How many expected people needed until 3 share a birthday?

The expected time to the first triple collision is $$\mathbb{E}(T) = \int_0^\infty \left(1+{x\over M}+{x^2\over2M^2}\right)^M \,e^{-x}\,dx.\tag1$$

In my answer here, I derived the formula for double collisions, and the extension to triple collisions is straightforward. From equation (1), we see that $\mathbb{E}(T)$ grows like a multiple of $M^{2/3}$.

I turned the page and noticed that in section 15.3 of Problems and Snapshots from the World of Probability by Blom, Holst, and Sandell, the authors give the precise asymptotic result $$\mathbb{E}(T)\approx 6^{1/3}\,\Gamma(4/3)\, M^{2/3}\approx 1.6226\,M^{2/3}.$$