Finding $ \int_{-\pi /2}^{\pi /2} \frac{\log (1 + b \sin x)}{\sin x}\,\mathrm dx$ given $|b|<1$

Find $$\int_{-\pi/2}^{\pi/2}\frac{\log(1+b\sin x)}{\sin x}\,\mathrm dx$$given that $|b|<1$.

I split the integral into$$I=\int_0^{\pi/2}f(x)\,\mathrm dx+\int_{-\pi/2}^0f(x)\,\mathrm dx$$ For the second term made the substitution $x =-t$ and further solved $I$ to get$$I=\int_{0} ^{\pi / 2} \frac{\log \frac{1 + b \sin x}{1- b \sin x}}{\sin x}\,\mathrm dx$$

I do not know how to proceed further. The answer is $\pi \arcsin b$.

Let $I(b)=\int_{-\pi/2}^{\pi/2}\frac{\log(1+b\sin(x))}{\sin(x)}\,dx$. Differentiating reveals

$$\begin{align} I'(b)&=\int_{-\pi/2}^{\pi/2} \frac1{1+b\sin(x)}\,dx\\\\ &=2\left(\frac{\arctan\left(\sqrt{\frac{1+b}{1-b}}\right)+\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)}{\sqrt{1-b^2}}\right)\\\\ &=\frac{\pi}{\sqrt{1-b^2}}\tag1 \end{align}$$

Using $I(0)=0$ and integrating $(1)$ yields

$$I(b)=\pi\arcsin(b)$$

This is not an answer but a curiosity

Using the tangent half-angle substitution $x=2\tan^{-1}(t)$, we have $$I=\int_0^{\frac \pi 2}\csc (x) \log \left(\frac{1+b \sin (x)}{1-b \sin (x)}\right)=\int_0^1\frac{1}{t}\log \left(1+\frac{4 b t}{t^2-2 b t+1}\right)\,dt$$ Using Taylor series

$$\frac{1}{t}\log \left(1+\frac{4 b t}{t^2-2 b t+1}\right)=2\sum_{n=0}^\infty \frac {a_+^{2 n+1}+a_-^{2 n+1}}{2n+1} t^{2n}\quad \text{where}\quad \color{blue}{a_\pm=b\pm\sqrt{b^2-1}}$$

$$I=2\sum_{n=0}^\infty \frac {a_+^{2 n+1}+a_-^{2 n+1}}{(2n+1)^2}$$ $$I=\color{blue}{\frac 12\Bigg[a_+ \Phi \left(a_+^2,2,\frac{1}{2}\right)+a_- \Phi \left(a_-^2,2,\frac{1}{2}\right)\Bigg]}$$ where appears the Lerch transcendent function.

I have not be able to simplify this expression