This Integral is really Hot!

Let $f(x) = \dfrac{2^x \tan^2(x)}{1+2^x}$ and $g(x) = \dfrac{3^x \tan^4(x)}{1+3^x}$ .

Now:

$H = \int_{-1}^{1} ( f(x) + g(x) ) \ dx = \int_{-1}^{1} ( f(-1+1-x) + g(-1+1-x) )\ dx = \int_{-1}^{1} ( f(-x) + g(-x) ) \ dx$

$\Rightarrow H = \int_{-1}^{1} \left ( \dfrac{2^x \tan^2(x)}{1+2^x} + \dfrac{3^x \tan^4(x)}{1+3^x} \right)\ dx = \int_{-1}^{1} \left ( \dfrac{\tan^2(x)}{1+2^x} + \dfrac{\tan^4(x)}{1+3^x} \right)\ dx$

$\Rightarrow 2H = \int_{-1}^{1} \left( \tan^2(x) + \tan^4(x) \right)\ dx = \int_{-1}^{1} \tan^2(x)\sec^2(x)\ dx$

$\Rightarrow H = \int_0^1 \tan^2(x)\sec^2(x)\ dx$

Let $u = \tan(x) \Rightarrow du = \sec^2(x)dx$

When $x=0, u=0 \text{ and when } x=1, u=\tan(1)$ . So:

$H = \int_0^{\tan(1)} u^2 \ du= \left. \dfrac{u^3}{3} \right|_0^{\tan(1)} = \dfrac{\tan^3(1)}{3}$

So, $A=0, B=1, C=3, D=3, A+B+C+D = \boxed{7}$ .

Note that:

$\begin{aligned} I_1 & = \int_{-1}^1 \frac {2^x \tan^2 x}{1+2^x} dx & \small {\color{#3D99F6}\text{Using identity }\int_a^b f(x) \ dx = \int_a^b f(a+b -x) \ dx} \\ & = \frac 12 \int_{-1}^1\left( \frac {2^x \tan^2 x}{1+2^x} + \frac {2^{-x} \tan^2 x}{1+2^{-x}} \right) dx \\ & = \frac 12 \int_{-1}^1 \left( \frac {2^x \tan^2 x}{1+2^x} + \frac {\tan^2 x}{2^x+1} \right) dx \\ & = \frac 12 \int_{-1}^1 \tan^2 x \ dx \end{aligned}$

Similarly, we have $\displaystyle I_2 = \int_{-1}^1 \frac {3^x \tan^4 x}{1+3^x} dx = \frac 12 \int_{-1}^1 \tan^4 x \ dx$

Therefore,

$\begin{aligned} I & = \int_{-1}^1\left( \frac {2^x \tan^2 x}{1+2^x} + \frac {3^x \tan^4 x}{1+3^x} \right) dx \\ & = \frac 12 \int_{-1}^1 (\tan^2 x + \tan^4 x) \ dx \\ & = \frac 12 \int_{-1}^1 \tan^2 x \sec^2 x \ dx \\ & = \frac 16 \tan^3 x \ \bigg|_{-1}^1 \\ & = \frac {\tan^3 (1)}3 \end{aligned}$

$\implies A+B+C+D = 0+1+3+3 = \boxed{7}$