Integral of $2^{2^{2^x}}$?

Question

$$\int2^{2^{2^x}}~\mathrm{d}x$$

Derivative is $\ln^3(2)2^{2^x+x+2^{2^x}}$.

So no substitution technique can be used. So please guide, I am confused.

Is this elliptic?

Answer 1

$\int2^{2^{2^x}}~dx$

$=\int e^{2^{2^x}\ln2}~dx$

$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{2^{2^xn}\ln^n2}{n!}\right)dx$

$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{e^{2^xn\ln2}\ln^n2}{n!}\right)dx$

$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\ln^n2}{n!}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty\dfrac{2^{kx}n^k\ln^{n+k}2}{n!k!}\right)dx$

$=\sum\limits_{n=0}^\infty\dfrac{x\ln^n2}{n!}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty\dfrac{2^{kx}n^k\ln^{n+k}2}{n!k!k\ln2}+C$

$=2x+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty\dfrac{2^{kx}n^k\ln^{n+k}2}{n!k!k\ln2}+C$