The OP clarified in a comment that he examines the standard bivariate normal distribution, with means and variances fixed to zero and unity correspondingly,
$$f(x,y) = \frac{1}{2 \pi \sqrt{1-\rho^2}}
\exp\left\{-\frac{x^2 +y^2 -2\rho xy}{2(1-\rho^2)}\right\} $$
In turn, this makes the distribution a member of the curved exponential family, and, as I have shown in my answer to this post the maximum likelihood estimator for $\rho$ in such a case does not equal the sample correlation coefficient. Specifically the sample correlation coefficient is
$$\tilde r = \frac 1n\sum_{i=1}^nx_iy_i$$
Denoting $\hat \rho$ the mle for $\rho$ and $(1/n)\sum_{i=1}^n(x_i^2 +y_i^2) = (1/n)S_2$, to be the sum of the sample variances of $X$ and $Y$, we obtain
$$\hat \rho: \hat \rho^3 -\tilde r \hat \rho^2 + \big[(1/n)S_2-1\big]\hat \rho -\tilde r=0$$
$$\Rightarrow \hat \rho\Big(\hat \rho^2 -\tilde r \hat \rho + \big[(1/n)S_2-1\big] \Big) = \tilde r$$
Doing the algebra, it is not difficult to conclude that we will obtain $\hat \rho = \tilde r$ if,and only if, $(1/n)S_2 =2$, i.e. only if it so happens that the sum of sample variances equals the sum of true variances.
So in general for finite samples,
$$\hat \rho \neq \tilde r$$
Both remain consistent, but this alone does not imply that the asymptotic distribution of the sample correlation coefficient will attain the Cramer-Rao bound, which is the one found by the OP. And it doesn't.
Best Answer
To my best memory, I've never come across a formal definition for this in a statistical text, but I think you can stitch one together from a few contextual readings. Start with Bayesian Data Analysis, p. 261:
The obstacle is generally the marginal likelihood, the denominator on the right-hand side of Bayes' rule, which could involve an integral that cannot be analytically expressed. For a more I think you'll find wiki's article on closed-form expression helpful for context (emphasis mine):
If you read on, you'll see a table of classes of expressions, and "Analytic expressions" includes several involved in the normalizing constants of exponential family distributions. E.g., the gamma function in the gamma distribution, and the Bessel function in the von-Mises Fisher.
Meaning, we're willing to admit at least these into our definition of "tractability." (There may be other distributions that involve the classes of operations classified as "analytic expressions"; I confess I'm not familiar.)