3. Given $\log _{a} b+3 \log _{b} a=\frac{13}{2}$. When $a>b>1$, the value of $\frac{a+b^{4}}{a^{2}+b^{2}}$ is ( ).
(A) 13
(B) 4
(C) 2
(D) 1
3. D.

Let $\log _{a} b=t$. Then $t+\frac{3}{t}=\frac{13}{2} \Rightarrow t=\frac{1}{2}$ or 6 $\Rightarrow b=\sqrt{a}$ or $a^{6}$ (discard). Hence $\frac{a+b^{4}}{a^{2}+b^{2}}=\frac{a+a^{2}}{a^{2}+a}=1$.