How to solve $\int \frac{a^x + b^x}{c^x + d^x}\:dx$

Question

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?

Answer 1

Identical to Kemono Chen's comment, write $$I=\int\frac{b^x}{c^x+d^x}\,dx=\int \frac{\left(\frac{b}{d}\right)^x}{\left(\frac{c}{d}\right)^x+1}\,dx$$ and get $$I=\frac{\left(\frac{b}{d}\right)^x }{\log \left(\frac{b}{d}\right)}\,\, _2F_1\left(1,\frac{\log \left(\frac{b}{d}\right)}{\log \left(\frac{c}{d}\right)};\frac{\log \left(\frac{b}{d}\right)}{\log \left(\frac{c}{d}\right)}+1;-\left(\frac{c}{d}\right)^x\right)$$ where appears the Gaussian or ordinary hypergeometric function.

Answer 2

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 $$\int \frac{\gamma^{2 x} dx}{\gamma^x + 1} = \frac{1}{\log \gamma}(\gamma^x - \log(\gamma^x + 1)) + C .$$

Answer 3

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

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

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

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