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I have the following two functions:

a[x_] := w''[x]^2
b[x_] := 1/L*\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(\(-L\)/2\), \(L/2\)]\(1/
     2\ \(w'\)[x]^2 \[DifferentialD]x\)\)

Then I calculate:

1/2*\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(\(-H\)/2\), \(H/2\)]\(
\*SubsuperscriptBox[\(\[Integral]\), \(\(-W\)/2\), \(W/2\)]\(
\*SubsuperscriptBox[\(\[Integral]\), \(\(-L\)/2\), \(L/2\)]
\*SubscriptBox[\(\[Sigma]\), \(0\)] \((\(-z\)\ *a[x] + 
        b[x])\)\ \[DifferentialD]x\ \[DifferentialD]y\ \
\[DifferentialD]z\)\)\) + Y/2*\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(\(-H\)/2\), \(H/2\)]\(
\*SubsuperscriptBox[\(\[Integral]\), \(\(-W\)/2\), \(W/2\)]\(
\*SubsuperscriptBox[\(\[Integral]\), \(\(-L\)/2\), \(L/2\)]
\*SuperscriptBox[\((\(-z\)\ *a[x] + 
         b[x])\), \(2\)] \[DifferentialD]x\ \[DifferentialD]y\ \(\(\
\[DifferentialD]\)\(z\)\(\ \)\)\)\)\)

The result is this:

(1/(2 L))H W Integrate[1/2 Derivative[1][w][x]^2, {x, -(L/2), L/2}, 
   Assumptions -> (-(H/2) < z < H/2 || -(H/2) > z > H/2) && (-(W/2) < 
        y < W/2 || -(W/2) > y > W/2) && (Im[L] != 0 || -(L/2) < x < L/
        2 || L/2 < x < -(L/2))] Subscript[\[Sigma], 0] + 
 1/2 Y ((1/(L^2))
    H W Integrate[1/2 Derivative[1][w][x]^2, {x, -(L/2), L/2}, 
      Assumptions -> (-(H/2) < z < H/2 || -(H/2) > z > H/
           2) && (-(W/2) < y < W/2 || -(W/2) > y > W/2) && (Im[L] != 
           0 || -(L/2) < x < L/2 || L/2 < x < -(L/2))]^2 + 
    1/12 H^3 W (w^\[Prime]\[Prime])[x]^4)

How can I get this result with all those assumptions ? all variables are real and positive.

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