Beautiful closed form!

My approach

Note that given summation can be further simplified to $\sum_{n=0}^{\infty} \left(\dfrac{(-1)^n}{(2n+1)} + \dfrac{(-1)^n}{(3m+1)(3m+2)} +\dfrac{(-1)^n}{(4m+1)(4m+2)(4m+3)}\right)$ Now let us make individual evaluation of the each sum. $\Phi_{1} =\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)} =\int_{0}^{1} \dfrac{1}{1+x^2}\,dx =\dfrac{\pi}{4}$ Similarly, $\begin{aligned} \Phi_{2} & = \sum_ {n=0}^{\infty} \dfrac{(-1)^n}{(3m+1)(3m+2)} \\&= \sum_{n=0}^{\infty}\left(\dfrac{(-1)^n}{(3m+1)} -\dfrac{(-1)^n}{(3m+2)}\right)\\& =\sum_{n=0}^{\infty}\left(\dfrac{(-1)^n}{n+1}-\dfrac{1}{3}\dfrac{(-1)^n}{n+1}\right)= \dfrac{2}{3}\sum_{n=0}^{\infty} \dfrac{(-1)^n}{n+1} = \dfrac{2}{3}\int_{0}^{1}\dfrac{1}{1+x}\,dx = \dfrac{2}{3}\log2 \end{aligned}$ Finally, $\Phi_{3} = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(4n+1)(4n+2)(4n+3)}$ Since $\dfrac{(-1)^n}{(4n+1)(4n+2)(4n+3)} = \dfrac{(-1)^n}{2}\left(\dfrac{1}{4n+1} -\dfrac{2}{4n+2} +\dfrac{1}{4n+3}\right)$ Now note that above decompose portion can be further written as $\begin{aligned} \Phi_3& = \dfrac{1}{2}\sum_{n=0}^{\infty} \int_{0}^{1} \left( x^{4n} + x^{4n+2}\right)\,dx-\dfrac{1}{2}\sum_{n=0}^{\infty} \dfrac{(-1)^n}{2n+1}\\& =\dfrac{1}{2}\int_{0}^{1}\left(\dfrac{1}{1+x^4}+\dfrac{x^2}{1+x^4}\right)\,dx-\dfrac{1}{2}\left(\sum_{n=0}^{\infty}\dfrac{(-1)^n}{2n+1}\right) \\& = \dfrac{1}{2}\int_{0}^{1}\left(\dfrac{1+x^2}{1+x^4}-\dfrac{1}{(1+x^2)}\right)\,dx\\& = \dfrac{I}{2}-\dfrac{\pi}{2\cdot 4} = \dfrac{\pi}{4\sqrt 2} -\dfrac{\pi}{8}\end{aligned}$ Thus we have then $\Phi_1 +\Phi_2 +\Phi_3 = \dfrac{2}{3}\log2 +\dfrac{\pi}{8}(1+\sqrt 2)$ Making $a+b+c+d=14$ .

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Note $I = \int\dfrac{1+x^2}{1+x^4} \,dx=\dfrac{1}{2}\int_{0}^{1}\dfrac{2(x^2+1)}{x^4+1}\,dx$ since $x^4+1= (x^2+\sqrt 2 x+1)(x^2-\sqrt 2 x+1)$ Hence $\begin{aligned} I & = \int\dfrac{1}{2}\left(\dfrac{1}{x^2+\sqrt 2x+1} +\dfrac{1}{x^2-\sqrt 2x +1}\right)\,dx\\&= \dfrac{\sqrt2 }{2}\left(\tan ^{-1} (x\sqrt 2+1)+\tan ^{-1}(x\sqrt 2 -1)\right)+C \end{aligned}$ setting the limit gives $= \dfrac{1}{\sqrt 2} \tan ^{-1}\left(\dfrac{\sqrt 2 x}{1- x^2}\right) =\dfrac{\pi}{2\sqrt 2}$