Answer
$$\frac{{2{{\tan }^{3/2}}x}}{3} + \frac{{2{{\tan }^{7/2}}x}}{7} + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {\tan x} {{\sec }^4}x} dx \cr
& {\text{property }}\sqrt a = {a^{1/2}} \cr
& = \int {{{\tan }^{1/2}}x{{\sec }^4}x} dx \cr
& {\text{split exponent of }}{\sec ^4}x \cr
& = \int {{{\tan }^{1/2}}x{{\sec }^2}x{{\sec }^2}x} dx \cr
& {\text{pythagorean identity 1 + ta}}{{\text{n}}^2}\theta = {\sec ^2}\theta \cr
& = \int {{{\tan }^{1/2}}x\left( {1 + {{\tan }^2}x} \right){{\sec }^2}x} dx \cr
& = \int {\left( {{{\tan }^{1/2}}x + {{\tan }^{5/2}}x} \right){{\sec }^2}x} dx \cr
& {\text{substitute }}u = \tan x,{\text{ }}du = {\sec ^2}xdx \cr
& = \int {\left( {{u^{1/2}} + {u^{5/2}}} \right)du} \cr
& {\text{find the antiderivative by the power rule}} \cr
& = \frac{{2{u^{3/2}}}}{3} + \frac{{2{u^{7/2}}}}{7} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \tan x \cr
& = \frac{{2{{\tan }^{3/2}}x}}{3} + \frac{{2{{\tan }^{7/2}}x}}{7} + C \cr} $$
