For $h>0$, I want to evaluate the integral
$$I = \int_{-\infty}^\infty \frac{x^2 \sinh^2x}{(\cosh x)^{4h+2}} dx.$$
Mathematica says that the answer is
$$16^{h-1} \left(\frac{\, _4F_3(2 h,2 h,2 h,4 h+2;2 h+1,2 h+1,2
h+1;-1)}{h^3}-\frac{16 \, _4F_3(2 h+1,2 h+1,2 h+1,4 h+2;2 h+2,2 h+2,2
h+2;-1)}{(2 h+1)^3}+\frac{\, _4F_3(2 h+2,2 h+2,2 h+2,4 h+2;2 h+3,2 h+3,2
h+3;-1)}{(h+1)^3}\right).$$
Can I simplify the above answer?
The motivation/plausibility for the simplification is as follows. For a similar integral
$$I' = \int_{-\infty}^\infty \frac{\sinh^2x}{(\cosh x)^{4h+2}} dx,$$
Mathematica gives an answer
$$\frac{\, _2F_1(1,-2 h-1;2 h+1;-1)-\frac{\sqrt{\pi } h \Gamma (2 h+1)}{\Gamma
\left(2 h+\frac{3}{2}\right)}-1}{2 h},$$
but a much simpler answer is available (see Two integrals whose answer involves gamma function):
$$ \frac{\sqrt\pi \Gamma(2h)}{2\Gamma\left(2h + \frac32\right)}.$$
A simplification of the hypergeometric function occurred in $I'$, and I suspect that a similar thing may be also possible for $I$.
Best Answer
Define the function $\mathcal{I}:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ via the improper integral
$$\mathcal{I}{\left(p\right)}:=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{x^{2}\sinh^{2}{\left(x\right)}}{\left[\cosh{\left(x\right)}\right]^{2p+2}}.\tag{1}$$
Suppose $p\in\mathbb{R}_{>0}$. We begin by using an inverse hyperbolic substitution to rewrite the integral as
$$\begin{align} \mathcal{I}{\left(p\right)} &=\int_{-\infty}^{\infty}\mathrm{d}\xi\,\frac{\xi^{2}\sinh^{2}{\left(\xi\right)}}{\left[\cosh{\left(\xi\right)}\right]^{2p+2}}\\ &=2\int_{0}^{\infty}\mathrm{d}\xi\,\frac{\xi^{2}\sinh^{2}{\left(\xi\right)}}{\left[\cosh{\left(\xi\right)}\right]^{2p+2}};~~~\small{\text{even symmetry}}\\ &=2\int_{0}^{\infty}\mathrm{d}\xi\,\xi^{2}\tanh^{2}{\left(\xi\right)}\left[\operatorname{sech}^{2}{\left(\xi\right)}\right]^{p}\\ &=2\int_{0}^{\infty}\mathrm{d}\xi\,\xi^{2}\tanh^{2}{\left(\xi\right)}\left[1-\tanh^{2}{\left(\xi\right)}\right]^{p}\\ &=2\int_{0}^{1}\mathrm{d}y\,y^{2}\left(1-y^{2}\right)^{p-1}\left[\operatorname{artanh}{\left(y\right)}\right]^{2};~~~\small{\left[\xi=\operatorname{artanh}{\left(y\right)}\right]}\\ &=2\int_{0}^{1}\mathrm{d}y\,y^{2}\left[\left(1-y\right)\left(1+y\right)\right]^{p-1}\left[\frac12\ln{\left(\frac{1+y}{1-y}\right)}\right]^{2}\\ &=\frac12\int_{0}^{1}\mathrm{d}y\,\left(\frac{1-y}{1+y}\right)^{p-1}\left(1+y\right)^{2p-2}y^{2}\ln^{2}{\left(\frac{1-y}{1+y}\right)}\\ &=\frac12\int_{1}^{0}\mathrm{d}x\,\frac{(-2)}{\left(1+x\right)^{2}}x^{p-1}\left(\frac{2}{1+x}\right)^{2p-2}\left(\frac{1-x}{1+x}\right)^{2}\ln^{2}{\left(x\right)};~~~\small{\left[y=\frac{1-x}{1+x}\right]}\\ &=2^{2p-2}\int_{0}^{1}\mathrm{d}x\,\frac{x^{p-1}\left(1-x\right)^{2}}{\left(1+x\right)^{2p+2}}\ln^{2}{\left(x\right)}.\tag{2}\\ \end{align}$$
Next, substituting $\frac{x}{1+x}=y$ and taking taking advantage of symmetry about the unit interval, we find
$$\begin{align} \mathcal{I}{\left(p\right)} &=2^{2p-2}\int_{0}^{1}\mathrm{d}x\,\frac{x^{p-1}\left(1-x\right)^{2}}{\left(1+x\right)^{2p+2}}\ln^{2}{\left(x\right)}\\ &=2^{2p-2}\int_{0}^{1}\mathrm{d}x\,\left(x^{p-1}-2x^{p}+x^{p+1}\right)\left(\frac{1}{1+x}\right)^{2p+2}\ln^{2}{\left(x\right)}\\ &=2^{2p-2}\int_{0}^{\frac12}\mathrm{d}y\,\frac{1}{\left(1-y\right)^{2}}\left[\left(\frac{y}{1-y}\right)^{p-1}-2\left(\frac{y}{1-y}\right)^{p}+\left(\frac{y}{1-y}\right)^{p+1}\right]\\ &~~~~~\times\left(1-y\right)^{2p+2}\ln^{2}{\left(\frac{y}{1-y}\right)};~~~\small{\left[x=\frac{y}{1-y}\right]}\\ &=2^{2p-2}\int_{0}^{\frac12}\mathrm{d}y\,\left[\left(1-y\right)^{2}-2y\left(1-y\right)+y^{2}\right]y^{p-1}\left(1-y\right)^{p-1}\ln^{2}{\left(\frac{y}{1-y}\right)}\\ &=2^{2p-2}\int_{0}^{\frac12}\mathrm{d}y\,\left(1-2y\right)^{2}y^{p-1}\left(1-y\right)^{p-1}\ln^{2}{\left(\frac{y}{1-y}\right)}\\ &=2^{2p-2}\int_{\frac12}^{1}\mathrm{d}t\,\left(1-2t\right)^{2}t^{p-1}\left(1-t\right)^{p-1}\ln^{2}{\left(\frac{t}{1-t}\right)};~~~\small{\left[y=1-t\right]}\\ &=2^{2p-3}\int_{0}^{1}\mathrm{d}t\,\left(1-2t\right)^{2}t^{p-1}\left(1-t\right)^{p-1}\ln^{2}{\left(\frac{t}{1-t}\right)}.\tag{3}\\ \end{align}$$
Define the function $\mathcal{J}:\mathbb{R}_{>0}^{2}\rightarrow\mathbb{R}$ via the definite integral
$$\mathcal{J}{\left(a,b\right)}:=\int_{0}^{1}\mathrm{d}t\,t^{a-1}\left(1-t\right)^{b-1}\ln^{2}{\left(\frac{t}{1-t}\right)}.\tag{4}$$
Given $\left(a,b\right)\in\mathbb{R}_{>0}^{2}$, we can calculate $\mathcal{J}$ in terms of derivatives of the beta function:
$$\begin{align} \mathcal{J}{\left(a,b\right)} &=\int_{0}^{1}\mathrm{d}t\,t^{a-1}\left(1-t\right)^{b-1}\ln^{2}{\left(\frac{t}{1-t}\right)}\\ &=\int_{0}^{1}\mathrm{d}t\,t^{a-1}\left(1-t\right)^{b-1}\left[\ln{\left(t\right)}-\ln{\left(1-t\right)}\right]^{2}\\ &=\int_{0}^{1}\mathrm{d}t\,t^{a-1}\left(1-t\right)^{b-1}\left[\ln^{2}{\left(t\right)}-2\ln{\left(t\right)}\ln{\left(1-t\right)}+\ln^{2}{\left(1-t\right)}\right]\\ &=\int_{0}^{1}\mathrm{d}t\,t^{a-1}\left(1-t\right)^{b-1}\ln^{2}{\left(t\right)}-2\int_{0}^{1}\mathrm{d}t\,t^{a-1}\left(1-t\right)^{b-1}\ln{\left(t\right)}\ln{\left(1-t\right)}\\ &~~~~~+\int_{0}^{1}\mathrm{d}t\,t^{a-1}\left(1-t\right)^{b-1}\ln^{2}{\left(1-t\right)}\\ &=\frac{\partial^{2}}{\partial a^{2}}\int_{0}^{1}\mathrm{d}t\,t^{a-1}\left(1-t\right)^{b-1}-2\frac{\partial^{2}}{\partial a\partial b}\int_{0}^{1}\mathrm{d}t\,t^{a-1}\left(1-t\right)^{b-1}\\ &~~~~~+\frac{\partial^{2}}{\partial b^{2}}\int_{0}^{1}\mathrm{d}t\,t^{a-1}\left(1-t\right)^{b-1}\\ &=\frac{\partial^{2}}{\partial a^{2}}\operatorname{B}{\left(a,b\right)}-2\frac{\partial^{2}}{\partial a\partial b}\operatorname{B}{\left(a,b\right)}+\frac{\partial^{2}}{\partial b^{2}}\operatorname{B}{\left(a,b\right)}\\ &=\left[\left(\psi{\left(a\right)}-\psi{\left(a+b\right)}\right)^{2}+\psi_{1}{\left(a\right)}-\psi_{1}{\left(a+b\right)}\right]\operatorname{B}{\left(a,b\right)}\\ &~~~~~-2\left[\left(\psi{\left(a\right)}-\psi{\left(a+b\right)}\right)\left(\psi{\left(b\right)}-\psi{\left(a+b\right)}\right)-\psi_{1}{\left(a+b\right)}\right]\operatorname{B}{\left(a,b\right)}\\ &~~~~~+\left[\left(\psi{\left(b\right)}-\psi{\left(a+b\right)}\right)^{2}+\psi_{1}{\left(b\right)}-\psi_{1}{\left(a+b\right)}\right]\operatorname{B}{\left(a,b\right)}\\ &=\left[\left(\psi{\left(a\right)}-\psi{\left(b\right)}\right)^{2}+\psi_{1}{\left(a\right)}+\psi_{1}{\left(b\right)}\right]\operatorname{B}{\left(a,b\right)}.\tag{5}\\ \end{align}$$
Continuing from where we left off in $(2)$, we obtain the following closed form expression for $\mathcal{I}$:
$$\begin{align} \mathcal{I}{\left(p\right)} &=2^{2p-3}\int_{0}^{1}\mathrm{d}t\,\left(1-2t\right)^{2}t^{p-1}\left(1-t\right)^{p-1}\ln^{2}{\left(\frac{t}{1-t}\right)}\\ &=2^{2p-3}\int_{0}^{1}\mathrm{d}t\,\left[t^{2}-2t\left(1-t\right)+\left(1-t\right)^{2}\right]t^{p-1}\left(1-t\right)^{p-1}\ln^{2}{\left(\frac{t}{1-t}\right)}\\ &=2^{2p-3}\left[\mathcal{J}{\left(p+2,p\right)}-2\mathcal{J}{\left(p+1,p+1\right)}+\mathcal{J}{\left(p,p+2\right)}\right]\\ &=2^{2p-2}\left[\mathcal{J}{\left(p+2,p\right)}-\mathcal{J}{\left(p+1,p+1\right)}\right]\\ &=2^{2p-2}\left[\left(\psi{\left(p+2\right)}-\psi{\left(p\right)}\right)^{2}+\psi_{1}{\left(p+2\right)}+\psi_{1}{\left(p\right)}\right]\operatorname{B}{\left(p+2,p\right)}\\ &~~~~~-2^{2p-2}\left[2\psi_{1}{\left(p+1\right)}\right]\operatorname{B}{\left(p+1,p+1\right)}\\ &=2^{2p-2}\left[\left(\frac{1}{p+1}+\frac{1}{p}\right)^{2}-\frac{1}{\left(p+1\right)^{2}}-\frac{1}{p^{2}}+2\psi_{1}{\left(p\right)}\right]\operatorname{B}{\left(p+2,p\right)}\\ &~~~~~+2^{2p-2}\left[\frac{2}{p^{2}}-2\psi_{1}{\left(p\right)}\right]\operatorname{B}{\left(p+1,p+1\right)}\\ &=2^{2p-1}\left[\frac{1}{p\left(p+1\right)}+\psi_{1}{\left(p\right)}\right]\operatorname{B}{\left(p+2,p\right)}\\ &~~~~~+2^{2p-1}\left[\frac{1}{p^{2}}-\psi_{1}{\left(p\right)}\right]\operatorname{B}{\left(p+1,p+1\right)}\\ &=2^{2p-1}\left[\frac{1}{p\left(p+1\right)}+\psi_{1}{\left(p\right)}\right]\frac{\left(p+1\right)}{2\left(2p+1\right)}\operatorname{B}{\left(p,p\right)}\\ &~~~~~+2^{2p-1}\left[\frac{1}{p^{2}}-\psi_{1}{\left(p\right)}\right]\frac{p}{2\left(2p+1\right)}\operatorname{B}{\left(p,p\right)}\\ &=\frac{2^{2p-2}}{\left(2p+1\right)}\left[\frac{2}{p}+\psi_{1}{\left(p\right)}\right]\operatorname{B}{\left(p,p\right)}\\ &=\frac{1}{2\left(2p+1\right)}\left[\frac{2}{p}+\psi_{1}{\left(p\right)}\right]\operatorname{B}{\left(p,\frac12\right)}.\blacksquare\\ \end{align}$$