How to Evaluate $\int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4}dx$?

Question

How to evaluate

$$ \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx \approx -0.162858 \tag{1}$$

The integral arises in the computation of

$$\left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)\left(\sum_{n=1}^{\infty} \frac{(-1)^n}{4n-1}\right)$$ as

$$ \scriptsize{\frac{\pi \ln(2)}{4\sqrt{2}} + \frac{\ln(2) \ln(3-2\sqrt{2})}{4\sqrt{2}} -3\ln(2) + \frac{\pi}{2}= \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx + \int_{0}^{1} \frac{x^{1/4}}{2(1+x)}\left(\tan^{-1}(x^{1/4}) - \tanh^{-1}(x^{1/4}) \right)}dx $$

A similar Integral

$$ \int_{0}^{1}\left( \frac{x^2 \ln(2)}{x^4-1} - \frac{x^2 \ln(1+x^4)}{x^4-1}\right)dx = C-\frac{\pi^2}{16}+\frac{\ln^2(\sqrt{2}-1)}{4}+\frac{\pi \ln (\sqrt{2}-1)}{4} \tag{2} $$

Where $ C $ = Catalan Constant

Unfortunately the same techniques I used to evaluate $(2)$ have not worked for $(1)$. I know only for integration - By parts, U-Sub, and using Taylor Series as well as Mathematica.

Q = Is there a closed form for Integral $(1)$?

EDIT

$$ (1) = \frac{-\pi^2}{8\sqrt{2}} + \frac{\pi \ln(8)}{4\sqrt{2}} +\frac{\ln(8) \ln(3-2\sqrt{2})}{4\sqrt{2}} -\frac{\pi \ln(3-2\sqrt{2})}{8\sqrt{2}} + 4\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} $$

$$\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} = \frac{1}{64}\left(\psi^{(1)}\left(\frac{7}{8}\right)-\psi^{(1)}\left(\frac{3}{8}\right)\right) + W $$

Where $W$ is some value.

Answer 1

$$I=\int_0^1 \frac{x^2\ln(1-x^4)}{1+x^4}dx=\frac{5\pi^2}{48\sqrt 2}+\frac{\pi\ln 2}{4\sqrt 2}+\frac{\pi\ln(\delta_S)}{2\sqrt 2}+\frac{\ln^2(\delta_S)}{2\sqrt 2}-\frac{\ln 2 \ln(\delta_S)}{2\sqrt 2 }$$ $$+\frac{3\operatorname{Li}_2\left(-(\delta_S)^2\right)}{8\sqrt 2}-\frac{3\operatorname{Li}_2\left(-1/(\delta_S)^2\right)}{8\sqrt 2}+\frac{\operatorname{Li}_2\left(1/(\delta_S)^4\right)}{8\sqrt 2}+\frac{\operatorname{Ti}_2\left(-\delta_S\right)}{\sqrt 2}-\frac{\operatorname{Ti}_2\left(1/\delta_S\right)}{\sqrt 2}$$ Where $\delta_S=1+\sqrt 2\, $ is the silver ratio, $\operatorname{Li}_2(x)$ is the dilogarithm and $\operatorname{Ti}_{2}(x)$ is the inverse tangent integral.

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To obtain this closed form we'll start by moving the bounds from $(0,1)$ to $(0,\infty)$. That's simply because afterwards the plan is to differentiate under the integral sign and usually having the bounds as $(0,\infty)$ simplifies the final result a lot when dealing with rational functions. $$I=\frac12\int_{-1}^1\frac{x^2\ln(1-x^4)}{1+x^4}dx\overset{x\to\frac{1-x}{1+x}}=\frac12\int_0^\infty \frac{(1-x)^2}{1+6x^2+x^4}\ln\left(\frac{8x(1+x^2)}{(1+x)^4}\right)dx$$ $$=\frac{3\pi \ln 2}{4\sqrt 2}-\frac{3\ln 2\ln(1+\sqrt 2)}{2\sqrt 2}+\frac12X-2Y$$ $$X=\int_0^\infty \frac{(1-x)^2\ln\left(1+x^2\right)}{1+6x^2+x^4}dx;\quad Y=\int_0^\infty \frac{(1-x)^2\ln\left(1+x\right)}{1+6x^2+x^4}dx$$ Also one integral from above vanished via the substitution $x\to \frac{1}{x}$: $$\int_0^\infty \frac{(1-x)^2\ln x}{1+6x^2+x^4}dx=0$$ As mentioned above, to evaluate $X$ and $Y$ we'll start by differentiating under the integral sign and even though the rational functions from above are somewhat more complicated than if we kept the original ones, we expect to have an easier time when integrating back.

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$$X(a)=\int_0^\infty\frac{(1-x)^2\ln\left(1+ax^2\right)}{1+6x^2+x^4}dx\Rightarrow X'(a)=\int_0^\infty \frac{(1-x)^2x^2}{(1+6x^2+x^4)(1+ax^2)}dx$$ $$=\frac{2}{1-6a+a^2}\int_0^\infty\left(\frac{ax}{1+ax^2}-\frac{ax+x^3}{1+6x^2+x^4}\right)dx$$ $$+\frac{1-a}{1-6a+a^2}\int_0^\infty\frac{1}{1+ax^2}dx-\frac{1}{1-6a+a^2}\int_0^\infty\frac{1-a-(a-5)x^2}{1+6x^2+x^4}dx$$ $$=\frac{\ln a}{1-6a+a^2}-\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{a-3}{1-6a+a^2}+\frac{\pi}{2\sqrt a}\frac{1-a}{1-6a+a^2}+\frac{\pi}{2\sqrt 2}\frac{a-3}{1-6a+a^2}$$ $$\Rightarrow X=\int_0^1\left(\frac{\ln a}{1-6a+a^2}-\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{a-3}{1-6a+a^2}\right)da+\frac{\pi\ln 2}{2\sqrt 2}+\frac{\pi\ln (1+\sqrt 2)}{2\sqrt 2}$$

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$$Y(a)=\int_0^\infty\frac{(1-x)^2\ln\left(1+ax\right)}{1+6x^2+x^4}dx\Rightarrow Y'(a)=\int_0^\infty \frac{(1-x)^2x}{(1+6x^2+x^4)(1+ax)}dx$$ $$=\int_0^\infty\left(\frac{(1+5a^2-2a^3)x+(1+a)^2x^3}{(1+6a^2+a^4)(1+6x^2+x^4)}-\frac{a(1+a)^2}{(1+6a^2+a^4)(1+ax)}\right)dx$$ $$+\int_0^\infty \left(\frac{a(1+a)^2-(2-5a-a^3)x^2}{(1+6a^2+a^4)(1+6x^2+x^4)}\right)dx$$ $$=-\frac{(1+a)^2\ln a}{1+6a^2+a^4}-\frac{a(3-a+a^2)+1}{1+6a^2+a^4}\frac{\ln(1+\sqrt 2)}{\sqrt 2}+\frac{\pi}{2\sqrt 2}\frac{a(3+a+a^2)-1}{1+6a^2+a^4}$$ $$\Rightarrow Y=-\int_0^1\frac{(1+a)^2\ln a}{1+6a^2+a^4}da-\frac{\pi\ln(1+\sqrt 2)}{8\sqrt 2}-\frac{3\ln 2\ln(1+\sqrt 2)}{4\sqrt 2}-\frac{\pi^2}{16\sqrt 2}+\frac{3\pi\ln 2}{8\sqrt 2}$$

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Let's reorder things a little, just to see what we have so far (combining with the $X$ and $Y$ integrals): $$I=\frac{\pi^2}{8\sqrt 2}+\frac{\pi\ln 2}{4\sqrt 2}+\frac{\pi\ln(1+\sqrt 2)}{2\sqrt 2}+2\int_0^1\frac{(1+x)^2\ln x}{1+6x^2+x^4}dx$$ $$+\frac12\int_0^1\left(\frac{\ln x}{1-6x+x^2}+\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{3-x}{1-6x+x^2}\right)dx$$

So all that's left is to evaluate the remaining two integrals. For the first one we have:

$$\int_0^1\frac{(1+x)^2\ln x}{1+6x^2+x^4}dx=\int_0^1\left(\frac{1}{2\sqrt 2}\frac{(1+\sqrt 2)-x}{(1+\sqrt 2)^2+x^2}-\frac{1}{2\sqrt 2}\frac{(1-\sqrt 2)-x}{(1-\sqrt 2)^2+x^2}\right)\ln x \, dx$$ $$\small =\frac{1}{8\sqrt 2}\left(\operatorname{Li}_2\left(-(1+\sqrt 2)^2\right)-\operatorname{Li}_2\left(-(1-\sqrt 2)^2\right)\right)+\frac{1}{2\sqrt 2}\left(\operatorname{Ti}_2\left(-(1+\sqrt 2)\right)-\operatorname{Ti}_2\left(-(1-\sqrt 2)\right)\right)$$ The result from above follows using: $$\int_0^1 \frac{x\ln x}{a^2+x^2}dx\overset{x^2\to x}=\frac14\int_0^1 \frac{\ln x}{a^2+x}dx=\frac14\operatorname{Li}_2\left(-\frac{1}{a^2}\right)$$ $$\int_0^1 \frac{a\ln x}{a^2+x^2}dx\overset{x\to ax}=\int_0^\frac1a\frac{\ln a +\ln x}{1+x^2}dx\overset{IBP}=-\operatorname{Ti}_2\left(\frac1a\right)$$

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The second one is a little troublesome since we must keep the terms togheter to avoid divergence issues. So we'll do a couple substitutions first: $$\int_0^1\left(\frac{\ln x}{1-6x+x^2}+\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{3-x}{1-6x+x^2}\right)dx$$ $$\overset{x\to\frac{1-x}{1+x}}=\int_0^1\left(\frac12\frac{1}{2x^2-1}\ln\left(\frac{1-x}{1+x}\right)+\frac{\ln(1+\sqrt 2)}{\sqrt 2}\left(\frac{2x}{2x^2-1}-\frac{1}{x+1}\right)\right)dx$$ $$=-\frac{\ln 2\ln(1+\sqrt 2)}{\sqrt 2}+\int_0^1\left(\frac12\frac{1}{2x^2-1}\ln\left(\frac{1-x}{1+x}\right)+\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{2x}{2x^2-1}\right)dx$$ $$\overset{\sqrt 2 x\to x}=-\frac{\ln 2\ln(1+\sqrt 2)}{\sqrt 2}-\frac{1}{\sqrt 2}\int_0^\sqrt 2\left(\frac12\frac{1}{1-x^2}\ln\left(\frac{\sqrt 2 -x}{\sqrt 2 + x}\right)+\frac{x}{1-x^2}\ln(1+\sqrt 2)\right)$$ $$\overset{x\to\frac{1-x}{1+x}}=\frac{\ln^2(1+\sqrt 2)}{\sqrt 2}-\frac{\ln 2 \ln(1+\sqrt 2)}{\sqrt 2 }-\frac{1}{4\sqrt 2}\int_{-(1-\sqrt 2)^2}^1\ln\left(\frac{1+(1+\sqrt 2)^2x}{1+\frac{x}{(1+\sqrt 2)^2}}\right)\frac{dx}{x}$$ $$\scriptsize{=\frac{\ln^2(1+\sqrt 2)}{\sqrt 2}-\frac{\ln 2 \ln(1+\sqrt 2)}{\sqrt 2 }-\frac{1}{4\sqrt 2}\left(\frac{\pi^2}{6}-\operatorname{Li}_2\left(-(1+\sqrt 2)^2\right)+\operatorname{Li}_2\left(-(1-\sqrt 2)^2\right)-\operatorname{Li}_2\left((1-\sqrt 2)^4\right)\right)}$$ The last integral was split into two and the result follows by using the definition of the dilogarithm.
Finally putting everything togheter gives the announced closed form.

Answer 2

I think (at least for the time being) tha we need to combine two things $$\frac {x^2}{x^4+1}=\frac{1}{2 \left(x^2+i\right)}+\frac{1}{2 \left(x^2-i\right)},$$ $$\log(1-x^4)=\log(1+x^2)+\log(1-x^2)$$ Each of the four integrals $$\int_0^1\frac{\log \left(1\pm x^2\right)}{2 \left(x^2\pm i\right)}\,dx$$ shows a complex expression involving a bunch of polylogarithms of complex arguments and, for sure, quite many terms invoving the roots of unity.

I obtained the final expression but I cannot simplify it. For sure, the numerical value is $$I=-0.16285829160764898119151607224709931337723009033493\cdots$$

I hope and wish that this could help other users.

Edit

Consider $$I(a)=\int_{0}^{1} x^2\frac{ \log \left(1-a^4 x^4\right)}{x^4+1}\,dx$$ $$I'(a)=-\int_{0}^{1}\frac{4 a^3 x^6}{\left(x^4+1\right) \left(1-a^4 x^4\right)}\,dx$$ $$I'(a)=\frac{a^3 \left(\pi +\log \left(3-2 \sqrt{2}\right)\right)}{\sqrt{2} \left(a^4+1\right)}+2\frac{\tan ^{-1}(a)-\tanh ^{-1}(a)}{a^4+1}$$ Integrating, we have $$\frac{\log (2) \left(\pi +\log \left(3-2 \sqrt{2}\right)\right)}{4 \sqrt{2}}+2\int_0^1 \frac{\tan ^{-1}(a)-\tanh ^{-1}(a)}{a^4+1}\,da$$ Now $$\tan ^{-1}(a)-\tanh ^{-1}(a)=-2\sum_{n=0}^\infty \frac{a^{4 n+3}}{4 n+3}$$ $$J=\int_0^1 \frac{\tan ^{-1}(a)-\tanh ^{-1}(a)}{a^4+1}\,da=-2\sum_{n=0}^\infty \frac 1{4 n+3}\int_0^1 \frac{a^{4 n+3}}{a^4+1}\,da$$ All of that leads for the integral $$I(1)=\frac{\log (2) \left(\pi +\log \left(3-2 \sqrt{2}\right)\right)}{4 \sqrt{2}}-\frac 12\sum_{n=0}^\infty \frac{H_{\frac{n}{2}}-H_{\frac{n-1}{2}}}{4 n+3}$$ I have not been able to make the result of the summation explicit. For its evaluation, we can make $$\sum_{n=0}^\infty \frac{H_{\frac{n}{2}}-H_{\frac{n-1}{2}}}{4 n+3}=\sum_{n=0}^p \frac{H_{\frac{n}{2}}-H_{\frac{n-1}{2}}}{4 n+3}+\sum_{n=p+1}^\infty \frac{H_{\frac{n}{2}}-H_{\frac{n-1}{2}}}{4 n+3}$$ and approximate the last summnand by $$\frac{H_{\frac{n}{2}}-H_{\frac{n-1}{2}}}{4 n+3}=\sum_{k=2}^? \frac {c_k}{4^{k-1}\, n^k}$$ where the first $c_k$'s are $$\{1,-5,15,-29,87,-773,2319,27859,-83577,-3812501,11437503\}$$ Using only what is given here, some results for the summation as a function of $p$ $$\left( \begin{array}{cc} p & \text{summation} \\ 5 & -0.16285829044608650070 \\ 10 & -0.16285829160689532584 \\ 15 & -0.16285829160764112453 \\ 20 & -0.16285829160764869429 \\ 25 & -0.16285829160764895979 \\ 30 & -0.16285829160764897866 \\ 35 & -0.16285829160764898078 \\ 40 & -0.16285829160764898111 \\ 45 & -0.16285829160764898117 \\ 50 & -0.16285829160764898119 \end{array} \right)$$

Answer 3

By expanding the denominator and integrating term-by-term we obtain: $$ -\sum _{n=0}^{\infty } \frac{(-1)^n }{4 n+3}H_{n+\frac{3}{4}}, $$ where $H_n$ is the $n$th harmonic number, in the formula above--the harmonic number of fractional order.

Not sure if it is useful, but the sum can be performed with the help of, e.g., Mathematica yielding a relatively simple expression: $$ \tfrac{1}{3} \Gamma \big(\tfrac{7}{4}\big)\, {}_2F_1^{(0,0,1,0)}\!\left(\tfrac{3}{4},1,\tfrac{7}{4},-1\right)-\frac{\gamma}{4 \sqrt{2}}\left[\pi -2\log(\sqrt{2}-1)\right]. $$ The superscript means that derivative is taken with respect to the parameter $c$ of the regularized hypergeometric function ${}_2F_1(a,b,c,z)$.

Answer 4

Proof:

Let $$ I=\int_0^1\frac{x^2 \log(1-x^4)}{1+x^4}\,\mathrm dx. $$ Expanding the factor of $(1+x^4)^{-1}$ and integrating termwise yields $$ I=-\frac{1}{4}\sum_{k=0}^\infty \frac{H_{k+3/4}}{k+3/4}(-1)^k. $$ Working with the properties of the Pochhammer symbol we may write $$ I=\frac{1}{3}\partial_c F\left({1,3/4 \atop c};-1\right)\bigg|_{c=7/4}-\frac{H_{3/4}}{3} F\left({1,3/4 \atop 7/4};-1\right). $$ We need to find an explicit expression for the remaining derivative. Using this relationship we have $$ F\left({1,3/4 \atop c};-1\right)=2^{-3/4}(c-1)\left(-\frac{1}{2}\right)^{1-c}\operatorname{B}_{-1}(c-1,7/4-c), $$ where $\operatorname{B}_z(a,b)$ is the incomplete beta function. It follows that $$ \partial_c F\left({1,3/4 \atop c};-1\right)\bigg|_{c=7/4}% =\frac{1}{4}(-1)^{3/4} (3\pi+\mathrm i(4+\log 8))\operatorname{B}_{-1}(3/4,0) -\frac{3}{4}(-1)^{1/4}\partial_c\operatorname{B}_{-1}(c-1,7/4-c)|_{c=7/4}, $$ where $$ \partial_c\operatorname{B}_{-1}(c-1,7/4-c)|_{c=7/4}=\partial_a\operatorname{B}_{-1}(a,b)|_{a=3/4, b=0}-\partial_b\operatorname{B}_{-1}(a,b)|_{a=3/4, b=0}. $$ We can evaluate the remaining derivatives using the derivatives of $\operatorname{B}_z(a,b)$ w.r.t. $a$ and $b$ here. We have for the first term $$ \partial_a\operatorname{B}_{-1}(a,b)|_{a=3/4, b=0}= \frac{1}{4} (-1)^{3/4} \left(\psi ^{(1)}\left(\frac{7}{8}\right)-\psi ^{(1)}\left(\frac{3}{8}\right)\right)+i \pi B_{-1}\left(\frac{3}{4},0\right). $$ The second term is more complicated as it has poles of the form $b^{-2}$ as $b\to 0$. The derivatives gives $$ \partial_b\operatorname{B}_{-1}(a,b)|_{a=3/4, b=0}=\lim_{b\to 0}\left( \frac{2^b}{b^2}{_3F_2}\left({\frac{1}{4},b,b\atop b+1,b+1};2\right)+\left(\psi ^{(0)}(b)-\psi^{(0)}\left(b+\frac{3}{4}\right)\right) \operatorname B\left(\frac{3}{4},b\right)-\log (2)\operatorname{B}_2\left(b,\frac{3}{4}\right)\right). $$ We make the following observations as $b\to 0$: $$ \left(\psi ^{(0)}(b)-\psi ^{(0)}\left(b+\frac{3}{4}\right)\right) B\left(\frac{3}{4},b\right)\sim -\frac{1}{b^2}+4 C-\frac{7 \pi ^2}{24}+\frac{\log ^2(8)}{2}-\frac{1}{2} \pi \log (8)+\mathcal O(b), $$ and $$ \frac{2^b}{b^2}{_3F_2}\left({1/4,b,b\atop b+1,b+1};2\right)\sim \frac{1}{b^2}+\frac{\log 2}{b}+\frac{1}{2}\log^2 2+\frac{1}{2}{_4F_3}\left({1,1,1,5/4\atop 2,2,2};2\right)+\mathcal O(b), $$ and $$ \log (2)\operatorname{B}_2\left(b,\frac{3}{4}\right)\sim\frac{\log 2}{b}+\log^22+\frac{\log 2}{2}{_3F_2}\left({1,1,5/4\atop 2,2};2\right)+\mathcal O(b). $$ These three results then give us $$ \partial_b\operatorname{B}_{-1}(a,b)|_{a=3/4, b=0}= 4\log^22-\frac{3}{2}\pi\log 2+4 C-\frac{7 \pi ^2}{24} +\frac{1}{2}{_4F_3}\left({1,1,1,5/4\atop 2,2,2};2\right) -\frac{\log 2}{2}{_3F_2}\left({1,1,5/4\atop 2,2};2\right). $$ Bringing all results together gives a final expression for $I$.

Edit:

Much of the special functions in this solution reduce to elementary functions by way of the differentiation formula \begin{align}\left(z\frac{d}{dz}+\beta_k-1\right){}_pF_q\biggl[ \begin{array}{c}\alpha_1,\ldots,\alpha_p \\ \beta_1,\ldots,\beta_k,\ldots,\beta_q\end{array};z\biggr]&=\\ =\left(\beta_k-1\right) {}_pF_q\biggl[ \begin{array}{c}\alpha_1,\ldots,\alpha_p \\ \beta_1,\ldots,\beta_k-1,\ldots,\beta_q\end{array};z\biggr]&. \tag{$\spadesuit$}\end{align}

For example, this formula gives $$ (z\partial_z+1){_3F_2}\left({1,1,5/4\atop 2,2};z\right)={F}\left({1,5/4\atop 2};z\right)=\frac{4}{z}((1-z)^{-1/4}-1). $$ Letting $y(z)={_3F_2}(1,1,5/4; 2,2;z)$ the above formula is written as $$ z y^\prime+y=\frac{4}{z}((1-z)^{-1/4}-1),\quad y(0)=1. $$ Solving the differential equation we find $$ y(z)=\frac{2}{z}\left((-2-2 i) \log \left(-\sqrt[4]{1-z}+i\right)-(2-2 i) \log \left(\sqrt[4]{1-z}+i\right)-4 \log \left(\sqrt[4]{1-z}+1\right)-(1-2 i) \pi +\log (64)\right). $$ Substituting $z=2$ and simplifying then gives the result $$ {_3F_2}\left({1,1,5/4\atop 2,2};2\right)=-\mathrm i\pi+4\log 2+(1-i) \log (3-2 \sqrt{2}). $$ I suspect my solution for $I$ here can be reduced down to to elementary functions without the need to beta and hypergeometric functions.

Answer 5

This may be counterproductive but we can obtain another form from @yarchik's solution. $$ \begin{aligned} S% &=-\sum _{n=0}^{\infty } \frac{(-1)^n }{4n+3}H_{n+\frac{3}{4}},\\ &=-\frac{1}{4}\sum _{n=0}^{\infty } \frac{(-1)^n }{n+3/4}H_{n+\frac{3}{4}},\\ &=-\frac{1}{4}\int_0^1x^{-1/4}\sum _{n=0}^{\infty } (-x)^nH_{n+\frac{3}{4}}\,\mathrm dx,\\ \end{aligned} $$ The remaining series can be solved exactly yielding $$ \begin{aligned} S% &=\frac{1}{7}\int_0^1\frac{x^{3/4}}{1+x}F\left({1,7/4\atop 11/4};-x\right)\,\mathrm dx-\frac{H_{3/4}}{4}\int_0^1\frac{x^{-1/4}}{1+x}\,\mathrm dx,\\ &=\frac{1}{7}\int_0^1\frac{x^{3/4}}{1+x}F\left({1,7/4\atop 11/4};-x\right)\,\mathrm dx-\frac{H_{3/4}(-1)^{1/4}}{2}(\operatorname{arctan}((-1)^{1/4})- \operatorname{arctanh}((-1)^{1/4})), \end{aligned} $$ The hypergeometric function in the integral is zero-balanced and may reduce further.