I want to calculate the convolution of a function $f(t)$ with two other functions $M_1(t)$ and $M_2(t)$, i.e. $f(t)*M_1(t)*M_2(t)$. The functions are as defined below where $u(t)$ is the step function. $F$, $R$, $\theta_s$, $T_v$, $T_1$, and $T_2$ are constants, and $T_v>T_1>T_2$.
$$ f(t) = \begin{cases} 0 & \quad \text{, }t<0 \\ -F\sin(\frac{F}{R}t+\theta_s) & \quad \text{, } 0\leq t< T_v\\ 0 & \quad \text{, } t\geq T_v \end{cases} $$
$$ M_1(t) = \frac{u(t)-u(t-T_1)}{T_1} $$
$$ M_2(t) = \frac{u(t)-u(t-T_2)}{T_2} $$
From the solution obtained by assigning random values to the constants and using MATLAB's discrete convolution function conv.m, I know the solution will be split into 7 piecewise functions on $[0,T_2)$, $[T_2,T_1)$, $[T_1,T_1+T_2)$, $[T_1+T_2,T_v)$, $[T_v,T_v+T_2)$, $[T_v+T_2,T_v+T_1)$, and $[T_v+T_1,T_v+T_1+T_2)$. The function at $(-\infty,0)$ and $[T_v+T_1+T_2,\infty)$ will be zero.
I thought it would be easier to solve $f(t)*M_1(t)*M_2(t)$ by solving it as $(f(t)*M_1(t))*M_2(t)$ but I am stuck after I obtained $f(t)*M_1(t)$. I calculated $f(t)*M_1(t)$ to be as shown below:
$$ f(t)*M_1(t) = \int\limits_0^t f(t-\tau)M_1(t) \mathrm{d}\tau = \begin{cases} 0 & \quad \text{, } t<0 \\ \frac{R}{T_1} \big[\cos(\frac{F}{R}t+\theta_s)- \cos(\theta_s) \big] & \quad \text{, }0\leq t< T_1 \\ \frac{R}{T_1} \big[\cos(\frac{F}{R}t+\theta_s)- \cos(\frac{F}{R}(t-T_1)+\theta_s) \big] & \quad \text{, } T_1\leq t< T_v\\ \frac{R}{T_1} \big[\cos(\frac{F}{R}T_v+\theta_s)- \cos(\frac{F}{R}(t-T_1)+\theta_s) \big] & \quad \text{, } T_v\leq t< T_v+T_1\\ 0 & \quad \text{, } t\geq T_v+T_1 \end{cases} $$
In particular, I am having difficulty determining the definite integrals and the integration bounds needed to solve the pieces of $(f(t)*M_1(t))*M_2(t)$ on $[T_1,T_1+T_2)$, $[T_v,T_v+T_2)$, and $[T_v+T_1,T_v+T_1+T_2)$. Can someone help explain to me how to solve this convolution for these specific pieces?