Rewrite the integral and apply integration by parts.

Let $I = \int \sec^3 x \, dx$.
We can write $\sec^3 x = \sec x \cdot \sec^2 x$.
Use integration by parts, $\int u \, dv = uv - \int v \, du$.
Let $u = \sec x$ and $dv = \sec^2 x \, dx$.
Then $du = \sec x \tan x \, dx$ and $v = \tan x$.

$$I = \sec x \tan x - \int \tan x (\sec x \tan x) \, dx$$ $$I = \sec x \tan x - \int \sec x \tan^2 x \, dx$$

Substitute this identity into the integral:

$$I = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx$$ $$I = \sec x \tan x - \int (\sec^3 x - \sec x) \, dx$$ $$I = \sec x \tan x - \int \sec^3 x \, dx + \int \sec x \, dx$$

Notice that the original integral $I = \int \sec^3 x \, dx$ appears on the right side.

$$I = \sec x \tan x - I + \int \sec x \, dx$$

Add $I$ to both sides:

$$2I = \sec x \tan x + \int \sec x \, dx$$

The integral of $\sec x$ is $\ln|\sec x + \tan x|$.

$$2I = \sec x \tan x + \ln|\sec x + \tan x| + C'$$

Divide by 2 to solve for $I$:

$$I = \frac{1}{2} (\sec x \tan x + \ln|\sec x + \tan x|) + C$$

where $C = \frac{C'}{2}$ is the constant of integration.

The final answer is $\frac{1}{2} (\sec x \tan x + \ln|\sec x + \tan x|) + C$.