Contour Integration with $\sec{(\sqrt{1-x^2})}$

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I've found that

$$\int_{-\infty}^{\infty} \sec{\left(\sqrt{1-x^2}\right)}\,dx = \pi^2\sum_{k=-\infty}^{\infty} \frac{(-1)^k (2k+1)}{\sqrt{\pi^2(2k+1)^2-4}} \approx 4.89477$$

by way of just playing around with some contours, in particular, a semicircle avoiding the branch (to which I defined to just be $[-1,1]$), then summing the infinite residues along the imaginary axis.

Is there any way to exploit a branch (dog-bone?) in a similiar to compute a series representation of

$$\int_{-1}^{1} \sec{\left(\sqrt{1-x^2}\right)}\,dx \, ?$$

Perhaps one way to make some progress is via a trig substitution?