Define the function $\mathcal{I}:\mathbb{R}_{\ge0}\rightarrow\mathbb{R}$ via the definite integral
$$\mathcal{I}{\left(a\right)}:=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(a+\sin{\left(\varphi\right)}\right)}.$$
The $a\ge1$ case has already been addressed elsewhere, so this answer will just consider the $0<a<1$ case.
Given $a\in(0,1)$ and setting $\arcsin{\left(a\right)}=:\alpha\in\left(0,\frac{\pi}{2}\right)$, we find
$$\begin{align}
\mathcal{I}{\left(a\right)}
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(a+\sin{\left(\varphi\right)}\right)}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(\sin{\left(\alpha\right)}+\sin{\left(\varphi\right)}\right)}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\alpha+\varphi}{2}\right)}\cos{\left(\frac{\alpha-\varphi}{2}\right)}\right)}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\left[-\ln{\left(2\right)}+\ln{\left(2\sin{\left(\frac{\alpha+\varphi}{2}\right)}\right)}+\ln{\left(2\cos{\left(\frac{\alpha-\varphi}{2}\right)}\right)}\right]\\
&=-\frac{\pi}{2}\ln{\left(2\right)}+\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi+\alpha}{2}\right)}\right)}+\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(2\cos{\left(\frac{\varphi-\alpha}{2}\right)}\right)}\\
&=-\frac{\pi}{2}\ln{\left(2\right)}+\int_{\alpha}^{\frac{\pi}{2}+\alpha}\mathrm{d}\vartheta\,\ln{\left(2\sin{\left(\frac{\vartheta}{2}\right)}\right)};~~~\small{\left[\varphi=\vartheta-\alpha\right]}\\
&~~~~~+\int_{-\alpha}^{\frac{\pi}{2}-\alpha}\mathrm{d}\omega\,\ln{\left(2\cos{\left(\frac{\omega}{2}\right)}\right)};~~~\small{\left[\varphi=\omega+\alpha\right]}\\
&=-\frac{\pi}{2}\ln{\left(2\right)}+\int_{\alpha}^{\frac{\pi}{2}+\alpha}\mathrm{d}\vartheta\,\ln{\left(2\sin{\left(\frac{\vartheta}{2}\right)}\right)}\\
&~~~~~+\int_{\frac{\pi}{2}+\alpha}^{\pi+\alpha}\mathrm{d}\vartheta\,\ln{\left(2\sin{\left(\frac{\vartheta}{2}\right)}\right)};~~~\small{\left[\omega=\pi-\vartheta\right]}\\
&=-\frac{\pi}{2}\ln{\left(2\right)}+\int_{\alpha}^{\pi+\alpha}\mathrm{d}\vartheta\,\ln{\left(2\sin{\left(\frac{\vartheta}{2}\right)}\right)}\\
&=-\frac{\pi}{2}\ln{\left(2\right)}+\operatorname{Cl}_{2}{\left(\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi+\alpha\right)},\\
\end{align}$$
where the Clausen function is given by the integral
$$\operatorname{Cl}_{2}{\left(\theta\right)}=-\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|};~~~\small{\theta\in\mathbb{R}}.$$
Basic properties of the Clausen function include:
$$\operatorname{Cl}_{2}{\left(m\pi\right)}=0;~~~\small{m\in\mathbb{Z}},$$
$$\operatorname{Cl}_{2}{\left(-\theta\right)}=-\operatorname{Cl}_{2}{\left(\theta\right)};~~~\small{\theta\in\mathbb{R}},$$
$$\operatorname{Cl}_{2}{\left(\theta+2m\pi\right)}=\operatorname{Cl}_{2}{\left(\theta\right)};~~~\small{m\in\mathbb{Z}\land\theta\in\mathbb{R}}.$$
Recall the definition of the so-called inverse tangent integral function:
$$\operatorname{Ti}_{2}{\left(x\right)}:=\int_{0}^{x}\mathrm{d}t\,\frac{\arctan{\left(t\right)}}{t};~~~\small{x\in\mathbb{R}}.$$
For $\theta\in\left(0,\frac{\pi}{2}\right)$, the inverse tangent integral $\operatorname{Ti}_{2}{\left(\tan{\left(\theta\right)}\right)}$ has the following closed form in terms of Clausen functions:
$$\begin{align}
\operatorname{Ti}_{2}{\left(\tan{\left(\theta\right)}\right)}
&=\int_{0}^{\tan{\left(\theta\right)}}\mathrm{d}t\,\frac{\arctan{\left(t\right)}}{t}\\
&=\int_{0}^{\theta}\mathrm{d}\vartheta\,\frac{\vartheta\sec^{2}{\left(\vartheta\right)}}{\tan{\left(\vartheta\right)}};~~~\small{\left[\arctan{\left(t\right)}=\vartheta\right]}\\
&=\int_{0}^{\theta}\mathrm{d}\vartheta\,\vartheta\csc{\left(\vartheta\right)}\sec{\left(\vartheta\right)}\\
&=\theta\ln{\left(\tan{\left(\theta\right)}\right)}-\int_{0}^{\theta}\mathrm{d}\vartheta\,\ln{\left(\tan{\left(\vartheta\right)}\right)};~~~\small{I.B.P.}\\
&=\theta\ln{\left(\tan{\left(\theta\right)}\right)}-\int_{0}^{2\theta}\mathrm{d}\varphi\,\frac12\ln{\left(\tan{\left(\frac{\varphi}{2}\right)}\right)};~~~\small{\left[\vartheta=\frac{\varphi}{2}\right]}\\
&=\theta\ln{\left(\tan{\left(\theta\right)}\right)}-\frac12\int_{0}^{2\theta}\mathrm{d}\varphi\,\left[\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)}-\ln{\left(2\cos{\left(\frac{\varphi}{2}\right)}\right)}\right]\\
&=\theta\ln{\left(\tan{\left(\theta\right)}\right)}+\frac12\left[-\int_{0}^{2\theta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)}+\int_{0}^{2\theta}\mathrm{d}\varphi\,\ln{\left(2\cos{\left(\frac{\varphi}{2}\right)}\right)}\right]\\
&=\theta\ln{\left(\tan{\left(\theta\right)}\right)}+\frac12\left[\operatorname{Cl}_{2}{\left(2\theta\right)}+\int_{0}^{2\theta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\pi-\varphi}{2}\right)}\right)}\right]\\
&=\theta\ln{\left(\tan{\left(\theta\right)}\right)}+\frac12\left[\operatorname{Cl}_{2}{\left(2\theta\right)}-\int_{\pi}^{\pi-2\theta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)}\right];~~~\small{\left[\varphi\mapsto\pi-\varphi\right]}\\
&=\theta\ln{\left(\tan{\left(\theta\right)}\right)}+\frac12\left[\operatorname{Cl}_{2}{\left(2\theta\right)}+\operatorname{Cl}_{2}{\left(\pi-2\theta\right)}-\operatorname{Cl}_{2}{\left(\pi\right)}\right]\\
&=\theta\ln{\left(\tan{\left(\theta\right)}\right)}+\frac12\operatorname{Cl}_{2}{\left(2\theta\right)}+\frac12\operatorname{Cl}_{2}{\left(\pi-2\theta\right)}.\\
\end{align}$$
We can use this relation to rewrite the expression for $\mathcal{I}$ obtained above as
$$\begin{align}
\mathcal{I}{\left(a\right)}
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(a+\sin{\left(\varphi\right)}\right)}\\
&=-\frac{\pi}{2}\ln{\left(2\right)}+\operatorname{Cl}_{2}{\left(\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi+\alpha\right)}\\
&=-\frac{\pi}{2}\ln{\left(2\right)}+\operatorname{Cl}_{2}{\left(\alpha\right)}+\operatorname{Cl}_{2}{\left(-\pi-\alpha\right)}\\
&=-\frac{\pi}{2}\ln{\left(2\right)}+\operatorname{Cl}_{2}{\left(\alpha\right)}+\operatorname{Cl}_{2}{\left(\pi-\alpha\right)}\\
&=-\frac{\pi}{2}\ln{\left(2\right)}-\alpha\ln{\left(\tan{\left(\frac{\alpha}{2}\right)}\right)}+2\operatorname{Ti}_{2}{\left(\tan{\left(\frac{\alpha}{2}\right)}\right)}\\
&=-\frac{\pi}{2}\ln{\left(2\right)}-\arcsin{\left(a\right)}\ln{\left(\frac{a}{1+\sqrt{1-a^{2}}}\right)}+2\operatorname{Ti}_{2}{\left(\frac{a}{1+\sqrt{1-a^{2}}}\right)}.\blacksquare\\
\end{align}$$