calculus - Integrate $int_0^1 frac{ln(1+x^a)}{1+x}, dx$

I have recently met with this integral:
$$\int_0^1 \frac{\ln(1+x^a)}{1+x}\, dx$$

I want to evaluate it in a closed form, if possible.

1st functional equation:

$\displaystyle f(a)=\ln^2 2-f\left ( \frac{1}{a} \right )$ since:
$$\begin{aligned}
f(a)=\int_{0}^{1}\frac{\ln (1+x^a)}{1+x}\,dx &= \int_{0}^{1}\ln \left ( 1+x^a \right )\left ( \ln (1+x) \right )'\,dx\\
&= \ln^2 2 - \int_{0}^{1}\frac{ax^{a-1}\ln (1+x)}{1+x^a}\,dx\\
&\overset{y=x^a}{=\! =\! =\!}\ln^2 2 - \int_{0}^{1}\frac{a y^{(a-1)/a}\ln \left ( 1+y^{1/a} \right )}{1+y}\frac{1}{a}y^{(1-a)/a}\,dy \\

&= \ln^2 2 -f\left ( \frac{1}{a} \right )
\end{aligned}$$

2nd functional equation:

$\displaystyle f(a)=-\frac{a\pi^2}{12}+f(-a)$ since:

$$\begin{aligned}
f(a)=\int_{0}^{1}\frac{\ln (1+x^a)}{1+x}\,dx &=\int_{0}^{1}\frac{\ln \left ( x^a\left ( 1+x^{-a} \right ) \right )}{1+x}\,dx \\
&= a\int_{0}^{1}\frac{\ln x}{1+x}\,dx+\int_{0}^{1}\frac{\ln (1+x^{-a})}{1+x}\,dx\\

&= a\int_{0}^{1}\frac{\ln x}{1+x}\,dx+f(-a)\\
&= a\int_{0}^{1}\ln x \sum_{n=0}^{\infty}(-1)^n x^n \,dx +f(-a)\\
&= a\sum_{n=0}^{\infty}(-1)^n \int_{0}^{1}\ln x \cdot x^n \,dx +f(-a)\\
&=\cdots\\
&=-\frac{a\pi^2}{12}+f(-a)
\end{aligned}$$

What I did try was:

1st way

$$\begin{aligned}
\int_{0}^{1}\frac{\log(1+x^a)}{1+x}\,dx &=\int_{0}^{1}\log(1+x^a)\sum_{n=0}^{\infty}(-1)^n x^n\,dx \\
&= \sum_{n=0}^{\infty}(-1)^n \int_{0}^{1}\log(1+x^a)x^n\,dx\\
&= \sum_{n=0}^{\infty}(-1)^n \{ \left[ \frac{x^{n+1}\log(1+x^a)}{n+1} \right]_0^1- \\
& \quad \quad \quad \quad \quad \frac{a}{n+1}\int_{0}^{1}\frac{x^{n+1}x^{a-1}}{1+x^a}\,dx \}
\end{aligned}$$

2nd way
I tried IBP but I get to an unpleasant integral of the form $\displaystyle \int_{0}^{1}\frac{\ln(1+x)x^{a-1}}{1+x^a}\,dx$.

3nd way
It was just an idea ... expand the nominator into Taylor Series , swip integration and summation and get into a digamma form . This way suggests that a closed form is far away ... since we are dealing with digammas here.

The result I got was:
$$\begin{aligned}
\int_{0}^{1}\frac{1}{1+x}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}x^{ak} &=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}\frac{x^{ak}}{1+x}dx \\
&= 1/2\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\left[\psi\left(\frac{ak}{2}+1\right)-\psi\left(\frac{ak}{2}+1/2\right)\right] \\
\end{aligned}$$
and I can't go on.