# 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

• W|A doesn't give a nice result. wolframalpha.com/input/… – Pranav Arora Jun 21 '14 at 4:34
• Not that it means much, but a bit of fiddling did not yield anything useful. If it were a definite integral over certain nice intervals, more tools would be available. – André Nicolas Jun 21 '14 at 4:40
• Not sure how you are moving to the last step there. Wouldn't the last integrand just be $t^{4/3}(1+t^3)^{1/3}$? – ClassicStyle Jun 21 '14 at 4:48

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

$$\mathcal{I} = \frac{12}{7} t^{\frac73} {}_2F_1\left( -\frac13, \frac79 ; \frac{16}{9}; -t^3 \right) = \frac{12}{7} x^{\frac{7}{12}} {}_2F_1\left( -\frac13, \frac79 ; \frac{16}{9}; -x^{\frac34} \right)$$

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int{\root[3]{x + \root[4]{x}} \over \root{x}}\,\dd x:\ {\large ?}}$

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

\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}

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 $0 \leq a \leq 2$.