how to use integral3 without loops

I am trying to run a code that uses integral3 but this seems to be taking so long. Is there a way to use vectors in integral3 rather than loops?

 L = 1:1:5;
 num = numel(L);
 eff = zeros(1, num);

 extcoef = [3 5 600 4 3 699 200 22 22 55 33]; 
 alp = numel(extcoef);

 for i = 1:num
    pl = L(i);

    Q = 0;

    for ii = 1:alp
        alpha = extcoef(ii);

        % define the function
        fun = @(x,y,z) (2*pi*pl)^-1*sin(x).*(exp(-alpha.*...
               (pl-y)./(sin(x).*sin(z)))+ exp(-alpha.*...
                   (pl+y)./(sin(x).*sin(z))));

        % integrate using triple integral
        q = integral3(fun,0.73,pi/2,0,pl,0,pi);
        Q = Q + q;
    end

    eff(i) = Q;

 end

 plot(L, eff);

thank you for your help regarding this..

[Roger Stafford]

You can easily obtain the definite integral of your function with respect to y from y = 0 to y = pl as an explicit function of just x and z. The resulting function will not be very much more complicated than your present function. I suggest you do that and then your problem becomes that of numerically solving a double integral of the new function with respect to x and z which should be much, much faster. For this latter you can use either 'quad2d', 'dblquad', or 'integral2', whichever proves to be faster.

[Roger Stafford]

Just to encourage you to avoid time-consuming triple integration, here is the function, g(x,z), you would use as integrand in a double integration with respect to x and z:

 g = @(x,z) sin(x).^2.*sin(z)/(2*alpha*pi*pl).*...
               (1-exp(-2*alpha*pl./(sin(x).*sin(z))));

This is obtained by integrating your 'fun' with respect to y from 0 to pl. As you see, it is even simpler than 'fun'. You should always use explicit solutions like this from calculus when available in order to minimize numerical processing.

[Ping]

Thank you Roger for this suggestion.