Speed up numeric integral

Hi,everyone.I got a problem of doing numeric integral on the matrix H (which is at the bottom part of my code).

The code works fine but I would like to speed it up? Am I doing something stupid to make it slow?Thanks for your help.

clear

L=0.1;
section=50;
a=L./2;
b=L./section;
v=3e8;
f1=1.2e9;
f2=4.8e9;
f3=7.5e9;
f4=10e9;
w1=(2.*pi.*f1);
w2=(2.*pi.*f2);
w3=(2.*pi.*f3);
w4=(2.*pi.*f4);
Z01=50;
Z02=75;
a0=(log(Z02./Z01))./(2..*L);
T=a;
Ub=a;
Lb=-a;
ub=a;
lb=-a;
k=5;

syms x y w m

%%% Revised Algorithm

l=0:(2.*L)./(2.*k):L  % Approximate sections dvided

% Approximate terms
sec=numel(l)-1;

Z1=[50 55 60 65 70 80 75];    % Set the impedance of the end of each sections

ac=10;                  % Choose the accuracy of dividing a section

for s=1:1:sec

    c=(l(s+1)-l(s))./ac; 

    app_cos=0.5./ac.*log(Z1(s+1)./Z1(s)).*sin(ac./2.*(w./v).*c).*cos(((ac+1)./2).*(w./v).*c)./sin(((c.*w)./(2.*v)));

    app_sin=0.5./ac.*log(Z1(s+1)./Z1(s)).*sin(ac./2.*(w./v).*c).*sin(((ac+1)./2).*(w./v).*c)./sin(((c.*w)./(2.*v)));

    amp=0.75;

    Ga(s)=(app_cos-(1i.*amp.*app_sin));

end

    Ta=(1-(abs(Ga(1:1:end)).^2));

%%% Turn closed loop terms into matrix presentation

for t=1:1:sec

    for r=1:1:sec

        E(t,r)=Ga(t).*Ga(r);

    end

end

E=triu(E,1);

for t=1:1:sec

    for r=1:1:sec
    
        D(t,r)=abs(E(r,t));

    end

end

C1=sum(D);

% Construct approximate loop terms 

for t=1:1:sec

    D1(t)=prod(Ta(1:t))./(1+(sum(C1(1:t))));

end

D=[1 D1(2:end)];

Dm=transpose(D).*D;

% Cmn

m=transpose(1:5);

n=1:5;

g=(v.^2.*w.^2.*cos((2.*w.*(- y + x))./v).*cos(20.*sym(pi).*m.*x).*cos(20.*sym(pi).*n.*y) + 10.*sym(pi).*m.*v.^3.*w.*sin(20.*sym(pi).*m.*x).*sin((2.*w.*(- y + x))./v).*cos(20.*sym(pi).*n.*y) - 10.*sym(pi).*n.*v.^3.*w.*sin(20.*sym(pi).*n.*y).*sin((2.*w.*(- y + x))./v).*cos(20.*sym(pi).*m.*x) + 100.*m.*n.*v.^4.*sym(pi).^2.*sin(20.*sym(pi).*m.*x).*sin(20.*sym(pi).*n.*y).*cos((2.*w.*(- y + x))./v))./(4.*(w.^2 - 100.*sym(pi).^2.*m.^2.*v.^2).*(w.^2 - 100.*sym(pi).^2.*n.^2.*v.^2));

for p=1:numel(l)-1

    x1=l(p);

    x2=l(p+1);

    for q=1:numel(l)-1

        y1=l(q);

        y2=l(p+1);

        Ht{p,q}=matlabFunction(((subs(g,[x y],[x2 y2])-subs(g,[x y],[x1 y1]))*Dm(p,q)));

    end

end

Hi=Ht{1};

for m=2:1:numel(Ht)

    Hi=Hi+vpa(Ht{m},3);

end

H=integral(Hi,w1,w2,"ArrayValued",true)

[Les Beckham]

Using symbolic variables/functions is much slower than doing regular numeric calculations. Is there a reason that you are using symbolics?

[Guan Hao]

@Les Beckham Yes,there are three variables (x,y,w) in my function g as you can see.I've try integral3,but it was super slow.That's why I want to symbolic integral x and y first, and numeric integral w to obtain the result.

[Torsten]

@Guan Hao

Do you see your mistake ?

syms x y 
x1 = 1;
x2 = 2;
y1 = 4;
y2 = 6;
g = 4*x*y;
G = int(int(g,x),y)

G = 

int(int(g,x,x1,x2),y,y1,y2)

ans = 

60

int(int(g,y,y1,y2),x,x1,x2)

ans = 

60

subs(G,[x y],[x2,y2])-subs(G,[x y],[x1,y1])

ans = 

128

[Guan Hao]

@Torsten Oh.Thanks for pointing that out.