I have recently found this property for the integral of a matrix exponential with a sum on the upper bound, valid for any scalars $a$ and $b$:
$$\int_0^{a+b}e^{As}ds = \int_0^ae^{As}ds + \int_0^be^{As}ds\big(A\int_0^ae^{As}ds + I\big)$$
I was not able to prove it myself. Any advice would be very appreciated! Thanks.
I was able to prove it, so I will post the solution in case anyone finds it useful. Actually, it is pretty simple, using the fact that
$$A\int_0^xe^{As}ds = e^{Ax} - I.\tag{1}\label{1}$$
By directly applying (\ref{1}) and splitting the exponent we obtain
$$A\int_0^{a+b} e^{As}ds = e^{A(a+b)} - I = e^{Ab}e^{Aa} - I.\tag{2}\label{2}$$
By inverting (\ref{1}) we can rewrite the matrix exponential as
$$e^{Ab}e^{Aa} - I = \big(e^{Ab}-I+I\big)\big(e^{Aa}-I+I\big) - I = \big( A\int_0^{b} e^{As}ds + I\big)\big( A\int_0^{a} e^{As}ds + I\big) - I.\tag{3}\label{3}$$
Putting together (\ref{2}) and (\ref{3}) we get
$$A\int_0^{a+b} e^{As}ds = A\big[\int_0^ae^{As}ds + \int_0^be^{As}ds\big(A\int_0^ae^{As}ds + I\big)\big].$$
Assuming that $A$ is invertible we get the desired property. $\square$