Computing $\sum _{k=1}^{\infty } \frac{\Gamma \left(\frac{k}{2}+1\right)}{k^2 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}$ in closed form

Writing $\dfrac{\Gamma\left(\frac{k}{2}+1\right)}{\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}$ as an integral and exchanging summation and integration, we get \begin{align} \sum^\infty_{k=1}\frac{\Gamma\left(\frac{k}{2}+1\right)}{k^2\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)} &=\frac{2}{\sqrt{\pi}}\int^1_0\frac{x{\rm Li}_2(x)}{\sqrt{1-x^2}}\ {\rm d}x\tag1\\ &=-\frac{2}{\sqrt{\pi}}\int^1_0\frac{\sqrt{1-x^2}\ln(1-x)}{x}\ {\rm d}x\tag2\\ &=-\frac{1}{\sqrt{\pi}}\int^\pi_0\frac{\cos^2{x}\ln(1-\sin{x})}{\sin{x}}\ {\rm d}x\tag3\\ &=-\frac{1}{\sqrt{\pi}}\left[\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n}\mathcal{A}_n+2\sum^\infty_{n=0}\frac{(-1)^{n-1}}{2n+1}\mathcal{B}_n\right]\tag4\\ \end{align} where \begin{align} \mathcal{A}_n =\int^\pi_0\frac{\cos^2{x}}{\sin{x}}(\cos(2nx)-1)\ {\rm d}x \ \ \ , \ \ \ \mathcal{B}_n =\int^\pi_0\frac{\cos^2{x}}{\sin{x}}\sin((2n+1)x)\ {\rm d}x\\ \end{align} Using simple trigonometric identities, it is not hard to see that, for $n\in\mathbb{N}$, \begin{align} \mathcal{A}_n-\mathcal{A}_{n-1}&=-\ \frac{1}{2n-3}-\frac{2}{2n-1}-\frac{1}{2n+1}\tag5\\ \mathcal{B}_n-\mathcal{B}_{n-1}&=0\tag6 \end{align} Thus we may obtain the closed forms for both sequences. \begin{align} \mathcal{A}_n=2H_n-4H_{2n}+\frac{1}{2n-1}-\frac{1}{2n+1}+2\ \ \ , \ \ \ \mathcal{B}_n=\pi-\frac{\pi}{2}\delta_{n0}\tag7 \end{align} Using the Taylor series for $\ln(1-x)$ and $\arctan{x}$, as well as the well-known identities \begin{align} \sum^\infty_{n=1}\frac{(-1)^{n-1}H_n}{n}&=\frac{\pi^2}{12}-\frac{1}{2}\ln^2{2}\tag8\\ \sum^\infty_{n=1}\frac{(-1)^{n-1}H_{2n}}{n}&=\frac{5\pi^2}{48}-\frac{1}{4}\ln^2{2}\tag9 \end{align} gives us \begin{align} \sum^\infty_{k=1}\frac{\Gamma\left(\frac{k}{2}+1\right)}{k^2\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)} &=-\frac{1}{\sqrt{\pi}}\left[2\left(\frac{\pi^2}{12}-\frac{1}{2}\ln^2{2}\right)-4\left(\frac{5\pi^2}{48}-\frac{1}{4}\ln^2{2}\right)+2-2\pi\left(\frac{\pi}{4}\right)+2\left(\frac{\pi}{2}\right)\right]\\ &=\frac{3\pi^2-4\pi-8}{4\sqrt{\pi}}\tag{10} \end{align} as the closed form.

Explanation:
$(1)$: Write $\displaystyle\frac{\Gamma\left(\frac{k}{2}+1\right)}{\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}=\frac{2}{\sqrt{\pi}}\int^1_0\frac{x^{k+1}}{\sqrt{1-x^2}}\ {\rm d}x$.
$(2)$: Integrated by parts.
$(3)$: Substitute $x\mapsto\sin{x}$ then $x\mapsto\pi-x$.
$(4)$: Use the fact that $\ln(2-2\sin{x})=2\mathrm{Re}\ln(1+ie^{ix})$ then expand the $\mathrm{RHS}$.
$(5)$: Write $\displaystyle\mathcal{A}_n-\mathcal{A}_{n-1}=-\int^\pi_0(1+\cos(2x))\sin((2n-1)x)\ {\rm d}x$.
$(6)$: Write $\displaystyle\mathcal{B}_n-\mathcal{B}_{n-1}=\int^\pi_0(1+\cos(2x))\cos(2nx)\ {\rm d}x$.
$(7)$: Sum up $(5)$ and $(6)$.
$(8),(9)$: Let $z=-1$, $z=i$ in $\displaystyle\sum^\infty_{n=1}\frac{H_n}{n}z^n={\rm Li}_2(z)+\frac{1}{2}\ln^2(1-z)$.
$(10)$: Apply $(7)$, $(8)$, $(9)$ to $(4)$.

Let's write this as a sum of two hopefully simpler sums. $$ \sum_{k=1}^\infty\frac{\Gamma(\frac k2+1)}{k^2\Gamma(\frac k2+\frac32)} =\sum_{k=1}^\infty\frac{\Gamma(k+\frac12)}{(2k-1)^2\Gamma(k+1)} +\sum_{k=1}^\infty\frac{\Gamma(k+1)}{4k^2\Gamma(k+\frac32)}\tag{1} $$ For integer $k$, we have $$ \Gamma(k+\tfrac12)=\frac{\sqrt\pi}{4^k}\binom{2k}{k}k!\tag{2} $$ We will proceed by computing particular series.

The First Sum on the Right of $\boldsymbol{(1)}$

Identity $(2)$ says that $$ \sum_{k=1}^\infty\frac{\Gamma(k+\frac12)}{(2k-1)^2\Gamma(k+1)} =\sqrt\pi\sum_{k=1}^\infty\frac{\binom{2k}{k}}{4^k(2k-1)^2}\tag{3} $$ Using the Extended Binomial Theorem, we get $$ \sum_{k=0}^\infty\frac{\binom{2k}{k}}{4^k}x^{2k}=(1-x^2)^{-1/2}\tag{4} $$ Therefore, letting $x=\sin(u)$ and noting that $\int\frac{\mathrm{d}u}{1+\cos(u)}=\frac{\sin(u)}{1+\cos(u)}$, we get $$ \begin{align} \sum_{k=1}^\infty\frac{\binom{2k}{k}}{4^k}\frac{x^{2k-1}}{2k-1} &=\int\frac{(1-x^2)^{-1/2}-1}{x^2}\,\mathrm{d}x\\ &=\int\frac{\mathrm{d}u}{1+\cos(u)}\\ &=\frac{\sin(u)}{1+\cos(u)}\\ &=\frac{x}{1+\sqrt{1-x^2}}\tag{5} \end{align} $$ and $$ \begin{align} \sum_{k=1}^\infty\frac{\binom{2k}{k}}{4^k}\frac{x^{2k-1}}{(2k-1)^2} &=\int\frac1{1+\sqrt{1-x^2}}\mathrm{d}x\\ &=\int\frac{\cos(u)}{1+\cos(u)}\,\mathrm{d}u\\ &=u-\frac{\sin(u)}{1+\cos(u)}\\ &=\sin^{-1}(x)-\frac{x}{1+\sqrt{1-x^2}}\tag{6} \end{align} $$ Putting together $(3)$ and $(6)$ yields $$ \sum_{k=1}^\infty\frac{\Gamma(k+\frac12)}{(2k-1)^2\Gamma(k+1)}=\sqrt\pi\left(\frac\pi2-1\right)\tag{7} $$

The Second Sum on the Right of $\boldsymbol{(1)}$

Identity $(2)$ says that $$ \sum_{k=1}^\infty\frac{\Gamma(k+1)}{4k^2\Gamma(k+\frac32)} =\frac1{2\sqrt\pi}\sum_{k=1}^\infty\frac{4^k}{k^2(2k+1)\binom{2k}{k}}\tag{8} $$ Equation $(2)$ from this answer says that $$ \sum_{k=0}^\infty\frac{4^kx^{2k}}{\binom{2k}{k}} =\frac1{1-x^2}\left[1+\sqrt{\frac{x^2}{1-x^2}}\sin^{-1}(x)\right]\tag{9} $$ Therefore, letting $x=\sin(u)$, $$ \begin{align} \sum_{k=0}^\infty\frac{4^kx^{2k+1}}{(2k+1)\binom{2k}{k}} &=\int\frac1{1-x^2}\left[1+\sqrt{\frac{x^2}{1-x^2}}\sin^{-1}(x)\right]\,\mathrm{d}x\\ &=\int\sec(u)(1+\tan(u)u)\,\mathrm{d}u\\ &=u\sec(u)\\ &=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}\tag{10} \end{align} $$ and $$ \begin{align} \sum_{k=1}^\infty\frac{4^kx^{2k}}{2k(2k+1)\binom{2k}{k}} &=\int\frac{\sin^{-1}(x)-x\sqrt{1-x^2}}{x^2\sqrt{1-x^2}}\,\mathrm{d}x\\ &=\int\frac{u-\sin(u)\cos(u)}{\sin^2(u)}\,\mathrm{d}u\\ &=1-u\cot(u)\\ &=1-\frac{\sin^{-1}(x)\sqrt{1-x^2}}{x}\tag{11} \end{align} $$ and $$ \begin{align} \sum_{k=1}^\infty\frac{4^kx^{2k}}{4k^2(2k+1)\binom{2k}{k}} &=\int\frac{x-\sin^{-1}(x)\sqrt{1-x^2}}{x^2}\,\mathrm{d}x\\ &=\int\frac{\sin(u)\cos(u)-u\cos^2(u)}{\sin^2(u)}\,\mathrm{d}u\\ &=\frac12u^2+u\cot(u)-1\\ &=\frac12\sin^{-1}(x)^2+\frac{\sin^{-1}(x)\sqrt{1-x^2}}{x}-1\tag{12} \end{align} $$ Combining $(8)$ and $(12)$ yields $$ \sum_{k=1}^\infty\frac{\Gamma(k+1)}{4k^2\Gamma(k+\frac32)} =\frac2{\sqrt\pi}\left(\frac{\pi^2}8-1\right)\tag{13} $$

Putting together $(1)$, $(7)$, and $(13)$ gives $$ \bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^\infty\frac{\Gamma(\frac k2+1)}{k^2\Gamma(\frac k2+\frac32)} =\frac1{4\sqrt\pi}\left(3\pi^2-4\pi-8\right)}\tag{14} $$

Using some results from my other answer, we can compute the second sum requested.

Similar to my other answer, write $$ \sum_{k=1}^\infty\frac{\Gamma(\frac k2+1)}{k^3\Gamma(\frac k2+\frac32)} =\sum_{k=1}^\infty\frac{\Gamma(k+\frac12)}{(2k-1)^3\Gamma(k+1)} +\sum_{k=1}^\infty\frac{\Gamma(k+1)}{8k^3\Gamma(k+\frac32)}\tag{1} $$

First Sum on the Right of $\boldsymbol{(1)}$

Using $(2)$ from my other answer, we get $$ \sum_{k=1}^\infty\frac{\Gamma(k+\frac12)}{(2k-1)^3\Gamma(k+1)} =\sqrt\pi\sum_{k=1}^\infty\frac{\binom{2k}{k}}{4^k(2k-1)^3}\tag{2} $$

Starting from $(6)$ in my other answer, we get $$ \begin{align} \sum_{k=1}^\infty\frac{\binom{2k}{k}}{4^k(2k-1)^3} &=\int_0^1\left(\frac{\sin^{-1}(x)}x-\frac1{1+\sqrt{1-x^2}}\right)\,\mathrm{d}x\\ &=\int_0^{\pi/2}\left(u\cot(u)-\frac{\cos(u)}{1+\cos(u)}\right)\,\mathrm{d}u\\ &=\left[u\log(\sin(u))\vphantom{\int}\right]_0^{\pi/2}-\int_0^{\pi/2}\log(\sin(u))\,\mathrm{d}u-\left[u-\frac{\sin(u)}{1+\cos(u)}\vphantom{\int}\right]_0^{\pi/2}\\ &=\frac\pi2\log(2)+1-\frac\pi2\tag{3} \end{align} $$ Combining $(2)$ and $(3)$, $$ \sum_{k=1}^\infty\frac{\Gamma(k+\frac12)}{(2k-1)^3\Gamma(k+1)} =\sqrt\pi\left(\frac\pi2\log(2)+1-\frac\pi2\right)\tag{4} $$

Second Sum on the Right of $\boldsymbol{(1)}$

Using $(2)$ from my other answer, we get $$ \sum_{k=1}^\infty\frac{\Gamma(k+1)}{8k^3\Gamma(k+\frac32)} =\frac2{\sqrt\pi}\sum_{k=1}^\infty\frac{4^k}{8k^3(2k+1)\binom{2k}{k}}\tag{5} $$ Starting from $(12)$ in my other answer, we get $$ \begin{align} \sum_{k=1}^\infty\frac{4^k}{8k^3(2k+1)\binom{2k}{k}} &=\int_0^1\left(\frac{\sin^{-1}(x)^2}{2x}-\frac{x-\sin^{-1}(x)\sqrt{1-x^2}}{x^2}\right)\,\mathrm{d}x\\ &=\int_0^{\pi/2}\left(\frac{u^2\cos(u)}{2\sin(u)}-\frac{\sin(u)\cos(u)-u\cos^2(u)}{\sin^2(u)}\right)\,\mathrm{d}u\\ &=\small\left[\frac{u^2}2\log(\sin(u))\right]_0^{\pi/2}-\int_0^{\pi/2}u\log(\sin(u))\,\mathrm{d}u-\left[\frac12u^2+u\cot(u)-1\right]_0^{\pi/2}\\ &=\frac{\pi^2}8\log(2)-\frac7{16}\zeta(3)-\frac{\pi^2}8+1\tag{6} \end{align} $$ Combining $(5)$ and $(6)$, $$ \sum_{k=1}^\infty\frac{\Gamma(k+1)}{8k^3\Gamma(k+\frac32)} =\frac1{8\sqrt\pi}\left(2\pi^2(\log(2)-1)-7\zeta(3)+16\right)\tag{7} $$

Putting together $(1)$, $(4)$, and $(7)$ yields $$ \bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^\infty\frac{\Gamma(\frac k2+1)}{k^3\Gamma(\frac k2+\frac32)} =\frac1{8\sqrt\pi}\left(6\pi^2(\log(2)-1)-7\zeta(3)+8\pi+16\right)}\tag{8} $$

Integrals Involving $\boldsymbol{\log(\sin(u))}$ Used Above

As shown in $(1)$ from this answer: $$ \log(\sin(u))=-\log(2)+\sum_{k=1}^\infty(-1)^{k-1}\frac{\cos(2ku)}{k}\tag{9} $$ For $k\in\mathbb{Z}$ and $k\ne0$, we have $$ \int_0^{\pi/2}\cos(2ku)\,\mathrm{d}u=0\tag{10} $$ and integration by parts gives $$ \begin{align} \int_0^{\pi/2}u\cos(2ku)\,\mathrm{d}u &=-\frac1{2k}\int_0^{\pi/2}\sin(2ku)\,\mathrm{d}u\\ &=\frac1{4k^2}(1-\cos(k\pi))\\ &=\frac{[k\text{ is odd}]}{2k^2}\tag{11} \end{align} $$ It follows that $$ \int_0^{\pi/2}\log(\sin(u))\,\mathrm{d}u=-\frac\pi2\log(2)\tag{12} $$ and $$ \int_0^{\pi/2}u\log(\sin(u))\,\mathrm{d}u=-\frac{\pi^2}8\log(2)+\frac7{16}\zeta(3)\tag{13} $$