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I am reading a physics paper in which I got stuck trying to reproduce some results. Since it seems to be strictly a question of mathematics (integration specifically) I decided to post it here. In summary, we have two variables, call them $chi$ and $h$, which are related according to the relation

$$
frac{dchi}{dh} = sqrt{frac{1+(1+6xi)xi h^2/M^2}{(1+xi h^2/M^2)^2}},
$$

where $xi gg 1$ and $M$ are both fixed parameters. I would like to integrate and solve $chi$ in terms of $h$. The answer is given without any more detail and supposed to be

$$
frac{sqrt{xi}}{M}chi = sqrt{1+6xi}sinh^{-1}{(sqrt{1+6xi},psi)} - sqrt{6xi}sinh^{-1}left(frac{sqrt{6xi}psi}{sqrt{1+psi^2}}right).
$$

where $psi(h) = sqrt{xi}h/M$. My attempt so far, looking at the answer for inspiration is of course start with the given substitution so that I have

$$
frac{sqrt{xi}}{M}chi = int frac{sqrt{1+(1+6xi)psi^2}}{1+psi^2}dpsi. qquad qquad (*)
$$

The first and only thing I have come up until now is integration by parts. I thought for instance I could take

$$
dv = sqrt{1+apsi^2} dpsi qquad text{where} qquad a = 1+6xi \
Rightarrow v = frac{1}{2}psi sqrt{1+apsi^2} + frac{sinh^{-1}(sqrt{a}psi)}{2sqrt{a}}
$$

where the second term kind of looks like something I am looking for. Then for $u$, I would take $u = 1/(1+psi^2)$. However, when I carry, I get to a point where the integration seems hopeless, unless some magical cancelation happens. I discarted the opposite choice for $u$ and $dv$, since I would end up with $tan^{-1}{psi}$ terms, and that doesn't look right. I was hoping you could advice me on how to approach this integral $(*)$, since I don't know if I quit too soon and should persevere with this approach or if I am hitting a dead end. The answer is supposed to be the general solution (i.e., to my understanding there's no physics in the way in justifying any step or approximation), so I would like the opinion of someone more experienced at taming integrals. As always, thank you so much for your time.

First let us denote: $$I=int frac{sqrt{1+ax^2}}{1+x^2}dx$$
Our goal is first to get rid of the square root. With $displaystyle{x=frac{tan t}{sqrt a}Rightarrow dx=frac{1}{sqrt a}sec^2 t dt} $ we get:
$$require{cancel}I=frac{1}{sqrt a}int frac{sqrt{1+cancel a frac{tan^2 t}{cancel a}}}{1+frac{tan^2 t}{a}}sec^2 t dt=sqrt a int frac{sec t}{a+tan^2 t}sec^2 t dt$$
$$=sqrt a int frac{1}{cos^3 t} frac{1}{a+frac{sin^2 t}{cos^2 t}}dt=sqrt a intfrac{1}{cos t}cdotfrac{1}{acos^2 t+sin^2 t}dt$$
$$=sqrt a int frac{cos t}{1-sin^2 t}cdot frac{1}{a(1-sin^2 t)+sin^2 t}dtoverset{sin t=y}=sqrt a int frac{1}{1-y^2}frac{1}{a(1-y^2)+y^2}dy$$
$$=sqrt aint frac{1}{y^2-1}dx-sqrt aint frac{1}{y^2+frac{a}{1-a}}dy $$
$$=frac{sqrt a}{2}lnleft(frac{y-1}{y+1}right)-sqrt{1-a} arctanleft(ysqrt{frac{1-a}{a}}right)+C, quad y=sin(arctan(sqrt a x))$$

As given by a CAS (and almost the same as given by David G. Stork in comments), the result, for the integral in title, is
$$int frac{sqrt{1+ax^2}}{1+x^2},dx=frac{2 sqrt{(1-a) a} sinh ^{-1}left(sqrt{a} xright)-i (a-1) log
left(frac{(2 a-1) x^2-2 i sqrt{1-a}, x sqrt{a x^2+1}+1}{x^2+1}right)}{2
sqrt{1-a}}$$ which is a real which seems to be difficult to simplify.

Using $a=1+6xi$ and assuming $xi >0$ and simplifying, this would give for
$$ I=int frac{sqrt{1+(1+6xi)x^2}}{1+x^2},dx$$
$$I=sqrt{1+6 xi } sinh ^{-1}left(xsqrt{1+6 xi} right)-sqrt{frac{3xi}{2}}
log left(frac{x left(2 sqrt{6xi} sqrt{1+ (1+ 6 xi)
x^2}+(1+12 xi) xright)+1}{x^2+1}right)$$ and the logarithm can be transformed in a $sinh(.)$