Let $a$, $b$ and $c$ be positive real scalars such that $a>b,c$ and consider the following integral $$ \int_{t_0}^{t_1} \log\left(\frac{a-c}{a-b\sin(\tau)}\right)\,\mathrm{d}\tau. $$
My question. Does there exist some sort of nice closed-form expression of the above integral? If so, does there also exist a "nice" closed-form expression of the following multiple integral $$ \int_{t_0}^{t_1} \int_{\tau_{1,0}}^{\tau_1} \cdots \int_{\tau_{k-1,0}}^{\tau_{k-1}} \log\left(\frac{a-c}{a-b\sin(\tau_k)}\right)\,\mathrm{d}\tau_k \cdots\,\mathrm{d}\tau_2\,\mathrm{d}\tau_1 \ \ \ ? $$
Considering $$I=\int \log\left(\frac{a-c}{a-b\sin(\tau)}\right)\,d\tau$$ use the usual tangent half-angle substitution to get $$\frac 12I =\int \frac{\log \left( (a-c)\left(t^2+1\right)\right)}{t^2+1}\,dt-\int\frac{\log \left(a t^2-2 b t+a\right)}{t^2+1}\,dt$$
Looking at @Maxim's answer to this question,and quoting
"Factoring the polynomials and formally expanding the logarithm reduces the indefinite integral to a sum of integrals of the form" $$\int \frac {\log (u)} {u + a}\, du = \log (u)\, \log \left( 1 + \frac u a \right) + \operatorname{Li}_2 \left( - \frac u a \right)$$