I need to check whether my evaluation of $\displaystyle \int \dfrac {\mathrm d x} {\sqrt {a x^2 + b x + c} }$ is correct.
This is what I have evaluated.
First we note that the integral is defined only when $a x^2 + b x + c > 0$, so it will be assumed throughout that this is the case.
I have also omitted the constant of integration out of typical childish laziness.
a) For $a > 0$ we have the following:
$\displaystyle \int \dfrac {\mathrm d x} {\sqrt {a x^2 + b x + c} } = \begin{cases} \dfrac 1 {\sqrt a} \dfrac {|2 a x + b|} {2 a x + b} \cosh^{-1} \left({\dfrac {|2 a x + b|} {\sqrt {b^2 - 4 a c} } }\right) & : b^2 - 4 a c > 0 \\ \dfrac 1 {\sqrt a} \sinh^{-1} \left({\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } }\right) & : b^2 - 4 a c < 0 \\ \dfrac 1 {\sqrt a} \ln |2 a x + b| + C & : b^2 - 4 a c = 0 \end{cases}$
b) For $a < 0$ we have:
$\displaystyle \int \dfrac {\mathrm d x} {\sqrt {a x^2 + b x + c} } = \dfrac {-1} {\sqrt {-a} } \arcsin \left({\dfrac {2 a x + b} {\sqrt {|b^2 - 4 a c|} } }\right)$
defined only when $b^2 - 4 a c \ne 0$.
The technique I used was to complete the square in $a x^2 + b x + c$ to get $\dfrac {(2 a x + b)^2 - (b^2 - 4 a c)} {4 a}$ and use that as the basis for a primitive of the form:
$\displaystyle \frac 1 A \int \dfrac {\mathrm d z} {\pm z^2 \pm D^2}$
where $z = 2 a x + b$ and $D^2 = b^2 - 4 a c$ and $\dfrac 1 A$ is the constant that arises during the substitution.
Having evaluated both cases arising from different discriminant signs when $a < 0$, the result I obtained seemed to indicate that they could both be merged by using a modulus operation.
I also noted that when $a < 0$ the expression $a^2 + b x + c$ appears always negative when $b^2 - 4 a c$, hence the lack of definition of the primitive when those two conditions hold simultaneously.
I have been working with respect to Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.280$.
The latter appears to be incomplete, giving merely:
$\displaystyle \int \dfrac {\mathrm d x} {\sqrt {a x^2 + b x + c} } = \begin{cases} \dfrac 1 {\sqrt a} \ln (2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b) \\ -\dfrac 1 {\sqrt {-a} } \sin^{-1} \left({\dfrac {2 a x + b} {\sqrt {b^2 - 4 a c} } }\right) \text { or } \dfrac 1 {\sqrt a} \sinh^{-1} \left({\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } }\right) \end{cases}$
without providing the conditions under which the various results hold.
He also appears to have glossed over the cases arising from different discriminant signs when $a < 0$ in the arcsin expression.
(I established to my satisfaction that the $\ln$ expression given by Spiegel is actually equivalent to the $\cosh^{-1}$ expression, give or take a constant that can be subsumed into the arbitrary constant of integration, but again, I may have made a mistake on that, so checking of my work would be greatly appreciated here.)