Rewrite the second factor in the numerator as
[tex]2x^2+6x+1=2(x+2)^2-2(x+2)-3[/tex]
Then in the entire integrand, set [tex]x+2=\sqrt3\sec t[/tex], so that [tex]\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt[/tex]. The integral is then equivalent to
[tex]\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt[/tex]
[tex]=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt[/tex]
[tex]=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt[/tex]
[tex]=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt[/tex]
Note that by letting [tex]x+2=\sqrt3\sec t[/tex], we are enforcing an invertible substitution which would make it so that [tex]t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}[/tex] requires [tex]0\le t<\dfrac\pi2[/tex] or [tex]\dfrac\pi2<t\le\pi[/tex]. However, [tex]\tan t[/tex] is positive over this first interval and negative over the second, so we can't ignore the absolute value.
So let's just assume the integral is being taken over a domain on which [tex]\tan t>0[/tex] so that [tex]|\tan t|=\tan t[/tex]. This allows us to write
[tex]=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt[/tex]
[tex]=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt[/tex]
We can show pretty easily that
[tex]\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C[/tex]
[tex]\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C[/tex]
[tex]\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C[/tex]
[tex]\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C[/tex]
which means the integral above becomes
[tex]=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C[/tex]
[tex]=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C[/tex]
Back-substituting to get this in terms of [tex]x[/tex] is a bit of a nightmare, but you'll find that, since [tex]t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}[/tex], we get
[tex]\sec t=\dfrac{x+2}{\sqrt3}[/tex]
[tex]\sec^2t=\dfrac{(x+2)^2}3[/tex]
[tex]\tan t=\sqrt{\dfrac{x^2+4x+1}3}[/tex]
[tex]\cot t=\sqrt{\dfrac3{x^2+4x+1}}[/tex]
[tex]\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}[/tex]
[tex]\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}[/tex]
etc.
