Solving integral of $e^\sqrt{x^2 + c}$ I am trying to solve an integral:
$$
\int {a e^{\dfrac{\sqrt{x^2 + b}}{c}} }dx
$$
Where a, b and c are all constants and positive.
My math is very rusty, so I don't even know if this is possible.
An approximation would also do.
I think it can also be expressed as:
$$
\int {a e^{\sqrt{dx^2 + e}} }dx
$$
If you can share any workings out that would be helpful, as I really want to relearn this stuff and be able to do it myself.
 A: $$I=\int ae^{\frac{\sqrt{x^2+c}}{b}}dx=a\int e^{\sqrt{(x/b)^2+c/b^2}}dx$$
now if we let $u=x/b$ then $dx=bdu$ and to simplify we will call $\alpha^2=c/b^2$ and so:
$$I=ab\int e^{\sqrt{u^2+\alpha^2}}du$$
Now this is not a very nice integral that I am struggling to find an elementary derivative for but we could try and substitute again:
$v^2=u^2+\alpha^2\Rightarrow 2vdv=2udu=2\sqrt{v^2-\alpha^2}du\Rightarrow du=\frac{vdv}{\sqrt{v^2-\alpha^2}}$ and so:
$$I=ab\int\frac{ve^v}{\sqrt{v^2-\alpha^2}}dv$$
now using IBP we get:
$$I=ab\sqrt{v^2-\alpha^2}e^v-\int\sqrt{v^2-\alpha^2}e^vdv$$
which still does not give an elementrary antiderivative but maybe you could see where a taylor series gets you?
A: Usually, integrands of the form $e^{f(x)}$ don't have a closed-form antiderivative and this case is no exception.
A simple approximation is by ignoring $b$,
$$\int ae^{\sqrt{x^2+b}/c}dx\approx \int ae^{x/c}dx=ace^{x/c}+C.$$
Unfortunately, neither $e^{\sqrt{x^2+b}/c}-e^x$ nor $e^{(\sqrt{x^2+b}-x)/c}$ seem to easily lead to improvements of this approximation.
