Yes I know I've asked this question before, but this time it's more specific.
I am still trying to get to grips with the calculation of $\int \dfrac {\mathrm d x} {\sqrt {(a x + b) (p x + q)} }$ (Schaum's Mathematical Handbook of Formulas and Tables, 1968).
This is for when $a p < 0$.
The answer is: $$\dfrac 2 {\sqrt {-a p} } \arctan \left({\sqrt {\dfrac {-p (a x + b) } {a (p x + q) } } }\right) + C$$
My working is using Euler's Substitution of the Third Kind.
$$\sqrt {a x^2 + b x + c} = \sqrt {a (x - \alpha) (x - \beta) } = (x - \alpha) t$$
First express the integrand as:
$$\int \frac {\mathrm d x} {\sqrt {(a x + b) (p x + q) } }= \int \frac {\mathrm d x} {\sqrt {a p \left({x - \left({-\frac b a}\right) }\right) \left({x - \left({-\frac q p}\right) }\right) } }$$
so that $a \gets a p$, $\alpha \gets -\dfrac b a$, $\beta \gets -\dfrac q p$.
Then:
$$x = \dfrac {a p \left({-\frac q p}\right) - \left({-\frac b a}\right) t^2} {a p - t^2} = \dfrac {-a q + \left({\frac b a}\right) t^2} {a p - t^2} = \dfrac {a^2 q - b t^2} {a (t^2 - a p) }$$
This leads to:
$$x - \alpha = \dfrac {a (\alpha - \beta) } {t^2 - a} = \dfrac {a p \left({\left({-\frac b a}\right) - \left({-\frac q p}\right) }\right) } {t^2 - a p}$$
substituting $a \gets a p$, $\alpha \gets -\dfrac b a$, $\beta \gets -\dfrac q p$.
Thus: $$x - \alpha = \dfrac {a q - p b} {t^2 - a p}$$
Then:
$$\dfrac {\mathrm d x} {\mathrm d t} = \dfrac {2 t a p \left({\left({-\frac q p}\right) - \left({-\frac b a}\right) }\right) } {(a p - t^2)^2} = \dfrac {2 t a (b p - a q) } {(t^2 - a p)^2}$$
Then after some algebra:
$$t = \pm a \sqrt {\dfrac {p x + q} {a x + b} }$$
(Not sure what to do about the $\pm$, I'll figure it out later.)
Putting it all together:
$$\int \frac {\mathrm d x} {\sqrt {(a x + b) (p x + q) } } = \int \frac {\mathrm d t} {\left({\dfrac {a q - p b} {t^2 - a p} }\right) t} \dfrac {2 t a (b p - a q) } {(t^2 - a p)^2} = -2 a \int \dfrac {\mathrm d t} {(t^2 - a p) }$$
Hence as $a p < 0$ by hypothesis:
$$\int \frac {\mathrm d x} {\sqrt {(a x + b) (p x + q) } } = -2 a \int \dfrac {\mathrm d t} {(t^2 + (-a p)) } = -2 a \arctan \dfrac t {\sqrt {-a p} } + C$$
which leads to: $$= -2 a \arctan {\dfrac {\pm a \sqrt {\frac {p x + q} {a x + b} } } {\sqrt {-a p} } } + C = -2 a \arctan {\sqrt {-\dfrac a p} \dfrac {\sqrt {p x + q} } {\sqrt {a x + b} } } + C$$
(The $\pm$ still needs to be resolved, clearly -- not done that.)
Clearly I'm on the right lines, but the details do not match what we're aiming for.
Can anyone put their finger on what I have done wrong?
Many thanks
The following calculation is off
$$\dfrac {\mathrm d x} {\mathrm d t} = \dfrac {2 t a p \left({\left({-\frac q p}\right) - \left({-\frac b a}\right) }\right) } {(a p - t^2)^2} = \dfrac {2 t a (b p - a q) } {(t^2 - a p)^2}$$ Instead, it should be $$\dfrac {\mathrm d x} {\mathrm d t} = \dfrac {2 t (b p - a q) } {(t^2 - a p)^2}$$ Then \begin{align} \int \frac {\mathrm d x} {\sqrt {(a x + b) (p x + q) } } &= -2 \int \dfrac {\mathrm d t} {(t^2 - a p) } =-\frac2{\sqrt{-ap}}\tan^{-1}\frac t{\sqrt{-ap}}\\ &= -\frac2{\sqrt{-ap}}\tan^{-1}\frac {a \sqrt {(a x + b) (p x + q) } }{\sqrt{-ap}\ (ax+b)}+C\\ &= \frac2{\sqrt{-ap}} \tan^{-1} {\sqrt {\dfrac {-p (a x + b) } {a (p x + q) } } } -\frac\pi2+ C \end{align}