The negative binomial distribition parametrized by mean and size can be given by
$$ \DeclareMathOperator{\P}{\mathbb{P}}
   \P (X=k) = \binom{k+m-1}{k}\left( \frac{m}{m+\mu} \right)^m \left( \frac{\mu}{m+\mu} \right)^k   
$$
for the outcome $k$ a nonnegative integer, and $\mu>0$ the mean, $m>0$ the size. I will do the calculations by maple.

The Fisher information matrix (of size $2\times 2$) has components 
$I_{\mu\mu}, I_{\mu m} \text{ and } I_{\mu m}$ given by
$$ \DeclareMathOperator{\E}{\mathbb{E}}
  I_{ij}=-\E\left\{ \frac{\partial^2}{\partial \theta_i \partial\theta_j}\log f(X;\theta)|\theta \right\}
$$ where here $\theta=(\mu, m)$. Then we (maple code at the end of post)  find 
$$
  I_{\mu\mu}=\frac{m}{(m+\mu)\mu}
$$
The diagonal term is simplest, it reduces to zero! That is the beauty of the mean parametrization
$$
I_{\mu m}=0
$$
For the last term, I cannot find a simple closed form. The result will involve a digamma function written $\Psi$ (log derivative of gamma function) and result will be a somewhat complex infinite series, which must be evaluated numerically:
$$
I_{mm}=\sum_{k=0}^\infty \binom{k+m-1}{k}\left\{ -m^{m-1}\mu^k (m+\mu)^{-m-2-k} \left( m(m+\mu)^2 \Psi(k+m) -m(m+\mu)^2 \Psi(m) +mk+\mu^2 \right) 
\right\}
$$
or maybe some other form could be found (later, the NIST handbook of mathematical functions has some nice integral formulas for $\Psi$ which could be useful). But, maybe numerical integration is not so much better than numerical summation?

Below some maple code (and output):

    f :=  binomial(k+m-1,k)*(m/(m+mu))^m * (mu/(m+mu))^k
                                                m         k
                                        /  m   \  /  mu  \ 
            f := binomial(k + m - 1, k) |------|  |------| 
                                        \m + mu/  \m + mu/ 
    lf := ln( binomial(k+m-1,k) ) + m*ln( m/(m+mu) ) + k* ln( mu/(m+mu) ) assuming m>0,mu>0;
                                            /  m   \       /  mu  \
     lf := ln(binomial(k + m - 1, k)) + m ln|------| + k ln|------|
                                            \m + mu/       \m + mu/
    
    simplify( -sum(f*diff(lf,mu,mu),k=0..infinity ) ) assuming m>0,mu>0;
                                   m     
                              -----------
                              (m + mu) mu
    
    simplify( -sum(f*diff(lf,mu,m),k=0..infinity ) ) assuming m>0,mu>0;
                                   0
    
    simplify( -sum(f*diff(lf,m,m),k=0..infinity ) ) assuming m>0,mu>0;
     infinity                                                         
      -----                                                           
       \                                                              
        )     (m - 1)         (-m - 2 - k)   k                        
    -  /     m        (m + mu)             mu  binomial(k + m - 1, k) 
      -----                                                           
      k = 0                                                           
    
      /          2                           2                     2\
      \m (m + mu)  Psi(1, k + m) - m (m + mu)  Psi(1, m) + m k + mu /