Too long for a comment: Your formula
$$\frac{1}{c^2}L(cB)=\frac{1}{c^2}\left(F(cB(T))-\frac{c^2}{2}\int_0^T f'(cB(t))dt-\frac12\int_0^T f^2(cB(t))dt\right):=\frac{1}{c^2}L^c(cB).$$
I cannot reproduce.

When I integrate $\int_0^Tf(cB(t))\,d(cB(t))$ by parts I get
\begin{align}
&\int_0^Tf(cB(t))\,d(cB(t))=-\int_0^TcB(t)\,df(cB(t))+\Big\langle f(cB),cB\Big\rangle_T\\
&\stackrel{\text{Ito's f.}}{=}-\int_0^TcB(t)\,f'(cB(t))\,d(cB(t))-\frac{1}{2}\int_0^TcB(t)\,f''(cB(t))\,c^2\,dt+\Big\langle f(cB),cB\Big\rangle_T\,.
\end{align}
Using Ito's formula we also get
$$
\Big\langle f(cB),cB\Big\rangle_T=\int_0^Tf'(cB(t))\,dt\,.
$$
Putting this together it looks quite different from your $L(cB(t)).$