Are there "one way" integrals? If we suppose that we can start with any function we like, can we work "backwards" and differentiate the function to create an integral that is hard to solve?
To define the question better, let's say we start with a function of our choosing, $f(x)$.  We can then differentiate the function with respect to $x$ do get $g(x)$:
$$g(x) = f'(x)$$
This, in turn, implies, under appropriate conditions, that the integral of $g(x)$ is $f(x)$:
$$\int_a^b { g(x) dx } = [f(x)]_a^b$$
I'm wondering what conditions are appropriate to allow one to easily get a $g(x)$ and $f(x)$ that assure that $f(x)$ can't be easily found from $g(x)$.
SUMMARY OF THE QUESTION
Can we get a function, $g(x)$, that is hard to integrate, yet we know the solution to?  It's important that no one else should be able to find the solution, $f(x)$, given only $g(x)$.  Please help!
POSSIBLE EXAMPLE
This question/integral seems like it has some potential.
DEFINITION OF HARDNESS
The solution to the definite integral can be returned with the most $n$ significant digits correct.  Then it is hard to do this if the time it takes is an exponential number of this time.
In other words, if we get the first $n$ digits correct, it would take roughly $O(e^n)$ seconds to do it.

Here is an example (by Harold Davenport) of a function that is hard to integrate:
$$\int{2t^6+4t^5+7t^4-3t^3-t^2-8t-8\over
(2t^2-1)^2\sqrt{t^4+4t^3+2t^2+1}}\,dt\ .$$ 
The primitive is given by (check it!)
$$\eqalign{&-2\log(t^2+4t+2)-\log(\sqrt2+2t)\cr&+ \log\left(\sqrt2
t+2\sqrt2-2\sqrt{t^4+4t^3+2t^2+1}\,\right)\cr
&-5\log(2t^2-1)-5\log\left(2\sqrt2 t+4\sqrt2 -t^2-4t-6\right) \cr&+
5\log\left(\bigl(4\sqrt2+19\bigr)\sqrt{t^4+4t^3+2t^2+1}\right.
\cr  &\qquad\qquad\left. - 16\sqrt2 t^2 -8\sqrt2 t +6\sqrt2
-29t^2-38t+5\right)\cr}$$ 
$$\eqalign{ &+
2\log\left(\bigl(10\sqrt2+17\bigr)\sqrt{t^4+4t^3+2t^2+1}\right.\cr
&\qquad\qquad\left.+4\sqrt2 t^2+16\sqrt2 t -2\sqrt2-11t^2-44t-39\right)
\cr &+ {1\over2}\log\left(\bigl(731\sqrt2
t+71492\sqrt2-70030t-141522\bigr) \sqrt{t^4+4t^3+2t^2+1}
\right.\cr &\qquad\qquad-40597\sqrt2t^3-174520\sqrt2t^2-122871\sqrt2
t+50648\sqrt2\cr&\qquad\qquad\left.+90874t^3+
403722t^2+ 272622t-61070 \vphantom{\sqrt{t^4}}\right)\cr & + 
{(2t+1)\sqrt{t^4+4t^3+2t^2+1}\over 4t^2-2}\ .\cr}$$
