Let $f(x)$ have the expansion $f(x)=x+c_3 x^3+c_4 x^4 +\cdots$, then define $y=x/a$ and you have $$I=\int_0^{\infty}{dx}\ln \bigg(1+\exp\left(-\frac{f(x)}{a}\right)\bigg)$$ $$=\int_0^\infty \left[a\ln \left(1+e^{-y}\right)-\frac{a^3 c_3 y^3}{e^y+1}-\frac{a^4 c_4 y^4}{e^y+1}+{\cal O}(a^5)\right]\,dy$$ $$=\frac{1}{12} \pi ^2 a-\frac{7}{120} \pi ^4 c_3 a^3-\frac{45}{2} \zeta (5)c_4 a^4+{\cal O}(a^5).$$

Let $f(x)$ have the expansion $f(x)=x+c_3 x^3+c_4 x^4 +\cdots$, then define $y=x/a$ and you have $$I=\int_0^{\infty}{dx}\ln \bigg(1+\exp\left(-\frac{f(x)}{a}\right)\bigg)$$ $$=\int_0^\infty \left[a\ln \left(1+e^{-y}\right)-\frac{a^3 c_3 y^3}{e^y+1}-\frac{a^4 c_4 y^4}{e^y+1}+{\cal O}(a^5)\right]\,dy$$ $$=\frac{1}{12} \pi ^2 a-\frac{7}{120} \pi ^4 c_3 a^3-\frac{45}{2} \zeta (5)c_4 a^4+{\cal O}(a^5).$$ The term of order $a^p$ has coefficient $-c_p\left(1-2^{-p}\right)p! \zeta (p+1)$, when $c_p$ is the coefficient of order $x^p$ in the expansion of $f(x)$.

Let $f(x)=x+c_3 x^3+c_4 y^4 +\cdots$, then define $y=x/a$ and you have $$I=\int_0^{\infty}{dx}\ln \bigg(1+\exp\left(-\frac{f(x)}{a}\right)\bigg)$$ $$=\int_0^\infty \left[a\ln \left(1+e^{-y}\right)-\frac{a^3 c_3 y^3}{e^y+1}-\frac{a^4 c_4 y^4}{e^y+1}+{\cal O}(a^5)\right]\,dy$$ $$=\frac{1}{12} \pi ^2 a-\frac{7}{120} \pi ^4 c_3 a^3-\frac{45}{2} \zeta (5)c_4 a^4+{\cal O}(a^5).$$

Let $f(x)$ have the expansion $f(x)=x+c_3 x^3+c_4 x^4 +\cdots$, then define $y=x/a$ and you have $$I=\int_0^{\infty}{dx}\ln \bigg(1+\exp\left(-\frac{f(x)}{a}\right)\bigg)$$ $$=\int_0^\infty \left[a\ln \left(1+e^{-y}\right)-\frac{a^3 c_3 y^3}{e^y+1}-\frac{a^4 c_4 y^4}{e^y+1}+{\cal O}(a^5)\right]\,dy$$ $$=\frac{1}{12} \pi ^2 a-\frac{7}{120} \pi ^4 c_3 a^3-\frac{45}{2} \zeta (5)c_4 a^4+{\cal O}(a^5).$$