Integral ${\large\int}_0^1\frac{dx}{(1+x^{\sqrt2})^{\sqrt2}}$ Mathematica claims that
$${\large\int}_0^1\!\!\frac{dx}{(1+x^{\sqrt2})^{\sqrt2}}=\frac{\sqrt\pi}{2^{\sqrt2}\sqrt2}\cdot\frac{\Gamma\left(\frac1{\sqrt2}\right)}{\Gamma\left(\frac12+\frac1{\sqrt2}\right)},\tag{$\diamond$}$$
and it also confirms numerically.
How can we prove $(\diamond)$?
This result seems interesting, because no such nice answer seems to occur for other algebraic powers, except trivial cases, when the antiderivative is an elementary function, e.g. 
$${\large\int}\frac{dx}{(1+x^\alpha)^\alpha}=\\
\small\frac{x\left(x^\alpha+1\right)^{1-\alpha}}{18}\left[\vphantom{\large|}\left(15(\alpha-1)+4\left(5(\alpha+1)+\left(6\alpha+(\alpha-3)x^{2\alpha}+2(\alpha+2)x^\alpha-3\right)x^\alpha\right)x^\alpha\right)x^\alpha+18\right],$$
where $\alpha=3+\sqrt{10}$.
 A: Hint. One may observe that
$$
\begin{align}
\int_0^\infty\frac{dx}{(1+x^a)^a}&=\int_0^1\frac{dx}{(1+x^a)^a}+\int_1^\infty\frac{dx}{(1+x^a)^a}
\\\\&=\int_0^1\frac{dx}{(1+x^a)^a}+\int_0^1\frac{dx}{\left(1+\frac1{x^a}\right)^a\:x^2} \quad (x \to 1/x)
\\\\&=\int_0^1\frac{dx}{(1+x^a)^a}+\int_0^1\frac{x^{a^2-2}dx}{(1+x^a)^a}
\\\\&=\int_0^1\frac{1+x^{a^2-2}}{(1+x^a)^a}\:dx
\end{align}
$$ giving, for $a:=\sqrt{2}$, that
$$
\int_0^1\frac{dx}{(1+x^{\sqrt2})^{\sqrt2}}=\frac12\int_0^\infty\frac{dx}{(1+x^{\sqrt2})^{\sqrt2}}=\frac{\Gamma\left(1+\frac{1}{\sqrt2}\right) \Gamma\left({\sqrt2}-\frac{1}{\sqrt2}\right)}{2\:\Gamma\left(\sqrt2\right)}
$$ by using the Euler beta result with the change of variable $t=\dfrac1{(1+x^{\sqrt2})^{\sqrt2}}$ in the latter integral.
A: Using Euler's integral representation of the Gaussian hypergeometric function, and assuming $a>0$, we get
$$ \begin{align} \int_{0}^{1} \frac{dx}{(1+x^{a})^{a}} &= \frac{1}{a} \int_{0}^{1} \frac{u^{1/a-1}}{(1+u)^{a}} \, du \\ &= \frac{1}{a} \int_{0}^{1} u^{1/a-1} (1-u)^{(1+1/a)-1/a-1}(1+u)^{-a} \, du \\ &= \frac{1}{a} \, B \left(\frac{1}{a}, 1 \right) {}_2F_{1} \left(a, \frac{1}{a}; 1+ \frac{1}{a}; -1 \right) \\&= {}_2F_{1} \left(a, \frac{1}{a}; 1+ \frac{1}{a}; -1 \right). \end{align}$$
In the case $a=\sqrt{2}$, we can apply Kummer's theorem since $1+\sqrt{2} - \frac{1}{\sqrt{2}}= 1+ \frac{1}{\sqrt{2}}$.
$$\begin{align} \int_{0}^{1} \frac{dx}{(1+x^{\sqrt{2}})^{\sqrt{2}}} &=  \frac{\Gamma\left(1+ \frac{1}{\sqrt{2}}\right) \Gamma\left(1+\frac{1}{\sqrt{2}}\right)}{\Gamma\left(1+ \sqrt{2}\right) \Gamma\left(1\right)} \\ &= \frac{\Gamma \left(1+ \frac{1}{\sqrt{2}} \right)^{2}}{\Gamma(1+\sqrt{2})}\\ &= \frac{1}{2 \sqrt{2}} \frac{\Gamma\left(\frac{1}{\sqrt{2}}\right)^{2}}{\Gamma(\sqrt{2})} \\ &\approx 0.6604707226 \end{align}$$
To get it in the form given by Mathematica, apply the duplication formula to $\Gamma \left(\frac{1}{\sqrt{2}} \right)$.
