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:
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 )

2nd functional equation:

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

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)\\

What I did try was:

1st way
\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 \}

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:
\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] \\
and I can’t go on.

Solutions Collecting From Web of "Integrate $\int_0^1 \frac{\ln(1+x^a)}{1+x}\, dx$"

Another series approach might offer a hint:

$$\int_0^1 x^pdp=\frac{1}{p+1}$$

$$\int_0^1 x^{ak} x^ldp=\frac{1}{ak+l+1}$$

$$\int_0^1 \frac{\ln(1+x^a)}{1+x}dx=\sum_{k=1}^{\infty} \sum_{l=0}^{\infty} \frac{(-1)^{k+l+1}}{k(ak+l+1)}=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \Phi(-1,1,ak+1)$$

Here $\Phi$ is Lerch Trancendent. I only consider $a>0$, otherwise this series might diverge. But really, we only need the case $a>1$ as you showed.

The double sum, despite being unwieldy, offers some insight into $f(a)$. For example, we can guess that for $a \to +\infty$ the asymptotic is $f(a) \to \frac{\pi^2}{12a}$:

$$\frac{1}{a} \sum_{k=1}^{\infty} \sum_{l=1}^{\infty} \frac{(-1)^{k+l}}{k^2+kl/a} \asymp^{a \to +\infty} \frac{1}{a} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2}=\frac{\pi^2}{12a}$$

This is not a proof! But the asymptotic is right, as is shown in this answer for a related integral.

I really suggest you read the links provided in the comments to find out about possible closed forms, but just in case I aslo compile the simplest cases with closed forms here:

a & f(a) \\
-2 & \dfrac{7 \pi ^2}{48}+\dfrac{3 \ln ^2 2}{4} \\
-1 & \dfrac{ \pi ^2}{12}+\dfrac{\ln ^2 2}{2}\\
-\dfrac{1}{2} & \dfrac{\pi ^2}{16}+\dfrac{\ln ^2 2}{4}\\
0 & \ln ^2 2 \\
\dfrac{1}{2} & \dfrac{\pi ^2}{48}+\dfrac{\ln ^2 2}{4} \\
1 & \dfrac{\ln ^2 2}{2} \\
2 & -\dfrac{\pi ^2}{48}+\dfrac{3 \ln ^2 2}{4}\\

See also this answer for a more comprehensive approach