Evaluation of $\int \frac{\sqrt[3]{x+\sqrt[4]{x}}}{\sqrt{x}}dx$ Evaluation of $\displaystyle \int \frac{\sqrt[3]{x+\sqrt[4]{x}}}{\sqrt{x}}dx$
$\bf{My\; Try::}$ Let $x=t^4\;,$ Then $dx = 4t^3dt$
So Integral is $\displaystyle \int\frac{\sqrt[3]{t^4+t}}{t^2} \cdot 4t^3dt$
So Integral is $\displaystyle 4\int t^{\frac{7}{3}}\cdot (1+t^{-3})^{\frac{1}{3}}$
Now How can i solve after that
Help me
Thanks
 A: Let $\mathcal{I}$ be the integral. You can actually evaluate it using the substitution $x = t^4$.
$$\mathcal{I} = 
\int\frac{\sqrt[3]{x + \sqrt[4]{x}}}{\sqrt{x}}dx
= \int\frac{\sqrt[3]{(t^3 + 1)t}}{t^2}4t^3dt
= 4\int\sqrt[3]{1+t^3}t^{4/3}dt
$$
For $|x| < 1$, we can expand the integrand at RHS using following expansion
$$\frac{1}{(1-t)^\gamma} = \sum_{k=0}^\infty \frac{(\gamma)_k}{k!} t^k$$
where $(\gamma)_k = \gamma(\gamma+1)\cdots(\gamma+k-1)$ is the rising Pochhammer symbol. This gives us
$$
\mathcal{I} 
= 4\int\left(\sum_{k=0}^\infty \frac{(-1)^k (-\frac13)_k}{k!}t^{3k}\right)t^{4/3}dt
= 4 \sum_{k=0}^\infty \frac{(-1)^k (-\frac13)_k}{k!}\frac{t^{3k+7/3}}{3k+7/3}
$$
Using another identity
$$\frac{(\gamma)_k}{(\gamma+1)_k} = \frac{\gamma}{\gamma+k}$$
We can transform above expression to
$$\mathcal{I} 
= \frac{12}{7} t^{\frac73}
\sum_{k=0}^\infty \frac{(-\frac13)_k}{k!}\frac{(\frac79)_k}{(\frac{16}{9})_k}(-t^3)^k
$$
The expansion in the right is that for a 
hypergeometric function ${}_2F_1$. As a result, up to an integration constant, we have
$$\mathcal{I}
= \frac{12}{7} t^{\frac73} {}_2F_1\left( -\frac13, \frac79 ; \frac{16}{9}; -t^3 \right)
= \frac{12}{7} x^{\frac{7}{12}} {}_2F_1\left( -\frac13, \frac79 ; \frac{16}{9}; -x^{\frac34} \right)$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int{\root[3]{x + \root[4]{x}} \over \root{x}}\,\dd x:\ {\large ?}}$

$$
\mbox{Lets consider}\quad\fermi\pars{x}\equiv
\int_{0}^{x}{\root[3]{t + \root[4]{t}} \over \root{t}}\,\dd t\,,\qquad x > 0
$$

\begin{align}
\fermi\pars{x}&=
\int_{0}^{x^{1/4}}{\root[3]{t^{4} + t} \over t^{2}}\,4t^{3}\,\dd t
=4\int_{0}^{x^{1/4}}\root[3]{t^{3} + 1}t^{4/3}\,\dd t
\\[3mm]&=4\int_{0}^{x^{3/4}}\pars{1 + t}^{1/3}t^{4/9}\,{1 \over 3}\,t^{-2/3}\dd t
={4 \over 3}\int_{0}^{x^{3/4}}\pars{1 + t}^{1/3}t^{-2/9}\dd t
\end{align}

Set $\ds{\xi \equiv {1 \over 1 + t}\quad\imp\quad t = {1 \over \xi} - 1}$:
  \begin{align}
\fermi\pars{x}&={4 \over 3}\int_{1}^{1/\pars{1 + x^{3/4}}}
\xi^{-1/3}\pars{{1 \over \xi} - 1}^{-2/9}\pars{-\,{\dd\xi \over \xi^{2}}}
\\[3mm]&={4 \over 3}\int_{1/\pars{1 + x^{3/4}}}^{1}
\xi^{-19/9}\pars{1 - \xi}^{-2/9}\,\dd\xi
\end{align}

The final result can be expressed in terms of the
Generalized Incomplete Beta Function
$\ds{{\rm B}\pars{z_{1},z_{2},a,b}}$:
\begin{align}
\fermi\pars{x}\equiv
\color{#66f}{\large\int_{0}^{x}{\root[3]{t + \root[4]{t}} \over \root{t}}\,\dd t
={4 \over 3}\,
{\rm B}\pars{{1 \over 1 + x^{3/4}},1,-\,{10 \over 9},{7 \over 9}}}
\end{align}
A: As Pranav Arora showed, there is no nice answer to this antiderivative. Personally, the only way I can think about it is a Taylor expansion of the integrand followed by a term by term integration.
As, you did, starting with $x=t^4$,we have $$\frac{\sqrt[3]{x+\sqrt[4]{x}}}{\sqrt{x}}=t^{-\frac{5}{3}} \sqrt[3] {1+t^3}=t^{-\frac{5}{3}} \Big(1+\frac{t^3}{3}-\frac{t^6}{9}+\frac{5 t^9}{81}-\frac{10
   t^{12}}{243}+O\left(t^{13}\right)\Big)$$
Then, replacing $t$ by $\sqrt[4] x$, we can find  that $$\frac{\sqrt[3]{x+\sqrt[4]{x}}}{\sqrt{x}}=\frac{1}{x^{5/12}}+\frac{\sqrt[3]{x}}{3}-\frac{x^{13/12}}{9}+\frac{5
   x^{11/6}}{81}-\frac{10 x^{31/12}}{243}+\frac{22 x^{10/3}}{729}-\frac{154
   x^{49/12}}{6561}+\frac{374 x^{29/6}}{19683}+O\left(x^{61/12}\right)$$ Now integration leads to 
$$\int \frac{\sqrt[3]{x+\sqrt[4]{x}}}{\sqrt{x}}dx=\frac{12 x^{7/12}}{7}+\frac{x^{4/3}}{4}-\frac{4 x^{25/12}}{75}+\frac{10
   x^{17/6}}{459}-\frac{40 x^{43/12}}{3483}+\frac{22 x^{13/3}}{3159}-\frac{616
   x^{61/12}}{133407}+\frac{748 x^{35/6}}{229635}+O\left(x^{73/12}\right)$$
If we compare the exact and approximate solutions for $$\int_0^a \frac{\sqrt[3]{x+\sqrt[4]{x}}}{\sqrt{x}}dx$$ they match quite well for $0 \leq a \leq 2$.
