What's a good approximation to the zeros of the cosine integral?

If you look here, you will see that the zeros are given by $$a+\frac{1}{a}-\frac{16}{3\, a^3}+\frac{1673}{15\, a^5}-\frac{507746}{105 \, a^7}+O\left(\frac{1}{a^9}\right)$$ where $a=k \pi$ for the cosine integral and $a=\left(k+\frac 12\right)\pi$ for the sine integral.

If you want a more compact form, you could use $$a+\frac{240 a^2+3739}{a(240 a^2+5019)}$$ I built it from the above.