Integral $\int \frac1{\sqrt {(a x + b) (p x + q)}}dx$

Question

I am working on this:

$$\int \dfrac 1 {\sqrt {(a x + b) (p x + q)} } \ \mathrm d x$$

valid where $(a x + b)(p x + q) > 0$.

I am specifically interested in the case where $\dfrac {b p - a q} p < 0$.

I do this:

Let $u = \sqrt {a x + b}$

$\leadsto x = \dfrac {u^2 - b} a$

$\leadsto \sqrt {p x + q} = \sqrt {p \left({\dfrac {u^2 - b} a}\right) + q}$

$= \sqrt {\dfrac {p \left({u^2 - b}\right) + a q} a}$

$= \sqrt {\dfrac {p u^2 - b p + a q} a}$

$= \sqrt {\dfrac p a} \sqrt {u^2 - \left({\dfrac {b p - a q} p}\right) }$

Then by a standard result:

$\displaystyle \int \frac {\mathrm d x} {\sqrt {(a x + b) (p x + q) } } = \int \frac {2 u \mathrm d u} {a \sqrt {\frac p a} \sqrt {u^2 - \left({\frac {b p - a q} p}\right) } u}$

$= \dfrac 2 {\sqrt {a p} } \int \dfrac {\mathrm d u} {\sqrt {u^2 - \left({\dfrac {b p - a q} p}\right) } }$

Now let $\dfrac {b p - a q} p < 0$.

Let $c^2 = -\dfrac {b p - a q} p$.

This gives us:

$\displaystyle \int \dfrac 1 {\sqrt {(a x + b) (p x + q)} } \ \mathrm d x = \frac 2 {\sqrt {a p} } \int \frac {\mathrm d u} {\sqrt {u^2 + c^2} }$

$= \dfrac 2 {\sqrt {a p} } \sinh^{-1} {\dfrac u c} + C$ (standard result)

$= \dfrac 2 {\sqrt {a p} } \sinh^{-1} \dfrac {\sqrt {a x + b} } {\sqrt {\dfrac {a q - b p} p} } + C$

(because $\dfrac {b p - a q} p = -\dfrac {a q - b p} p$)

$= \dfrac 2 {\sqrt {a p} } \sinh^{-1} \sqrt {\dfrac {p (a x + b) } {a q - b p} } + C$

But when I look in the book (in this case Murray Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), I see this as result $14.120$:

$$\dfrac 2 {\sqrt {-a p} } \arctan \sqrt {\dfrac {-p (a x + b) } {a (p x + q)} } $$

EDIT 4-Jun-2023: Corrected denominator of argument of arctangent to $a (p x + q)$ from $a q - b p$.

Can't see where I'm doing it wrong.

Answer 1

Amplified comment: "Convert an $\arcsin$ to an $\arctan$."
I learned this long ago because certain computer languages had only $\arctan$ as built-in function, and you had to use formulas if you wanted $\arcsin, \arccos$, etc. (BASIC had ATN arctangent, but none of the other "arc" trig functions)

So here it is:

$$ \arcsin t = 2 \arctan\left(\frac{t}{1+\sqrt{1-t^2}}\right) $$

Here are the graphs.

Identical