How to find inverse of $y = x C_1 + \tan^{-1}(x + rh) + C_2$? I have a function as below for which inputs are x, and rh. $C_i$ are constants.
$$
y = x C_1 + \tan^{-1}(x + rh) + C_2  \tag{1}
$$
Now, given y, I need to find x. I could not find any relation in arctan that could help me here. Kindly help. 
Background: I am trying to reverse the stull's formula to find dry bulb temperature, given wet bulb temperature and relative humidity. Eq(1) is simplified form of stull's formula. 
If I take $u = y - C_2, v = x, D = rh$ then I get, via tangent simplification, 
$$
u = \dfrac{v(1+C_1) + D}{1 - vC_1(v - D)}  \tag{2}
$$
where I am still lost, because I need to find $v$ for given $u$, but (2) is other way round.
 A: If $C_1$ and $C_2$ are known:
We have $y-xC_1-C_2=\arctan(x+rh)$. Now using the fact that $\tan(\arctan x) = x$ for all real $x$, we can simplify to reach a form where we can apply a Taylor approximation:
Since you are given $y$ and $C_2$, you only have something in the form $\tan(xa+b)$. Using the fact that $\tan(-x) = -\tan(x)$ and the addition formula for tangent:
$$x = \tan(y-xC_1 - C_2) +rh$$
$$= -\tan(-y+xC_1+C_2) + rh$$
$$= -\tan\big((-y+C_2) + xC_1 \big) + rh$$
and now you can use the Taylor series approximation given by Wolfram Alpha:
$$x \approx \tan b + rh + a \sec^2 b + a^2 \tan b \sec^2 b + O(a^3)$$
A: Since, from comments, you seem to be concerned by a rather narrow range for $T$, what you could do is to expand the rhs of the equation
$$y = C_1 T + \tan^{-1}(T + rh) + C_2$$ as a Taylor series at the midpoint (say at $T=T_{mid}$).
This would give
$$y=\left(C_1T_{mid} +\tan ^{-1}(T_{mid}+rh)+C_2\right)+(T-T_{mid})
   \left(\frac{1}{(T_{mid}+rh)^2+1}+C_1\right)+O\left((T-T_{mid})^2\right)$$
