#int (7x^2-3)^(3/2)*dx#
After using #x=sqrt(3/7)*secu# and #dx=sqrt(3/7)*secu*tanu*du# transforms, this integral became,
#int [7*3/7(secu)^2-3]^(3/2)*sqrt(3/7)*secu*tanu*du#
=#int [3(secu)^2-3]^(3/2)*sqrt(3/7)*secu*tanu*du#
=#int [3(tanu)^2]^(3/2)*sqrt(3/7)*secu*tanu*du#
=#int [3sqrt3(tanu)^3*sqrt(3/7)*secu*tanu*du#
=#int [9/sqrt7*(tanu)^4*secu*du#
Now, I solved #int (tanu)^4*secu*du# integral,
#int (tanu)^4*secu*du#
=#int [(secu)^2-1]^2*secu*du#
=#int (secu)^5*du-2int (secu)^3*du+int secu*du#
After using reduction formula for #int (secu)^n*du#,
#int (secu)^n*du=1/(n-1)*(secu)^(n-2)*tanu+(n-2)/(n-1)*int (secu)^(n-2)*du#
#int (tanu)^4*secu*du#
=#1/4*(secu)^3*tanu+3/4int (secu)^3*du-2int (secu)^3*du+int secu*du#
=#1/4*(secu)^3*tanu-5/4int (secu)^3*du+int secu*du#
=#1/4*(secu)^3*tanu-5/8*secu*tanu-5/8int secu*du+int secu*du#
=#1/4*(secu)^3*tanu-5/8*secu*tanu+3/8int secu*du#
=#1/4*(secu)^3*tanu-5/8*secu*tanu+3/8ln(secu+tanu)+sqrt7/9*C1#
Hence,
=#int 9/sqrt7*(tanu)^4*secu*du#
=#9/(4sqrt7)*(secu)^3*tanu#-#45/(8sqrt7)*secu*tanu#+#27/(8sqrt7)*ln(secu+tanu)+C1#
After using #x=sqrt(3/7)*secu#, #secu=sqrt(7/3)*x# and #tanu=sqrt(7x^2-3)/sqrt3# inverse transformations, I found
#int (7x^2-3)^(3/2)*dx#
=#9/(4sqrt7)*7sqrt7/9*x^3*sqrt(7x^2-3)#-#45/(8sqrt7)*sqrt7/3*xsqrt(7x^2-3#+#27/(8sqrt7)*ln(sqrt(7/3)*x+sqrt(7x^2-3)/sqrt3)+C1#
=#7/4x^3*sqrt(7x^2-3)#-#15/8xsqrt(7x^2-3#+#(27sqrt7)/56*ln(xsqrt7+sqrt(7x^2-3))+C#
Note: #C=C1-(27sqrt7)/112*Ln3#
