= Solution
Put $d=\sqrt{a^2-b^2-c^2}>0$ and $h=b-ic$. The denominator of the complex <rational function> becomes
$$
P(z)=\frac h2z^2+az+\frac{b+ic}{2}.
$$
The <quadratic formula> gives its two zeros, hence the two <poles>:
$$
\boxed{z_{\rm in}=\frac{-a+d}{b-ic},\qquad z_{\rm out}=\frac{-a-d}{b-ic}.}
$$
Their product is $(b+ic)/(b-ic)$, so $|z_{\rm in}z_{\rm out}|=1$. To locate them one also needs to exclude the possibility of both lying on the <unit circle>. Here their moduli are explicit:
$$
|z_{\rm in}|=\frac{a-d}{\sqrt{b^2+c^2}}=\sqrt{\frac{a-d}{a+d}}<1,\qquad
|z_{\rm out}|=\sqrt{\frac{a+d}{a-d}}>1.
$$
Both are <simple poles>: $P'(z)=hz+a$ takes the nonzero values $d$ and $-d$ at them. In particular their <residues> are $1/d$ and $-1/d$.
For the real <integral>, use the <change of variables> $z=e^{i\theta}$. As $\theta$ runs from $-\pi$ to $\pi$, this traverses the <unit circle> once counterclockwise, and
$$
d\theta=\frac{dz}{iz},\quad\cos\theta=\frac{z+z^{-1}}2,\quad\sin\theta=\frac{z-z^{-1}}{2i}.
$$
Thus $I=(1/i)\oint_{|z|=1}f(z)dz$. Only $z_{\rm in}$ is enclosed. The <residue theorem>, or <Cauchy integral formula> after subtracting its simple-pole part, gives
$$
\boxed{I=\frac1i\frac{2\pi i}{d}=\frac{2\pi}{\sqrt{a^2-b^2-c^2}}.}
$$
The real denominator is positive because $a>\sqrt{b^2+c^2}$. As an independent check, write $b\cos\theta+c\sin\theta=\sqrt{b^2+c^2}\cos(\theta-\phi)$; a period shift reduces the answer to the existing <reciprocal affine sine integral>.
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