\[ y'(x)=\frac {2 x F(y(x)+\log (2 x+1))+F(y(x)+\log (2 x+1))-2}{2 x+1} \] ✓ Mathematica : cpu = 0.351902 (sec), leaf count = 117
DSolve[Derivative[1][y][x] == (-2 + F[Log[1 + 2*x] + y[x]] + 2*x*F[Log[1 + 2*x] + y[x]])/(1 + 2*x),y[x],x]
\[\text {Solve}\left [\int _1^{y(x)}-\frac {F(K[2]+\log (2 x+1)) \int _1^x-\frac {2 F'(K[2]+\log (2 K[1]+1))}{F(K[2]+\log (2 K[1]+1))^2 (2 K[1]+1)}dK[1]-1}{F(K[2]+\log (2 x+1))}dK[2]+\int _1^x\left (\frac {2}{F(\log (2 K[1]+1)+y(x)) (2 K[1]+1)}-1\right )dK[1]=c_1,y(x)\right ]\] ✓ Maple : cpu = 0.155 (sec), leaf count = 27
dsolve(diff(y(x),x) = 1/(2*x+1)*(2*F(y(x)+ln(2*x+1))*x+F(y(x)+ln(2*x+1))-2),y(x))
\[y \left (x \right ) = -\ln \left (2 x +1\right )+\RootOf \left (-x +\int _{}^{\textit {\_Z}}\frac {1}{F \left (\textit {\_a} \right )}d \textit {\_a} +c_{1}\right )\]