2*x + Log[4]) - E^(E^x^2 + x^2)*
x^2*(2*x + Log[4]))*Log[x + Log[2]]^2)/((E^E^x^2 + x)^2*((5 - 2*x + 10*E^(E^x^2 + x^2)*x)/(E^E^x^2 + x) - ((1
+ 2*E^(E^x^2 + x^2)*x)*(5*E^E^x^2 + (5 - x)*x))/(E^E^x^2 + x)^2)*(x + Log[2])))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2326

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]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 e^{\frac {10 e^{e^{x^2}}-2 (-5+x) x}{e^{e^{x^2}}+x}} \log (x+\log (2)) \left (\left (e^{e^{x^2}}+x\right )^2+x \left (-x (x+\log (2))-e^{e^{x^2}} (2 x+\log (4))+e^{e^{x^2}+x^2} x^2 (2 x+\log (4))\right ) \log (x+\log (2))\right )}{\left (e^{e^{x^2}}+x\right )^2 (x+\log (2))} \, dx \\ & = 2 \int \frac {e^{\frac {10 e^{e^{x^2}}-2 (-5+x) x}{e^{e^{x^2}}+x}} \log (x+\log (2)) \left (\left (e^{e^{x^2}}+x\right )^2+x \left (-x (x+\log (2))-e^{e^{x^2}} (2 x+\log (4))+e^{e^{x^2}+x^2} x^2 (2 x+\log (4))\right ) \log (x+\log (2))\right )}{\left (e^{e^{x^2}}+x\right )^2 (x+\log (2))} \, dx \\ & = -\frac {e^{\frac {2 \left (5 e^{e^{x^2}}+(5-x) x\right )}{e^{e^{x^2}}+x}} x \left (x (x+\log (2))+e^{e^{x^2}} (2 x+\log (4))-e^{e^{x^2}+x^2} x^2 (2 x+\log (4))\right ) \log ^2(x+\log (2))}{\left (e^{e^{x^2}}+x\right )^2 \left (\frac {5-2 x+10 e^{e^{x^2}+x^2} x}{e^{e^{x^2}}+x}-\frac {\left (1+2 e^{e^{x^2}+x^2} x\right ) \left (5 e^{e^{x^2}}+(5-x) x\right )}{\left (e^{e^{x^2}}+x\right )^2}\right ) (x+\log (2))} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17

\[\ln \left (\ln \left (2\right )+x \right )^{2} {\mathrm e}^{-\frac {2 \left (-5 \,{\mathrm e}^{{\mathrm e}^{x^{2}}}+x^{2}-5 x \right )}{{\mathrm e}^{{\mathrm e}^{x^{2}}}+x}}\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-\frac {2 \left (-5 e^{e^{x^2}}-5 x+x^2\right )}{e^{e^{x^2}}+x}} \left (2 e^{2 e^{x^2}} \log (x+\log (2))+2 x^2 \log (x+\log (2))+\left (-2 x^3-2 x^2 \log (2)\right ) \log ^2(x+\log (2))+e^{e^{x^2}} \left (4 x \log (x+\log (2))+\left (-4 x^2-4 x \log (2)+e^{x^2} \left (4 x^4+4 x^3 \log (2)\right )\right ) \log ^2(x+\log (2))\right )\right )}{x^3+x^2 \log (2)+e^{2 e^{x^2}} (x+\log (2))+e^{e^{x^2}} \left (2 x^2+2 x \log (2)\right )} \, dx=e^{\left (-\frac {2 \, {\left (x^{2} - 5 \, x - 5 \, e^{\left (e^{\left (x^{2}\right )}\right )}\right )}}{x + e^{\left (e^{\left (x^{2}\right )}\right )}}\right )} \log \left (x + \log \left (2\right )\right )^{2} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (27) = 54\).

Time = 0.36 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.47 \[ \int \frac {e^{-\frac {2 \left (-5 e^{e^{x^2}}-5 x+x^2\right )}{e^{e^{x^2}}+x}} \left (2 e^{2 e^{x^2}} \log (x+\log (2))+2 x^2 \log (x+\log (2))+\left (-2 x^3-2 x^2 \log (2)\right ) \log ^2(x+\log (2))+e^{e^{x^2}} \left (4 x \log (x+\log (2))+\left (-4 x^2-4 x \log (2)+e^{x^2} \left (4 x^4+4 x^3 \log (2)\right )\right ) \log ^2(x+\log (2))\right )\right )}{x^3+x^2 \log (2)+e^{2 e^{x^2}} (x+\log (2))+e^{e^{x^2}} \left (2 x^2+2 x \log (2)\right )} \, dx=e^{\left (-x^{2} + \frac {x^{3} + x^{2} e^{\left (e^{\left (x^{2}\right )}\right )} - 2 \, x^{2} + x e^{\left (x^{2}\right )} + 10 \, x + e^{\left (x^{2} + e^{\left (x^{2}\right )}\right )} + 10 \, e^{\left (e^{\left (x^{2}\right )}\right )}}{x + e^{\left (e^{\left (x^{2}\right )}\right )}} - e^{\left (x^{2}\right )}\right )} \log \left (x + \log \left (2\right )\right )^{2} \]

[In]

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

[Out]

e^(-x^2 + (x^3 + x^2*e^(e^(x^2)) - 2*x^2 + x*e^(x^2) + 10*x + e^(x^2 + e^(x^2)) + 10*e^(e^(x^2)))/(x + e^(e^(x
^2))) - e^(x^2))*log(x + log(2))^2

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-\frac {2 \left (-5 e^{e^{x^2}}-5 x+x^2\right )}{e^{e^{x^2}}+x}} \left (2 e^{2 e^{x^2}} \log (x+\log (2))+2 x^2 \log (x+\log (2))+\left (-2 x^3-2 x^2 \log (2)\right ) \log ^2(x+\log (2))+e^{e^{x^2}} \left (4 x \log (x+\log (2))+\left (-4 x^2-4 x \log (2)+e^{x^2} \left (4 x^4+4 x^3 \log (2)\right )\right ) \log ^2(x+\log (2))\right )\right )}{x^3+x^2 \log (2)+e^{2 e^{x^2}} (x+\log (2))+e^{e^{x^2}} \left (2 x^2+2 x \log (2)\right )} \, dx=\int -\frac {{\mathrm {e}}^{\frac {2\,\left (5\,x+5\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}-x^2\right )}{x+{\mathrm {e}}^{{\mathrm {e}}^{x^2}}}}\,\left ({\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\left ({\ln \left (x+\ln \left (2\right )\right )}^2\,\left (4\,x\,\ln \left (2\right )-{\mathrm {e}}^{x^2}\,\left (4\,x^4+4\,\ln \left (2\right )\,x^3\right )+4\,x^2\right )-4\,x\,\ln \left (x+\ln \left (2\right )\right )\right )-2\,{\mathrm {e}}^{2\,{\mathrm {e}}^{x^2}}\,\ln \left (x+\ln \left (2\right )\right )+{\ln \left (x+\ln \left (2\right )\right )}^2\,\left (2\,x^3+2\,\ln \left (2\right )\,x^2\right )-2\,x^2\,\ln \left (x+\ln \left (2\right )\right )\right )}{x^2\,\ln \left (2\right )+{\mathrm {e}}^{2\,{\mathrm {e}}^{x^2}}\,\left (x+\ln \left (2\right )\right )+x^3+{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\left (2\,x^2+2\,\ln \left (2\right )\,x\right )} \,d x \]

[In]

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

[Out]

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