Instead of forcing Mathematica to arrive at the desired form, we can analyse the expression. Let us first get rid of the square root, which clearly cannot be simplified:

sqrt = Sqrt[1 + (3/2 - Sqrt[2]) (1 + U^2)]

Then we look at the remaining part:

expr2 = expr*sqrt /. {U -> Sqrt[x]};
FunctionPoles[expr2, x]
 (* {} *)

Thus, expr2 is a polynomial. Next we find the roots of the numerator and denominator

Solve[Numerator[expr3] == 0, x] // FullSimplify
Solve[Denominator[expr3] == 0, x] // FullSimplify
 
(* {{x -> -7 - 4 Sqrt[2]}, {x -> -7 - 4 Sqrt[2]}, {x -> -6 - 
    4 Sqrt[2]}, {x -> -6 - 4 Sqrt[2]}, {x -> -6 - 
    4 Sqrt[2]}, {x -> -6 - 4 Sqrt[2]}, {x -> -3 - 2 Sqrt[2] + (
    2 (2 + Sqrt[2]))/c}, {x -> -3 - 2 Sqrt[2] + (2 (2 + Sqrt[2]))/c}} *)

(* {{x -> -7 - 4 Sqrt[2]}, {x -> -7 - 4 Sqrt[2]}, {x -> -6 - 
    4 Sqrt[2]}, {x -> -6 - 4 Sqrt[2]}, {x -> -6 - 
    4 Sqrt[2]}, {x -> -6 - 4 Sqrt[2]}, {x -> -3 - 2 Sqrt[2] + (
    2 (2 + Sqrt[2]))/c}} *)

This leads to a lot cancellations, which can be automated.

Instead of forcing Mathematica to arrive at the desired form, we can analyse the expression. Let us first get rid of the square root, which clearly cannot be simplified:

sqrt = Sqrt[1 + (3/2 - Sqrt[2]) (1 + U^2)]

Then we look at the remaining part:

expr2 = expr*sqrt /. {U -> Sqrt[x]};
FunctionPoles[expr2, x]
 (* {} *)

Thus, expr2 is a polynomial, which can be simplified with methods for polynomial functions:

expr3 = PolynomialQuotient[Numerator[expr2], Denominator[expr2], x] //
   FullSimplify
(* 1/4 r (-4 + c (2 + Sqrt[2] - (-2 + Sqrt[2]) x)) *)

And the final simplified expression recovered as

exprSimple = (expr3 /. {x -> U^2})/sqrt

$$\frac{r \left(c \left(-\left(\left(\sqrt{2}-2\right) U^2\right)+\sqrt{2}+2\right)-4\right)}{4 \sqrt{\left(\frac{3}{2}-\sqrt{2}\right) \left(U^2+1\right)+1}}$$