The following reduces the equation to a quartic. Let $b=2 a d, x = a(y - d)$, then:
$$\frac{y+d}{\sqrt{(y+d)^2 + 1}} - \frac{y-d}{\sqrt{(y-d)^2 + 1}} = c$$
Squaring:
$$\frac{(y+d)^2}{(y+d)^2 + 1} + \frac{(y-d)^2}{(y-d)^2 + 1} - 2 \;\frac{y^2-d^2}{\sqrt{(y+d)^2 + 1}\sqrt{(y-d)^2 + 1}} = c^2$$
Eliminating denominators:
$$(y+d)^2\big((y-d)^2 + 1\big) + (y-d)^2\big((y+d)^2 + 1\big) -c^2 \big((y+d)^2 + 1\big) \big((y-d)^2 + 1\big)
= 2 (y^2-d^2)\sqrt{\big((y+d)^2 + 1\big)\big((y-d)^2 + 1\big)}$$
Collecting and rearranging:
$$(2-c^2)(y^2-d^2)^2 + 2(1-c^2)(y^2+d^2) -c^2 = 2(y^2-d^2)\sqrt{(y^2-d^2)^2 + 2(y^2+d^2)+1}$$
Squaring again gives a quartic in $z = y^2$, which then could technically be solved in radicals.