Given integral
#\int x^3\sqrt{-x^2+8x-9}\ dx#
#=\int x^3\sqrt{-(x^2-8x+16)+16-9}\ dx#
#=\int x^3\sqrt{7-(x-4)^2}\ dx#
Let #x-4=\sqrt7\sin\theta\implies dx=\sqrt7\cos\theta\ d\theta#
#=\int (\sqrt7\sin\theta+4)^3\sqrt{7-7\sin^2\theta}\sqrt7\cos\theta\ d\theta#
#=\int (\sqrt7\sin\theta+4)^3(\sqrt7\cos\theta)\sqrt7\cos\theta\ d\theta#
#=7\int (7\sqrt7\sin^3\theta+84\sin^2\theta+48\sqrt7\sin\theta+64)\cos^2\theta\ d\theta#
#=7\int (7\sqrt7\sin^3\theta\cos^2\theta+84\sin^2\theta\cos^2\theta+48\sqrt7\sin\theta\cos^2\theta+64\cos^2\theta)\ d\theta#
#=7\int (7\sqrt7(1-\cos^2\theta)\sin\theta\cos^2\theta+21(2\sin\theta\cos\theta)^2+48\sqrt7\cos^2\theta \sin\theta+64({1+\cos2\theta}/{2}))\ d\theta#
#=49\sqrt7\int (\cos^2\theta-\cos^4\theta)\sin\theta\ d\theta+147\int \sin^2 2\theta\ d\theta+336\sqrt7\int \cos^2\theta\ sin\theta\ d\theta+224\int (1+\cos2\theta)\ d\theta#
#=49\sqrt7\int (\cos^2\theta-\cos^4\theta)\sin\theta\ d\theta+147\int \sin^2 2\theta\ d\theta+336\sqrt7\int \cos^2\theta\ sin\theta\ d\theta+224\int (1+\cos2\theta)\ d\theta#
#=49\sqrt7\int (\cos^4\theta-\cos^2\theta)(-\sin\theta\ d\theta)+147\int \frac{1-\cos 4\theta}{2}\ d\theta-336\sqrt7\int \cos^2\theta(-\ sin\theta\ d\theta)+224\int (1+\cos2\theta)\ d\theta#
#=49\sqrt7\int (\cos^4\theta-\cos^2\theta)\ \ d(\cos\theta)+147/2\int (1-\cos 4\theta)\ d\theta-336\sqrt7\int \cos^2\theta\ \ d(\ cos\theta)+224\int \ d\theta+112\int \cos2\theta\ d(2\theta)#
#=49\sqrt7(\cos^5\theta/5-\cos^3\theta/3)+147/2\theta-147/2{\sin 4\theta}/4-336\sqrt7\cos^3\theta/3+224\theta+112\sin2\theta\+C#
Hope you can solve further by substituting back #\sin\theta={x-4}/\sqrt7#
