#int x^4*sqrt(x^2-4)*dx#
After using #x=2secu# and #dx=2secu*tanu*du# transforms, this integral became,
#int (2secu)^4*sqrt[(2secu)^2-4]*2secu*tanu*du#
=#int 32(secu)^5*tanu*sqrt[4(tanu)^2]*du#
=#int 64(secu)^5*(tanu)^2*du#
=#int 64(secu)^5*[(secu)^2-1]*du#
=#int 64(secu)^7*du#-#int 64(secu)^5*du#
After using #int (secu)^n*du#=#1/(n-1)*(secu)^(n-2)*tanu#+#(n-2)/(n-1)*int (secu)^n*du# reduction formula,
=#int 64(secu)^7*du#-#int 64(secu)^5*du#
=#64*[1/6*(secu)^5*tanu+5/6*int (secu)^5*du#]-#int 64(secu)^5*du#
=#32/3*(secu)^5*tanu+160/3*int (secu)^5*du#-#int 64(secu)^5*du#
=#32/3*(secu)^5*tanu-32/3*int (secu)^5*du#
=#32/3*(secu)^5*tanu#-#32/3*[1/4*(secu)^3*tanu#+#3/4*int (secu)^3*du#]
=#32/3*(secu)^5*tanu#-#8/3*(secu)^3*tanu#-#8int (secu)^3*du#
=#32/3*(secu)^5*tanu#-#8/3*(secu)^3*tanu#-#8*[1/2*secu*tanu#+#1/2*int secu*du#]
=#32/3*(secu)^5*tanu#-#8/3*(secu)^3*tanu#-#4secu*tanu#-#4int secu*du#
=#32/3*(secu)^5*tanu#-#8/3*(secu)^3*tanu#-#4secu*tanu#-#4ln(secu+tanu)+C1#
After using #x=2secu#, #secu=x/2# and #tanu=sqrt(x^2-4)/2# inverse transformations, I found
#int x^4*sqrt(x^2-4)*dx#
=#(1/6*x^5-1/6*x^3-x)*sqrt(x^2-4)#-#4ln[(x+sqrt(x^2-4))/2]+C1#
=#1/6*(x^5-x^3-6x)*sqrt(x^2-4)#-#4ln(x+sqrt(x^2-4))+C#
Note: #C=C1+4Ln2#
