[mlir][math] Implement alternative decomposition for tanh (#85025)

The previous implementation decomposes `tanh(x)` into
`(exp(2x) - 1)/(exp(2x)+1), x < 0`
`(1 - exp(-2x))/(1 + exp(-2x)), x >= 0`
This is fine as it avoids overflow with the exponential, but the whole
decomposition is computed for both cases unconditionally, then the
result is chosen based off the sign of the input. This results in doing
two expensive `exp` computations.

The proposed change avoids doing the whole computation twice by
exploiting the reflection symmetry `tanh(-x) = -tanh(x)`. We can
"normalize" the input to be positive by setting `y = sign(x) * x`, where
the sign of `x` is computed as `sign(x) = (float)(x > 0) * (-2) + 1`.
Then compute `z = tanh(y)` with the decomposition above for `x >=0` and
"denormalize" the result `z * sign(x)` to retain the sign. The reason it
is done this way is that it is very amenable to vectorization.

This method trades the duplicate decomposition computations (which takes
5 instructions including an extra expensive `exp` and `div`) for 4 cheap
instructions to compute the signs value
1. `arith.cmpf` (which is a pre-existing instruction in the previous
impl)
2. `arith.sitofp`
3. `arith.mulf`
4. `arith.addf`

and 1 more instruction to get the right sign in the result
5. `arith.mulf`. Moreover, numerically, this implementation will yield
the exact same results as the previous implementation.

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