This is not an answer; rather, it is an extended comment on Terry's comments on Zurab's answer.
Here's an explicit counterexample to Popov's formula: Let
$$
iA = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad
B = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} .
$$
Then the integrand equals
$$
\begin{pmatrix} e^t & 0 \\ 0 & e^{-t} \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} e^{-t} & 0 \\ 0 & e^t \end{pmatrix}
= \begin{pmatrix} 0 & -e^{2t} \\ e^{-2t} & 0 \end{pmatrix} \equiv F(t) .
$$
However, the matrix $C\equiv iA+B$ from Popov's formula equals
$$
C = \begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix} ,
$$
so satisfies $C^2=0$, hence $e^{CT} = 1+CT$, and I think it's already very plausible that we cannot possibly have $e^{CT}e^{-iAT}$ equal to the exponential of the LHS. However, I also computed the LHS explicitly and found
$$
\exp \left( \int_0^T F(t)\, dt \right) = \begin{pmatrix} \cos\mu & (a/\mu)\sin\mu\\
-(\mu/a)\sin\mu & \cos\mu\end{pmatrix}, \quad a =-e^T\sinh T, \: \mu =\sqrt{\frac{\cosh 2T - 1}{2}} .
$$
Update: I think I've successfully read Popov's mind now: He's really claiming that (notation as above, $F$ is the integrand)
$$
Te^{\int_0^T F} = e^{(iA+B)T}e^{-iAT} \quad\quad (1)
$$
where (the first) $T$ is the time-ordering "operator" (see my comment above). More succinctly (and precisely), we define the LHS as the fundamental matrix of $Y'=YF$. Then (1) holds trivially (both sides solve the same IVP). While $\int F$ does make an appearance on the LHS, this is a purely notational sleight of hand and the formula really seems useless for the problem at hand.