Let's try a close variant : differentiate relatively to $d$ instead of $a$ at the start (to avoid the additional factor $(x-d)$ in the integral and complications in the final integration) :
$$I\left(d\right)=\int_{-\infty}^{\infty}e^{-b^{2}\left(x-c\right)^{2}}\ \mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x$$
$$\frac{\mathrm{d}I\left(d\right)}{\mathrm{d}d}=\frac {-2a}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-b^{2}(x-c)^{2}-a^{2}(x-d)^{2}}\,\mathrm{d}x$$
$$\frac{\mathrm{d}I\left(d\right)}{\mathrm{d}d}=\frac {-2a}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-b^{2}(x-c)^{2}-a^{2}(x-d)^{2}}\,\mathrm{d}x$$
$$\frac{\mathrm{d}I\left(d\right)}{\mathrm{d}d}=\frac {-2a}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-\left(a^2+b^2\right)\left(x-\frac{b^2c+a^2d}{a^2+b^2}\right)^2+\frac{\left(b^2c+a^2d\right)^2}{a^2+b^2}-\left(b^2c^2+a^2d^2\right)}\,\mathrm{d}x$$
$$\frac{\mathrm{d}I\left(d\right)}{\mathrm{d}d}=\frac {-2a}{\sqrt{\pi}}e^{\frac{\left(b^2c+a^2d\right)^2}{a^2+b^2}-\left(b^2c^2+a^2d^2\right)}\int_{-\infty}^{\infty}e^{-\left(a^2+b^2\right)y^2}\,\mathrm{d}y$$
$$\frac{\mathrm{d}I\left(d\right)}{\mathrm{d}d}=\frac {-2a}{\sqrt{\pi}}e^{\frac{-a^2b^2\left(c-d\right)^2}{a^2+b^2}}\frac{\sqrt{\pi}}{\sqrt{a^2+b^2}}=-2a\frac{e^{\frac{-a^2b^2\left(c-d\right)^2}{a^2+b^2}}}{\sqrt{a^2+b^2}}$$
At this point we have to integrate again relatively to $d$ to get (up to a function $C$ independent of $d$) :
$$I\left(d\right)=\frac {\sqrt{\pi}}b\operatorname{erf}\left(\frac{ab(c-d)}{\sqrt{a^2+b^2}}\right)+C(a,b,c)$$
(Alpha integration check : note that the denominator is $b$ and not $\sqrt{b}$ nor my earlier $ab$ as I checked numerically!)
After that you'll just have to prove that $C(a,b,c)\equiv 0$
Note that the integration relatively to $d$ seems more straightforward than relatively to $a$ in your case (I'm not saying it can't be done your way!).
Hoping it clarified things a little even if it didn't answer your question,
This one is a cakewalk. Just use the definition $$K(1/2)=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-(1/2)\sin^{2}x}}=\frac{\sqrt{2}}{2}\int_{0}^{\pi}\frac{dx}{\sqrt{2-\sin^{2}x}}$$ and then put $t=\tan (x/2)$ so that $dx=2/(1+t^2)\,dt$ and $$2-\sin^{2}x=2-\frac{4t^2}{(1+t^2)^{2}}=2\cdot\frac{1+t^{4}}{(1+t^{2})^{2}}$$ and hence we have $$K(1/2)=\frac{1}{\sqrt{2}}\int_{0}^{\infty}\frac{1+t^{2}}{\sqrt{2}\sqrt{1+t^{4}}}\cdot\frac{2}{1+t^2}\,dt=\int_{0}^{\infty}\frac{dt}{\sqrt{1+t^{4}}}$$ Similarly put $\tan t=x^2$ to get $$\frac{1}{2}\int_{0}^{\pi/2}\frac{dt}{\sqrt{\sin t\cos t}} =\int_{0}^{\infty} \frac{dx} {\sqrt{1+x^{4}}}$$ This avoids the more complicated approaches from the theory of elliptic and theta functions and AGM.
Best Answer
Let $\mathcal{I}$ denote the value of the definite integral
$$\mathcal{I}:=\int_{0}^{1}\mathrm{d}u\,\sqrt{\frac{u^{2}-2-2\sqrt{u^{4}-u^{2}+1}}{4u^{6}-8u^{4}+8u^{2}-4}}\approx1.5436866339.$$
Note: the denominator of the radicand of the outer square root has the factorization
$$\begin{align} 4u^{6}-8u^{4}+8u^{2}-4 &=4t^{3}-8t^{2}+8t-4;~~~\small{\left[u^{2}=t\right]}\\ &=4\left(t^{3}-2t^{2}+2t-1\right)\\ &=4\left(t-1\right)\left(t^{2}-t+1\right).\\ \end{align}$$
Using the substitution $u^{2}=t$, the integral $\mathcal{I}$ can be rewritten as
$$\begin{align} \mathcal{I} &=\int_{0}^{1}\mathrm{d}u\,\sqrt{\frac{u^{2}-2-2\sqrt{u^{4}-u^{2}+1}}{4u^{6}-8u^{4}+8u^{2}-4}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{1}{2\sqrt{t}}\sqrt{\frac{t-2-2\sqrt{t^{2}-t+1}}{4t^{3}-8t^{2}+8t-4}};~~~\small{\left[u=\sqrt{t}\right]}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{1}{2\sqrt{t}}\sqrt{\frac{t-2-2\sqrt{t^{2}-t+1}}{4\left(t-1\right)\left(t^{2}-t+1\right)}}\\ &=\frac14\int_{0}^{1}\mathrm{d}t\,\sqrt{\frac{2-t+2\sqrt{t^{2}-t+1}}{t\left(1-t\right)\left(t^{2}-t+1\right)}}.\\ \end{align}$$
Now the inner radical is just a square root of a quadratic function, which suggests that the integral might be further simplified using an appropriate Euler substitution.
Consider a substitution given implicitly by the relation
$$\sqrt{t^{2}-t+1}=t+x.$$
Solving for $t$, we obtain $t=\frac{1-x^{2}}{1+2x}$. The integral $\mathcal{I}$ is then transformed to
$$\begin{align} \mathcal{I} &=\frac14\int_{0}^{1}\mathrm{d}t\,\sqrt{\frac{2-t+2\sqrt{t^{2}-t+1}}{t\left(1-t\right)\left(t^{2}-t+1\right)}}\\ &=\frac14\int_{1}^{0}\mathrm{d}x\,\frac{(-2)\left(1+x+x^{2}\right)}{\left(1+2x\right)^{2}}\sqrt{\frac{3\left(1+x\right)}{\left(1-x\right)}\cdot\frac{\left(1+2x\right)}{x\left(2+x\right)}\cdot\frac{\left(1+2x\right)^{2}}{\left(1+x+x^{2}\right)^{2}}};~~~\small{\left[t=\frac{1-x^{2}}{1+2x}\right]}\\ &=\frac12\int_{0}^{1}\mathrm{d}x\,\sqrt{\frac{3\left(1+x\right)}{x\left(1-x\right)\left(2+x\right)\left(1+2x\right)}}\\ &=\frac{\sqrt{3}}{2}\int_{0}^{1}\mathrm{d}x\,\frac{\left(1+x\right)}{\sqrt{x\left(1-x\right)\left(1+x\right)\left(1+2x\right)\left(2+x\right)}}.\\ \end{align}$$
Next, look what happens when we transform the integral using the linear fractional transformation $x=\frac{1-y}{1+y}$:
$$\begin{align} \mathcal{I} &=\frac{\sqrt{3}}{2}\int_{0}^{1}\mathrm{d}x\,\frac{\left(1+x\right)}{\sqrt{x\left(1-x\right)\left(1+x\right)\left(1+2x\right)\left(2+x\right)}}\\ &=\frac{\sqrt{3}}{2}\int_{0}^{1}\mathrm{d}x\,\frac{\left(1+x\right)}{\sqrt{\left(1+x\right)^{6}\left(\frac{1-x}{1+x}\right)\left(\frac{x}{1+x}\right)\left(\frac{1}{1+x}\right)\left(\frac{1+2x}{1+x}\right)\left(\frac{2+x}{1+x}\right)}}\\ &=\frac{\sqrt{3}}{2}\int_{0}^{1}\mathrm{d}x\,\frac{\left(\frac{1}{1+x}\right)^{2}}{\sqrt{\left(\frac{1-x}{1+x}\right)\left(\frac{x}{1+x}\right)\left(\frac{1}{1+x}\right)\left(\frac{1+2x}{1+x}\right)\left(\frac{2+x}{1+x}\right)}}\\ &=\frac{\sqrt{3}}{2}\int_{1}^{0}\mathrm{d}y\,\frac{(-2)}{\left(1+y\right)^{2}}\cdot\frac{\left(\frac{1+y}{2}\right)^{2}}{\sqrt{y\left(\frac{1-y}{2}\right)\left(\frac{1+y}{2}\right)\left(\frac{3-y}{2}\right)\left(\frac{3+y}{2}\right)}};~~~\small{\left[x=\frac{1-y}{1+y}\right]}\\ &=\sqrt{3}\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{y\left(1-y\right)\left(1+y\right)\left(3-y\right)\left(3+y\right)}}\\ &=\sqrt{3}\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{y\left(1-y^{2}\right)\left(9-y^{2}\right)}}.\\ \end{align}$$
Recalling Euler's integral formula for the Gauss hypergeometric function: for real argument and parameters,
$$\int_{0}^{1}\mathrm{d}t\,\frac{t^{b-1}\left(1-t\right)^{c-b-1}}{\left(1-zt\right)^{a}}=\operatorname{B}{\left(b,c-b\right)}\,{_2F_1}{\left(a,b;c;z\right)};~~~\small{\left(a,b,c,z\right)\in\mathbb{R}^{4}\land0<b<c\land z<1},$$
(where $\operatorname{B}$ here denotes the usual beta function), we arrive at the following representation for $\mathcal{I}$ as a particular value of ${_2F_1}$:
$$\begin{align} \mathcal{I} &=\sqrt{3}\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{y\left(1-y^{2}\right)\left(9-y^{2}\right)}}\\ &=\frac{1}{\sqrt{3}}\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{y}\sqrt{1-y^{2}}\sqrt{1-\frac19y^{2}}}\\ &=\frac{1}{\sqrt{3}}\int_{0}^{1}\mathrm{d}t\,\frac{1}{2\sqrt{t}}\cdot\frac{1}{\sqrt[4]{t}\sqrt{1-t}\sqrt{1-\frac19t}};~~~\small{\left[y=\sqrt{t}\right]}\\ &=\frac{1}{2\sqrt{3}}\int_{0}^{1}\mathrm{d}t\,\frac{t^{b-1}\left(1-t\right)^{c-b-1}}{\left(1-zt\right)^{a}};~~~\small{\left[a:=\frac12,b:=\frac14,c:=\frac34,z:=\frac19\right]}\\ &=\frac{1}{2\sqrt{3}}\operatorname{B}{\left(b,c-b\right)}\,{_2F_1}{\left(a,b;c;z\right)}\\ &=\frac{1}{2\sqrt{3}}\operatorname{B}{\left(\frac14,\frac12\right)}\,{_2F_1}{\left(\frac12,\frac14;\frac34;\frac19\right)}.\\ \end{align}$$
Given $\left(a,b,z\right)\in\mathbb{R}_{>0}\times\mathbb{R}_{>0}\times\left(0,1\right)$, the Gauss hypergeometric function obeys the following two functional relations:
$${_2F_1}{\left(a,b;2b;z\right)}=\left(\frac{1+\sqrt{1-z}}{2}\right)^{-2a}\,{_2F_1}{\left(a,a-b+\frac12;b+\frac12;\left(\frac{1-\sqrt{1-z}}{1+\sqrt{1-z}}\right)^{2}\right)};~~~\small{b<a+\frac12},$$
and
$$\begin{align} {_2F_1}{\left(a,b;\frac12;z\right)} &=\frac{\Gamma{\left(a+\frac12\right)}\,\Gamma{\left(b+\frac12\right)}}{2\,\Gamma{\left(\frac12\right)}\,\Gamma{\left(a+b+\frac12\right)}}\bigg{[}{_2F_1}{\left(2a,2b;a+b+\frac12;\frac{1-\sqrt{z}}{2}\right)}\\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+{_2F_1}{\left(2a,2b;a+b+\frac12;\frac{1+\sqrt{z}}{2}\right)}\bigg{]}.\\ \end{align}$$
The following pair of identities are then immediate corollaries of the pair above by setting $b=a$: for $0<a\land0<z<1$,
$${_2F_1}{\left(a,\frac12;a+\frac12;z^{2}\right)}=\left(1+z\right)^{-2a}\,{_2F_1}{\left(a,a;2a;\frac{4z}{\left(1+z\right)^{2}}\right)},$$
and
$$\begin{align} {_2F_1}{\left(a,a;\frac12;z\right)} &=\frac{\left[\Gamma{\left(a+\frac12\right)}\right]^{2}}{2\,\Gamma{\left(\frac12\right)}\,\Gamma{\left(2a+\frac12\right)}}\bigg{[}{_2F_1}{\left(2a,2a;2a+\frac12;\frac{1-\sqrt{z}}{2}\right)}\\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+{_2F_1}{\left(2a,2a;2a+\frac12;\frac{1+\sqrt{z}}{2}\right)}\bigg{]}.\\ \end{align}$$
Continuing with our main evaluation of the integral $\mathcal{I}$, the quadratic transformations of ${_2F_1}$ given above allow us to reduce our integral to standard complete elliptic integrals:
$$\begin{align} \mathcal{I} &=\frac{1}{2\sqrt{3}}\operatorname{B}{\left(\frac14,\frac12\right)}\,{_2F_1}{\left(\frac12,\frac14;\frac34;\frac19\right)}\\ &=\frac{1}{2\sqrt{3}}\operatorname{B}{\left(\frac14,\frac12\right)}\,{_2F_1}{\left(\frac14,\frac12;\frac34;\frac19\right)}\\ &=\frac{1}{2\sqrt{3}}\operatorname{B}{\left(\frac14,\frac12\right)}\,\frac{\sqrt{3}}{2}\,{_2F_1}{\left(\frac14,\frac14;\frac12;\frac34\right)}\\ &=\frac{1}{2\sqrt{3}}\operatorname{B}{\left(\frac14,\frac12\right)}\,\frac{\sqrt{3}}{2}\cdot\frac{\left[\Gamma{\left(\frac34\right)}\right]^{2}}{2\,\Gamma{\left(\frac12\right)}}\bigg{[}{_2F_1}{\left(\frac12,\frac12;1;\frac{2-\sqrt{3}}{4}\right)}+{_2F_1}{\left(\frac12,\frac12;1;\frac{2+\sqrt{3}}{4}\right)}\bigg{]}\\ &=\frac{\Gamma{\left(\frac14\right)}\,\Gamma{\left(\frac34\right)}}{8}\bigg{[}{_2F_1}{\left(\frac12,\frac12;1;\frac{2-\sqrt{3}}{4}\right)}+{_2F_1}{\left(\frac12,\frac12;1;\frac{2+\sqrt{3}}{4}\right)}\bigg{]}\\ &=\frac{\sqrt{2}\,\pi}{8}\bigg{[}{_2F_1}{\left(\frac12,\frac12;1;\frac{2-\sqrt{3}}{4}\right)}+{_2F_1}{\left(\frac12,\frac12;1;\frac{2+\sqrt{3}}{4}\right)}\bigg{]}\\ &=\frac{1}{2\sqrt{2}}\bigg{[}K{\left(\frac{\sqrt{2-\sqrt{3}}}{2}\right)}+K{\left(\frac{\sqrt{2+\sqrt{3}}}{2}\right)}\bigg{]}\\ &=\frac{1}{2\sqrt{2}}\bigg{[}K{\left(\frac{\sqrt{2-\sqrt{3}}}{2}\right)}+K^{\prime}{\left(\frac{\sqrt{2-\sqrt{3}}}{2}\right)}\bigg{]}\\ &=\frac{1}{2\sqrt{2}}\bigg{[}K{\left(\sin{\frac{\pi}{12}}\right)}+K^{\prime}{\left(\sin{\frac{\pi}{12}}\right)}\bigg{]}.\\ \end{align}$$
where here $K{(k)}$ is the complete elliptic integral of the first kind defined as a function of elliptic modulus $k$ by
$$K{(k)}:=\int_{0}^{1}\mathrm{d}x\,\frac{1}{\sqrt{(1-x^{2})(1-k^{2}x^{2})}};~~~\small{-1<k<1},$$
and $K^{\prime}{(k)}$ is the complementary complete elliptic integral of the first kind and is defined in terms of $K$ by
$$K^{\prime}{(k)}:=K{\left(\sqrt{1-k^{2}}\right)}.$$
We can complete our calculation by recognizing that the modulus $k=\sin{\frac{\pi}{12}}$ is in fact the third elliptic integral singular value, $k_{3}$.
Finally, we obtain:
$$\begin{align} \mathcal{I} &=\frac{1}{2\sqrt{2}}\bigg{[}K{\left(\sin{\frac{\pi}{12}}\right)}+K^{\prime}{\left(\sin{\frac{\pi}{12}}\right)}\bigg{]}\\ &=\frac{1}{2\sqrt{2}}\bigg{[}K{\left(k_{3}\right)}+K^{\prime}{\left(k_{3}\right)}\bigg{]}\\ &=\frac{1}{2\sqrt{2}}\bigg{[}1+\frac{K^{\prime}{\left(k_{3}\right)}}{K{\left(k_{3}\right)}}\bigg{]}K{\left(k_{3}\right)}\\ &=\frac{1+\sqrt{3}}{2\sqrt{2}}\,K{\left(k_{3}\right)}\\ &=\frac{1+\sqrt{3}}{2\sqrt{2}}\cdot\frac{\sqrt[4]{3}}{6}\operatorname{B}{\left(\frac12,\frac16\right)}\\ &=\frac{1+\sqrt{3}}{2^{5/2}\,3^{3/4}}\cdot\frac{\Gamma{\left(\frac12\right)}\,\Gamma{\left(\frac16\right)}}{\Gamma{\left(\frac23\right)}}\\ &=\frac{1+\sqrt{3}}{2^{11/6}\,3^{3/4}}\cdot\frac{\pi\,\Gamma{\left(\frac13\right)}}{\left[\Gamma{\left(\frac23\right)}\right]^{2}}\\ &=\frac{\sqrt{3+2\sqrt{3}}}{2^{10/3}}\cdot\frac{\left[\Gamma{\left(\frac13\right)}\right]^{3}}{\pi}.\blacksquare\\ \end{align}$$