Given real parameters $\left(\alpha,\beta,\gamma,\delta\right)\in\mathbb{R}^{4}$ such that $0<\alpha<\delta$ and real arguments $\left(x,y\right)\in\left(-\infty,1\right)^{2}$, we can express the Appell $F_{1}$ function via the integral representation
$$\begin{align}
F_{1}{\left(\alpha;\beta,\gamma;\delta;x,y\right)}
&=\frac{1}{\operatorname{B}{\left(\alpha,\delta-\alpha\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{t^{\alpha-1}\left(1-t\right)^{\delta-\alpha-1}}{\left(1-xt\right)^{\beta}\left(1-yt\right)^{\gamma}}.\\
\end{align}$$
Starting from the integral representation of the $F_{1}$ function for the particular set of parameters that we're interested in, we obtain an integral of a simple algebraic function with elementary antiderivative: for any fixed but arbitrary $\left(x,y\right)\in\left(-\infty,1\right)^{2}$,
$$\begin{align}
F_{1}{\left(1;1,\frac12;2;x,y\right)}
&=\frac{1}{\operatorname{B}{\left(1,1\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\left(1-xt\right)\sqrt{1-yt}}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{\left(1-xt\right)\sqrt{1-yt}}\\
&=\int_{1}^{0}\mathrm{d}u\,\frac{\left(-1\right)\left(1-x\right)}{\left(1-xu\right)^{2}}\cdot\frac{\left(1-xu\right)\sqrt{1-xu}}{\left(1-x\right)\sqrt{\left(1-y\right)-\left(x-y\right)u}};~~~\small{\left[t=\frac{1-u}{1-xu}\right]}\\
&=\frac{1}{\sqrt{1-y}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-xu\right)\left[1-\left(\frac{x-y}{1-y}\right)u\right]}}.\\
\end{align}$$
Suppose $a\in\left(0,1\right)$ and $x<1\land x\neq0$. Setting $y=ax$, we then find
$$\begin{align}
F_{1}{\left(1;1,\frac12;2;x,ax\right)}
&=\frac{1}{\sqrt{1-ax}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-xu\right)\left[1-\left(\frac{x-ax}{1-ax}\right)u\right]}}\\
&=\frac{1}{\sqrt{1-ax}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-xu\right)\left[1-\left(\frac{1-a}{1-ax}\right)xu\right]}}\\
&=\frac{1}{x\sqrt{1-ax}}\int_{0}^{x}\mathrm{d}v\,\frac{1}{\sqrt{\left(1-v\right)\left[1-\left(\frac{1-a}{1-ax}\right)v\right]}};~~~\small{\left[u=\frac{v}{x}\right]}\\
&=\frac{1}{x\sqrt{1-ax}}\int_{1-x}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[1-\left(\frac{1-a}{1-ax}\right)\left(1-w\right)\right]}};~~~\small{\left[v=1-w\right]}\\
&=\frac{1}{x}\int_{1-x}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[\left(1-ax\right)-\left(1-a\right)\left(1-w\right)\right]}}\\
&=\frac{1}{x}\int_{1-x}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[a\left(1-x\right)+\left(1-a\right)w\right]}}\\
&=\frac{1}{x}\int_{0}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[a\left(1-x\right)+\left(1-a\right)w\right]}}\\
&~~~~~-\frac{1}{x}\int_{0}^{1-x}\mathrm{d}w\,\frac{1}{\sqrt{w\left[a\left(1-x\right)+\left(1-a\right)w\right]}}\\
&=\frac{1}{x}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t\right)\left[\left(1-ax\right)-\left(1-a\right)t\right]}};~~~\small{\left[w=1-t\right]}\\
&~~~~~-\frac{1}{x}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{t\left[a+\left(1-a\right)t\right]}};~~~\small{\left[w=\left(1-x\right)t\right]}\\
&=\frac{1}{x\sqrt{1-ax}}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t\right)\left[1-\left(\frac{1-a}{1-ax}\right)t\right]}}\\
&~~~~~-\frac{1}{x}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-u\right)\left[1-\left(1-a\right)u\right]}};~~~\small{\left[t=1-u\right]}\\
&=\frac{2}{x\sqrt{1-ax}}\,{_2F_1}{\left(\frac12,1;\frac32;\frac{1-a}{1-ax}\right)}-\frac{2}{x}\,{_2F_1}{\left(\frac12,1;\frac32;1-a\right)},\\
\end{align}$$
where in the last line above we've used the Euler integral representation formula to express the remaining integrals in terms of the ${_2F_1}$ function:
$$\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta-1}\left(1-t\right)^{\gamma-\beta-1}}{\left(1-zt\right)^{\alpha}}=\operatorname{B}{\left(\beta,\gamma-\beta\right)}\,{_2F_1}{\left(\alpha,\beta;\gamma;z\right)};~~~\small{z<1\land0<\beta<\gamma}.$$
$$\tag*{$\blacksquare$}$$
tl;dr: Wolfram is giving the general solution that works even when $n$ isn't an integer.
Long version:
Let's look at the definition of $_2F_1$:
$$
_2F_1(a,b;c;x) = \sum_{k=0}^\infty \frac{\Gamma(a+k)\Gamma(b+k)\Gamma(c)}{\Gamma(a)\Gamma(b)\Gamma(c+k)}\frac{x^k}{\Gamma(k+1)}
$$
Plugging in $a = 1$, $b = n+1$, $c = n+2$ gives
\begin{align}
-\frac{x^{n+1}}{n+1}\,_2F_1(1,n+1;n+2;x) = &-\frac{x^{n+1}}{n+1}\sum_{k=0}^\infty \frac{\Gamma(1+k)\Gamma(n+1+k)\Gamma(n+2)}{\Gamma(1)\Gamma(n+1)\Gamma(n+2+k)}\frac{x^k}{\Gamma(k+1)}
\\= &-\frac{x^{n+1}}{n+1}\sum_{k=0}^\infty \frac{n+1}{n+1+k}x^k
\\ = &-\sum_{k=0}^\infty \frac{x^{n+1+k}}{n+1+k}.
\end{align}
Additionally using $\ln(1-x) = -\sum_{j=1}^\infty x^j/j$, Wolfram is saying
$$
\int\frac{x^n-1}{x-1}dx = \sum_{j=1}^\infty\frac{x^j}{j} -\sum_{k=0}^\infty \frac{x^{n+1+k}}{n+1+k}.
$$
Now at this point you may have noticed that the second sum will cancel out every term of the first sum beyond $j = n$. This is true, but only if $n$ is an integer. Wolfram never assumes any information you don't give it, so it gives the answer in terms of the hypergeometric function. If $n$ is an integer, the cancellation I mentioned happens, and you get
$$
\int\frac{x^n-1}{x-1}dx = \sum_{j=1}^n\frac{x^j}{j}
$$
as expected.
Best Answer
Consider the integral in the form $$ I = \int_{-\infty}^{\infty} (1 + p \, t^2) \, {}_{2}F_{1}\left(a, b; c; \frac{q^2 \, t^2}{4 \, \alpha \, (1 + p \, t^2)} \right) \, dt. $$ Converting the ${}_{2}F_{1}$ into its series form then $$ I = \sum_{n=0}^{\infty} \frac{(a)_{n} \, (b)_{n}}{n! \, (c)_{n}} \, \left(\frac{q^2}{4 \, \alpha}\right)^n \, J_{n} $$ where $J_{n}$ is the integral $$ J_{n} = \int_{-\infty}^{\infty} \frac{t^{2 n} \, dt}{(1 + p \, t^2)^{n-1}}. $$ Since \begin{align} \int_{-\infty}^{\infty} f(t^2) \, dt &= \int_{-\infty}^{0} f(t^2) \, dt + \int_{0}^{\infty} f(t^2) \, dt \\ &= 2 \, \int_{0}^{\infty} f(t^2) \, dt \hspace{5mm} \text{let} \, t = -u \, \text{in the first integral} \\ &= \int_{0}^{\infty} f(u) \, u^{-1/2} \, du \hspace{5mm} \text{where} \, t = \sqrt{u} \, \text{was used}. \end{align} With this then $J_{n}$ becomes $$ J_{n} = \int_{0}^{\infty} u^{n-1/2} \, (1 + p \, u)^{1-n} \, du.$$ Comparing this to the hypergeometric function ${}_{2}F_{1}$ integral form, namely, $$ \int_{0}^{\infty} t^{c-b-1} \, (1+t)^{c-a} \, (1-z+t)^{-a} \, dt = \frac{\Gamma(b) \, \Gamma(c-b)}{\Gamma(c)} \, {}_{2}F_{1}(a, b; c; z).$$ In the case $c = a$ this reduces to \begin{align} \int_{0}^{\infty} t^{a-b-1} \, (1-z+t)^{-a} \, dt &= \frac{\Gamma(b) \, \Gamma(a-b)}{\Gamma(a)} \, {}_{2}F_{1}(a, b; a; z) \\ &= \frac{\Gamma(b) \, \Gamma(a-b)}{\Gamma(a)} \, (1-z)^{-a}. \end{align} Now, \begin{align} J_{n} &= \int_{0}^{\infty} u^{n-1/2} \, (1 + p \, u)^{1-n} \, du \\ &= p^{1-n} \, \int_{0}^{\infty} u^{n-1/2} \, \left(\frac{1}{p} + t\right)^{-(n-1)} \, du \\ &= p^{1-n} \, \frac{\Gamma\left(- \frac{3}{2}\right) \, \Gamma\left(n + \frac{1}{2}\right)}{\Gamma(n-1)} \, \left(\frac{1}{p}\right)^{3/2} \\ &= \frac{\Gamma\left(- \frac{3}{2}\right) \, \Gamma\left(\frac{1}{2}\right)}{\sqrt{p}} \, \frac{\left(\frac{1}{2}\right)_{n}}{(n-2)! \, p^{n}}. \end{align} Returning to $I$: \begin{align} I &= \sum_{n=0}^{\infty} \frac{(a)_{n} \, (b)_{n}}{n! \, (c)_{n}} \, \left(\frac{q^2}{4 \, \alpha}\right)^n \, J_{n} \\ &= \frac{\Gamma\left(- \frac{3}{2}\right) \, \Gamma\left(\frac{1}{2}\right)}{\sqrt{p}} \, \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n} \, (a)_{n} \, (b)_{n}}{(n-2)! \, (1)_{n} \, (c)_{n}} \, \left(\frac{q^2}{4 \, \alpha \, p}\right)^n \\ &= \frac{\Gamma\left(- \frac{3}{2}\right) \, \Gamma\left(\frac{1}{2}\right)}{\sqrt{p}} \, \sum_{n=2}^{\infty} \frac{\left(\frac{1}{2}\right)_{n} \, (a)_{n} \, (b)_{n}}{(n-2)! \, (1)_{n} \, (c)_{n}} \, \left(\frac{q^2}{4 \, \alpha \, p}\right)^n \\ &= \frac{\Gamma\left(- \frac{3}{2}\right) \, \Gamma\left(\frac{1}{2}\right)}{\sqrt{p}} \, \frac{\left(\frac{1}{2}\right)_{2} \, (a)_{2} \, (b)_{2}}{(1)_{2} \, (c)_{2}} \, \sum_{n=0}^{\infty} \frac{\left(\frac{5}{2}\right)_{n} \, (a+2)_{n} \, (b+2)_{n}}{n! \, (3)_{n} \, (c+2)_{n}} \, \left(\frac{q^2}{4 \, \alpha \, p}\right)^{n+2} \\ I &= \frac{\pi \, (a)_{2} \, (b)_{2}}{2 \, (c)_{2}} \, \left(\frac{q^2}{4 \, \alpha \, p} \right)^{2} \, {}_{3}F_{2}\left(\frac{5}{2}, a+2, b+2; 3, c+2; \frac{q^2}{4 \, \alpha \, p} \right). \end{align} Associating the constants will lead to the desired result.
For the case of $c = \frac{1}{2}$ the given result has a nice reduction as seen by \begin{align} I_{c = 1/2} &= \int_{-\infty}^{\infty} (1 + p \, t^2) \, {}_{2}F_{1}\left(a, b; \frac{1}{2}; \frac{q^2 \, t^2}{4 \, \alpha \, (1 + p \, t^2)} \right) \, dt \\ &= \frac{\pi \, (a)_{2} \, (b)_{2} \, q^4}{24 \, (\alpha \, p)^2} \, {}_{2}F_{1}\left(a+2, b+2; 3; \frac{q^2}{4 \, \alpha \, p} \right). \end{align}