How to find laplace transform of $\,\sinh(ct)\int_a^te^{au}\sinh(bu)\,du$

Question

How to find laplace transform of $$\sinh(ct)\int_a^te^{au}\sinh(bu)\,du.$$

I tried to integrate inner function and then do it, but it became way more tedious. So I was thinking there should be good way out for this. Thanks for helping

Answer 1

\begin{align} &\mathcal{L}\left( \sinh(ct) \int_a^te^{au}\sinh(bu)\,\text{d}u \right)\\ =&\,\mathcal{L}\left( \sinh(ct) \frac {e^{a^2} (b \cosh(ab) - a\sinh(ab)) + e^{at} (a\sinh(bt) - b\cosh(bt))} {(a-b)(a+b)} \right) \\ =&\,\mathcal{L}\Bigg( -\frac{a e^{a^2} \sinh(ab) \sinh(ct)}{(a-b)(a+b)} +\frac{e^{a^2} b \cosh(ab) \sinh(ct)}{(a-b)(a+b)} \\ &\qquad+\frac{a e^{at} \sinh(bt) \sinh(ct)}{(a-b)(a+b)} -\frac{b e^{at} \cosh(bt) \sinh(ct)}{(a-b)(a+b)} \Bigg)\\ =&\,-\frac{a e^{a^2} \sinh(ab)}{(a-b)(a+b)} \frac{c}{s^2-c^2}+\frac{e^{a^2} b \cosh(ab)}{(a-b)(a+b)} \frac{c}{s^2-c^2}\\ &\,+\frac{a e^{at}}{(a-b)(a+b)} \frac{2 b c s} {b^4 - 2b^2 \left( c^2+s^2 \right) + \left(c^2-s^2\right)^2}\\ &\,-\frac{b e^{at}}{(a-b)(a+b)} \frac{1}{4} \left( \frac{1}{-b-c+s} +\frac{1}{b-c+s} -\frac{1}{b+c+s} +\frac{1}{b-c-s} \right) \end{align}