# Get Integral Result without Assumptions

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.