Answer
$\frac{1}{{15}}{\left( {16 + {x^2}} \right)^{3/2}}\left( {3{x^2} - 32} \right) + C$
Work Step by Step
$$\eqalign{
& \int {{x^3}\sqrt {16 + {x^2}} } dx \cr
& {\text{From the triangle}} \cr
& {\text{tan}}\theta = \frac{x}{4},{\text{ }}x = 4\tan \theta ,{\text{ }}dx = 4{\sec ^2}\theta d\theta \cr
& \sec \theta = \frac{{\sqrt {16 + {x^2}} }}{4},{\text{ }}\sqrt {16 + {x^2}} = 4\sec \theta \cr
& {\text{Substituting}}{\text{, we obtain}} \cr
& \int {{x^3}\sqrt {16 + {x^2}} } dx = \int {{{\left( {4\tan \theta } \right)}^3}\left( {4\sec \theta } \right)\left( {4{{\sec }^2}\theta } \right)} d\theta \cr
& = \int {{4^5}{{\tan }^3}\theta {{\sec }^3}\theta } d\theta \cr
& = {4^5}\int {{{\tan }^2}\theta {{\sec }^2}\theta \sec \theta \tan \theta } d\theta \cr
& {\text{Use the pythagorean identity ta}}{{\text{n}}^2}\theta = {\sec ^2}\theta - 1 \cr
& = {4^5}\int {\left( {{{\sec }^2}\theta - 1} \right){{\sec }^2}\theta \sec \theta \tan \theta } d\theta \cr
& = {4^5}\int {\left( {{{\sec }^4}\theta - {{\sec }^2}\theta } \right)\sec \theta \tan \theta } d\theta \cr
& {\text{Integrating}} \cr
& = {4^5}\left( {\frac{{{{\sec }^5}\theta }}{5} - \frac{{{{\sec }^3}\theta }}{3}} \right) + C \cr
& = \frac{{{4^5}}}{{15}}{\sec ^3}\theta \left( {3{{\sec }^2}\theta - 5} \right) + C \cr
& {\text{Where sec}}\theta = \frac{{\sqrt {16 + {x^2}} }}{4} \cr
& = \frac{{{4^5}}}{{15}}{\left( {\frac{{\sqrt {16 + {x^2}} }}{4}} \right)^3}\left( {3{{\left( {\frac{{\sqrt {16 + {x^2}} }}{4}} \right)}^2} - 5} \right) + C \cr
& {\text{Simplifying}} \cr
& = \frac{{{4^5}}}{{15}}\left( {\frac{{{{\left( {16 + {x^2}} \right)}^{3/2}}}}{{{4^3}}}} \right)\left( {3\left( {\frac{{\left( {16 + {x^2}} \right)}}{{16}}} \right) - 5} \right) + C \cr
& = \frac{{16}}{{15}}{\left( {16 + {x^2}} \right)^{3/2}}\left( {\frac{{48 + 3{x^2} - 80}}{{16}}} \right) + C \cr
& = \frac{1}{{15}}{\left( {16 + {x^2}} \right)^{3/2}}\left( {3{x^2} - 32} \right) + C \cr} $$
