In a manuscript I am currently reading, the authors propose a modified softplus function
$$g(a)=\frac{\log\left(2^a +1 \right)}{\log(2)}$$
for some $a \in \mathbb{R}$. The authors then claim that if $a$ is a polynomial, e.g. $a(x)=c_0 + c_1x+c_2x^2$, then the following integral has a closed-form solution:
$$\int_0^tg\left( \partial a(t) \right)\,dt$$
where $\partial$ denotes the derivative. Naively plugging in $g$ and then $\partial a(t)$, I obtain:
$$\int_0^t\frac{\log\left(2^{\partial a(t)} +1 \right)}{\log(2)} \,dt =\int_0^t\frac{\log\left(2^{c_1+2c_2t} +1 \right)}{\log(2)} \,dt$$
WolframAlpha informs me that this still has a closed form solution, but if we move one order higher, the integral seems to be indefinite. I assume there is some simplification or trick which I don't see to solve the integral more easily. Would you have any idea what this might be?
For the indefinite integral $$\int\frac{\log\left(2^{c_1+2c_2t} +1 \right)}{\log(2)}\, dt,$$ assuming that $\log$ is the natural logarithm, we can use the substitution \begin{align*} u &= -2^{c_{1} + 2c_{2}t}\\ du &= -2c_{2}\log(2)2^{c_{1} + 2c_{2}t} \,dt = 2c_{2}\log(2)u\,dt. \end{align*}
This gives us \begin{align*} \int \frac{\log\left(2^{c_1+2c_2t} +1 \right)}{\log(2)}\, dt &= \int\frac{\log(1-u)}{2c_{2}\log^{2}(2)u}\,du\\[5pt] &= \frac{1}{2c_{2}\log^{2}(2)}\int\frac{\log(1-u)}{u}\,du\\[5pt] &=-\frac{1}{2c_{2}\log^{2}(2)}\operatorname{Li_{2}}(u)+C\\[5pt] &=-\frac{\operatorname{Li_{2}}(-2^{c_{1} + 2c_{2}t})}{2c_{2}\log^{2}(2)}+C. \end{align*} Here $\operatorname{Li_2}$ is the dilogarithm given by $$\operatorname{Li_2}(z) = -\int_{0}^{z}\frac{\ln(1-u)}{u}\,du.$$ So, for the definite integral: $$\boxed{\int_0^x\frac{\log\left(2^{c_1+2c_2t} +1 \right)}{\log(2)}\, dt = \frac{\operatorname{Li_{2}}(-2^{c_{1}}) - \operatorname{Li_{2}}(-2^{c_{1} + 2c_{2}x})}{2c_{2}\log^{2}(2)}.}$$