Cfrgtkky

As an extension of this question how would we address integrals of the form:

$$int frac{a^x + b^x}{c^x + d^x}:dx$$

Where $a,b,c,d in mathbb{R}^{+}$ are the values are distinct.

Does anyone have any starting points?

By multiplying through by $frac{d^{-x}}{d^{-x}}$, we may assume w.l.o.g. that $d = 1$ and so evaluate
$$int frac{alpha^x + beta^x}{gamma^x + 1} dx .$$

In the special case $a d - b c = 0$ ($alpha = beta gamma$), the integrand simplifies to an exponential function. But generically, we may as well split the integrand into its summands, and so it's enough to evaluate $$int frac{alpha^x dx}{gamma^x + 1}. $$

For $alpha = 1$, an elementary integration gives $$boxed{int frac{dx}{gamma^x + 1} = x - log_{gamma}(gamma^x + 1) + C} .$$

For $alpha neq 1$, substituting $u = alpha^x, du = alpha^x log alpha$ gives (as was essentially observed in the comments)
$$boxed{int frac{alpha^x dx}{gamma^x + 1}
= frac{1}{log alpha} int frac{du}{1 + u^{lambda}} = {}_2 F_1(1, lambda^{-1}; 1 + lambda^{-1}; -u^{lambda}) u + C , quad lambda := log_{alpha} gamma},$$
where ${}_2 F_1$ is Gauss' hypergeometric function.

For generic $lambda$ this expression is the best we can do, but if $lambda$ is rational, say, $lambda = frac{m}{n}$, so that $alpha^m = gamma^n$, the substitution $u = v^n, du = n v^{n - 1} dv$ rationalizes the integral, making it amenable to other familiar techniques:
$$int frac{du}{1 + u^{lambda}} = n int frac{v^{n - 1} dv}{1 + v^m} .$$

For example, for $lambda = 2$ ($gamma = alpha^2$), the integral is
$$int frac{alpha^x dx}{alpha^{2 x} + 1} = frac{1}{log alpha} int frac{du}{1 + u^2} = frac{1}{log alpha} arctan (x log alpha) + C ,$$
and for $lambda = frac{1}{2}$ ($alpha = gamma^2$), the integral is
$$int frac{gamma^{2 x} dx}{gamma^x + 1} = frac{1}{log gamma}(gamma^x - log(gamma^x + 1)) + C .$$

Identical to Kemono Chen's comment, write
$$I=intfrac{b^x}{c^x+d^x},dx=int frac{left(frac{b}{d}right)^x}{left(frac{c}{d}right)^x+1},dx$$ and get
$$I=frac{left(frac{b}{d}right)^x }{log
left(frac{b}{d}right)},, _2F_1left(1,frac{log
left(frac{b}{d}right)}{log left(frac{c}{d}right)};frac{log
left(frac{b}{d}right)}{log
left(frac{c}{d}right)}+1;-left(frac{c}{d}right)^xright)$$ where appears the Gaussian or ordinary hypergeometric function.

$int{frac{a^x+b^x}{c^x+d^x}}dx=int{frac{a^x+b^x}{c^x}frac{1}{1+(d/c)^x}}dx$

$=int{((a/c)^x+(b/c)^x)(1-(d/c)^x+(d/c)^{2x}- ...)}dx$

$=int(((a/c)^x-(ad/c^2)^x+....)+((b/c)^x-(bd/c^2)^x+....))dx$

Which can probably be integrated term-wise into an infinite sum