Integrating $\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}$ 
I'm struggling with the integral, $$\int\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}dx.$$

I tried it as follows:
Substituting $x-2 = \frac1t \implies dx = \frac{-dt}{t^2}.$
$$\therefore \int\frac{dx}{(x-2)^4 \sqrt{x^2 + 6x + 2}} = \int \frac{- dt}{\frac{t^2}{t^4} \sqrt{(\frac1t + 2)^2 + 6 (\frac1t + 2) + 2}} = \int \frac{-t^3}{\sqrt{18t^2 + 10t + 1}}\ dt$$
How to continue from here?

Substitute $y=x-2$ to rewrite the integral as
$$I_4= \int\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}dx
=\int \frac{1}{y^4 \sqrt{y^2 + 10y + 18}}dy $$
and then integrate by parts to get the reduction formula
$$18(n-1)I_n=-\frac{\sqrt{y^2 + 10y + 18}}{y^{n-1}}-5(2n-3)I_{n-1}-(n-2)I_{n-2}
$$
Apply the formula three times to reduce it to
$$I_4=\frac1{54} \left(-\frac1{y^3} + \frac{25}{36y^2}-\frac{101}{216y}\right)\sqrt{y^2 + 10y + 18}-\frac{355}{11664}I_1
$$
where
$$I_1= \int\frac{1}{y \sqrt{y^2 + 10y + 18}}dy
=-\frac1{\sqrt{18}}\tanh^{-1}\frac{\sqrt{18}\sqrt{y^2 + 10y + 18} }{5y+18}
$$
