[Largedisplaystyle int_0^1frac{ln^3(1+x)\,ln^2x}xmathrm{d}x
]
(Largemathbf{Solution:})
I will be using the following results:
[2sum^infty_{n=1}frac{H_n}{n^q}=(q+2)zeta(q+1)-sum^{q-2}_{j=1}zeta(j+1)zeta(q-j)
]
[sum^infty_{n=1}frac{H_n}{n^22^n}=zeta(3)-frac{pi^2}{12}ln{2}
]
[sum^infty_{n=1}frac{H_n}{n^32^n}={
m Li}_4left(frac{1}{2}
ight)+frac{pi^4}{720}-frac{1}{8}zeta(3)ln{2}+frac{1}{24}ln^4{2}
]
[egin{align*}
sum^infty_{n=1}frac{H_n}{n^42^n}
=&2{
m Li}_5left(frac{1}{2}
ight)+frac{1}{32}zeta(5)+{
m Li}_4left(frac{1}{2}
ight)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}
end{align*}]
Using (I) to denote the integral in question,
[egin{align*}
mathcal{I}
&=-int^1_0frac{ln^3{x}ln^2(1+x)}{1+x}{
m d}x\&=-int^2_1frac{ln^2{x}ln^3(x-1)}{x}{
m d}x\
&=underbrace{-int^1_frac{1}{2}frac{ln^2{x}ln^3(1-x)}{x}{
m d}x}_{mathcal{I}_1}underbrace{+3int^1_{frac{1}{2}}frac{ln^3{x}ln^2(1-x)}{x}}_{mathcal{I}_2}underbrace{-3int^1_{frac{1}{2}}frac{ln^4{x}ln(1-x)}{x}{
m 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}{
m d}x
]
On the other hand, (xmapsto1-x) yields
[mathcal{I}_1=-int^frac{1}{2}_0frac{ln^3{x}ln^2(1-x)}{1-x}{
m d}x
]
Combining these two equalities, we have
[egin{align*}
mathcal{I}_1
&=frac{1}{6}ln^6{2}-frac{1}{2}int^1_0frac{ln^3{x}ln^2(1-x)}{1-x}{
m d}x\&=frac{1}{6}ln^6{2}-frac{1}{2}frac{partial^5eta}{partial a^3partial b^2}(1,0^+)\
&=frac{1}{6}ln^6{2}-frac{1}{2}left[frac{1}{b}+mathcal{O}(1)
ight]left[left(12zeta^2(3)-frac{23pi^6}{1260}
ight)b+mathcal{O}(b^2)
ight]_{b=0}\
&=frac{23pi^6}{2520}-6zeta^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),
[egin{align*}
mathcal{I}_2&=6sum^infty_{n=1}frac{H_n}{n+1}int^1_frac{1}{2}x^nln^3{x} {
m d}x\&=6sum^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}}
ight)\
&=color{Magenta}{-sum^infty_{n=1}frac{36H_n}{(n+1)^5}}\, color{Black}{+}\, color{Orange}{sum^infty_{n=1}frac{36H_n}{(n+1)^52^{n+1}}}\, color{Black}{+}\, color{Green}{sum^infty_{n=1}frac{36ln{2}H_n}{(n+1)^42^{n+1}}}\, color{Black}{+}\, color{Cyan}{sum^infty_{n=1}frac{18ln^2{2}H_n}{(n+1)^32^{n+1}}}\&~~~+color{Purple}{sum^infty_{n=1}frac{6ln^3{2}H_n}{(n+1)^22^{n+1}}}\
&=color{Magenta}{-frac{pi^6}{35}+18zeta^2(3)}+color{Orange}{sum^infty_{n=1}frac{36H_n}{n^52^{n}}-36{
m Li}_6left(frac{1}{2}
ight)}+color{Green}{36{
m Li}_5left(frac{1}{2}
ight)ln{2}+frac{9}{8}zeta(5)ln{2}}\
&~~~+color{Green}{36{
m Li}_4left(frac{1}{2}
ight)ln^2{2}-frac{pi^4}{20}ln^2{2}+18zeta(3)ln^3{2}-3pi^2zeta(3)ln{2}-pi^2ln^4{2}+frac{9}{10}ln^6{2}}\
&~~~+color{Cyan}{frac{pi^4}{40}ln^2{2}-frac{9}{4}zeta(3)ln^3{2}+frac{3}{4}ln^6{2}}+color{Purple}{frac{3}{4}zeta(3)ln^3{2}-ln^6{2}}\
&=sum^infty_{n=1}frac{36H_n}{n^52^{n}}-36{
m Li}_6left(frac{1}{2}
ight)-frac{pi^6}{35}+36{
m Li}_5left(frac{1}{2}
ight)ln{2}+frac{9}{8}zeta(5)ln{2}+36{
m Li}_4left(frac{1}{2}
ight)ln^2{2}\
&~~~-frac{pi^4}{40}ln^2{2}+18zeta^2(3)-3pi^2zeta(3)ln{2}+frac{33}{2}zeta(3)ln^3{2}-pi^2ln^4{2}+frac{13}{20}ln^6{2}
end{align*}]
For (mathcal{I}_3),
[egin{align*}
mathcal{I}_3
=&3sum^infty_{n=1}frac{1}{n}int^1_frac{1}{2}x^{n-1}ln^4{x} {
m d}x\
=&3sum^infty_{n=1}frac{1}{n}frac{partial^4}{partial n^4}left(frac{1}{n}-frac{1}{n2^n}
ight)\
=&sum^infty_{n=1}left(frac{72}{n^6}-frac{72}{n^62^n}-frac{72ln{2}}{n^52^n}-frac{36ln^2{2}}{n^42^n}-frac{12ln^3{2}}{n^32^n}-frac{3ln^4{2}}{n^22^n}
ight)\
=&-72{
m Li}_6left(frac{1}{2}
ight)+frac{8pi^6}{105}-72{
m Li}_5left(frac{1}{2}
ight)ln{2}-36{
m Li}_4left(frac{1}{2}
ight)ln^2{2}\&-frac{21}{2}zeta(3)ln^3{2}+frac{3pi^2}{4}ln^4{2}-frac{1}{2}ln^6{2}
end{align*}]
Thus
[oxed{egin{align*}
int_0^1frac{ln^3(1+x)\,ln^2x}xmathrm{d}x
=&~color{blue}{36sum^infty_{n=1}frac{H_n}{n^52^n}-108{
m Li}_6left(frac{1}{2}
ight)+frac{143pi^6}{2520}-36{
m Li}_5left(frac{1}{2}
ight)ln{2}}\
&color{blue}{+\, frac{9}{8}zeta(5)ln{2}-frac{pi^4}{40}ln^2{2}+12zeta^2(3)-3pi^2zeta(3)ln{2}}\
&color{blue}{+\, 6zeta(3)ln^3{2}-frac{pi^2}{4}ln^4{2}+frac{3}{20}ln^6{2}}
end{align*}}]
We note that
[egin{align*}
zeta(ar{5},1)
=&frac{1}{24}int^1_0frac{ln^4{x}ln(1+x)}{1+x}{
m d}x
=frac{1}{24}int^2_1frac{ln{x}ln^4(x-1)}{x}{
m d}x\
=&-frac{1}{24}int^1_frac{1}{2}frac{ln{x}ln^4(1-x)}{x}{
m d}x+frac{1}{6}int^1_frac{1}{2}frac{ln^2{x}ln^3(1-x)}{x}{
m d}x-frac{1}{4}int^1_frac{1}{2}frac{ln^3{x}ln^2(1-x)}{x}{
m d}x\
&+frac{1}{6}int^1_frac{1}{2}frac{ln^4{x}ln(1-x)}{x}{
m d}x+frac{1}{144}ln^6{2}\
=&underbrace{-frac{1}{24}int^frac{1}{2}_0frac{ln^4{x}ln(1-x)}{1-x}{
m d}x}_{mathcal{J}}-3sum^infty_{n=1}frac{H_n}{n^52^n}+7{
m Li}_6left(frac{1}{2}
ight)-frac{17pi^6}{5040}+{
m Li}_5left(frac{1}{2}
ight)ln{2}\
&-frac{3}{32}zeta(5)ln{2}-{
m Li}_4left(frac{1}{2}
ight)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,
[egin{align*}
mathcal{J}=&frac{1}{24}sum^infty_{n=1}H_nfrac{partial^4}{partial n^4}left(frac{1}{(n+1)2^{n+1}}
ight)=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}-{
m Li}_6left(frac{1}{2}
ight)+{
m Li}_5left(frac{1}{2}
ight)ln{2}+frac{1}{32}zeta(5)ln{2}+{
m Li}_4left(frac{1}{2}
ight)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}-{
m Li}_6left(frac{1}{2}
ight)+{
m Li}_5left(frac{1}{2}
ight)ln{2}+frac{1}{32}zeta(5)ln{2}+{
m Li}_4left(frac{1}{2}
ight)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(ar{5},1)) as
[egin{align*}
zeta(ar{5},1)
=&-2sum^infty_{n=1}frac{H_n}{n^52^n}+6{
m Li}_6left(frac{1}{2}
ight)-frac{17pi^6}{5040}+2{
m Li}_5left(frac{1}{2}
ight)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
[egin{align*}
sum^infty_{n=1}frac{H_n}{n^52^n}
=&3{
m Li}_6left(frac{1}{2}
ight)-frac{1}{2}zeta(ar{5},1)-frac{17pi^6}{10080}+{
m Li}_5left(frac{1}{2}
ight)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(ar{5},1))
[Largeoxed{displaystyle int_0^1frac{ln^3(1+x)\,ln^2x}xmathrm{d}x=color{blue}{-frac{pi^6}{252}-18zeta(ar{5},1)+3zeta^2(3)}}
]