What is a closed form for ${\large\int}_0^1\frac{\ln^3(1+x)\,\ln^2x}xdx$?

Solution 1:

I will be using the following results: $$2\sum^\infty_{n=1}\frac{H_n}{n^q}=(q+2)\zeta(q+1)-\sum^{q-2}_{j=1}\zeta(j+1)\zeta(q-j)\tag1$$ $$\sum^\infty_{n=1}\frac{H_n}{n^22^n}=\zeta(3)-\frac{\pi^2}{12}\ln{2}\tag2$$ $$\sum^\infty_{n=1}\frac{H_n}{n^32^n}={\rm Li}_4\left(\tfrac{1}{2}\right)+\frac{\pi^4}{720}-\frac{1}{8}\zeta(3)\ln{2}+\frac{1}{24}\ln^4{2}\tag3$$ \begin{align} \sum^\infty_{n=1}\frac{H_n}{n^42^n} =&2{\rm Li}_5\left(\tfrac{1}{2}\right)+\frac{1}{32}\zeta(5)+{\rm Li}_4\left(\tfrac{1}{2}\right)\ln{2}-\frac{\pi^4}{720}\ln{2}+\frac{1}{2}\zeta(3)\ln^2{2}\\&-\frac{\pi^2}{12}\zeta(3)-\frac{\pi^2}{36}\ln^3{2}+\frac{1}{40}\ln^5{2}\tag4 \end{align} Proofs of $(1)$, $(2)$ and $(4)$ can be found here, here and here respectively. Unfortunately, there has not been a mathematically sound proof of $(3)$ on MSE as of now.

Using $\mathcal{I}$ to denote the integral in question, \begin{align} \mathcal{I} &=-\int^1_0\frac{\ln^3{x}\ln^2(1+x)}{1+x}{\rm d}x\\ &=-\int^2_1\frac{\ln^2{x}\ln^3(x-1)}{x}{\rm d}x\\ &=\underbrace{-\int^1_\frac{1}{2}\frac{\ln^2{x}\ln^3(1-x)}{x}{\rm d}x}_{\mathcal{I}_1}\underbrace{+3\int^1_{\frac{1}{2}}\frac{\ln^3{x}\ln^2(1-x)}{x}}_{\mathcal{I}_2}\underbrace{-3\int^1_{\frac{1}{2}}\frac{\ln^4{x}\ln(1-x)}{x}{\rm d}x}_{\mathcal{I}_3}-\frac{1}{6}\ln^6{2} \end{align} For $\mathcal{I}_1$, integration by parts gives $$\mathcal{I}_1=\frac{1}{3}\ln^6{2}-\int^1_\frac{1}{2}\frac{\ln^3{x}\ln^2(1-x)}{1-x}{\rm d}x$$ On the other hand, $x\mapsto1-x$ yields $$\mathcal{I}_1=-\int^\frac{1}{2}_0\frac{\ln^3{x}\ln^2(1-x)}{1-x}{\rm d}x$$ Combining these two equalities, we have \begin{align} \mathcal{I}_1 &=\frac{1}{6}\ln^6{2}-\frac{1}{2}\int^1_0\frac{\ln^3{x}\ln^2(1-x)}{1-x}{\rm d}x\\ &=\frac{1}{6}\ln^6{2}-\frac{1}{2}\frac{\partial^5\beta}{\partial a^3\partial b^2}(1,0^+)\\ &=\frac{1}{6}\ln^6{2}-\frac{1}{2}\left[\frac{1}{b}+\mathcal{O}(1)\right]\left[\left(12\zeta^2(3)-\frac{23\pi^6}{1260}\right)b+\mathcal{O}(b^2)\right]_{b=0}\\ &=\frac{23\pi^6}{2520}-6\zeta^2(3)+\frac{1}{6}\ln^6{2} \end{align} Even with the help of Wolfram Alpha, evaluating that fifth derivative was horribly unpleasant to say the least. As for $\mathcal{I}_2$, \begin{align} \mathcal{I}_2 =&6\sum^\infty_{n=1}\frac{H_n}{n+1}\int^1_\frac{1}{2}x^n\ln^3{x}\ {\rm d}x\\ =&6\sum^\infty_{n=1}\frac{H_n}{n+1}\frac{\partial^3}{\partial n^3}\left(\frac{1}{n+1}-\frac{1}{(n+1)2^{n+1}}\right)\\ =&\color{#E2062C}{-\sum^\infty_{n=1}\frac{36H_n}{(n+1)^5}}+\color{#FF4F00}{\sum^\infty_{n=1}\frac{36H_n}{(n+1)^52^{n+1}}}+\color{#00A000}{\sum^\infty_{n=1}\frac{36\ln{2}H_n}{(n+1)^42^{n+1}}}+\color{#21ABCD}{\sum^\infty_{n=1}\frac{18\ln^2{2}H_n}{(n+1)^32^{n+1}}}\\&+\color{#6F00FF}{\sum^\infty_{n=1}\frac{6\ln^3{2}H_n}{(n+1)^22^{n+1}}}\\ =&\color{#E2062C}{-\frac{\pi^6}{35}+18\zeta^2(3)}+\color{#FF4F00}{\sum^\infty_{n=1}\frac{36H_n}{n^52^{n}}-36{\rm Li}_6\left(\tfrac{1}{2}\right)}+\color{#00A000}{36{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}+\frac{9}{8}\zeta(5)\ln{2}}\\ &+\color{#00A000}{36{\rm Li}_4\left(\tfrac{1}{2}\right)\ln^2{2}-\frac{\pi^4}{20}\ln^2{2}+18\zeta(3)\ln^3{2}-3\pi^2\zeta(3)\ln{2}-\pi^2\ln^4{2}+\frac{9}{10}\ln^6{2}}\\ &+\color{#21ABCD}{\frac{\pi^4}{40}\ln^2{2}-\frac{9}{4}\zeta(3)\ln^3{2}+\frac{3}{4}\ln^6{2}}+\color{#6F00FF}{\frac{3}{4}\zeta(3)\ln^3{2}-\ln^6{2}}\\ =&\sum^\infty_{n=1}\frac{36H_n}{n^52^{n}}-36{\rm Li}_6\left(\tfrac{1}{2}\right)-\frac{\pi^6}{35}+36{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}+\frac{9}{8}\zeta(5)\ln{2}+36{\rm Li}_4\left(\tfrac{1}{2}\right)\ln^2{2}\\ &-\frac{\pi^4}{40}\ln^2{2}+18\zeta^2(3)-3\pi^2\zeta(3)\ln{2}+\frac{33}{2}\zeta(3)\ln^3{2}-\pi^2\ln^4{2}+\frac{13}{20}\ln^6{2} \end{align} For $\mathcal{I}_3$, \begin{align} \mathcal{I}_3 =&3\sum^\infty_{n=1}\frac{1}{n}\int^1_\frac{1}{2}x^{n-1}\ln^4{x}\ {\rm d}x\\ =&3\sum^\infty_{n=1}\frac{1}{n}\frac{\partial^4}{\partial n^4}\left(\frac{1}{n}-\frac{1}{n2^n}\right)\\ =&\sum^\infty_{n=1}\left(\frac{72}{n^6}-\frac{72}{n^62^n}-\frac{72\ln{2}}{n^52^n}-\frac{36\ln^2{2}}{n^42^n}-\frac{12\ln^3{2}}{n^32^n}-\frac{3\ln^4{2}}{n^22^n}\right)\\ =&-72{\rm Li}_6\left(\tfrac{1}{2}\right)+\frac{8\pi^6}{105}-72{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}-36{\rm Li}_4\left(\tfrac{1}{2}\right)\ln^2{2}\\&-\frac{21}{2}\zeta(3)\ln^3{2}+\frac{3\pi^2}{4}\ln^4{2}-\frac{1}{2}\ln^6{2} \end{align} Thus \begin{align} \color{#BF00FF}{\mathcal{I} =}&\color{#BF00FF}{36\sum^\infty_{n=1}\frac{H_n}{n^52^n}-108{\rm Li}_6\left(\tfrac{1}{2}\right)+\frac{143\pi^6}{2520}-36{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}+\frac{9}{8}\zeta(5)\ln{2}-\frac{\pi^4}{40}\ln^2{2}}\\&\color{#BF00FF}{+12\zeta^2(3)-3\pi^2\zeta(3)\ln{2}+6\zeta(3)\ln^3{2}-\frac{\pi^2}{4}\ln^4{2}+\frac{3}{20}\ln^6{2}} \end{align} We note that \begin{align} \zeta(\bar{5},1) =&\frac{1}{24}\int^1_0\frac{\ln^4{x}\ln(1+x)}{1+x}{\rm d}x\\ =&\frac{1}{24}\int^2_1\frac{\ln{x}\ln^4(x-1)}{x}{\rm d}x\\ =&-\frac{1}{24}\int^1_\frac{1}{2}\frac{\ln{x}\ln^4(1-x)}{x}{\rm d}x+\frac{1}{6}\int^1_\frac{1}{2}\frac{\ln^2{x}\ln^3(1-x)}{x}{\rm d}x-\frac{1}{4}\int^1_\frac{1}{2}\frac{\ln^3{x}\ln^2(1-x)}{x}{\rm d}x\\ &+\frac{1}{6}\int^1_\frac{1}{2}\frac{\ln^4{x}\ln(1-x)}{x}{\rm d}x+\frac{1}{144}\ln^6{2}\\ =&\underbrace{-\frac{1}{24}\int^\frac{1}{2}_0\frac{\ln^4{x}\ln(1-x)}{1-x}{\rm d}x}_{\mathcal{J}}-3\sum^\infty_{n=1}\frac{H_n}{n^52^n}+7{\rm Li}_6\left(\tfrac{1}{2}\right)-\frac{17\pi^6}{5040}+{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}\\ &-\frac{3}{32}\zeta(5)\ln{2}-{\rm Li}_4\left(\tfrac{1}{2}\right)\ln^2{2}+\frac{\pi^4}{480}\ln^2{2}-\frac{1}{2}\zeta^2(3)+\frac{\pi^2}{4}\zeta(3)\ln{2}-\frac{19}{24}\zeta(3)\ln^3{2}\\ &+\frac{\pi^2}{24}\ln^4{2}-\frac{17}{360}\ln^6{2} \end{align} since we have already derived the values of the last three integrals. For the remaining integral, \begin{align} \mathcal{J} =&\frac{1}{24}\sum^\infty_{n=1}H_n\frac{\partial^4}{\partial n^4}\left(\frac{1}{(n+1)2^{n+1}}\right)\\ =&\sum^\infty_{n=1}\frac{H_n}{(n+1)^52^{n+1}}+\sum^\infty_{n=1}\frac{\ln{2}H_n}{(n+1)^42^{n+1}}+\sum^\infty_{n=1}\frac{\ln^2{2}H_n}{2(n+1)^32^{n+1}}+\sum^\infty_{n=1}\frac{\ln^3{2}H_n}{6(n+1)^22^{n+1}}\\ &+\sum^\infty_{n=1}\frac{\ln^4{2}H_n}{24(n+1)2^{n+1}}\\ =&\sum^\infty_{n=1}\frac{H_n}{n^52^n}-{\rm Li}_6\left(\tfrac{1}{2}\right)+{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}+\frac{1}{32}\zeta(5)\ln{2}+{\rm Li}_4\left(\tfrac{1}{2}\right)\ln^2{2}-\frac{\pi^4}{720}\ln^2{2}\\ &+\frac{1}{2}\zeta(3)\ln^3{2}-\frac{\pi^2}{12}\zeta(3)\ln{2}-\frac{\pi^2}{36}\ln^4{2}+\frac{1}{40}\ln^6{2}+\frac{\pi^4}{1440}\ln^2{2}-\frac{1}{16}\zeta(3)\ln^3{2}\\&+\frac{1}{48}\ln^6{2}+\frac{1}{48}\zeta(3)\ln^3{2}-\frac{1}{36}\ln^6{2}+\frac{1}{48}\ln^6{2}\\ =&\sum^\infty_{n=1}\frac{H_n}{n^52^n}-{\rm Li}_6\left(\tfrac{1}{2}\right)+{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}+\frac{1}{32}\zeta(5)\ln{2}+{\rm Li}_4\left(\tfrac{1}{2}\right)\ln^2{2}-\frac{\pi^4}{1440}\ln^2{2}\\ &+\frac{11}{24}\zeta(3)\ln^3{2}-\frac{\pi^2}{12}\zeta(3)\ln{2}-\frac{\pi^2}{36}\ln^4{2}+\frac{7}{180}\ln^6{2}\\ \end{align} Hence we can express $\zeta(\bar{5},1)$ as \begin{align} \zeta(\bar{5},1) =&-2\sum^\infty_{n=1}\frac{H_n}{n^52^n}+6{\rm Li}_6\left(\tfrac{1}{2}\right)-\frac{17\pi^6}{5040}+2{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}-\frac{1}{16}\zeta(5)\ln{2}+\frac{\pi^4}{720}\ln^2{2}\\ &-\frac{1}{2}\zeta^2(3)-\frac{1}{3}\zeta(3)\ln^3{2}+\frac{\pi^2}{6}\zeta(3)\ln{2}+\frac{\pi^2}{72}\ln^4{2}-\frac{1}{120}\ln^6{2} \end{align} This implies that \begin{align} \sum^\infty_{n=1}\frac{H_n}{n^52^n} =&3{\rm Li}_6\left(\tfrac{1}{2}\right)-\frac{1}{2}\zeta(\bar{5},1)-\frac{17\pi^6}{10080}+{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}-\frac{1}{32}\zeta(5)\ln{2}+\frac{\pi^4}{1440}\ln^2{2}\\ &-\frac{1}{4}\zeta^2(3)-\frac{1}{6}\zeta(3)\ln^3{2}+\frac{\pi^2}{12}\zeta(3)\ln{2}+\frac{\pi^2}{144}\ln^4{2}-\frac{1}{240}\ln^6{2} \end{align} Plucking this back into the original integral, we get another form in terms of $\zeta(\bar{5},1)$ \begin{align} \color{#BF00FF}{\mathcal{I} =}&\color{#BF00FF}{-\frac{\pi^6}{252}-18\zeta(\bar{5},1)+3\zeta^2(3)} \end{align} This is as close to a "closed form" as I can get. The sheer number of cancellations involved in the last step makes me think that my answer could be roundabout and inefficient. Note that no known simple closed form for $\zeta(\bar{5},1)$ exists, implying that closed forms for higher power integrals are unlikely to exist as well.

Solution 2:

This is not going to be a complete answer but since this kind of approach has not been presented here yet and since I believe that it can be brought to a successful completion given enough time under disposal( which I lack now) I present the approach now. Denote: \begin{eqnarray} {\mathcal I}^{(2,3)} := \int\limits_0^1 \frac{\log(\xi)^2 \log(1+\xi)^3}{\xi} d\xi \end{eqnarray} Then we have: \begin{eqnarray} &&{\mathcal I}^{(2,3)} = \left. \frac{\partial^2}{\partial \theta_1^2} \frac{\partial^3}{\partial \theta_2^3} \int\limits_0^1 \xi^{\theta_1-1} (1+\xi)^{\theta_2} d\xi \right|_{\theta_1=\theta_2=0} \\ &&= \left. \frac{\partial^2}{\partial \theta_1^2} \frac{\partial^3}{\partial \theta_2^3} \left[\sum\limits_{l=0}^\infty \frac{ (\theta_2)_{(l)} }{ \theta_1^{(l+1)} } \cdot 2^{\theta_2-l} (-1)^l \right] \right|_{\theta_1=\theta_2=0}\\ &&= \sum\limits_{l=1}^\infty \left(\log(2)^2 + \frac{\log(4)}{l} + \frac{2}{l^2} + [H_l]^2 - H_l^{(2)}-\frac{2}{l} H_l - 2 \log(2) H_l \right)\cdot \\ &&\left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l\cdot 2^l} \right) \end{eqnarray} The first line is straightforward. In the second line we computed the integral in question by integrating by parts. Finally in the last line we computed the partial derivatives using Higher order derivatives of the binomial factor and the chain rule. Now, the sums look scary but it appears that those sums have actually a much simpler integrals representation than the original integral we want to compute. As a matter of fact the following holds: \begin{eqnarray} \sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l} \right) x^l = -\int\limits_0^1 \frac{x}{1-\xi x} \cdot [\log(1-\xi)]^3 d\xi \end{eqnarray} Using the generation function above we compute the harmonic sums in question. We have: \begin{eqnarray} \sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l} \right) \cdot \frac{1}{2^l} &=& \frac{21}{4} \zeta(4)\\ \sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l^2} \right) \cdot \frac{1}{2^l} &=& -\frac{3 \pi ^2 \zeta (3)}{8}+12 \zeta (5)-\frac{7}{120} \pi ^4 \log (2)\\ \sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l^3} \right) \cdot \frac{1}{2^l} &=& -\int\limits_0^1 \frac{1}{\xi} Li_2(\frac{\xi}{2}) \log(1-\xi)^3 d\xi\\ \sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l} \right) \cdot \frac{1}{2^l} \cdot H_l &=& -\frac{7}{8} \pi^2 \zeta(3) + \frac{279}{16} \zeta(5)\\ \sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l^2} \right) \cdot \frac{1}{2^l} \cdot H_l &=& \int\limits_{0}^1 \frac{Li_2(-\xi)}{\xi(1+\xi)} \cdot [\log(\frac{1-\xi}{1+\xi})]^3 d\xi\\ \sum\limits_{l=1}^\infty \left(\frac{[H_l]^3+3 H_l H_l^{(2)} + 2 H_l^{(3)}}{l^2} \right) \cdot \frac{1}{2^l} \cdot \left([H_l]^2-H_l^{(2)}\right) &=& -12 \left( \zeta(-4,1,1)-\zeta(4,-1,1)\right)- \frac{1}{8} \left(\pi^4 \log(2) + 14 \pi^2 \zeta(3) - 279 \zeta(5)\right) \end{eqnarray} It is clear that the remaining sums are more complicated and more time is required to bring this thread to completion. We will finish this work as soon as possible.

Solution 3:

Let us denote: \begin{equation} {\mathcal I}^{(3,2)}:=\int\limits_0^1 \frac{\log(1+x)^3}{x} \cdot [\log(x)]^2 dx \end{equation} We have: \begin{eqnarray} &&{\mathcal I}^{(3,2)}=\\ &&-\frac{2}{3} i \pi \left(-12 \text{Li}_5\left(\frac{1}{2}\right)-12 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+\frac{\pi ^2 \zeta (3)}{2}+\frac{99 \zeta (5)}{16}-\frac{21}{4} \zeta (3) \log ^2(2)-\frac{2 \log ^5(2)}{5}+\frac{1}{3} \pi ^2 \log ^3(2)\right)-\\ &&\frac{2}{3} \int\limits_0^1 \frac{6\left(\text{Li}_4(x+1)-\frac{\pi ^4}{90}\right)+3 \text{Li}_2(x+1) \log ^2(x+1)-6 \text{Li}_3(x+1) \log (x+1)}{x} \cdot \log(x)dx \end{eqnarray} In the above we used the knowledge of the anti-derivative of the fraction in the integrand and we integrated by parts once. Indeed we have: \begin{eqnarray} \int \frac{\log(1+x)^k}{x} dx = \sum\limits_{l=1}^{k+1} (-1)^l \binom{k}{l-1} (l-1)! Li_l(1+x) \log(1+x)^{k+1-l} \end{eqnarray} Now it is actually fairy easy to construct the anti-derivative of the fraction in the remaining integrand above and then to do another integration by parts.As a matter of fact we have: \begin{eqnarray} &&\int \frac{6\left(\text{Li}_4(x+1)-\frac{\pi ^4}{90}\right)+3 \text{Li}_2(x+1) \log ^2(x+1)-6 \text{Li}_3(x+1) \log (x+1)}{x} dx=\\ &&\log (-x) \left(6 \left(\text{Li}_4(x+1)-\frac{\pi ^4}{90}\right)+3 \text{Li}_2(x+1) \log ^2(x+1)-6 \text{Li}_3(x+1) \log (x+1)\right)+\\ &&3 \int \frac{Li_1(1+x)^2}{(1+x)} \cdot \log(1+x)^2 dx \end{eqnarray}

In doing that it turns out that the boundary term vanishes and then what we are left with are integrals of the kind $\int\limits_0^{1/2} \log(x)^p \log(1-x)^q/x dx$ for $p+q \le 5$. All those integrals have allready been dealt with and are expressed thorough poly-logarithms with one exception only , namely when $(p,q)=(3,2)$. In this case a new quantity ${\bf H}^{(1)}_5(1/2)$ enters the result. Then the final result reads: \begin{eqnarray} &&{\mathcal I}^{(3,2)}=\\ &&-108 \text{Li}_6\left(\frac{1}{2}\right)-36 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+\frac{429 \zeta (2)^3}{35}+12 \zeta (3)^2-\frac{3}{2} \zeta (2) \log ^4(2)+6 \zeta (3) \log ^3(2)-\frac{9}{10} \zeta (2)^2 \log ^2(2)-18 \zeta (3) \zeta (2) \log (2)+\frac{9}{8} \zeta (5) \log (2)+\frac{3 \log ^6(2)}{20} + 36 {\bf H}^{(1)}_5(1/2) \end{eqnarray} Below I include the Mathematica code that verifies the results:

M = 2000; Clear[H];
H[p_, q_, x_] := N[Sum[ HarmonicNumber[n, p]/n^q x^n, {n, 1, M}], 50];
k = 3;
NIntegrate[Log[1 + x]^k/x Log[x]^2, {x, 0, 1}, WorkingPrecision :> 30]

(*The border term is equal to Int Log[1+x]^3 Log[x]/x,{x,0,1}]*)
-2 I Pi/3 (Pi^2/3 Log[2]^3 - 2/5 Log[2]^5 + Pi^2/2 Zeta[3] + 
    99/16 Zeta[5] - 21/4 Zeta[3] Log[2]^2 - 
    12 PolyLog[4, 1/2] Log[2] - 12 PolyLog[5, 1/2]) - 
 2/3 NIntegrate[(3 Log[1 + x]^2 PolyLog[2, 1 + x] - 
      6 Log[1 + x] PolyLog[3, 1 + x] + 
      6 (-PolyLog[4, 1] + PolyLog[4, 1 + x])) Log[x]/x, {x, 0, 1}, 
   WorkingPrecision :> 30]

(3 Log[2]^6)/20 - 36 Log[2] PolyLog[5, 1/2] - 108 PolyLog[6, 1/2] + 
 6 Log[2]^3 Zeta[3] + 12 Zeta[3]^2 + 9/8 Log[2] Zeta[5] - 
 3/2 Log[2]^4 Zeta[2] - 18 Log[2] Zeta[3] Zeta[2] - 
 9/10 Log[2]^2 Zeta[2]^2 + (429 Zeta[2]^3)/35 + 36 H[1, 5, 1/2]

Solution 4:

partial solution

using the following identity: ( I can provide the proof if needed) $$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)$$

replace $x$ with $-x$ , then multiply both sides by $\ln^3x$ and integrate from $0$ to $1$, we have \begin{align} I&=\int_0^1\frac{\ln^2(1+x)\ln^3x}{1+x}\ dx=\sum_{n=1}^\infty (-1)^n\left(H_n^2-H_n^{(2)}\right)\int_0^1x^n\ln^3x\ dx\\ &=-6\sum_{n=1}^\infty \frac{(-1)^n}{(n+1)^4}\left(H_n^2-H_n^{(2)}\right)=6\sum_{n=1}^\infty \frac{(-1)^n}{n^4}\left(H_{n-1}^2-H_{n-1}^{(2)}\right)\\ &=6\sum_{n=1}^\infty \frac{(-1)^n}{n^4}\left(H_{n}^2-H_{n}^{(2)}-2\frac{H_n}{n}+\frac2{n^2}\right)\\ &=6\left(\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^4}-2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}-\frac{31}{16}\zeta(6)\right) \end{align}