It looks indeed simple however

Consider the following Maclaurin series:

$\begin{aligned} \frac 12 \ln \left(\frac {1+x}{1-x} \right) & = x + \frac {x^3}3 + \frac {x^5}5 + \frac {x^7}7 + \cdots \\ \frac x2 \ln \left(\frac {1+x}{1-x} \right) & = x^2 + \frac {x^4}3 + \frac {x^6}5 + \frac {x^8}7 + \cdots \\ \frac {ix}2 \ln \left(\frac {1+ix}{1-ix} \right) & = - x^2 + \frac {x^4}3 - \frac {x^6}5 + \frac {x^8}7 + \cdots \\ \frac x2 \ln \left(\frac {1+x}{1-x} \right) + \frac {ix}2 \ln \left(\frac {1+ix}{1-ix} \right) & = 2 \left(\frac {x^4}3 + \frac {x^8}7 + \frac {x^{12}}{11} + \frac {x^{16}}{15} + \cdots \right) \\ \frac x4 \left(\ln \left(\frac {1+x}{1-x} \right) + i \ln \left(\frac {1+ix}{1-ix} \right) \right) & = \frac {x^4}3 + \frac {x^8}7 + \frac {x^{12}}{11} + \frac {x^{16}}{15} + \cdots & \small \color{#3D99F6} \text{Put }x = i^\frac 12 \\ \frac {i^\frac 12}4 \left(\ln \left(\frac {1+i^\frac 12}{1-i^\frac 12} \right) + i \ln \left(\frac {1+i^\frac 32}{1-i^\frac 32} \right) \right) & = - \frac 13 + \frac 17 - \frac 1{11} + \frac 1{15} + \cdots \end{aligned}$

Therefore, we have:

$\begin{aligned} \sum_{n=0}^\infty \frac {(-1)^n}{(4n+3)} & = - \frac {i^\frac 12}4 \left(\ln \left(\frac {1+i^\frac 12}{1-i^\frac 12} \right) + i \ln \left(\frac {1+i^\frac 32}{1-i^\frac 32} \right) \right) = - \frac {e^{\frac \pi 4i}}4 \left(\ln \left(\frac {1+e^{\frac \pi 4i}}{1-e^{\frac \pi 4i}} \right) + i \ln \left(\frac {1+e^{\frac {3\pi}4i}}{1-e^{\frac {3\pi}4i}} \right) \right) \\ & = -\frac {e^{\frac \pi 4i}}4 \left(\ln \left(\frac {1+ \frac {1+i}{\sqrt 2}}{1- \frac {1+i}{\sqrt 2}} \right) + i \ln \left(\frac {1- \frac {1-i}{\sqrt 2}}{1+\frac {1-i}{\sqrt 2}} \right) \right) = - \frac {e^{\frac \pi 4i}}4 \left(\ln \left(\frac {\sqrt 2+1+i}{\sqrt 2-1-i} \right) + i \ln \left(\frac {\sqrt 2-1+i}{\sqrt 2+1-i} \right) \right) \\ & = -\frac {(1+i)\left(\ln i - \ln (\sqrt 2-1) + i (\ln (\sqrt 2 - 1) + \ln i) \right)}{4\sqrt 2} \\ & = \frac {\pi + 2\ln(\sqrt 2-1)}{4\sqrt 2} \end{aligned}$

Therefore, $a+b+c = 2+1+4 = \boxed 7$ .

My 1st approach Let's make some observation for the entries of $n=0,1,2,3\cdots$ . $\Phi=\sum_{n=0}^{\infty} \dfrac{(-1)^n}{4n+1} = \dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{11}-\dfrac{1}{15}+ \dfrac{1}{19}-\dfrac{1}{23} +\cdots$ Here now I will define two sums $\Phi_1$ and $\Phi_2$ such that $\Phi= \Phi_1+\Phi_2$ . So let's recall for $|x|< 1$ we have $\Phi_1=2\sum_{j=0}^{\infty} \dfrac{x^{2j+1}}{2j+1} = \ln\left(\dfrac{1+x}{1-x}\right)$ Now let me set $x = i^2e^{i\frac{\pi}{4}}$ and plug it to $\Phi_1$ . we can see $\Re\Phi_1$ $\begin{aligned} \Re \ln\left(\dfrac{1-e^{i\frac{\pi}{4}}}{1+e^{i\frac{\pi}{4}}}\right)&= \sqrt 2\left(-1 +\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}+\cdots \right)\\\Re\ln\left(\dfrac{\sqrt 2 -1 -i}{\sqrt 2 +1 +i}\right)& =\sqrt 2\left( -1 +\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}+\cdots\right )\\ \Phi_1& = \dfrac{1}{\sqrt 2} \ln(\sqrt 2-1)\end{aligned}$ One the similar fashion set $x=i^2 e^{i\frac{5\pi}{4}}$ for the next series namely $\Phi_2$ and we can observe that $\Im\Phi_2$ . $\begin{aligned} \Im \ln\left(\dfrac{1+e^{i\frac{\pi}{4}}}{1-e^{i\frac{\pi}{4}}}\right)& =\sqrt 2\left(1 +\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots \right)\\\Im\ln\left(\dfrac{\sqrt 2 -1 -i}{\sqrt 2 +1 +i}\right)& =\sqrt 2\left( 1 +\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots\right )\\ \Phi_1& = \dfrac{\pi}{2 \sqrt 2}\end{aligned}$ Now adding $\begin{aligned} \Phi_1+\Phi_2& = 2\left( \dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{11}-\dfrac{1}{15}+ \dfrac{1}{19}-\dfrac{1}{23} +\cdots \right)\\ \sum_{n=0}^{\infty}\dfrac{(-1)^n}{4n+3} & = \dfrac{1}{2}(\Phi_1+\Phi_2)= \dfrac{\pi+2\ln(\sqrt 2-1)}{4\sqrt 2}\end{aligned}$

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2nd approch

Let sum be $\Phi$ $\sum_{n=0}^{\infty}\dfrac{(-1)^n}{4n+3} = \sum_{n=0}^{\infty} \int_{0}^{1} x^{4n+2}\,dx = \int_{0}^{1}\dfrac{x^2}{1+x^4}\,dx$ Since $\int\dfrac{x^2}{1+x^4}\,dx = \dfrac{1}{2\sqrt 2}\tan^{-1}\left(\dfrac{x^2+1}{\sqrt 2 x}\right)+\dfrac{1}{4\sqrt2}\log\left(\dfrac{x^2+1-\sqrt 2 x}{x^2+1+\sqrt 2 x}\right)+ C$ setting the limit we obtain $\Phi =\dfrac{\pi +2\ln(\sqrt 2-1)}{4\sqrt 2}$

Note: $\begin{aligned} \int \dfrac{x^2}{1+x^4}\,dx & =\dfrac{1}{2}\int\left(\dfrac{x^2+1}{1+x^4}+ \dfrac{x^2-1}{1+x^4} \right)\,dx\\& = \dfrac{1}{2}\int\left(\dfrac{1+\frac{1}{x^2} }{x^2+\frac{1}{x^2} }+ \dfrac{1-\frac{1}{x^2} }{x^2+\frac{1}{x^2} } \right)\,dx \\ & =\dfrac{1}{2}\int\left(\dfrac{1+\frac{1}{x^2} }{\,(x-\frac{1}{x})^2 +2 }+ \dfrac{1-\frac{1}{x^2} }{{\,(x+\frac{1}{x})^2 -2} } \right)\,dx\end{aligned}$ Substitute $x -\frac{1}{x} = t \implies 1+\frac{1}{x^2}=\,dt$ and $x +\frac{1}{x} = u\implies 1-\frac{1}{x^2}=\,du$ . We have then $\begin{aligned} \int \dfrac{x^2}{1+x^4} & = \dfrac{1}{2}\int \left( \dfrac{\,dt}{ t^2 +2} +\dfrac{\,du}{u^2-2}\right) \\& = \left(\dfrac{1}{2\sqrt 2}\tan^{-1} \left(\dfrac{x^2-1}{2\sqrt x}\right)+\dfrac{1}{4\sqrt 2}\ln\left[\dfrac{x^2+1-\sqrt 2 x}{x^2+1+\sqrt 2 x}\right]\right)+C \end{aligned}$