Answer
$$\frac{1}{{15}}{\left( {{x^2} - 25} \right)^{3/2}}\left( {3{x^2} + 50} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{x^3}\sqrt {{x^2} - 25} } dx \cr
& {\text{Using substitution }}x = 5\sec \theta \cr
& x = 5\sec \theta ,{\text{ }}dx = 5\sec \theta \tan \theta d\theta \cr
& {\text{Substitute}} \cr
& = \int {{x^3}\sqrt {{x^2} - 25} } dx \cr
& = \int {{{\left( {5\sec \theta } \right)}^3}\sqrt {{{\left( {5\sec \theta } \right)}^2} - 25} } \left( {5\sec \theta \tan \theta } \right)d\theta \cr
& {\text{Simplifying}} \cr
& = \int {{5^3}{{\sec }^3}\theta \sqrt {25{{\sec }^2}\theta - 25} } \left( {5\sec \theta \tan \theta } \right)d\theta \cr
& = \int {{5^4}{{\sec }^4}\theta \sqrt {25{{\tan }^2}\theta } } \left( {5\sec \theta \tan \theta } \right)d\theta \cr
& = {5^5}\int {{{\sec }^4}\theta {{\tan }^2}\theta } d\theta \cr
& = {5^5}\int {{{\sec }^2}\theta {{\sec }^2}\theta {{\tan }^2}\theta } d\theta \cr
& {\text{Use the pythagorean identity }}{\tan ^2}\theta + 1 = {\sec ^2}\theta \cr
& = {5^5}\int {\left( {{{\tan }^2}\theta + 1} \right){{\sec }^2}\theta {{\tan }^2}\theta } d\theta \cr
& = {5^5}\int {\left( {{{\tan }^4}\theta + {{\tan }^2}\theta } \right){{\sec }^2}\theta } d\theta \cr
& {\text{Integrating}} \cr
& = {5^5}\left( {\frac{{{{\tan }^5}\theta }}{5}} \right) + {5^5}\left( {\frac{{{{\tan }^3}\theta }}{3}} \right) + C \cr
& = {5^4}{\tan ^5}\theta + \frac{{{5^5}}}{3}{\tan ^3}\theta + C \cr
& {\text{Where }}\sec \theta = \frac{x}{5},{\text{ }}\tan \theta = \frac{{\sqrt {{x^2} - 25} }}{5} \cr
& = {5^4}{\left( {\frac{{\sqrt {{x^2} - 25} }}{5}} \right)^5} + \frac{{{5^5}}}{3}{\left( {\frac{{\sqrt {{x^2} - 25} }}{5}} \right)^3} + C \cr
& {\text{Simplifying}} \cr
& = \frac{1}{5}{\left( {{x^2} - 25} \right)^{5/2}} + \frac{{25}}{3}{\left( {{x^2} - 25} \right)^{3/2}} + C \cr
& {\text{Factoring}} \cr
& \frac{1}{{15}}{\left( {{x^2} - 25} \right)^{3/2}}\left( {3{x^2} - 75 + 125} \right) + C \cr
& \frac{1}{{15}}{\left( {{x^2} - 25} \right)^{3/2}}\left( {3{x^2} + 50} \right) + C \cr} $$
