1. **State the problem:** We want to evaluate the integral $$\int 5 \tan^3(x) \sec(x) \, dx.$$\n\n2. **Recall relevant formulas and identities:**\n- We know that $\frac{d}{dx}(\sec x) = \sec x \tan x$.\n- Also, $\tan^2 x = \sec^2 x - 1$.\n\n3. **Rewrite the integral using $\tan^2 x = \sec^2 x - 1$: **\n$$\int 5 \tan^3(x) \sec(x) \, dx = 5 \int \tan^3(x) \sec(x) \, dx = 5 \int \tan(x) \tan^2(x) \sec(x) \, dx$$\n$$= 5 \int \tan(x) (\sec^2(x) - 1) \sec(x) \, dx = 5 \int \tan(x) (\sec^3(x) - \sec(x)) \, dx.$$\n\n4. **Distribute inside the integral:**\n$$= 5 \int \tan(x) \sec^3(x) \, dx - 5 \int \tan(x) \sec(x) \, dx.$$\n\n5. **Use substitution for the first integral:** Let $u = \sec x$, then $du = \sec x \tan x \, dx$.\nRewrite the first integral: $$\int \tan(x) \sec^3(x) \, dx = \int \sec^2(x) (\sec x \tan x \, dx) = \int u^2 \, du.$$\n\n6. **Integrate:**\n$$\int u^2 \, du = \frac{u^3}{3} + C = \frac{\sec^3 x}{3} + C.$$\n\n7. **For the second integral:** $$\int \tan(x) \sec(x) \, dx,$$ use substitution $v = \sec x$, $dv = \sec x \tan x \, dx$, so\n$$\int \tan(x) \sec(x) \, dx = \int dv = v + C = \sec x + C.$$\n\n8. **Combine results:**\n$$5 \left( \frac{\sec^3 x}{3} - \sec x \right) + C = \frac{5}{3} \sec^3 x - 5 \sec x + C.$$\n\n**Final answer:** $$\boxed{\int 5 \tan^3(x) \sec(x) \, dx = \frac{5}{3} \sec^3 x - 5 \sec x + C}.$$