Latest Edit
Great thanks to Quanto who gave a complete proof to the question by setting up a simple reduction formula on $I_n$ below:
\begin{align} \boxed{I_{n}
=\ \frac1{2(4n+1)}+ \frac1{2(4n)}+ \frac1{4(4n-1)}-\frac14I_{n-1}}
\end{align}
Let’s start with the easy one.
$$
\begin{aligned}I_2& =\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{\left[1+\tan \left(\frac{\pi}{4}-x\right)\right]^2} d x \\
& =\frac{1}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(1+\tan x)^2 d x \\
& =\frac{1}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(1+2 \tan x+\tan ^2 x\right) d x \\
& =\frac{1}{2} \int_0^{\frac{\pi}{4}}\left(1+\tan ^2 x\right) d x \quad \textrm{ (Properties of odd and even functions.)}
\\
& =\frac{1}{2} \int_0^{\frac{\pi}{4}} \sec ^2 x d x \\
& =\frac{1}{2}[\tan x]_0^{\frac{\pi}{4}} \\
& =\frac{1}{2}
\end{aligned}
$$
To proceed, I tried further with the same method of evaluating $I_2$ by putting $x\mapsto \frac{\pi}{4} -x$ which transforms
$$
\begin{aligned}
I_n & =\frac{1}{2^n} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(1+\tan x)^n d x \\
& =\frac{1}{2^n} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) \tan ^k x d x \\
& =\frac{1}{2^{n-1}} \sum_{k=0}^{\left[\frac{n}{2}\right]}\left(\begin{array}{l}
n \\
2 k
\end{array}\right) \int_0^{\frac{\pi}{4}} \tan ^{2 k} x d x \quad \textrm{ (Properties of odd and even functions.)}
\end{aligned}
$$
Using the reduction formula for $\int_0^{\frac{\pi}{4}} \tan ^{2 k} x d x,$ we can continue our evaluation on $I_n$ as below:
$$
\begin{aligned}
I_3=-\frac{\pi}{8}+\frac{3}{4},\quad I_4=\frac{2}{3} -\frac{\pi}{8}, \quad I_5=\frac{5}{12}-\frac{\pi}{16},
\end{aligned}
$$
For $n=6$, letting $x\mapsto \frac{\pi}{4}-x $ transforms the integral into
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{d x}{(1+\tan x)^6} & =\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left[1+\tan \left(\frac{\pi}{4}-x\right)\right]^6} \\
& =\frac{1}{64} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(1+\tan x)^6 d x\\\\
\end{aligned}
$$
Expanding the integrand into $3$ odd and $4$ even functions simplifies it into
$$
\begin{aligned}I_6&=\frac{1}{32} \int_0^{\frac{\pi}{4}}\left(1 +\tan ^6 x +15 \tan ^2 x+15 \tan ^4 x\right) d x\\&= \frac{1}{32} \int_0^{\frac{\pi}{4}}\left(1-\tan ^2 x+\tan ^4 x+15 \tan ^2 x\right) d(\tan x)\\& =\frac{1}{32}\left[\tan x+\frac{14 \tan ^3 x}{3}+\frac{\tan ^5 x}{5}\right]_0^{\frac{\pi}{4}}\\&=\frac{11}{60} \end{aligned}
$$
When evaluating $I_6$, I found the symmetry of the binomial coefficients in expansion, I can group the corresponding terms together and found that $I_n$ has no $\pi$ when $n\equiv 2 \quad \pmod 4$, i.e.
$$I_{4n+2}\in Q.$$
Proof:
$$
\begin{aligned}
I_{n} & =\frac{1}{2^{4 n+2}} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(1+\tan x)^{4 n+2} d x \\
& = \frac{1}{2^{4 n+2}} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sum_{k=0}^{2n+1} {4n+2\choose 2k} \tan ^k x d x\\
\end{aligned}
$$
Using the properties of odd and even function simplifies the integral
$$
I_{n}=\frac{1}{2^{4 n+1}} \sum_{k=0}^{2 n+1}\int_0^{\frac{\pi}{4}}\left(\begin{array}{c}
4 n+2 \\
2 k
\end{array}\right) \tan ^{2 k} x d x
$$
Grouping the symmetric terms gives
$$
I_{n}=\frac{1}{2^{4 n+1}} \sum_{k=0}^{n+1}\left(\begin{array}{c}
4 n+2 \\
2 k
\end{array}\right) \int_0^{\frac{\pi}{4}}
\left(\tan ^{2 k} x+\tan ^{4 n+2-2 k} x\right) d x
$$
$(-1)^k+(-1)^{2 n+1-k} =(-1)^k\left[1+(-1)^{2(n-k)+1}\right] =0 \Rightarrow 1+y| y^k+y^{2 n+1-k}. $
Therefore $\tan ^{2 k} x+\tan ^{4 n+2-2 k} x$ is divisible by $1+\tan^2x$ and hence
$$\tan ^{2 k} x+\tan ^{4 n+2-2 k} x=(1+\tan^2x)P_k(\tan x)$$
for some polynomial $P_k(y)$ with integer coefficients. $$
\begin{aligned}
I_{n}= & \frac{1}{2^{4 n+1}} \sum_{k=0}^{n+1}\left(\begin{array}{c}
4 n+2 \\
2 k
\end{array}\right) \int_0^{\frac{\pi}{4}} P_k(\tan x) d(\tan x) \\
= & \frac{1}{2^{4 n+1}} \sum_{k=0}^{n+1}\left(\begin{array}{c}
4 n+2 \\
2 k
\end{array}\right) Q_k(1) \in \mathbb{Q} \\
\end{aligned}
$$
where $ Q_k^{\prime}(y)=P_k(y).$
Is there any other simpler proof?
Comments and alternative methods are highly appreciated.
Best Answer
Here is a simpler proof. Utilize the reduction formula $$\int_0^{\frac\pi2}\frac1{(1+\tan x)^m}dx=K_m= \frac1{2(m-1)}+K_{m-1}-\frac12K_{m-2} $$ to reduce \begin{align} I_n=&\int_0^{\frac\pi2}\frac1{(1+\tan x)^{4n+2}}dx\\ =& \ \frac1{2(4n+1)}+ \frac1{2(4n)}+ \frac1{4(4n-1)}-\frac14I_{n-1} \end{align} Thus, the starting rational value $I_0= \int_0^{\frac\pi2}\frac1{(1+\tan x)^{2}}dx=\frac12 $ yields successively the rational values below $$I_1=\frac{11}{60},\>\>\> I_2=\frac{34}{315},\>\>\> I_3=\frac{27341}{360360},\>\>\> \cdots $$