Not a complete answer but some thoughts that might be of help in evaluating them possibly numerically.

Let $I(a,b,c) = \displaystyle \int_0^{\infty} \dfrac{x^{a+b-1} dx}{(1+x)^{b}(1+x^2)^{am/2}}$.

Setting $x = \tan(\theta)$. We then get that

$$I(a,b,c) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \sec^2(\theta) d \theta}{(1+\tan(\theta))^b \left(1+\tan^2(\theta) \right)^{am/2}}$$

$$I(a,b,c) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \sec^2(\theta) d \theta}{(1+\tan(\theta))^b \sec^{am}(\theta)}$$

$$I(a,b,c) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \cos^{am-2}(\theta) d \theta}{(1+\tan(\theta))^b}$$

Let us use the following short hand notation. $c = \cos(\theta)$ and $s = \sin(\theta)$. We then get that $$I(a,b,c) = \int_0^{\pi/2} \dfrac{s^{a+b-1} c^{am-a-1} d \theta}{(s+c)^b}$$ Hence, we are interested in evaluating integral of the form $$J(p,q,r) = \int_0^{\pi/2} \dfrac{s^p c^q d \theta}{(s+c)^r}$$ Next note that $$J(p,q,r) = J(q,p,r) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\star)$$ Next note that $$J(p,q,0) = \dfrac{\beta((p+1)/2,(q+1)/2)}2$$ Further

 $$J(p,q,r+2) = \int_0^{\pi/2} \dfrac{s^p c^q d \theta}{(s+c)^r (1+2sc)}$$ $$J(p,q,r+2) = \sum_{k=0}^{\infty} \int_0^{\pi/2} \dfrac{s^p c^q (-2sc)^k d \theta}{(s+c)^r}$$ $$J(p,q,r+2) = \sum_{k=0}^{\infty}(-2)^k J(p+k,q+k,r)$$

Hence, you can use this to compute $J(p,q,r)$ for all even $r$.

For odd $r$, the same idea as before works but you need to evaluate for $r=1$. For $r=1$, you could split the integral from $0$ to $\pi/4$ and $\pi/4$ to $\pi/2$ to write $\dfrac1{s+c}$ as a power series in $s$ and $c$ and evaluate them accordingly.

Few other relations, which might be of use. We have $$\lim_{p \to \infty} J(p,q,r) = \lim_{q \to \infty} J(p,q,r) = \lim_{r \to \infty} J(p,q,r) = 0$$ Note that $$J(p+2,q,r) = \int_0^{\pi/2} \dfrac{s^p c^q (1-c^2) d \theta}{(s+c)^r}$$ Hence, we get that $$J(p+2,q,r) + J(p,q+2,r) = J(p,q,r)$$ Setting $p=q$, and making use of $(\star)$, we get that $$J(p+2,p,r) = \dfrac{J(p,p,r)}2$$ Note that $$J(0,0,r) = \int_0^{\pi/2} \dfrac{d \theta}{(s+c)^r} = \int_0^{\pi/2} \dfrac1{2^{r/2}} \dfrac{d \theta}{\sin^r(\theta + \pi/4)}$$

Let $I(a,b,m) = \displaystyle \int_0^{\infty} \dfrac{x^{a+b-1} dx}{(1+x)^{b}(1+x^2)^{am/2}}$.

Setting $x = \tan(\theta)$. We then get that

$$I(a,b,m) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \sec^2(\theta) d \theta}{(1+\tan(\theta))^b \left(1+\tan^2(\theta) \right)^{am/2}}$$

$$I(a,b,m) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \sec^2(\theta) d \theta}{(1+\tan(\theta))^b \sec^{am}(\theta)}$$

$$I(a,b,m) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \cos^{am-2}(\theta) d \theta}{(1+\tan(\theta))^b}$$

Let us use the following short hand notation. $c = \cos(\theta)$ and $s = \sin(\theta)$. We then get that $$I(a,b,m) = \int_0^{\pi/2} \dfrac{s^{a+b-1} c^{am-a-1} d \theta}{(s+c)^b}$$ Hence, we are interested in evaluating integral of the form $$J(p,q,r) = \int_0^{\pi/2} \dfrac{s^p c^q d \theta}{(s+c)^r}$$ First some observations: $$J(p,q,r) = J(q,p,r) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\star)$$ $$J(p,q,0) = \dfrac{\beta((p+1)/2,(q+1)/2)}2$$ $$J(p,q,-1) = \int_0^{\pi/2} s^p c^q (s+c) d \theta = J(p+1,q,0) + J(p,q+1,0)$$ Further  $$J(p,q,r+2) = \int_0^{\pi/2} \dfrac{s^p c^q d \theta}{(s+c)^r (1+2sc)}$$ $$J(p,q,r+2) = \sum_{k=0}^{\infty} \int_0^{\pi/2} \dfrac{s^p c^q (-2sc)^k d \theta}{(s+c)^r}$$ $$J(p,q,r+2) = \sum_{k=0}^{\infty}(-2)^k J(p+k,q+k,r)$$

Hence, you can use this to compute $J(p,q,r)$ for all $r$ since these can be obtained as we know the values for $J(p,q,0)$ and $J(p,q,-1)$. (For even $r$, $J(p,q,r)$ will eventually depend on $J(p,q,0)$ and for odd $r$, $J(p,q,r)$ will eventually depend on $J(p,q,-1)$)

Few other relations, which might be of use. We have $$\lim_{p \to \infty} J(p,q,r) = \lim_{q \to \infty} J(p,q,r) = \lim_{r \to \infty} J(p,q,r) = 0$$ Note that $$J(p+2,q,r) = \int_0^{\pi/2} \dfrac{s^p c^q (1-c^2) d \theta}{(s+c)^r}$$ Hence, we get that $$J(p+2,q,r) + J(p,q+2,r) = J(p,q,r)$$ Setting $p=q$, and making use of $(\star)$, we get that $$J(p+2,p,r) = \dfrac{J(p,p,r)}2$$ Note that $$J(0,0,r) = \int_0^{\pi/2} \dfrac{d \theta}{(s+c)^r} = \int_0^{\pi/2} \dfrac1{2^{r/2}} \dfrac{d \theta}{\sin^r(\theta + \pi/4)}$$

Not quite an answer but some thoughts that might be of help in evaluating them possibly numerically.

Let $I(a,b,c) = \displaystyle \int_0^{\infty} \dfrac{x^{a+b-1} dx}{(1+x)^{b}(1+x^2)^{am/2}}$.

Setting $x = \tan(\theta)$. We then get that

$$I(a,b,c) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \sec^2(\theta) d \theta}{(1+\tan(\theta))^b \left(1+\tan^2(\theta) \right)^{am/2}}$$

$$I(a,b,c) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \sec^2(\theta) d \theta}{(1+\tan(\theta))^b \sec^{am}(\theta)}$$

$$I(a,b,c) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \cos^{am-2}(\theta) d \theta}{(1+\tan(\theta))^b}$$

Let us use the following short hand notation. $c = \cos(\theta)$ and $s = \sin(\theta)$. We then get that $$I(a,b,c) = \int_0^{\pi/2} \dfrac{s^{a+b-1} c^{am-a-1} d \theta}{(s+c)^b}$$ Hence, we are interested in evaluating integral of the form $$J(p,q,r) = \int_0^{\pi/2} \dfrac{s^p c^q d \theta}{(s+c)^r}$$ Next note that $$J(p,q,r) = J(q,p,r) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\star)$$ Next note that $$J(p,q,0) = \dfrac{\beta((p+1)/2,(q+1)/2)}2$$ Further

$$J(p,q,r+2) = \int_0^{\pi/2} \dfrac{s^p c^q d \theta}{(s+c)^r (1+2sc)}$$ $$J(p,q,r+2) = \sum_{k=0}^{\infty} \int_0^{\pi/2} \dfrac{s^p c^q (-2sc)^k d \theta}{(s+c)^r}$$ $$J(p,q,r+2) = \sum_{k=0}^{\infty}(-2)^k J(p+k,q+k,r)$$ \end{align} Hence, you can use this to compute $J(p,q,r)$ for all even $r$.

For odd $r$, the same idea as before works but you need to evaluate for $r=1$. For $r=1$, you could split the integral from $0$ to $\pi/4$ and $\pi/4$ to $\pi/2$ to write $\dfrac1{s+c}$ as a power series in $s$ and $c$ and evaluate them accordingly.

Few other relations, which might be of use. We have $$\lim_{p \to \infty} J(p,q,r) = \lim_{q \to \infty} J(p,q,r) = \lim_{r \to \infty} J(p,q,r) = 0$$ Note that $$J(p+2,q,r) = \int_0^{\pi/2} \dfrac{s^p c^q (1-c^2) d \theta}{(s+c)^r}$$ Hence, we get that $$J(p+2,q,r) + J(p,q+2,r) = J(p,q,r)$$ Setting $p=q$, and making use of $(\star)$, we get that $$J(p+2,p,r) = \dfrac{J(p,p,r)}2$$ Proceeding on similar lines, it is not hard to show that $$J(p+2k,p,r) = \dfrac{J(p,p,r)}{2^k}$$ Note that $$J(0,0,r) = \int_0^{\pi/2} \dfrac{d \theta}{(s+c)^r} = \int_0^{\pi/2} \dfrac1{2^{r/2}} \dfrac{d \theta}{\sin^r(\theta + \pi/4)}$$

Not a complete answer but some thoughts that might be of help in evaluating them possibly numerically.

Let $I(a,b,c) = \displaystyle \int_0^{\infty} \dfrac{x^{a+b-1} dx}{(1+x)^{b}(1+x^2)^{am/2}}$.

Setting $x = \tan(\theta)$. We then get that

$$I(a,b,c) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \sec^2(\theta) d \theta}{(1+\tan(\theta))^b \left(1+\tan^2(\theta) \right)^{am/2}}$$

$$I(a,b,c) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \sec^2(\theta) d \theta}{(1+\tan(\theta))^b \sec^{am}(\theta)}$$

$$I(a,b,c) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \cos^{am-2}(\theta) d \theta}{(1+\tan(\theta))^b}$$

Let us use the following short hand notation. $c = \cos(\theta)$ and $s = \sin(\theta)$. We then get that $$I(a,b,c) = \int_0^{\pi/2} \dfrac{s^{a+b-1} c^{am-a-1} d \theta}{(s+c)^b}$$ Hence, we are interested in evaluating integral of the form $$J(p,q,r) = \int_0^{\pi/2} \dfrac{s^p c^q d \theta}{(s+c)^r}$$ Next note that $$J(p,q,r) = J(q,p,r) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\star)$$ Next note that $$J(p,q,0) = \dfrac{\beta((p+1)/2,(q+1)/2)}2$$ Further

$$J(p,q,r+2) = \int_0^{\pi/2} \dfrac{s^p c^q d \theta}{(s+c)^r (1+2sc)}$$ $$J(p,q,r+2) = \sum_{k=0}^{\infty} \int_0^{\pi/2} \dfrac{s^p c^q (-2sc)^k d \theta}{(s+c)^r}$$ $$J(p,q,r+2) = \sum_{k=0}^{\infty}(-2)^k J(p+k,q+k,r)$$

Hence, you can use this to compute $J(p,q,r)$ for all even $r$.

For odd $r$, the same idea as before works but you need to evaluate for $r=1$. For $r=1$, you could split the integral from $0$ to $\pi/4$ and $\pi/4$ to $\pi/2$ to write $\dfrac1{s+c}$ as a power series in $s$ and $c$ and evaluate them accordingly.

Few other relations, which might be of use. We have $$\lim_{p \to \infty} J(p,q,r) = \lim_{q \to \infty} J(p,q,r) = \lim_{r \to \infty} J(p,q,r) = 0$$ Note that $$J(p+2,q,r) = \int_0^{\pi/2} \dfrac{s^p c^q (1-c^2) d \theta}{(s+c)^r}$$ Hence, we get that $$J(p+2,q,r) + J(p,q+2,r) = J(p,q,r)$$ Setting $p=q$, and making use of $(\star)$, we get that $$J(p+2,p,r) = \dfrac{J(p,p,r)}2$$ Note that $$J(0,0,r) = \int_0^{\pi/2} \dfrac{d \theta}{(s+c)^r} = \int_0^{\pi/2} \dfrac1{2^{r/2}} \dfrac{d \theta}{\sin^r(\theta + \pi/4)}$$