Trying to calculate the length of some geodesic I came across an expression of the form $$ \int \sqrt{f'(x)} \,\mathrm{d}x $$ which got me wondering if there is any way to simplify or calculate such integrals in general (where $f: \mathbb{R} \to \mathbb{R}$ is sufficiently regular).
EDIT: I'm looking for a simplification in terms on $f$. (Anti-)derivatives of $f$ and integral transforms of $f$ would be interesting too.
EDIT 2: The asymptotic expansion for the integral of a product $$ \int f g = \sum_{k=0}^{\infty} (-1)^k f^{(k)} g^{(-1-k)} $$
@AmbretteOrrisey has given in the comments actually looks quite interesting (even if it is not very useful). I guess something similar for my expression would be nice, too.
I think you're looking for some sort of analog to $$\frac {\mathrm d(\sqrt {f'(x)})} {\mathrm dx} = \frac {f''(x)} {2 \sqrt{f'(x)}} $$
But for $\int \sqrt{f'(x)} \mathrm dx$. As alluded to in the comment, you will probably be disappointed to know that there is no such connection. Unfortunately integrals rarely work out this nicely.
Indeed, $\int \sqrt{f(x)} \mathrm dx$ may be completely different from $\int f(x) \mathrm dx$. For example:
$$\int (1 - x^3) \mathrm dx$$
is elementary and a polynomial itself, but:
$$\int \sqrt{1 - x^3} \mathrm dx$$
can only be resolved in terms of the elliptic integrals. (but has nice definite integrals over eg. $[0,1]$ with the aid of the $\Gamma$ function)
It ultimately boils down to the properties of $f'$. Certain $f'$s will resolve nicely, for example squares of functions with known antiderivatives, or in general even powers of nice functions, (eg. $\sqrt{f'(x)} = \text{polynomial} \cdot \sin x$ which will fall out, eventually, by parts) others will have unwieldly elementary antiderivatives, (eg. reciprocals of very high-order polynomials, which will eventually decompose into partial fractions and admit elementary anti-derivatives) or none at all (eg. $\frac {\sin x} x$).
Predicting whether a function has an elementary antiderivative is a question in differential algebra and is not an easy question at all. I don't have the background to understand it, but there is an answer here that addresses the topic.