To the power 11

$\begin{aligned} I & = \int \frac {x^3+x}{(x^4+2x^2+3)^{11}} dx & \small \color{#3D99F6} \text {Let }u =x^4+2x^2+3, du = 4x^3+4x \ dx \\ &=\int \frac {du}{4u^{11}} \\ &= -\frac 1{40u^{10}}+C\\ &=-\frac 1{40(x^4+2x^2+3)^{10}} +C \end{aligned}$

$\implies a+b-c+d-e+f-g = 1+2-3+4-2-10+40=\boxed{32}$

there may be better ways to do this but right now I now only of the u substitution way so I came up with this solution.

Let $u = (x+1)$

notice $(x^4 + 2x^2 + 3)^{11}$ can be written as $((x+1)^2 + 2)^{11}$ .

also $x^3 +x$ can also be written as $x(x^2+1)$ .

also $du = dx$ so the integral can be written as $u / (u+2)^{11} * 1/2 * du$ .we take the $1/2$ out of the integral sign (constant).

now we again substitute $u^2 = v$ . also $dv = 2u du$ .

again the indefinite integral becomes $1/4$ (integral sign) $(v+2)^{-11} dv$ which is easy to evaluate and we get

$(v+2)^{-10} / -10$ $+$ $C_1$

now we back substitute $u$ and $x$ to get,

$(x^4 + 2x^2 + 3)^{-10}$ $/$ $-40$ $+$ $C_1$

compare co efficients and get the answer.