Evaluate $\int_{0}^{1}x\sqrt{1+x^{3}}\mathrm{d}x$ How to evaluate 
$$\displaystyle \int_{0}^{1}x\sqrt{1+x^{3}}\mathrm{d}x$$
I tried
$$\int_{0}^{1}x\sqrt{1+x^{3}}\mathrm{d}x=\frac{1}{3}\int_{0}^{1}x^{-\frac{1}{3}}\left ( 1+x \right )^{\frac{1}{2}}\mathrm{d}x$$
But I can't figure out how to go on.
 A: Let's see the more general form below
$$\int_{0}^{u}y^{b -1}\left ( u-y \right )^{c-b-1}\left ( y+\frac{u}{x} \right )^{-a}\,\mathrm{d}t$$
consider the hypergeometric functions $_{2}F_{1}$
$$_{2}F_{1}\left ( a,b;c;x \right )=\frac{1}{\mathrm{B}\left ( b,c-b \right )}\int_{0}^{1}t^{b-1}\left ( 1-t \right )^{c-b-1}\left ( 1-tx \right )^{-a}\,\mathrm{d}t$$
where
$$\mathrm{B}\left ( a,b \right )=\int_{0}^{1}t^{a-1}\left ( 1-t \right )^{b-1}\,\mathrm{d}t$$
then we get
$$\int_{0}^{1}t^{b}\left ( 1-t \right )^{c}\left ( 1-tx \right )^{a}\,\mathrm{d}t=\mathrm{B}\left ( b+1,c+1 \right )\, _{2}F_{1}\left (-a,b+1;b+c+2;x  \right )$$
do the substitution $y=tu~,~x\rightarrow -x$ ,
\begin{align*}
\int_{0}^{1}t^{b}\left ( 1-t \right )^{c}\left ( 1-tx \right )^{a}\,\mathrm{d}t&=\int_{0}^{1}\left ( \frac{y}{u} \right )^{b}\left ( 1-\frac{y}{u} \right )^{c}\left ( 1+\frac{yx}{u} \right )^{a}\frac{1}{u}\,\mathrm{d}y\\
&=\left ( \frac{u}{x} \right )^{-a}u^{-b-c-1}\int_{0}^{u}y^{b}\left ( u-y \right )^{c}\left ( y+\frac{u}{x} \right )^{a}\,\mathrm{d}y
\end{align*}
then let $b+1\rightarrow b~,~c+1\rightarrow c-b~,~a\rightarrow -a$ we have
$$\int_{0}^{u}y^{b -1}\left ( u-y \right )^{c-b-1}\left ( y+\frac{u}{x} \right )^{-a}\,\mathrm{d}t=\left ( \frac{u}{x} \right )^{a}u^{c-1}\mathrm{B}\left ( b,c-b \right )\, _{2}F_{1}\left ( a,b;c;-x \right )$$
so,make $u=1~,~x=1~,~b=\dfrac{2}{3}~,~c=\dfrac{5}{3}~,~a=-\dfrac{1}{2}$ and the answer will follow.
