Evaluating $\int_0^1 \sqrt{1 + x ^4 } \, d x$

$$\int_{0}^{1}\sqrt{\,1 + x^{4}\,}\,\,\mathrm{d}x$$

I used substitution of tanx=z but it was not fruitful. Then i used $(x-1/x)= z$ and
$(x)^2-1/(x)^2=z$ but no helpful expression was derived.
I also used property $\int_0^a f(a-x)=\int_0^a f(x)$

Solutions Collecting From Web of "Evaluating $\int_0^1 \sqrt{1 + x ^4 } \, d x$"

We can do better than hypergeometric function and elliptic integral:
$$\color{blue}{\int_0^1 {\sqrt {1 + {x^4}} dx} = \frac{{\sqrt 2 }}{3} + \frac{{{\Gamma ^2}(\frac{1}{4})}}{{12\sqrt \pi }}}$$

Firstly, integration by part gives
$$\int_0^1 {\sqrt {1 + {x^4}} dx} = \sqrt 2 – 2\int_0^1 {\frac{{{x^4}}}{{\sqrt {1 + {x^4}} }}dx} = \sqrt 2 – 2\int_0^1 {\left( {\sqrt {1 + {x^4}} – \frac{1}{{\sqrt {1 + {x^4}} }}} \right)dx}$$
Hence
$$\int_0^1 {\sqrt {1 + {x^4}} dx} = \frac{{\sqrt 2 }}{3} + \frac{2}{3}\int_0^1 {\frac{1}{{\sqrt {1 + {x^4}} }}dx}$$
Making $x=1/u$ in the last integral gives
$$\int_0^1 {\frac{1}{{\sqrt {1 + {x^4}} }}dx} = \frac{1}{2}\int_0^\infty {\frac{1}{{\sqrt {1 + {x^4}} }}} dx = \frac{1}{{8\sqrt \pi }}{\Gamma ^2}(\frac{1}{4})$$
which can be evaluated by using some formula for beta function.

Consider the ${}_{2}F_{1}$ hypergeometric integral form given by
$${}_{2}F_{1}(a, b; c; x) = \frac{\Gamma(c)}{\Gamma(b) \, \Gamma(c-b)} \, \int_{0}^{1} t^{b-1} \, (1-t)^{c-b-1} \, (1-x \, t)^{-a} \, dt$$
leads to, with $a=-1/2$, $b=1/4$, $c=5/4$, $x=-1$,
$${}_{2}F_{1}\left(-\frac{1}{2}, \frac{1}{4}; \frac{5}{4}; -1\right) = \frac{1}{4} \, \int_{0}^{1} t^{-3/4} \, \sqrt{1+ t} \, dt.$$
Now let $t = x^{4}$ to obtain
$$\int_{0}^{1} \sqrt{1 + x^4} \, dx = {}_{2}F_{1}\left(-\frac{1}{2}, \frac{1}{4}; \frac{5}{4}; -1\right) = 1.08943…$$

Another non-elementary answer, from Maple, is
$$\int_0^1 \sqrt{1+x^4}\; dx = \frac{\sqrt {2}+{\it EllipticK} \left( 1/\sqrt {2} \right)}{3}$$