When is ${\large\int}\frac{dx}{\left(1+x^a\right)^a}$ an elementary function? Consider the following indefinite integral:
$$F_a(x)=\int\frac{dx}{\left(1+x^a\right)^a}.$$
Here $a\in\mathbb R$ is a parameter, and $x>0$ is a variable. For what values of the parameter $a$ the function $F_a(x)$ is an elementary function of the variable $x$?
Here are some examples (in each case there is implicitly an additive constant of integration).

$$F_1(x)=\int\frac{dx}{\left(1+x\right)}=\ln(1+x)$$

$$F_{-1}(x)=\int\frac{dx}{\left(1+x^{-1}\right)^{-1}}=x+\ln x$$

$$F_2(x)=\int\frac{dx}{\left(1+x^2\right)^2}=\frac12\left(\frac{x}{x^2+1}+\arctan(x)\right)$$

$$F_{1/2}(x)=\int\frac{dx}{\left(1+x^{1/2}\right)^{1/2}}=\frac43\left(x^{1/2}-2\right) \left(1+x^{1/2}\right)^{1/2}$$

$$F_{1/3}(x)=\int\frac{dx}{\left(1+x^{1/3}\right)^{1/3}}=\frac9{40}\left(9-6x^{1/3}+5x^{2/3}\right)\left(1+x^{1/3}\right)^{2/3}$$

$$F_{1+\sqrt2}(x)={\large\int}\frac{dx}{\left(1+x^{1+\sqrt2}\right)^{1+\sqrt2}}=\left(x+\frac{x^{2+\sqrt2}}{\sqrt2}\right)\left(1+x^{1+\sqrt2}\right)^{-\sqrt2}$$

$$F_{1-\sqrt2}(x)={\large\int}\frac{dx}{\left(1+x^{1-\sqrt2}\right)^{1-\sqrt2}}=\left(x-\frac{x^{2-\sqrt{2}}}{\sqrt{2}}\right) \left(1+x^{1-\sqrt{2}}\right)^{\sqrt{2}}$$

$$F_{(1+\sqrt5)/2}(x)={\large\int}\frac{dx}{\left(1+x^{(1+\sqrt5)/2}\right)^{(1+\sqrt5)/2}}=x \left(1+x^{(1+\sqrt5)/2}\right)^{(1-\sqrt5)/2}$$

$$F_{(1-\sqrt5)/2}(x)={\large\int}\frac{dx}{\left(1+x^{(1-\sqrt5)/2}\right)^{(1-\sqrt5)/2}}=x\left(1+x^{(1-\sqrt{5})/2}\right)^{(1+\sqrt{5})/2}$$

All these results can be easily verified by differentiation. For the general case, Mathematica gives the result in terms of the hypergeometric function:
$$F_a(x)=\int\frac{dx}{\left(1+x^a\right)^a}=x\cdot{_2F_1}\left(\frac1a,a;1+\frac1a;-x^a\right).$$
For what values of the parameter $a$ does this hypergeometric function reduce to an elementary function?
 A: I can give partial answers to your question : 
1) $a\in \mathbb{Z}$ 
Every rational function has elementary antiderivative.
2) $1/a\in\mathbb{Z}$. 
Let $y=x^{1/n}$. Then the integral becomes
$$ n\int y^{n-1}(1+y)^{-1/n}dy$$
which can be solved by $n-1$ times of integration by parts. 
3) $a-1/a=n\in\mathbb{Z}$
Let $u=x^{a}$. Then the integral becomes
$$ \int u^{a-n-1}(1+u)^{-a}du=\int u^{-n-1}\left(\frac{1+u}{u}\right)^{a}du$$
If $n\geq 1$, use substitution $t=(u+1)/u$ which gives 
$$ -\int t^{a}(t-1)^{n-1} dt$$
and if $n\leq -1$, use $t=u/(u+1)$ then
$$ \int t^{n-1-a}(1-t)^{-n-1} dt$$
where both can be solved by integration by parts. 
4) For $a \in \mathbb{Q}$, Chebyshev's theorem states that if $a, b\in\mathbb{R}$ and $p, q, r\in \mathbb{Q}$, then  $\int x^{p}(a+bx^{r})^{q} dx$ can be expressed as elementary function iff at least one of $(p+1)/r, q, (p+1)/r+q$ is integer. Applying for  our case, only for $a\in \mathbb{Z}$ or $1/a\in \mathbb{Z}$, $F_{a}(x)$ can be expressed as elementary function. 
A: Using integration by parts leads to \begin{align}\int\frac{dx}{(1+x^a)^a}&=\frac x{(1+x^a)^a}+\int\frac{a^2x^adx}{(1+x^a)^{a+1}}\\
&=\frac x{(1+x^a)^a}+\int a^2\left\{\frac1{(1+x^a)^a}-\frac1{(1+x^a)^{a+1}}\right\}dx\\
&=\frac x{(1-a^2)(1+x^a)^a}-\frac{a^2}{1-a^2}\int\frac{dx}{(1+x^a)^{a+1}}
\end{align}
so you can also consider integrals of the form $\int dx/(1+x^a)^{a+n}$. 
The substitution $u=1/x$ gives$$F_a(x)=\int\frac{-du}{u^2(1+u^{-a})^a}=-\int\frac{u^{a^2-2}du}{(1+u^a)^a}$$ $$\int\frac{dx}{(1+x^a)^{a+n}}=-\int\frac{u^{a^2+an-2}du}{(1+u^a)^{a+n}}$$which implies $a^2+an-1$ must be a multiple of $a$ (i.e. $a^2-1=k\cdot a$).
Subbing $u=x^a$ turns $F_a(x)$ into $$\frac1a\int\frac{u^{1/a-1}du}{(1+u)^a}$$which if elementary will only have roots, rational functions, and logarithms (specifically, logarithms will only appear when $a\in\mathbb{Z}$).
