<s>[INST] Analytic expression for the primitive of square root of a quadratic Can an analytic expression be given for 
$$\int \sqrt{ax^2 + bx +c} \, dx$$
I think substitution doesn't work in this case (I need to compute the integral $\int_0^t \ldots$).
 [/INST] When you see
$$
ax^2 + \underbrace{{}\quad bx\quad{}}_\text{1st-degree term} + c,
$$
it may help to remember that there is a standard technique in algebra for reducing problems involving quadratic polynomials with a first-degree term to problems involving quadratic polynomials with no first-degree term.  It's called "completing the square".  You write
$$
ax^2+bx+c = a\left( x^2 + \frac b a x\right) + c.
$$
Then you need to work on $\displaystyle x^2 + \frac b a x$.
Half of the coefficient of the first-degree term is $\dfrac{b}{2a}$.  If you square that and add it to this expression you're working on, you get a perfect square—i.e. something squared:
$$
\underbrace{x^2 + \frac b a x} \quad +\quad \frac{b^2}{4a^2} = \left( x + \frac{b}{2a} \right)^2.
$$
So
\begin{align}
ax^2+bx+c = a\left( x^2 + \frac b a x\right) + c & = a\left( x^2 + \frac b a x + \frac{b^2}{4a^2} \right) - a\left( \frac{b^2}{4a^2} \right) + c \\[12pt]
& = a\left( x+ \frac{b}{2a} \right)^2 + \frac{4ac-b^2}{4a} \\[12pt]
& au^2 + \text{constant}.
\end{align}
Let's call that last constant capital $C$, and later we'll recall that it's $\dfrac{4ac-b^2}{4a}$.
Then since $u= x + \dfrac{b}{2a}$, we have $du = dx$, and the integral becomes
$$
\int \sqrt{au^2+C}\,du.
$$
Now we'd like a "$1$" where $C$ is, so that we can apply trigonometric identities.  So do a bit of algebra:
$$
\int \sqrt{au^2+C}\,du = \int \sqrt{\frac{a}{C} u^2 + 1} \, du.
$$
We also need $(\text{something})^2+1$, in order to apply the identity involving $\tan^2\theta+1$.  So we write:
$$
\int\sqrt{\left(u\sqrt{\frac{a}{C}}\right)^2+1}\  du.
$$
Then we have
$$
\int \sqrt{w^2 + 1}\  du.
$$
Since $w=u\sqrt{\dfrac{a}{C}}$, we have $dw = du\sqrt{\dfrac{a}{C}}$, so $du = dw\sqrt{\dfrac{C}{a}}$.
Now we have
$$
\sqrt{\dfrac{C}{a}} \int \sqrt{w^2+1}\  dw.
$$
This is
$$
\sqrt{\dfrac{C}{a}} \int \sqrt{\tan^2\theta+1}\  \sec^2\theta\,d\theta.
$$
$$
= \sqrt{\dfrac{C}{a}} \int \sec^3\theta\,d\theta.
$$
In April 2007, I wrote this Wikipedia article, which has since been edited by a number of others, and by me, explaining how to treat that integral and why it matters.
Later note: The above works if $a$ and (capital) $C$ are positive.  This implies (among other things) that $b^2-4ac<0$, so the quadratic polynomial cannot be factored using real numbers.
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