Help solving another integral $\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$ $$\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$$
Complete the square?
$$\int \frac{1}{(2(x+1)^2-4)^\frac{3}{2}} \ dx$$
Not sure what do do next, or if I should try something else?
Big help if you can show as "step-by-step" possible as you can.
Thanks in advance!

@Chandru1:
I'm confused at what happens to the 2 that gets factored out when completing the square (the 2 in the denominator) because it looks like Arturo left it out and instead substituted: 
$\int \frac {1}{(2(x+1)^2-4)^\frac{3}{2}} \ dx$ 
$u = x+1$
 $du =dx$ 
which changes the integral to  $\int \frac {1}{(2(u^2-4))^\frac{3}{2}}$ 
Would I be best factoring it out as a 1/2 in front of the integral  or leaving it in and then using trig. substitution like this?  
$u= \sqrt{2}sec(t)$
 $du = \sqrt{2}sec(t)*tan(t) dt$ 
Which changes it to:
$\int \frac {\sqrt{2}sec(t)*tan(t)}{(2(2sec^2(t)-4))^\frac{3}{2}} \ dt$ = $\int \frac {\sqrt{2}sec(t)*tan(t)}{(2(tan^2(t)))^\frac{3}{2}} \ dt$ = $\int \frac {\sqrt{2}sec(t)*tan(t)}{(2tan^3(t))} \ dt$
My $\sqrt{2}$ factors out like yours, but I have no 1/8 so figure I factor it out as a 1/2 earlier like I thought?  But then the constant in front of my integral would be $\frac{\sqrt{2}}{2}$  and you have $\frac{\sqrt{2}}{8}$
I know what's wrong, just not how I went wrong.

Hint. $(2x^{2}+4x-2) = 2(x^{2}+2x +1) -4 = 2(x+1)^{2}-4$. Now subsitute $x+1 = \sqrt{2} \sec{t}$. Then you have $dx = \sqrt{2} \sec{t} \cdot \tan{t} \ dt$. Now, $$2 \cdot \Bigl[(x+1)^{2} -2 \Bigr]= 2 \cdot \Bigl[ 2 \cdot (\sec^{2}{t}-1)\Bigr] = 4 \cdot (\sec^{2}{t}-1)$$.
Now what will you get if you power them by $\frac{3}{2}$. You will get $4^{3/2} \cdot \Bigl[ \tan^{2}{t}\Bigr]^{3/2} = 8 \cdot \tan^{3}{t}$.
And try to evaluate the integral. Simplyfying you should get 
\begin{align*}
\int \frac{1}{(2x^{2}+4x-2)^{-\frac{3}{2}}} \ \textrm{dx} &= \frac{\sqrt{2}}{8} \int\frac{\sec{t} \cdot \tan{t}}{\tan^{3}{t}} \ \textrm{dt} \\ &= \frac{\sqrt{2}}{8} \int\frac{\cos{t}}{\sin^{2}{t}} \ \textrm{dt}
\end{align*}
