integration - Integrating the following $int sqrt{tan x+1},dx$

Question: Integrate the following, $$\int\sqrt{\tan x+1}\;dx.$$

Wolfram Alpha returns a non-elementary answer. Can someone please spot the mistake I have made here:

First consider this integral:

$$\int \frac{1}{(x+1)\sqrt{x+3}} \, dx = -\sqrt{2}\tanh^{-1}\frac{\sqrt{x+3}}{\sqrt{2}} + c$$

Wolfram Alpha confirms that result.

Then, we have

$$I=\int \sqrt{\tan x+1} \, dx, \quad \tan x=u+2, \quad dx=\frac{du}{\sec^{2}x}=\frac{du}{(u+2)^{2}-1}=\frac{dx}{(u+3)(u+1)}$$

So this transforms the integral to the first integral on this post, which we can evaluate. Then after evaluation and resubstitution I get:

$$I=-\sqrt{2}\tanh^{-1}\frac{\sqrt{\tan x+1}}{\sqrt{2}}+c$$

However differentiating this with Wolfram Alpha gives me a messy trigonometric expression which doesn't seem to be equal (I tested some values in both expressions and get different answers). I also estimated the area under the integral between some values and also obtained different answers using the closed form. Any ideas why?

EDIT: I used the wrong identity. Nevertheless, we can still use this method to integrate sqrt(tanhx integrals). E.g:
$$I=\int \sqrt{\tanh x+1} \, dx, \tanh x=u+2,\quad -dx = \frac{du}{\operatorname{sech}^2 x} = \frac{du}{(u+2)^2-1} = \frac{dx}{(u+3)(u+1)}$$

To obtain:
$\int \sqrt{\tanh x+1} \, dx = I=\sqrt{2}\tanh^{-1} \dfrac{\sqrt{\tanh x+1}}{\sqrt{2}}+c$