Using
$$\text{Li}_{2a+1}(x)-\text{Li}_{2a+1}(1/x)=\frac{i\,\pi\ln^{2a}(x)}{(2a)!}+2\sum_{k=0}^a \frac{\zeta(2a-2k)}{(2k+1)!}\ln^{2k+1}(x)\tag{1}$$
and
$$\int_0^1x^{n-1}\operatorname{Li}_a(x)\mathrm{d}x=(-1)^{a-1}\frac{H_n}{n^a}-\sum_{k=1}^{a-1}(-1)^k\frac{\zeta(a-k+1)}{n^k}\tag{2}$$
We have
$$\int_0^1\frac{\text{Li}_{2a+1}(x)}{1+x^2}dx=(2^{-4a-3}-2^{-2a-3})\pi\zeta(2a+1)+\frac{2a+1}{2}\beta(2a+2)$$ $$-\sum_{k=0}^a \zeta(2a-2k) \beta(2k+2);\tag{3}$$
$$\int_0^1\frac{\ln^{2a}(x)\ln(1-x)}{1+x^2}dx=\frac{(2a)!}{2}\ln(2)\beta(2a+1)-\frac{(2a+1)!}{2}\beta(2a+2)$$
$$+(2a)!\sum_{k=0}^a \zeta(2a-2k) \beta(2k+2);\tag{4}$$
$$\sum_{n=0}^\infty\frac{(-1)^n H_{2n+1}}{(2n+1)^{2a+1}}=(2^{-4a-3}-2^{-2a-3})\pi\zeta(2a+1)+\frac{2a+1}{2}\beta(2a+2)$$ $$-\sum_{k=0}^a \zeta(2a-2k) \beta(2k+2)+\sum_{k=1}^{2a}(-1)^k\zeta(2a-k+2)\beta(k).\tag{5}$$
Proof of $(1)$: Differentiating then integrating back, we have
$$\text{Li}_2(x)+\text{Li}_2(1/x)=-i\pi\ln(x)-\frac12\ln^2(x)+c$$
where $c=2\zeta(2)$ by setting $x=1$.
Keep dividing by $x$ then integrate, we get
$$\text{Li}_3(x)-\text{Li}_3(1/x)=-i\pi\frac{\ln^2(x)}{2}-\frac{\ln^3(x)}{3!}+2\zeta(2)\ln(x)$$
$$\text{Li}_4(x)+\text{Li}_4(1/x)=-i\pi\frac{\ln^3(x)}{3!}-\frac{\ln^4(x)}{4!}+\zeta(2)\ln^2(x)+2\zeta(4)$$
This pattern shows that
$$\text{Li}_{2a}(x)+\text{Li}_{2a}(1/x)=-\frac{i\,\pi\ln^{2a-1}(x)}{(2a-1)!}+2\sum_{k=0}^a \frac{\zeta(2a-2k)}{(2k)!}\ln^{2k}(x)$$
$$\text{Li}_{2a+1}(x)-\text{Li}_{2a+1}(1/x)=-\frac{i\,\pi\ln^{2a}(x)}{(2a)!}+2\sum_{k=0}^a \frac{\zeta(2a-2k)}{(2k+1)!}\ln^{2k+1}(x)$$
These two identities are good for $x\ge1$. To make them work for $x\le1$, let $x\to 1/x$, we get
$$\text{Li}_{2a}(x)+\text{Li}_{2a}(1/x)=\frac{i\,\pi\ln^{2a-1}(x)}{(2a-1)!}+2\sum_{k=0}^a \frac{\zeta(2a-2k)}{(2k)!}\ln^{2k}(x)$$
$$\text{Li}_{2a+1}(x)-\text{Li}_{2a+1}(1/x)=\frac{i\,\pi\ln^{2a}(x)}{(2a)!}+2\sum_{k=0}^a \frac{\zeta(2a-2k)}{(2k+1)!}\ln^{2k+1}(x)$$
Proof of $(2)$: Using the definition of the harmonic number,
$$H_n=\int_0^1\frac{1-x^n}{1-x}dx\overset{IBP}{=}-n\int_0^1 x^{n-1}\ln(1-x)dx$$
$$\overset{IBP}{=}n\left(\zeta(2)-n\int_0^1 x^{n-1}\text{Li}_2(x)dx\right)$$
$$\overset{IBP}{=}n\left(\zeta(2)-n\left(\zeta(3)-n\int_0^1 x^{n-1}\text{Li}_3(x)dx\right)\right).$$
Integrating by parts repeatedly gives $(2)$.
Proof of $(3)$:
$$\int_0^1\frac{\text{Li}_{2a+1}(x)}{1+x^2}dx=\int_0^\infty\frac{\text{Li}_{2a+1}(x)}{1+x^2}dx-\underbrace{\int_1^\infty\frac{\text{Li}_{2a+1}(x)}{1+x^2}dx}_{x\to 1/x}$$
$$=\int_0^\infty\frac{\text{Li}_{2a+1}(x)}{1+x^2}dx-\int_0^1\frac{\text{Li}_{2a+1}(1/x)}{1+x^2}dx$$ add the integral to both sides
$$2\int_0^1\frac{\text{Li}_{2a+1}(x)}{1+x^2}dx=\int_0^\infty\frac{\text{Li}_{2a+1}(x)}{1+x^2}dx+\int_0^1\frac{\text{Li}_{2a+1}(x)-\text{Li}_{2a+1}(1/x)}{1+x^2}dx$$ $$=I_1+I_2\tag{b}$$
For the first integral, write the integral form of the polylogarithm,
$$I_1=\int_0^\infty\frac{1}{1+x^2}\left(\frac1{(2a)!}\int_0^1 \frac{x\ln^{2a}(y)}{1-xy}dy\right)dx$$
$$=\frac1{(2a)!}\int_0^1 \ln^{2a}(y)\left(\int_0^\infty\frac{x}{(1+x^2)(1-xy)}dx\right)dy$$
$$=\frac1{(2a)!}\int_0^1 \ln^{2a}(y)\left(-\frac{\pi}{2}\frac{y}{1+y^2}-\frac{i\pi}{1+y^2}-\frac{\ln(y)}{1+y^2}\right)dy$$
$$=(4^{-2a-1}-4^{-a-1})\pi\zeta(2a+1)-i\pi\beta(2a+1)+(2a+1)\beta(2a+2).$$
For the second integral, by using $(1)$, we have
$$I_2=\frac{i\,\pi}{(2a)!}\int_0^1\frac{\ln^{2a}(x)}{1+x^2}dx+2\sum_{k=0}^a \frac{\zeta(2a-2k)}{(2k+1)!}\int_0^1\frac{\ln^{2k+1}(x)}{1+x^2}dx$$
$$=i\,\pi\beta(2a+1)-2\sum_{k=0}^a \zeta(2a-2k)\beta(2k+2).$$
Plugging $I_1$ and $I_2$ in $(b)$ completes the proof.
Proof of $(4)$: Again, using the integral form of the polylogarthm, we have
$$\int_0^1\frac{\text{Li}_{2a+1}(x)}{1+x^2}dx=\int_0^1\frac{1}{1+x^2}\left(\frac1{(2a)!}\int_0^1 \frac{x\ln^{2a}(y)}{1-xy}dy\right)dx$$
$$=\frac1{(2a)!}\int_0^1 \ln^{2a}(y)\left(\int_0^1\frac{x}{(1+x^2)(1-xy)}dx\right)dy$$
$$=\frac1{(2a)!}\int_0^1 \ln^{2a}(y)\left(\frac{\ln(2)}{2(1+y^2)}-\frac{\pi\,y}{4(1+y^2)}-\frac{\ln(1-y)}{1+y^2}\right)dy$$
$$=\frac12\ln(2)\beta(2a+1)+(2^{-4a-3}-2^{-2a-3})\pi\zeta(2a+1)-\frac1{(2a)!}\int_0^1\frac{\ln^{2a}(x)\ln(1-x)}{1+x^2}dx.$$
The integral on the LHS is given in $(2)$.
Proof of $(5)$:
$$\int_0^1\frac{\text{Li}_{2a+1}(x)}{1+x^2}dx=\sum_{n=0}^\infty (-1)^n\int_0^1 x^{2n} \text{Li}_{2a+1}(x) dx$$
use the result $(2)$
$$=\sum_{n=0}^\infty (-1)^n\left(\frac{H_{2n+1}}{(2n+1)^{2a+1}}-\sum_{k=1}^{2a}(-1)^k \frac{\zeta(2a-k+2)}{(2n+1)^k}\right)$$
$$=\sum_{n=0}^\infty\frac{(-1)^n H_{2n+1}}{(2n+1)^{2a+1}}-\sum_{k=1}^{2a} (-1)^k \zeta(2a-k+2)\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^k}$$
$$=\sum_{n=0}^\infty\frac{(-1)^n H_{2n+1}}{(2n+1)^{2a+1}}-\sum_{k=1}^{2a} (-1)^k \zeta(2a-k+2)\beta(k).$$
Question: Are the results $(3)$ to $(5)$ known in the literature? If so, any reference? Thanks.