Compute $\int\frac{e^{f(x)}}{f(x)}\,dx$ I am wondering if there is a non-elementary function in literature that describes the integration of $e^{f(x)}/f(x)$ with respect to $x$, where $f(x)$ is:
$$\sqrt{A + Bx + Cx^2}$$
or
$$\sqrt{(A-x)^2 + C}$$
Is it possible or will series expansion / Guassian quadrature or other numerical methods be necessary to estimate the integral?
Thank you,
F
 A: $\int\dfrac{e^{f(x)}}{f(x)}dx=\int\dfrac{1}{f(x)}\sum\limits_{n=0}^\infty\dfrac{(f(x))^n}{n!}dx=\int\sum\limits_{n=0}^\infty\dfrac{(f(x))^{n-1}}{n!}dx$
Then find $\int(f(x))^{n-1}~dx$ for any non-negative integral $n$ .
When $f(x)=\sqrt{A+Bx+Cx^2}$ and $f(x)=\sqrt{(A-x)^2+C}$ , $\int(f(x))^{n-1}~dx$ have close-forms for any non-negative integral $n$ .
A: First of all, note that substitution lets you do away with the $x - A$ part and so it suffices to consider antidifferentiating
$$f(x) := \frac{e^{\sqrt{x^2 + a}}}{\sqrt{x^2 + a}}$$
only. The series expansion approach leads us to consider
$$f(x) = \left(x \mapsto \frac{e^x}{x}\right)(\sqrt{x^2 + a})$$
where we've written it as an anonymous composition with $x \mapsto \frac{e^x}{x}$, and we have, using the exponential series,
$$\frac{e^x}{x} = \frac{1}{x} + \left(\sum_{n=0}^{\infty} \frac{x^n}{(n+1)!}\right)$$
so that
$$f(x) = \frac{1}{\sqrt{x^2 + a}} + \left(\sum_{n=0}^{\infty} \frac{(\sqrt{x^2 + a})^n}{(n + 1)!}\right)$$
which becomes
$$f(x) = \frac{1}{\sqrt{x^2 + a}} + \left(\sum_{n=0}^{\infty} \frac{(x^2 + a)^{n/2}}{(n + 1)!}\right)$$
Then
$$\int f(x)\ dx = \ln\left(\sqrt{x^2 + a} + x\right) + \left[\sum_{n=0}^{\infty} \frac{1}{(n + 1)!} \left(\int (x^2 + a)^{n/2}\ dx\right)\right]$$
so all we are left with to deal with now is
$$\int (x^2 + a)^{n/2}\ dx$$
Now we can go further with the binomial series
$$\begin{align}
(x^2 + a)^{n/2} &= \sum_{k=0}^{\infty} \binom{n/2}{k} x^{2k} a^{(n/2)-k}\\
&= \sum_{k=0}^{\infty} \binom{n/2}{k} x^{2k} a^{n/2} a^{-k}\\
&= a^{n/2} \sum_{k=0}^{\infty} \binom{n/2}{k} x^{2k} a^{-k}\\
&= a^{n/2} \sum_{k=0}^{\infty} \binom{n/2}{k} (x^2)^k a^{-k}\\
&= a^{n/2} \sum_{k=0}^{\infty} \binom{n/2}{k} \frac{(x^2)^k}{a^k}\\
&= a^{n/2} \sum_{k=0}^{\infty} \binom{n/2}{k} \left(\frac{x^2}{a}\right)^k
\end{align}$$
and we can use the "falling factorial power" $(a)_n$ to write this as
$$(x^2 + a)^{n/2} = a^{n/2} \sum_{k=0}^{\infty} \frac{(n/2)_k}{k!} \left(\frac{x^2}{a}\right)^k$$
which then converts to a "rising factorial power" $a^{(n)}$ by
$$\begin{align}
(x^2 + a)^{n/2} &= a^{n/2} \sum_{k=0}^{\infty} \frac{(-1)^k (n/2)^{(k)}}{k!} \left(\frac{x^2}{a}\right)^k\\
&= a^{n/2} \sum_{k=0}^{\infty} (n/2)^{(k)} \frac{\left(-\frac{x^2}{a}\right)^k}{k!}\end{align}$$
You should see that we're gunning for something here - because this poster smells hypergeometric series. Termwise integration yields
$$\int (x^2 + a)^{n/2}\ dx = x a^{n/2} \sum_{k=0}^{\infty} (n/2)^{(k)} \frac{\left(-\frac{x^2}{a}\right)^k}{(2k+1) k!}$$
Hence we can sum all this up to get
$$\begin{align}
\int f(x)\ dx &= \ln\left(\sqrt{x^2 + a} + x\right) + \left[\sum_{n=0}^{\infty} \frac{1}{(n+1)!} \left(x a^{n/2} \sum_{k=0}^{\infty} (n/2)^{(k)} \frac{\left(-\frac{x^2}{a}\right)^k}{(2k+1) k!}\right)\right]\\
&= \ln\left(\sqrt{x^2 + a} + x\right) + x \left[\sum_{n=0}^{\infty} \frac{a^{n/2}}{(n+1)!} \left(\sum_{k=0}^{\infty} (n/2)^{(k)} \frac{\left(-\frac{x^2}{a}\right)^k}{(2k+1) k!}\right)\right]\\
&= \ln\left(\sqrt{x^2 + a} + x\right) + x \left[\sum_{n=0}^{\infty} \frac{a^{n/2}}{(n+1) n!} \left(\sum_{k=0}^{\infty} (n/2)^{(k)} \frac{\left(-\frac{x^2}{a}\right)^k}{(2k+1) k!}\right)\right]\\
&= \ln\left(\sqrt{x^2 + a} + x\right) + x \left[\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{a^{n/2}}{(n+1) n!} (n/2)^{(k)} \frac{\left(-\frac{x^2}{a}\right)^k}{(2k+1) k!}\right]\\
&= \ln\left(\sqrt{x^2 + a} + x\right) + x \left[\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(n/2)^{(k)}}{(n+1)(2k+1)} \frac{a^{n/2} \left(-\frac{x^2}{a}\right)^k}{n! k!}\right]\\
&= \ln\left(\sqrt{x^2 + a} + x\right) + x \left[\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(n/2)^{(k)}}{(n+1)(2k+1)} \frac{\left(\sqrt{a}\right)^n \left(-\frac{x^2}{a}\right)^k}{n! k!}\right]\\ \end{align}$$
Now note it is easy to find from the definition of the rising factorial power that
$$n + 1 = \frac{2^{(n)}}{1^{(n)}}$$
$$2k + 1 = 2 \left(k + \frac{1}{2}\right) = \frac{\left(\frac{3}{2}\right)^{(k)}}{\left(\frac{1}{2}\right)^{(k)}}$$
Hence we get
$$\int f(x)\ dx = \ln\left(\sqrt{x^2 + a} + x\right) + x \left[\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{1^{(n)} \left(\frac{1}{2}\right)^{(k)} (n/2)^{(k)}}{2^{(n)} \left(\frac{3}{2}\right)^{(k)}} \frac{(\sqrt{a})^n \left(-\frac{x^2}{a}\right)^k}{n! k!}\right]$$
(plus a constant) Yet alas, it seems we can go no further. It is very close to a two-variable kind of hypergeometric series in $\sqrt{a}$ and $-\frac{x^2}{a}$, but there is the nasty cross-index rising factorial power factor
$$(n/2)^{(k)}$$
to deal with. And I don't think any of the canonical hypergeometric series/functions, despite there being lots of them - more than Wolfram knows, esp. the two-variables kinds - have a nasty little cross-term like this. Now there could be some transformation to deal with it that goes to a known form, but I don't know - if anyone has any pointers, feel free to comment.
