* Method 1
```
Integrate[
 Boole[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}](6 x^2 + 
    33 x*y + 42 y^2)^4, {x, -∞, ∞}, {y, -∞, ∞}]
```
> `4096/225`

* Method 2
```
Integrate[(6 x^2 + 33 x*y + 42 y^2)^4, {x, y} ∈ 
  ImplicitRegion[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}, {x, 
    y}]]
```
> `4096/225`

* Method 3
```
(* change of variables *)
sol = Solve[{u == 2 x + 7 y, v == 3 x + 6 y}, {x, y}][[1]];
(*  {x -> 1/9 (-6 u + 7 v), y -> 1/9 (3 u - 2 v)}  *)
expr = (6 x^2 + 33 x*y + 42 y^2)^4 /. sol // Simplify;
(* u^4 v^4 *)
jacobian = Grad[{x, y} /. sol, {u, v}] // Det // Abs
(* 1/9 *)
Integrate[expr*jacobian, {u, -2, 2}, {v, -2, 2}]
```
> `4096/225`

* Method 4
```
Factor[6 x^2 + 33 x*y + 42 y^2]
(* 3 (x + 2 y) (2 x + 7 y) *)
```
It means that the integrand is just `(u*v)^4` if we set `u=3(x+2y)` and `v=2x+7y`. The region became `-2<=u<=2` and `-2<=v<=2`.
To calculate `Grad[{x,y},{u,v}]//Det`, we just need to calculate `1/Grad[{u,v},{x,y}]//Det`.

```
1/(Grad[{3(x+2y),2x+7y}, {x, y}] // Det // Abs)
```
`1/9`

```
Integrate[(u*v)^4*1/9,{u,-2,2},{v,-2,2}]
```

> `4096/225`