My attempt:
$$\phi =\frac{1+\sqrt{5}}{2}, \; \bar{\phi }=\frac{1-\sqrt{5}}{2}\Rightarrow y^4-y^3+y^2-y+1=\left ( y^2-\phi y+1 \right )\left ( y^2-\bar{\phi }y+1 \right )$$
\begin{multline*}
\int\limits_{0}^{8}\frac{dx}{x^2+\sqrt[3]{x}}\overset{y=\sqrt[3]{x}}{=}\int\limits_{0}^{2}\frac{3y^2}{y^6+y}dy=\int\limits_{0}^{2}\frac{3y}{y^5+1}dy=\int\limits_{0}^{2}\frac{3ydy}{(y+1)(y^2-\phi y+1)(y^2-\bar{\phi }y+1)}=\\=3\int\limits_{0}^{2}\left ( \frac{A}{y+1}+\frac{By+C}{y^2-\phi y+1}+\frac{Dy+E}{y^2-\bar{\phi }y+1} \right )dy=\\=3\Bigg( A\ln(y+1)\Bigg|_0^{2}+\frac{B}{2}\ln(y^2-\phi y+1)\Bigg|_0^{2}+\left ( C-\frac{B}{2} \right )\int\limits_{0}^{2}\frac{dy}{y^2-\phi y+1}+\\+\frac{D}{2}\ln\left ( y^2-\bar{\phi }y+1 \right )\Bigg|_{0}^{2}+\left ( E-\frac{D}{2} \right )\int\limits_{0}^{2}\frac{dy}{y^2-\bar{\phi }+1} \Bigg)=\\=3\Bigg( A\ln 2+\frac{B}{2}\ln (5-2\phi )+\left ( C-\frac{B}{2} \right )\frac{2}{\sqrt{4-\phi ^2}}\arctan \left ( \frac{2y+\phi}{\sqrt{4-\phi^2}} \right )\Bigg|_0^{2}+\\+\frac{D}{2}\ln \left ( 5-2\bar{\phi} \right )+\left ( E-\frac{D}{2} \right )\frac{2}{\sqrt{4-\bar{\phi}^2}}\arctan \left ( \frac{2y+\bar{\phi}}{\sqrt{4-\bar{\phi}^2}} \right )\Bigg|_{0}^2 \Bigg)
\end{multline*}
Where
$$\left ( A,B,C,D,E \right )=\left ( -\frac{1}{5},\frac{1-\sqrt{5}}{10},\frac{1-\sqrt{5}}{10},\frac{1+\sqrt{5}}{10},\frac{1+\sqrt{5}}{10} \right )$$
Wolfram writes such an answer through a hypergeometric function:
$$\int\frac{dx}{x^2+\sqrt[3]{x}}=\frac{3}{2}x^{2/3}{}_2F_1\left ( \frac{2}{5},1;\frac{7}{5};-x^{5/3} \right )+C$$
How do you get the same answer and calculate a definite integral through this answer?
Best Answer
The question is - when i correctly get is - as follows. We have an integral on $[0,8]$, $$ J = \int_0^8 \frac1{x^2 + x^{1/3}}\;dx\ , $$ we have (A) one way to compute it explicitly in terms of elementary functions, and we have (B) a rather direct way to express a primitive of $x\to 1/(x^2 + x^{1/3})$ in terms of a special hypergeometric function, in essence some ${}_2F_1$-function.
The question is how can we go from the corresponding values of the obtained ${}_2F_1$-function to the computed result. The computed result is a complicated linear combination of values of $\log$ (evaluated at some complex numbers). And well, values of hypergeometric functions are not immediately such known $\log$-values (and linear combinations of them involving algebraic numbers as scalars).
I will do something, to realize a bridge between the two "worlds", (A) and (B), without going through the integral again, but do not expect a perfect term by term match of terms, terms from the given sum with terms from broken pieces of the hypergeometric series.
Note that the integral to be computed is a special incomplete beta value, see also Wolfram on the Incomplete Beta Function, so $$ \begin{aligned} B(z;a,b) &= \int_0^z u^{a-1}(1-u)^{b-1}\; du \\ &= \frac 1az^a\; {}_2F_1(a,1-b;\ a+1;\ z) \\ &= z^a\sum_{n\ge 0}\frac{(1-b)_n}{n!\;(a+n)}\; z^n \ , \\ &\qquad\text{ and on the other side} \\ B(z;a,b) &= \int_0^z u^{a-1}(1-u)^{b-1}\; du \\ &\qquad u=\frac v{1+v}=1-\frac1{1+v}\ ,\ v=\frac u{1-u}\ ,\ du=\frac{dv}{(1+v)^2} \\ &= \int_0^{z/(1-z)}\frac{v^{a-1}}{(1+v)^{a-1}}\cdot\frac 1{(1+v)^{b-1}}\; \frac{dv}{(1+v)^2} \\ &= \int_0^{z/(1-z)}\frac{v^{a-1}}{(1+v)^{a+b}}\;dv \end{aligned} $$
In our case: $$ \begin{aligned} J &= \int_0^8 \frac1{x^2 + x^{1/3}}\;dx \qquad(y=x^{1/3}\ ,\ x=y^3\ ,\ dx=3y^2\; dy) \\ &= \int_0^2 \frac1{y^6 + y}\;3y^2\; dy = 3\int_0^2 \frac y{y^5 + 1}\; dy\qquad(y^5=v) \\ &= \frac 35\int_0^{32}\frac{v^{-3/5}}{1+v}\; dv = \frac 35\int_0^{32}\frac{v^{2/5-1}}{(1+v)^{2/5+3/5}}\; dv \\ &= \frac 35 B\left(\frac {32}{33};\ \frac 25,\frac 35\right) \\ &= \frac 35 \cdot\frac 52 \cdot\left(\frac{32}{33}\right)^{2/5} \;{}_2F_1\left(\frac 25,\frac 25;\ \frac 75;\ \frac{32}{33}\right) \\ &= \frac 35 \cdot\frac 52 \cdot\left(\frac{32}{33}\right)^{2/5} \cdot\left(\frac{ 1}{33}\right)^{-2/5} \;{}_2F_1\left(\frac 25,1;\ \frac 75;\ -32\right) \\ &=6\cdot\;{}_2F_1\left(\frac 25,1;\ \frac 75;\ -32\right) \ . \end{aligned} $$ This is exactly the value of the primitive function delivered by Wolfram, mentioned in the question, taken between $0$ and $8$. So we have shown this relation. This is not answering the question, but it shows that Wolfram is not doing anything special, it just recognizes the given integral as an incomplete beta function.
To do something in the direction of connecting the "worlds" (A) and (B), we need a way to express $\displaystyle {}_2F_1\left(\frac 25,1;\ \frac 75;\ -32\right)$ as a sum of $\log$-values. (Without going back through $J$, but rather by reshaping the obtained hypergeometric function value.)
To do something, we use functional equations. Above, we have used one of the formulas (Pfaff) $$ \begin{aligned} {}_2F_1(a,b;\ c;\ z) &= (1-z)^{c-a-b}\cdot\; {}_2F_1(c-a,c-b;\ c;\ z) &&\text{(Euler)} \\ {}_2F_1(a,b;\ c;\ z) &= (1-z)^{-a}\cdot\; {}_2F_1\left(a,c-b;\ c;\ \frac z{z-1}\right) &&\text{(Pfaff)} \end{aligned} $$ and we went from the "small" argument $z=32/33\in(0,1)$ to the "beautiful" argument $z=-32$. If we want to use series, and we do want, this "beautiful" argument $z=-32$ is not really giving us convergent series. Going back to the "small" argument, as above or the other way around is not really helping us... (Applying twice Pfaff with a switch inside the $(a,b)$ argument we get Euler.) For instance $$ J = 6\cdot\;{}_2F_1\left(\frac 25,1;\ \frac 75;\ -32\right) = \frac 2{11}\cdot\;{}_2F_1\left(1,1;\ \frac 75;\ \frac{32}{33}\right) \ . $$ (And now what...?) So we use a general variable $z$ and analytic continuation to proceed. (Some monodromy may have to be controlled on the road.)
With a general argument $z$ we have $$ \begin{aligned} {}_2F_1\left(\frac 25,1;\ \frac 75;\ z\right) &= \sum_{n\ge 0} \frac{(2/5)_n\; (1)_n}{(7/5)_n\;\phantom{ (1)_n}} \cdot\frac 1{n!} \cdot z^n \\ &=1+\sum_{n\ge 1}\frac {\frac 25}{\frac 25+n}\cdot z^n \\ &=1+2\sum_{n\ge 1}\frac 1{5n+2}\cdot z^n \ . \end{aligned} $$ I hope the connection to $\log$-values is now clear. (We have to consider a linear combination of $\log$-values in $z$, $\zeta_5 z$, $\zeta_5^2 z$, $\zeta_5^3 z$, $\zeta_5^4 z$, here $\zeta_5$ being a complex primitive fifth root of the unity.) We still cannot plug in $z=-32$, but we can use the functional equation for $\log$ to go to values we can plug in.
Details can be given, but typing them would not add a substantial value in the direction of asking the stated question. To see that we (may) get the same value (explicitly or in a hidden way) could depend on the fact that computing the "general version" of $\frac J3=\int_0^2 \frac y{y^5 + 1}\; dy$ which is (analytic, so we restrict to a good region, and work with a complex $w$ with $|w|<1$) $$ \begin{aligned} \int_0^w \frac y{y^5 + 1}\; dy &= \int_0^w(y-y^6+y^{11}-y^{16}+\dots)\; dy \\ &= \frac 12w^2 - \frac 17 w^7 + \frac 1{12}w^{12} - \frac 1{17}w^{17} \pm\dots \end{aligned} $$