How can we show that $\int_{-\infty}^{+\infty}{b\over 1+2ax+(cx)^2}\cdot{\mathrm dx\over 1-2ax+(cx)^2}={\pi\over 2}?$ Consider this integral

$$\int_{-\infty}^{+\infty}{b\over 1+2ax+(cx)^2}\cdot{\mathrm dx\over 1-2ax+(cx)^2}={\pi\over 2}\tag1$$
  Where $a^2+b^2=c^2$; Pythagoras  theorem.

Making an attempt:
Complete the square of $(cx)^2+2ax+1=c^2\left(x+{a\over c^2}\right)^2+{c^2-a^2\over c^2}$, I can't see how this would be of any useful.
$1+2x^2(c^2-2a^2)+c^4x^4=(1+2ax+c^2x^2)(1-2ax+c^2x^2)$
I am even sure how to go about making an attempt to find the equivalent of $(1)$.
How may we go about and prove $(1)?$

Let $I$ denote the integral and write $(a, b) = (c\alpha, c\beta)$ so that $\alpha^2 + \beta^2 = 1$. Then by the substitution $x = 1/ct$, we have
\begin{align*}
I
= \int_{-\infty}^{\infty} \frac{b}{(1 + c^2x^2)^2 - 4a^2x^2} \, dx
&= \int_{-\infty}^{\infty} \frac{\beta}{(t + t^{-1})^2 - 4\alpha^2} \, dt \\
&= \int_{-\infty}^{\infty} \frac{\beta}{(t - t^{-1})^2 + 4\beta^2} \, dt.
\end{align*}
Now it follows from the Glasser's master theorem1) that
$$ I = \int_{-\infty}^{\infty} \frac{\beta}{u^2 + 4\beta^2} \, du = \left[ \frac{1}{2}\arctan\left(\frac{u}{2\beta}\right) \right]_{-\infty}^{\infty} = \frac{\pi}{2}. $$

$\text{1)}$ Notice that the following good old trick also works: substituting $t \mapsto 1/t$, we have
$$ \int_{0}^{\infty} \frac{\beta}{(t - t^{-1})^2 + 4\beta^2} \, dt
= \int_{0}^{\infty} \frac{t^{-2} \beta}{(t - t^{-1})^2 + 4\beta^2} \, dt. $$
Thus it follows that
$$ I
= \int_{0}^{\infty} \frac{(1 + t^{-2}) \beta}{(t - t^{-1})^2 + 4\beta^2} \, dt
= \int_{-\infty}^{\infty} \frac{\beta}{u^2 + 4\beta^2} \, du $$
where $u = t - 1/t$. The rest is the same.
