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just to follow up on Michael Renardy's suggestion: let's take $n$ even $>2$, then the integrand in $I_n(t)$ has the small-$t$ expansion [with $E_n(t)=a_0+a_1 t^2+a_2 t^4+ {\cal O}(t^6)$]:

$$K(t)=\frac{(t/h_2)^{2n}}{E_n(t)^2\sqrt{1-t^2}\sqrt{1-h_3^2(t/h_2)^2}}=c_0 (t/h_2)^{2n}+c_1 (t/h_2)^{2n+2}+c_2 (t/h_2)^{2n+4} + {\cal O}(t^{2n+6})$$

with coefficients

$$c_0=\frac{1}{a_0^2},\;\;c_1=\frac{a_0 (h_2^2+h_3^2)-4 a_1 h_2^2}{2 a_0^3},$$

$$c_2=\frac{ h_2^4\left(3 a_0^2-8 a_0 a_1-16a_0 a_2+24 a_1^2\right)+3 a_0^2 h_3^4+2 a_0 h_2^2 h_3^2 (a_0-4 a_1)}{8 a_0^4}$$

integration $I_n(\rho)=\int_0^{h_2/\rho}K(t)dt$ gives you the desired first three terms of the expansion,

$$I_n(\rho)=\frac{c_0 h_2 }{2n+1}(1/\rho)^{2n+1}+\frac{c_1 h_2}{2n+3}(1/\rho)^{2n+3}+\frac{c_2 h_2}{2n+5}(1/\rho)^{2n+5}+{\cal O}(1/\rho)^{2n+7}$$

just to follow up on Michael Renardy's suggestion: let's take $n$ even $>2$, then the integrand in $I_n(t)$ has the small-$t$ expansion [using $E_n(t)=a_0+a_1 t^2+a_2 t^4+ {\cal O}(t^6)$, with $n$-dependent coefficients $a_p$]:

$$K(t)=\frac{(t/h_2)^{2n}}{E_n(t)^2\sqrt{1-t^2}\sqrt{1-h_3^2(t/h_2)^2}}=c_0 (t/h_2)^{2n}+c_1 (t/h_2)^{2n+2}+c_2 (t/h_2)^{2n+4} + {\cal O}(t^{2n+6})$$

with coefficients

$$c_0=\frac{1}{a_0^2},\;\;c_1=\frac{a_0 (h_2^2+h_3^2)-4 a_1 h_2^2}{2 a_0^3},$$

$$c_2=\frac{ h_2^4\left(3 a_0^2-8 a_0 a_1-16a_0 a_2+24 a_1^2\right)+3 a_0^2 h_3^4+2 a_0 h_2^2 h_3^2 (a_0-4 a_1)}{8 a_0^4}$$

integration $I_n(\rho)=\int_0^{h_2/\rho}K(t)dt$ gives you the desired first three terms of the expansion for $n$ even,

$$I_n(\rho)=\frac{c_0 h_2 }{2n+1}(1/\rho)^{2n+1}+\frac{c_1 h_2}{2n+3}(1/\rho)^{2n+3}+\frac{c_2 h_2}{2n+5}(1/\rho)^{2n+5}+{\cal O}(1/\rho)^{2n+7}$$

for $n$ odd $>3$ we have instead $E_n(t)=t\times[a_0+a_1 t^2+a_2 t^4+ {\cal O}(t^6)]$ so the expansion becomes

$$I_n(\rho)=\frac{c_0/ h_2 }{2n-1}(1/\rho)^{2n-1}+\frac{c_1/ h_2}{2n+1}(1/\rho)^{2n+1}+\frac{c_2/ h_2}{2n+3}(1/\rho)^{2n+3}+{\cal O}(1/\rho)^{2n+5}$$

just to follow up on Michael Renardy's suggestion: let's take $n$ even $>2$, then the integrand in $I_n(t)$ has the small-$t$ expansion [with $E_n(t)=a_0+a_1 t^2+a_2 t^4+ {\cal O}(t^6)$]:

$$\frac{t^{2n}/h_2^{2n}}{(E_n(t))^2\sqrt{1-t^2}\sqrt{1-\frac{h_3^2}{h_2^2}t^2}}=c_0 t^{2n}+c_1 t^{2n+2}+c_2 t^{2n+4} + {\cal O}(t^{2n+6})$$

with coefficients

$$c_0=\frac{1}{h_2^{2 n}a_0^2},\;\;c_1=\frac{a_0 (h_2^2+h_3^2)-4 a_1 h_2^2}{2h_2^{2 n+2} a_0^3},$$

$$c_2=\frac{ h_2^4\left(3 a_0^2-8 a_0 (a_1+2 a_2)+24 a_1^2\right)+3 a_0^2 h_3^4+2 a_0 h_2^2 h_3^2 (a_0-4 a_1)}{8h_2^{2 n+4} a_0^4}$$

integration gives you the desired first three terms of the expansion,

$$I_n(\rho)=\frac{c_0}{2n+1}(h_2/\rho)^{2n+1}+\frac{c_1}{2n+3}(h_2/\rho)^{2n+3}+\frac{c_2}{2n+5}(h_2/\rho)^{2n+5}+{\cal O}(\rho^{-2n-7})$$

just to follow up on Michael Renardy's suggestion: let's take $n$ even $>2$, then the integrand in $I_n(t)$ has the small-$t$ expansion [with $E_n(t)=a_0+a_1 t^2+a_2 t^4+ {\cal O}(t^6)$]:

$$K(t)=\frac{(t/h_2)^{2n}}{E_n(t)^2\sqrt{1-t^2}\sqrt{1-h_3^2(t/h_2)^2}}=c_0 (t/h_2)^{2n}+c_1 (t/h_2)^{2n+2}+c_2 (t/h_2)^{2n+4} + {\cal O}(t^{2n+6})$$

with coefficients

$$c_0=\frac{1}{a_0^2},\;\;c_1=\frac{a_0 (h_2^2+h_3^2)-4 a_1 h_2^2}{2 a_0^3},$$

$$c_2=\frac{ h_2^4\left(3 a_0^2-8 a_0 a_1-16a_0 a_2+24 a_1^2\right)+3 a_0^2 h_3^4+2 a_0 h_2^2 h_3^2 (a_0-4 a_1)}{8 a_0^4}$$

integration $I_n(\rho)=\int_0^{h_2/\rho}K(t)dt$ gives you the desired first three terms of the expansion,

$$I_n(\rho)=\frac{c_0 h_2 }{2n+1}(1/\rho)^{2n+1}+\frac{c_1 h_2}{2n+3}(1/\rho)^{2n+3}+\frac{c_2 h_2}{2n+5}(1/\rho)^{2n+5}+{\cal O}(1/\rho)^{2n+7}$$

just to follow up on Michael Renardy's suggestion: let's take $n$ even $>2$, then the integrand in $I_n(t)$ has the small-$t$ expansion [with $E_n(t)=a_0+a_1 t^2+a_2 t^4+ {\cal O}(t^6)$]:

$$\frac{t^{2n}/h_2^{2n}}{(E_n(t))^2\sqrt{1-t^2}\sqrt{1-\frac{h_3^2}{h_2^2}t^2}}=c_0 t^{2n}+c_1 t^{2n+2}+c_2 t^{2n+4} + {\cal O}(t^{2n+6})$$

with coefficients

$$c_0=\frac{1}{h_2^{2 n}a_0^2},\;\;c_1=\frac{a_0 (h_2^2+h_3^2)-4 a_1 h_2^2}{2h_2^{2 (n+1)} a_0^3}$$

$$c_2=\frac{ h_2^4\left(3 a_0^2-8 a_0 (a_1+2 a_2)+24 a_1^2\right)+3 a_0^2 h_3^4+2 a_0 h_2^2 h_3^2 (a_0-4 a_1)}{8h_2^{2 (n+2)} a_0^4}$$

integration gives you the desired first few terms of the expansion,

$$I_n(\rho)=\frac{c_0}{2n+1}(h_2/\rho)^{2n+1}+\frac{c_1}{2n+3}(h_2/\rho)^{2n+3}+\frac{c_2}{2n+5}(h_2/\rho)^{2n+5}+{\cal O}(\rho^{-2n-7})$$

just to follow up on Michael Renardy's suggestion: let's take $n$ even $>2$, then the integrand in $I_n(t)$ has the small-$t$ expansion [with $E_n(t)=a_0+a_1 t^2+a_2 t^4+ {\cal O}(t^6)$]:

$$\frac{t^{2n}/h_2^{2n}}{(E_n(t))^2\sqrt{1-t^2}\sqrt{1-\frac{h_3^2}{h_2^2}t^2}}=c_0 t^{2n}+c_1 t^{2n+2}+c_2 t^{2n+4} + {\cal O}(t^{2n+6})$$

with coefficients

$$c_0=\frac{1}{h_2^{2 n}a_0^2},\;\;c_1=\frac{a_0 (h_2^2+h_3^2)-4 a_1 h_2^2}{2h_2^{2 n+2} a_0^3},$$

$$c_2=\frac{ h_2^4\left(3 a_0^2-8 a_0 (a_1+2 a_2)+24 a_1^2\right)+3 a_0^2 h_3^4+2 a_0 h_2^2 h_3^2 (a_0-4 a_1)}{8h_2^{2 n+4} a_0^4}$$

integration gives you the desired first three terms of the expansion,

$$I_n(\rho)=\frac{c_0}{2n+1}(h_2/\rho)^{2n+1}+\frac{c_1}{2n+3}(h_2/\rho)^{2n+3}+\frac{c_2}{2n+5}(h_2/\rho)^{2n+5}+{\cal O}(\rho^{-2n-7})$$