Latest Edit
Great thanks to @Quanto for his closed form for
$$\int_{0}^{\frac{\pi}{4}} \sec^{2n+1} x \ dx=
\frac1{2^{2n}}
\binom {2n}n \ln(\sqrt2+1) +\sum_{k=0}^{n-1} \binom {2n}k \frac{(\sqrt2+1)^{2(n-k)} – (\sqrt2-1)^{2(n-k)}}{2^{2n+1}(n-k)}, $$
by which the closed form for integrals with odd powers can be found as
$$\boxed{\quad \int_{0}^{\frac{\pi}{2}} \frac{d x}{(\sin x+\cos x)^{2 n+1}}\\=
\frac1{2^{3n-\frac{1}{2} }}
\binom {2n}n \ln(\sqrt2+1) +\sum_{k=0}^{n-1} \binom {2n}k \frac{(\sqrt2+1)^{2(n-k)} – (\sqrt2-1)^{2(n-k)}}{2^{3n+\frac{1}{2} }(n-k)} } $$
Inspired by the post, I firstly try to generalise it to even powers as
$$
I_{n}=\int_{0}^{\frac{\pi}{2}} \frac{d x}{(\sin x+\cos x)^{2 n}},
$$
where $n\in N.$
For simplicity, we express the denominator as cosine using compound angle formula.
$$
\begin{aligned}
I_{n} &=\int_{0}^{\frac{\pi}{2}} \frac{d x}{(\sin x+\cos x)^{2 n}} \\
&=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left[\sqrt{2} \cos \left(\frac{\pi}{4}-x\right)\right]^{2 n}} \\
& \stackrel{x\mapsto\frac{\pi}{2}-x}{=} \frac{1}{2^{n}}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\cos ^{2 n} x}\\ & \stackrel{Symm}{=} \frac{1}{2^{n-1}} \int_{0}^{\frac{\pi}{4}} \sec ^{2 n} x d x
\end{aligned}
$$
Letting $t= \tan x$ yields
$$
\begin{aligned}
I_{n} &= \frac{1}{2^{n-1}}\int_{0}^{1}\left(1+t^{2}\right)^{n-1} d t \\
&= \frac{1}{2^{n-1}} \sum_{k=0}^{n-1}\left(\begin{array}{c}
n-1 \\
k
\end{array}\right) \int_{0}^{1} t^{2 k} d t \\
&=\boxed{\frac{1}{2^{n-1}} \sum_{k=0}^{n-1}\left(\begin{array}{l}
n-1 \\
\quad k
\end{array}\right) \frac{1}{2 k+1}}
\end{aligned}
$$
Examples:
$$
I_{2} =\frac{1}{2}\left(1+\frac{1}{3}\right)=\frac{2}{3},I_{5}=\frac{83}{315}; I_{10}=\frac{26206}{230945};
$$
However, it is harder to find the integral of odd powers. $$
J_{n}=\int_{0}^{\frac{\pi}{2}} \frac{d x}{(\sin x+\cos x)^{2 n+1}}= \frac{1}{2^{n-\frac{1}{2} }} \int_{0}^{\frac{\pi}{4}} \sec ^{2 n+1} x d x
$$
where $n=0, 1, 2,…$
I had found a reduction formula for it as
$$
\boxed{J_{n+1}=\frac{1}{2 n}+\frac{2 n-1}{4 n} J_{n}}
$$
via the reduction formula for $$
\int_{0}^{\frac{\pi}{4}} \sec ^{2 n+1} x d x=\frac{1}{2 n}\left[2^{n-\frac{1}{2}}+(2 n-1) \int_{0}^{\frac{\pi}{4}} \sec ^{2 n-1} x d x\right]
$$
Examples
$$
\begin{aligned}
J_{0} &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \sec x d x \\
&=\sqrt{2}\left|\ln \left|\sec x+\tan x\right|\right]_{0} ^{\frac{\pi}{4}}\\
&=\sqrt{2} \ln (\sqrt{2}+1)
\end{aligned}
$$
\begin{align}
\begin{aligned}
J_{1} &=\frac{1}{2}+\frac{1}{4} \cdot \sqrt{2} \ln (\sqrt{2}+1) \\
&=\frac{1}{2}+\frac{1}{2 \sqrt{2}} \ln (\sqrt{2}+1)
\end{aligned}
\end{align}
$$
\begin{aligned}
J_{2} &=\frac{1}{4}+\frac{3}{8}\left[\frac{1}{2}+\frac{1}{2 \sqrt{2}} \ln (\sqrt{2}+1)\right] \\
&=\frac{7}{16}+\frac{3}{16 \sqrt{2}} \ln (\sqrt{2}+1)
\end{aligned}
$$
Although I can find the integral with odd powers one by one, I still can’t find its closed form. Your help will be highly appreciated.
Best Answer
For the odd case, substitute $\sec x= \cosh t$ to integrate \begin{align} &\int_{0}^{\frac{\pi}{4}} \sec^{2n+1} x \ d x =\int_0^{\ln(\sqrt2+1)}\cosh^{2n}t\ dt\\ =& \ \frac1{2^{2n}}\int_0^{\ln(\sqrt2+1)}\bigg[ \binom {2n}n+2 \sum_{k=0}^{n-1} \binom {2n}k \cosh2(n-k)t\bigg]\ dt\\ = &\ \frac1{2^{2n}} \binom {2n}n \ln(\sqrt2+1) +\sum_{k=0}^{n-1} \binom {2n}k \frac{(\sqrt2+1)^{2(n-k)} - (\sqrt2-1)^{2(n-k)}}{2^{2n+1}(n-k)} \\ \end{align}