What is the closed form of the value of $\int_0^\pi\frac{\mathrm dx}{\sqrt x+\sin x}$, if it exists?

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This integral came up in a review of determining convergence of improper integrals.

$$\int_0^\pi\frac{\mathrm dx}{\sqrt x+\sin x}$$

It's easy to show the integrand is bounded below by $\dfrac1{\sqrt x}$ over the same interval. This second integral converges to

$$\int_0^\pi\frac{\mathrm dx}{\sqrt x}=2\sqrt\pi$$

and so the first must converge.

What I'm interested in is determining whether the first integral has a closed form. The only method I can think of that might offer some hope to finding one would be to integrate the complex-valued function $f(z)=\dfrac1{\sqrt z+\sin z}$ along what I'll call a "barbell contour" similar to the one shown in this example on Wikipedia.

So the idea is to examine the component integrals,

$$\int_Cf(z)\,\mathrm dz=\left\{\int_{\gamma_1}+\int_{\epsilon_1}^{\pi-\epsilon_2}+\int_{\gamma_2}+\int_{\pi-\epsilon_2}^{\epsilon_1}\right\}f(z)\,\mathrm dz$$

where $\epsilon_1,\epsilon_2$ respectively denote the radii of the circular ends of the barbell centered at $z=0$ and $z=\pi$.

Now, my fleeting memory of complex analysis suggests to me that the integral over $\gamma_2$ will disappear as $\epsilon_2\to0$ as a result of the Cauchy-Goursat theorem since there are no singularities within this portion of the contour, but I'm not sure this is correct.

The rest eludes me, as the course I took never got as far as explaining branch points/cuts - though I think I have a basic understanding of what they are, not necessarily how to work around them - and warrants hefty review on my part. Any help on illuminating the procedure here is appreciated!

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I doubt that there is a closed form. Numerically the result is approximately $$2.40899073296947082180319435824$$ The Inverse Symbolic Calculator, Maple's "identify", and the OEIS get no hits.

Since $0 < \sin(x) < \sqrt{x}$ for $0 < x < \pi$, we can expand in a series

$$ \dfrac{1}{\sqrt{x} + \sin(x)} = \sum_{k=0}^\infty (-1)^k \frac{\sin^k x}{x^{(k+1)/2}} $$

Unfortunately the integrals of these terms from $0$ to $\pi$ are complicated expressions involving sine or cosine integrals ($\text{Si}$ or $\text{Ci}$ if $k$ is odd, Fresnel sine or cosine integrals if $k$ is even.