# how to integrate : $\int_{x^2}^{x^3}\frac{dt}{\sqrt{1+t^4}}$

Need to compute:

$$\int_{x^2}^{x^3}\frac{dt}{\sqrt{1+t^4}}$$

I tried to use partial fraction but got a messy algebra.

thanks.

#### Solutions Collecting From Web of "how to integrate : $\int_{x^2}^{x^3}\frac{dt}{\sqrt{1+t^4}}$"

If you only interested in the cases of real number $x$ :

Case $1$: $x\geq1$

Then $\int_{x^2}^{x^3}\dfrac{dt}{\sqrt{1+t^4}}$

$=\int_{x^2}^{x^3}\dfrac{dt}{t^2\sqrt{1+\dfrac{1}{t^4}}}$

$=\int_{x^2}^{x^3}\dfrac{1}{t^2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^{-4n}}{4^n(n!)^2}dt$

$=\int_{x^2}^{x^3}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^{-4n-2}}{4^n(n!)^2}dt$

$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^{-4n-1}}{4^n(n!)^2(-4n-1)}\right]_{x^2}^{x^3}$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(4n+1)x^{8n+2}}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(4n+1)x^{12n+3}}$

Case $2$: $-1\leq x\leq1$

Then $\int_{x^2}^{x^3}\dfrac{dt}{\sqrt{1+t^4}}$

$=\int_{x^2}^{x^3}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^{4n}}{4^n(n!)^2}dt$

$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^{4n+1}}{4^n(n!)^2(4n+1)}\right]_{x^2}^{x^3}$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{12n+3}}{4^n(n!)^2(4n+1)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{8n+2}}{4^n(n!)^2(4n+1)}$

Case $3$: $x\leq-1$

Then $\int_{x^2}^{x^3}\dfrac{dt}{\sqrt{1+t^4}}$

$=-\int_{x^3}^{x^2}\dfrac{dt}{\sqrt{1+t^4}}$

$=-\int_{x^3}^{-1}\dfrac{dt}{\sqrt{1+t^4}}-\int_{-1}^1\dfrac{dt}{\sqrt{1+t^4}}-\int_1^{x^2}\dfrac{dt}{\sqrt{1+t^4}}$

$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^{-4n-1}}{4^n(n!)^2(-4n-1)}\right]_{x^3}^{-1}-\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^{4n+1}}{4^n(n!)^2(4n+1)}\right]_{-1}^1-\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^{-4n-1}}{4^n(n!)^2(-4n-1)}\right]_1^{x^2}$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-12n-3}}{4^n(n!)^2(4n+1)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!(-1)^{-4n-1}}{4^n(n!)^2(4n+1)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(4n+1)}+\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!(-1)^{4n+1}}{4^n(n!)^2(4n+1)}+\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-8n-2}}{4^n(n!)^2(4n+1)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(4n+1)}$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(4n+1)x^{12n+3}}+\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(4n+1)x^{8n+2}}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{2^{2n-1}(n!)^2(4n+1)}$

For any $x \in [0,\infty]$, let $I(x)$ and $J(x)$ be the integrals
$$I(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}} \quad\text{ and }\quad J(x) = \int_{x^2}^{x^3} \frac{dt}{\sqrt{1+t^4}} = I(x^3) – I(x^2)$$
Notice under the substitution $t = \frac{1}{u}$,
$$\frac{dt}{\sqrt{1+t^4}} = – \frac{du}{\sqrt{1+u^4}}$$
We find $I(x)$ and $J(x)$ satisfy following identities:

$$I(x) + I(\frac{1}{x}) = I(\infty)\quad\text{ and }\quad J(x) = -J(\frac{1}{x})$$
This means we only need to figure out what $I(x)$ is for $0 < x < 1$.

Part 1 : As an Hypergeometric function

For $|t| < 1$, $\frac{1}{\sqrt{1+t^4}}$ has an absolute convergence power series expansion. We can integrate the expansion term by term to get

$$I(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}} = \int_0^x \left[ \sum_{k=0}^{\infty} \frac{(\frac12)_k}{k!}(-t^4)^k\right] dt = \sum_{k=0}^\infty \frac{(\frac12)_k}{k!}\frac{(-1)^k x^{4k+1}}{4k+1}$$
where $(\gamma)_k = \gamma(\gamma+1)\cdots(\gamma+k-1)$ is the rising
Pochhammer symbol. Notice

$$\frac{(\gamma)_k}{(\gamma+1)_k} = \frac{\gamma(\gamma+1)\cdots(\gamma+k-1)}{(\gamma+1)(\gamma+2)\cdots(\gamma+k)} = \frac{\gamma}{\gamma+k}$$
We have $\frac{1}{4k+1} = \frac{(\frac14)_k}{(\frac54)_k}$ and $I(x)$ becomes

$$I(x) = x \sum_{k=0}^\infty \frac{(\frac14)_k (\frac12)_k}{k!(\frac54)_k}(-x^4)^k$$
The expansion of the RHS is that for a
Hypergeometric function with
argument $-x^4$. More precisely,

$$\begin{cases} I(x) &= x\cdot{}_2F_1( \frac14, \frac12; \frac54 ; -x^4)\\ J(x) &= I(x^3) – I(x^2) = x^3\cdot{}_2F_1( \frac14, \frac12; \frac54 ; -x^{12}) – x^2\cdot{}_2F_1( \frac14, \frac12; \frac54 ; -x^8) \end{cases}$$

Even though it is sort of hard to extract analytic relations from hypergeometric functions,
there are highly efficient numeric routines around to evaluate them accurately. If what you
want is to evaluate it and get a number, this will be a good representation for $I(x)$ and hence $J(x)$.

On WolframAlpha, you can access the hypergeometric function though the function
$\bf\text{HypergeometricPFQ}$ and evaluate $I(x)$ using the expression:

$$\bf x*\text{HypergeometricPFQ}[\{1/4,1/2\},\{5/4\},-x^4]$$

Part 2: As an elliptic integral

As pointed out by others, the integrals can be recasted as elliptical integrals.
Introduce variable $c$ and $s$ such that

$$t^2 = \frac12 \left(\frac{1}{c^2} – c^2\right)\quad\text{ and }\quad s = \sqrt{1-c^2}$$

We have $\quad\displaystyle \sqrt{1+t^4} = \sqrt{ 1 + \left(\frac12 \left(\frac{1}{c^2} – c^2\right)\right)^2 } = \frac12 \left(\frac{1}{c^2} + c^2\right).$

This implies
$\quad\displaystyle 2tdt = dt^2 = -\left(\frac{1}{c^3} + c\right) dc = -2\sqrt{1+t^4}\frac{dc}{c}$ and as a result,
\begin{align} \frac{dt}{\sqrt{1+t^4}} &= -\frac{dc}{tc} = -\frac{cdc}{tc^2} = \frac{sds}{tc^2} = \frac{sds}{c\sqrt{\frac12\left(1-c^4\right)}} = \frac{ds}{c\sqrt{\frac12 (1+c^2)}}\\ & = \frac{ds}{\sqrt{(1-s^2)(1 – \frac12 s^2})} \end{align}
Notice $s^2 = 1 – c^2 = 1 – \left(\frac12(\frac{1}{c^2} + c^2) – \frac12(\frac{1}{c^2} – c^2)\right) = 1 + t^2 – \sqrt{1+t^4}$, we obtain
an alternate representation of $I(x)$:

$$I(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}} = F( \sqrt{1 + x^2 – \sqrt{1+x^4}} \;; \frac12 )$$

where $\quad\displaystyle F(y \;; m ) = \int_0^y \frac{ds}{\sqrt{(1-s^2)(1-ms^2)}}\quad$
is the Jacobi’s form of incomplete elliptic integral of the first kind.
Please note that there is more than one form/convention of elliptic integrals. A lot of authors use the same term “incomplete elliptic integral of the first kind” for following integral:

$$F( \phi \mid m ) = \int_0^\phi \frac{d\theta}{\sqrt{1-m\sin^2\theta}} = \int_0^{\sin\phi} \frac{ds}{\sqrt{(1-s^2)(1-ms^2)}} = F( \sin\phi \;; m)$$
On WolframAlpha, you can access the second form using the function $\bf \text{EllipticF}$ and
evaluate $I(x)$ using the expression:

$$\bf \text{EllipticF}[ \text{ArcSin}[ \text{Sqrt}[1+x^2-\text{Sqrt}[1+x^4]] ], m ]$$