Are there fast(er) ways to compute inverses of Hermitian matrices?

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I'm dealing with some Hermitian matrices valued in symbolic expressions. I need to be able to compute inverses of these matrices, but they seem to get big pretty fast. That is it would nice to be able to do this in a reasonable amount of time with 10x10 matrices at least, and hopefully larger than that.
I'm currently running a cell where the inverses of 2 10x10 matrices are being computed, and it's taken well over 10 minutes.

Are there ways to exploit the fact that these matrices are Hermitian to compute the inverses faster, and/or are there better ways to compute inverses of symbolic matrices than the standard .inverse()??

EDIT: I originally avoided an example, because their construction is somewhat convoluted as you'll see below.

First we have two variables which for which I want to run the larger program for various choices thereof.

p = 1, q= 5

X = {(i): var("X_{}".format(i), latex_name="X_{}") for i in range(2*p*q)}

e = {(i,j,k): var("e_{}{}{}".format(i,j,k), latex_name="e_{{{}{}{}}}".format(i,j,k)) for i in range(2) for j in range(q) for k in range((p+q))}

The following creates a list L such that L[i] is the collection of lists of length i whose elements are integers between 0 and q-1 in strictly increasing order. This is used throughout the larger program.

L = [[[]]]

LL = []
for i in range(q):
    LL.append([i])
L.append(LL)

if q>=2:
    for k in range(2,q+1):
        LLL = []
        for i in range(binomial(q, k-1)):
            for j in range(q):
                if L[k-1][i][len(L[k-1][i])-1]<j:
                    mm = []
                    for t in L[k-1][i]:
                        mm.append(t)
                    mm.append(j)
                    LLL.append(mm)
        L.append(LLL)

Here, a matrix is defined which we be used to construct the matrices I'm interested in.

HE = matrix(SR, q, q)

for i in range(q):
    for j in range(q):
        prod = 0
        for k in range(p):
            prod -= (e[(0, i, k)]*e[(1, j, k)])
        for l in range(p+q):
            if l>=p:
                prod += (e[(0, i, l)]*e[(0, j, l)])
        HE[i,j] = prod

h = {(i): var("h_{}".format(i)) for i in range(q+1)}

for i in range(q+1):
    h[i] = matrix(SR, binomial(q, i), binomial(q, i))

h[0][0,0] = 1

for i in range(1, q+1):
    for z in L[i]:
        for w in L[i]:
            h[i][L[i].index(z), L[i].index(w)] = HE[z, w].det()

Finally, we come to the inverse matrices I would like to compute. It is absolutely necessary that I have these inverse matrices.

hinv = {(i): var("hinv_{}".format(i)) for i in range(q+1)}

for i in range(q+1):
    hinv[i] = h[i].inverse()