Finding a closed form for $\int \frac{dx}{(1+x)(1+x^a)}$ if one exists. Find a closed form for this integral
$$\int \frac{dx}{(1+x)(1+x^a)}$$
This integral has the possibility of not having a closed form in which case can it be proven?

Feeble attempt so far:
$$\int \frac{dx}{(1+x)(1+x^a)}=\frac{\log(x+1)}{1+x^a}-\int\frac{ax^{a-1}\log(x+1)}{(x^a+1)^3}dx$$It is starting to feel analytical.

WA is not happy with it. No elementary function representation found it says. WA
 A: It is clearly known that this integral should have closed form when $a$ is a rational number.
Let $a=\dfrac{p}{q}$ , where $p\in\mathbb{Z}$ , $q\in\mathbb{Z}^+$ and $\text{gcd}(p,q)=1$ ,
Then $\int\dfrac{dx}{(1+x)(1+x^a)}=\int\dfrac{dx}{(1+x)(1+x^\frac{p}{q})}$
Let $u=x^\frac{1}{q}$ ,
Then $x=u^q$
$dx=qu^{q-1}~du$
$\therefore\int\dfrac{dx}{(1+x)(1+x^\frac{p}{q})}=\int\dfrac{qu^{q-1}}{(1+u^q)(1+u^p)}du$ , which is an integral of rational function and it should have closed form.
When $a$ is an irrational number, it is afraid that you can only solve this integral by these approaches:
When $|x|<1$ and $a>0$ ,
Then $\int\dfrac{dx}{(1+x)(1+x^a)}$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^n}{1+x^a}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty(-1)^{n+k}x^{n+ak}~dx$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{n+ak+1}}{n+ak+1}+C$
When $|x|>1$ and $a>0$ ,
Then $\int\dfrac{dx}{(1+x)(1+x^a)}$
$=\int\dfrac{dx}{x^{a+1}(1+x^{-1})(1+x^{-a})}$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{-n}}{x^{a+1}(1+x^{-a})}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{-n-ak}}{x^{a+1}}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty(-1)^{n+k}x^{-n-a(k+1)-1}~dx$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{-n-a(k+1)}}{-n-a(k+1)}+C$
$=-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}}{(n+a(k+1))x^{n+a(k+1)}}+C$
When $|x|<1$ and $a<0$ ,
Then $\int\dfrac{dx}{(1+x)(1+x^a)}$
$=\int\dfrac{dx}{x^a(1+x)(1+x^{-a})}$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^n}{x^a(1+x^{-a})}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{n-ak}}{x^a}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty(-1)^{n+k}x^{n-a(k+1)}~dx$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{n-a(k+1)+1}}{n-a(k+1)+1}+C$
When $|x|>1$ and $a<0$ ,
Then $\int\dfrac{dx}{(1+x)(1+x^a)}$
$=\int\dfrac{dx}{x(1+x^{-1})(1+x^a)}$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{-n}}{x(1+x^a)}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{-n+ak}}{x}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty(-1)^{n+k}x^{-n+ak-1}~dx$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{-n+ak}}{-n+ak}+C$
$=-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}}{(n-ak)x^{n-ak}}+C$
