If $$\int \sqrt{\sec 2x - 1} dx = \alpha \log_e \left|\cos 2x + \beta + \sqrt{\cos 2x\left(1 + \cos\frac{1}{\beta}x\right)}\right|$$ + constant, then $$\beta - \alpha$$ is equal to

$$I = \int \sqrt{\sec 2x - 1} \, dx = \int \sqrt{\frac{1 - \cos 2x}{\cos 2x}} \, dx = \int \sqrt{\frac{2\sin^2 x}{\cos 2x}} \, dx$$

Assuming $$\sin x > 0$$:

$$I = \sqrt{2} \int \frac{\sin x}{\sqrt{\cos 2x}} \, dx = \sqrt{2} \int \frac{\sin x}{\sqrt{2\cos^2 x - 1}} \, dx$$

Using substitution $$\cos x = t \implies -\sin x \, dx = dt$$:

$$I = -\sqrt{2} \int \frac{dt}{\sqrt{2t^2 - 1}} = -\int \frac{\sqrt{2} \, dt}{\sqrt{(\sqrt{2}t)^2 - 1}} = -\ln \left\vert{} \sqrt{2}t + \sqrt{2t^2 - 1} \right\vert{} + C$$

$$I = -\ln \left\vert{} \sqrt{2}\cos x + \sqrt{\cos 2x} \right\vert{} + C$$

$$I = -\frac{1}{2} \ln \left\vert{} \left( \sqrt{2}\cos x + \sqrt{\cos 2x} \right)^2 \right\vert{} + C$$

$$I = -\frac{1}{2} \ln \left\vert{} 2\cos^2 x + \cos 2x + 2\sqrt{2}\cos x\sqrt{\cos 2x} \right\vert{} + C$$

$$I = -\frac{1}{2} \ln \left\vert{} 2\cos 2x + 1 + 2\sqrt{\cos 2x} \sqrt{1 + \cos 2x} \right\vert{} + C$$

$$I = -\frac{1}{2} \ln \left\vert{} 2\left( \cos 2x + \frac{1}{2} + \sqrt{\cos 2x (1 + \cos 2x)} \right) \right\vert{} + C$$

$$I = -\frac{1}{2} \ln \left\vert{} \cos 2x + \frac{1}{2} + \sqrt{\cos 2x (1 + \cos 2x)} \right\vert{} + C'$$

$$\alpha = -\frac{1}{2}, \quad \beta = \frac{1}{2}$$

$$\beta - \alpha = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1$$