Indefinite integral of $\int \frac 1 x \operatorname{arsech} \frac x a \, \mathrm d x$ Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.668$ gives the indefinite integral of the area hyperbolic secant (that is, the "inverse" hyperbolic secant) as:
$$\int \frac 1 x \operatorname{arsech} \frac x a \, \mathrm d x = -\frac 1 2 \ln (a/x) \ln (4 a/x) - \frac {(x/a)^2} {2 \cdot 2 \cdot 2} - \frac {1 \cdot 3 (x/a)^4} {2 \cdot 4 \cdot 4 \cdot 4} - \cdots$$
This can be expressed as:
$$\int \frac 1 x \operatorname{arsech}  \frac x a \, \mathrm d x = -\frac 1 2 \ln \left({\frac a x}\right) \ln \left({\frac {4 a} x}\right) - \sum_{n \mathop \ge 0} \frac {(2 n)!} {2^{2 n} (n!)^2 (2 n)^2} \left({\frac x a}\right)^{2 n} + C$$
Getting nearly there is easy enough.
I take as my starting point the power series expansion of the area hyperbolic cosine of $\dfrac x a$:
$$\operatorname{arcosh} x = \ln \frac {2 x} a - \left({\sum_{n \mathop = 1}^\infty \frac {(2 n)!} {2^{2 n} (n!)^2 (2 n)} \left({\frac a x}\right)^{2 n} }\right)$$
(This is derived from the result in Spiegel, item $20.40$.)
From it I get the same for the area hyperbolic secant $\operatorname{arsech}$ of $\dfrac x a$, as it's just the $\operatorname{arcosh}$ of $\dfrac a x$:
$$\operatorname{arsech} x = \ln \frac {2 a} x - \left({\sum_{n \mathop = 1}^\infty \frac {(2 n)!} {2^{2 n} (n!)^2 (2 n)} \left({\frac x a}\right)^{2 n} }\right)$$
and integrate term by term (justified by Fubini's theorem, I believe) to get me eventually to:
$$\int \frac 1 x \operatorname{arsech} \frac x a \, \mathrm d x = -\frac 1 2 \ln^2 \left({\dfrac x {2 a} }\right) + \sum_{n \mathop = 0}^\infty \frac {(2 n)!} {2^{2 n} (n!)^2 (2 n)^2} \left({\frac x a}\right)^{2 n} + C$$
During the course of the above I inverted the reciprocal in the logarithm of the integrand to get it into a standard form for integration $\displaystyle \int \dfrac {\ln (c x)} x \mathrm d x = \dfrac {\ln^2 {c x} } 2$ which I think I got right.
The above is consistent with the result quoted in Schaum for the indefinite integral for the area hyperbolic cosine, where the logarithm term was left as a square.
So there are $2$ questions outstanding:

*

*How do you actually get from $\ln^2 \left({\dfrac x {2 a} }\right)$ to $\ln \left({\dfrac a x}\right) \ln \left({\dfrac {4 a} x}\right)$? I get that they will differ by a constant which can be subsumed into a constant of integration, but manipulation of $\ln^2 \left({\dfrac x {2 a} }\right)$ gets me only as far as $\left({\ln \left({\dfrac a x}\right)}\right)^2 + 2 \ln 2 \ln \dfrac a x + \left({\ln 2}\right)^2$ and at this point I can't see how to proceed. I can't reduce the $2 \ln 2 \ln \dfrac a x$ and get it to go the way I want it to.


*How did Spiegel ever get to that $\ln \left({\dfrac a x}\right) \ln \left({\dfrac {4 a} x}\right)$ term in the first place? His quoted result for the indefinite integral for the area hyperbolic cosine has that term in the $\ln^2$ form, the same as what I got for the area hyperbolic secant. That is, what direction could he have taken with his integration so as to land upon a result in that form? I can't see why he would deliberately manipulate it into that form, as he is happy enough to leave the indefinite integral for the area hyperbolic cosine in the $\ln^2$ form.

Since you are calculating indefinite integral, the inconsistency is obvious, the difference between two results must be a constant!

*

*The difference:

$$\ln \left(\frac{a}{x}\right) \ln \left(\frac{4 a}{x}\right)-\ln ^2\left(\frac{x}{2 a}\right)=
\ln \left(\frac{a}{x}\right) \left[\ln \left(\frac{a}{x}\right)+2 \log (2)\right]-\left[\ln \left(\frac{a}{x}\right)+\ln (2)\right]^2
=-\ln ^2(2)$$


*The reason: the real domain of ${\rm arcsech(x)}$ is $[0,1]$, thus when you change the operations of expansion and integral, you must be careful, i.e.
$$
\int\to\int_a^x
$$
In other words, you should have
$$
\int_a^x \frac{\ln (2 a)-\ln (z)}{z} \, dz=-\frac{1}{2} \ln \left(\frac{a}{x}\right) \ln \left(\frac{4 a}{x}\right)
$$
