The simplified answer is here.
Read below for the work.
Using the following identity:
#tan^2x + 1 = sec^2x#
the following substitution works out:
#x = 2tantheta#
#dx = 2sec^2thetad theta#
#x^3 = 8tan^3theta#
#sqrt(x^2 + 4) = 2sectheta#
Now we get:
#int 8tan^3theta * 2sectheta * 2sec^2thetad theta#
#= 32int tan^3theta sec^3thetad theta#
#= 32int tan^2thetatantheta sec^3thetad theta#
Now let's separate it so hopefully we get something we can do with u-substitution.
#= 32int (sec^2theta - 1)tantheta sec^3thetad theta#
#= 32int sec^5thetatantheta - sec^3thetatanthetad theta#
Notice how we can force this to look like #u^4du - u^2du#:
#= 32int (sectheta)^4secthetatantheta - (sectheta)^2secthetatanthetad theta#
Now, we can do another substitution:
#u = sectheta#
#du = secthetatanthetad theta#
This gives:
#= 32int u^4 - u^2 du#
#= 32(u^5/5 - u^3/3)#
#= 32(sec^5theta/5 - sec^3theta/3)#
Now, since #2sectheta = sqrt(x^2 + 4) -> sectheta = sqrt(x^2 + 4)/2#:
#= 32((x^2 + 4)^(5/2)/(2^5*5) - (x^2 + 4)^(3/2)/(2^3*3))#
For convenience I've purposefully multiplied the second fraction by #4/4#.
#= cancel(32)((x^2 + 4)^(5/2)/(cancel(32)*5) - (4(x^2 + 4)^(3/2))/(cancel((4*8))*3))#
#= color(green)((x^2 + 4)^(5/2)/(5) - (4(x^2 + 4)^(3/2))/(3) + C)#
If you want, you can stop here, but Wolfram Alpha gives you a different answer, so I'll show you how to get there too if you use that to check. Alternatively you could say "Is A = B" and it'll check for you, where A = one answer and B = the one you're checking.
Factor out #1/5#:
#= 1/5[(x^2 + 4)^(5/2) - (20(x^2 + 4)^(3/2))/(3)]#
Factor out #1/3#:
#= 1/15[3(x^2 + 4)^(5/2) - 20(x^2 + 4)^(3/2)]#
Factor out #(x^2 + 4)^(3/2)#:
#= 1/15(x^2 + 4)^(3/2)[3(x^2 + 4) - 20]#
Distribute and add:
#= 1/15(x^2 + 4)^(3/2)[3x^2 + 12 - 20]#
#= color(blue)(1/15(x^2 + 4)^(3/2)[3x^2 - 8] + C)#
There it is.
