Hm `is` is much faster than `identical`, even for tiny indices:

```
In [9]: i = Index([1, 2, 3])

In [10]: timeit i.identical(i)
1000000 loops, best of 3: 1.43 µs per loop

In [11]: timeit i is i
10000000 loops, best of 3: 60.2 ns per loop
```

Is there any way to try the `is` check and if that fails then use `identical`.