Evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}$ How to prove that 

$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$

where $H_n=1+\frac1{2}+\frac1{3}+...+\frac1{n}$ is the $n$th harmonic number.
This sum was proposed by Cornel and I solved it using integration but can we solve it using series manipulation? 
The integral representation of the sum is $\ \displaystyle\frac12\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx$ in case it is needed.
 A: \begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}&=\sum_{n=1}^\infty(-1)^{n-1} H_n\int_0^1\frac12x^{2n}\ln^2 x\ dx\\
&=-\frac12\int_0^1\ln^2x\sum_{n=1}^\infty(-x^2)H_n\\
&=\frac12\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx\tag{1}
\end{align}

\begin{align}
I&=\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx\\
&=\int_0^\infty\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx-\underbrace{\int_1^\infty\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx}_{\large x\mapsto1/x}\\
&=\underbrace{\int_0^\infty\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx}_{\large x^2\mapsto x}-I+2\int_0^1\frac{\ln^3x}{1+x^2}\ dx\\
2I=&\frac18\int_0^\infty\frac{\ln^2x\ln(1+x)}{\sqrt{x}(1+x)}\ dx+2(-6\beta(4))\\
I&=\frac1{16}\lim_{a\ \mapsto1/2\\b\ \mapsto1/2}\frac{-\partial^3}{\partial a^2\partial b}\text{B}(a,b)-6\beta(4)\\
&=\frac1{16}(14\pi\zeta(3)+2\pi^3\ln2)-6*\frac1{768}\left(\psi^{(3)}\left(\frac14\right)-8\pi^4\right)\\
&=\frac{7\pi}{8}\zeta(3)+\frac{\pi^3}{8}\ln2-\frac1{128}\left(\psi^{(3)}\left(\frac14\right)-8\pi^4\right)\tag{2}
\end{align}

Plugging $(2)$ in $(1)$ we get

$$\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$


Notes: 
$\displaystyle\beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}\ $ is the Dirichlet beta function and the value of $\beta(4)$ can be found here.
$\displaystyle\text{B}(a,b)=\int_0^\infty\frac{x^{a-1}}{(1+x)^{a+b}}\ dx$ is beta function.
A: $$\int_{0}^{\infty }\frac{ln(1+x^2)ln^2x}{1+x^2}dx\\
\\
let\ I(a)=\int_{0}^{\infty }\frac{ln^2(x)ln(1+a^2.x^2)}{1+x^2}\\
\\
\therefore I'(a)=\int_{0}^{\infty }\frac{2ax^2ln^2(x)}{(1+a^2x^2)(1+x^2)}dx=\frac{2a}{1-a^2}\int_{0}^{\infty }\frac{ln^2(x)}{1+a^2x^2}-\frac{ln^2}{1+x^2}dx\\
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let\ \ G=\int_{0}^{\infty }\frac{ln^2(x))}{1+a^2x^2}dx\ \ \ \ ,\ \ but \ we \ know\ \\
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G(a)=\int_{0}^{\infty }\frac{x^p}{(1+x^2)a^{p+1}}dx=\frac{\pi }{2}\frac{sec\frac{\pi p}{2}}{a^{p+1}}\\
\\
\therefore \frac{\partial^2 G(a)}{\partial^2 p}=\frac{\pi }{2}[tan(\frac{\pi p}{2})sec(\frac{\pi p}{2})+ln(a)sec(\frac{\pi p}{2})ln(a)a^{-p-1}]+\frac{1}{a^{p+1}}[\frac{\pi ^{2}}{4}tan^2(\frac{\pi p}{2})sec(\frac{\pi p}{2})+sec^3(\frac{\pi p}{2})-\frac{\pi }{2}tan(\frac{\pi p}{2})sec(\frac{\pi p}{2})ln(a)]\\$$
$$now\ \ take\ \ p=0\ \\
\\
\therefore \frac{\partial ^2 G(a)}{\partial p^2}_{p=0}=\frac{\pi }{2}[\frac{ln^2(a)}{a}+\frac{\pi ^{2}}{4a}]=\int_{0}^{\infty }\frac{ln^2(x)}{1+a^2x^2}dx\ ,\ \ \ \ take\ a=1\\
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\therefore \int_{0}^{1 }\frac{ln^2(x)dx}{1+x^2}=\frac{\pi ^{3}}{8}, \ \ \ \ \ now\ going\ to \ I\\
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\therefore I(a)=\int_{0}^{1}(\frac{ln^2(x)}{1+a^2x^2}-\frac{ln^2(x)}{1+x^2})dx=\frac{\pi ^{3}}{8}(\frac{1-a}{a})+\frac{\pi ln^2(a)}{2a}\\
\\
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\therefore I'(a)=\frac{2a}{1-a^2}(\frac{\pi ^{3}}{8}(\frac{1-a}{a})+\frac{\pi ln^2(a)}{2a})$$
$$\therefore I(1)=\frac{\pi ^{3}}{4}\int_{0}^{1}\frac{dx}{1+x}+\pi \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx\\
\\
\therefore \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx=\frac{1}{2}\int_{0}^{1}\frac{ln^2(x)}{1-x}dx+\frac{1}{2}\int_{0}^{1}\frac{ln^2(x)}{1+x}dx\\
\\
\therefore \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx=\frac{1}{2}[-Ln^2(x)ln(1-x)\tfrac{1}{0}+ln^2(x)ln(1+x)\tfrac{1}{0}+2\int_{0}^{1}\frac{ln(1-x)lnx}{x}dx-2\int_{0}^{1}\frac{ln(1+x)lnx}{x}dx]\\
\\
\\
\therefore \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx=\frac{7}{4}\zeta (3)\\
\\
\therefore I=\int_{0}^{\infty }\frac{ln^2(x)ln(1+x^2)}{1+x^2}dx=\frac{\pi ^{3}}{4}ln(2)+\frac{7\pi }{4}\zeta (3)$$
A: A nice generalization proved here:
$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{(2n+1)^{2a+1}}=(2a+1)\beta(2a+2)-\frac{\ln(2)|E_{2a}|}{(2a)!}\left(\frac{\pi}{2}\right)^{2a+1}$$
$$-\frac12\left(\frac{\pi}{2}\right)^{2a+1}\sum_{k=1}^{a} \frac{|E_{2a-2k}|}{(2a-2k)!}{\pi^{-2k}}(2^{2k+1}-1)\zeta(2k+1),$$
where $E_r$ is the Euler numbers.
Set $a=1,2,3 $ we get
$$\sum_{n=1}^\infty\frac{(-1)^n H_n}{(2n + 1)^3}=3\beta(4)-\frac{7\pi}{16}\zeta(3)-\frac{\pi^3}{16}\ln(2)$$
$$\sum_{n=1}^\infty\frac{(-1)^n H_n}{(2n + 1)^5}=5\beta(6)-\frac{31\pi}{64}\zeta(5)-\frac{7\pi^3}{128}\zeta(3)-\frac{5\pi^5}{768}\ln(2)$$
$$\sum_{n=1}^\infty\frac{(-1)^n H_n}{(2n + 1)^7}=7\beta(8)-\frac{127\pi}{256}\zeta(7)-\frac{31\pi^3}{512}\zeta(5)-\frac{35\pi^5}{6144}\zeta(3)-\frac{61\pi^7}{92160}\ln(2)$$
A: very nice solution Ali , 
this integral $$I=\int_{0}^{\infty }\frac{ln(1+x^2)ln^2x}{1+x^2}dx$$
i have another approach to evaluate 
i will post it
