# Integrals of Pairs of Products of Squares of Jacobi Elliptic Functions

We use combinations of the following to evaluate integrals in this post:

The order of presentation here is done to avoid look ahead bias, thus avoiding circularity when referencing previous results.

All results can be found in section 361 of Handbook of Elliptic Integrals for Engineers and Scientists by Byrd and Friedman. We use their equation numbering scheme.

Note that this post will be continuously updated as results are required for evaluating other integrals.

## BF 361.02

We begin with

\frac{\partial}{\partial u}\mathrm{cn} = -\,\mathrm{sn} \, \mathrm{dn}

Squaring both sides, integrating by parts, and then making substitutions for $$\mathrm{cn}^{2}$$, we have
\begin{align}
\int \mathrm{sn}^{2} \mathrm{dn}^{2} &= \int \left(\frac{\partial}{\partial u}\mathrm{cn}\right) \left(\frac{\partial}{\partial u}\mathrm{cn}\right) \\
&= -\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} + \int \mathrm{cn}^{2} \mathrm{dn}^{2} – k^{2}\int \mathrm{cn}^{2} \mathrm{sn}^{2} \\
&= -\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} + \int (1 – \,\mathrm{sn}^{2})\,\mathrm{dn}^{2} – \int (\mathrm{dn}^{2} – {k^{\prime}}^{2} ) \mathrm{sn}^{2} \\
&= \frac{1}{3 k^{2}}\left[{k^{\prime}}^{2}u + (2k^{2} – 1)E \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{align}

## BF 361.01

From BF 361.02 we have

\int \mathrm{sn}^{2} ({k^{\prime}}^{2} + k^{2}\mathrm{cn}^{2}) = \frac{1}{3 k^{2}}\left[{k^{\prime}}^{2}u + (2k^{2} – 1)E \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]

Rearranging yields

\int \mathrm{sn}^{2} \mathrm{cn}^{2} = \frac{1}{3 k^{4}}\left[(2 -\,k^{2})E \,-\,2{k^{\prime}}^{2}u \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]

## BF 361.03

From BF 361.02 we have

\int (1 \,-\,\mathrm{cn}^{2}) \mathrm{dn}^{2} = \frac{1}{3 k^{2}}\left[{k^{\prime}}^{2}u + (2k^{2} – 1)E \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]

Rearranging yields

\int \mathrm{dn}^{2} \mathrm{cn}^{2} = \frac{1}{3 k^{2}}\left[(1+k^{2})E \,-\,{k^{\prime}}^{2}u + k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]

## BF 361.05

Let

\mathrm{sc}^{2} \mathrm{dn}^{2} = (\mathrm{nc}^{2}-1)\mathrm{dn}^{2} = \mathrm{nc}^{2} \mathrm{dn}^{2} – \mathrm{dn}^{2} = \mathrm{dc}^{2} – \mathrm{dn}^{2}

And thus,

\int \mathrm{sc}^{2} \mathrm{dn}^{2} = \int \mathrm{dc}^{2} – \int \mathrm{dn}^{2} = u \,-\, 2E + \mathrm{dn}\,\mathrm{sc}

## BF 361.06

Let

\mathrm{cs}^{2} \mathrm{dn}^{2} = (\mathrm{ns}^{2}-1)\mathrm{dn}^{2} = \mathrm{ns}^{2} \mathrm{dn}^{2} – \mathrm{dn}^{2} = \mathrm{ds}^{2} – \mathrm{dn}^{2}

And thus,

\int \mathrm{cs}^{2} \mathrm{dn}^{2} = \int \mathrm{ds}^{2} – \int \mathrm{dn}^{2} = {k^{\prime}}^{2}u \,-\, 2E \,-\, \mathrm{dn}\,\mathrm{cs}

## BF 361.08

\begin{align}
\int \mathrm{ds}^{2} \mathrm{dn}^{2} &= \int (\mathrm{ns}^{2}-k^{2})\mathrm{dn}^{2} = \int \mathrm{ds}^{2} \,-\, k^{2} \int \mathrm{dn}^{2} \\
&= {k^{\prime}}^{2}u \,-\, (1+k^{2})E \,-\, \mathrm{dn}\,\mathrm{cs}
\end{align}

## BF 361.14

\int \mathrm{ds}^{2} \mathrm{dc}^{2} = \int (\mathrm{cs}^{2} + {k^{\prime}}^{2})\mathrm{dc}^{2} = 2{k^{\prime}}^{2}u \,-\, (1+{k^{\prime}}^{2})E + \mathrm{dn}({k^{\prime}}^{2}\mathrm{sc} \,-\, \mathrm{cs})

## BF 361.09

\int \mathrm{cs}^{2} \mathrm{cn}^{2} = \int (\mathrm{ns}^{2} \,-\, 1) \mathrm{cn}^{2} = \int (\mathrm{cs}^{2} \,-\, \mathrm{cn}^{2}) = \frac{1}{k^{2}} [(k^{2} + 1)E + {k^{\prime}}^{2}u \,-\, k^{2}\mathrm{dn}\,\mathrm{cs}]

## BF 361.10

\int \mathrm{ns}^{2} \mathrm{nc}^{2} = \int (\mathrm{cs}^{2} + 1) \mathrm{nc}^{2} = \int (\mathrm{ns}^{2} + \mathrm{nc}^{2}) = \frac{1}{{k^{\prime}}^{2}} [2{k^{\prime}}^{2}u \,-\, (1+{k^{\prime}}^{2})E + \mathrm{dn}(\mathrm{sc} \,-\, {k^{\prime}}^{2}\mathrm{cs})]

## BF 361.11

Let

\mathrm{ns}^{2} \mathrm{nd}^{2} = (\mathrm{ds}^{2} + k^{2})\mathrm{nd}^{2} = \mathrm{ns}^{2} + k^{2}\mathrm{nd}^{2}

\int \mathrm{ns}^{2} \mathrm{nd}^{2} = \int (\mathrm{ns}^{2} + k^{2}\mathrm{nd}^{2}) = \frac{1}{{k^{\prime}}^{2}} [{k^{\prime}}^{2}u + (k^{2} \,-\, {k^{\prime}}^{2})E \,-\, \mathrm{cn}({k^{\prime}}^{2}\mathrm{ds} + k^{4}\mathrm{sd})]

## BF 361.12

Let

\mathrm{nc}^{2} \mathrm{nd}^{2} = \frac{\mathrm{dc}^{2} \,-\, k^{2}}{{k^{\prime}}^{2}} \mathrm{nd}^{2} = \frac{\mathrm{nc}^{2}}{{k^{\prime}}^{2}} \,-\, \frac{k^{2} \mathrm{nd}^{2}}{{k^{\prime}}^{2}}

\int \mathrm{nc}^{2} \mathrm{nd}^{2} = \int \left( \frac{\mathrm{nc}^{2}}{{k^{\prime}}^{2}} \,-\, \frac{k^{2} \mathrm{nd}^{2}}{{k^{\prime}}^{2}} \right) = \frac{1}{{k^{\prime}}^{4}} [{k^{\prime}}^{2}u \,-\, (1+k^{2})E + \mathrm{sn} (\mathrm{dc} + k^{4}\mathrm{cd})]

## BF 361.20

\begin{align}
\int \mathrm{cs}^{2} \mathrm{nd}^{2} &= \int \mathrm{cs}^{2} (k^{2} \mathrm{sd}^{2} + 1) = \int (k^{2} \mathrm{cd}^{2} + \mathrm{cs}^{2}) \\
&= u \,-\, 2E + \mathrm{cn}(k^{2} \mathrm{sd} \,-\, \mathrm{ds})
\end{align}

## BF 361.31

\begin{align}
\int \mathrm{dn}^{2} \mathrm{dc}^{2} &= \int (k^{2}\mathrm{cn} + {k^{\prime}}^{2}) \mathrm{dc} = \int (k^{2}\mathrm{dn}^{2} + {k^{\prime}}^{2}) \mathrm{dc}^{2} \\
&= {k^{\prime}}^{2}u + (1 \,-\, 2{k^{\prime}}^{2})E + {k^{\prime}}^{2}\mathrm{dn}\,\mathrm{sc}
\end{align}

## BF 361.07

We begin with

\frac{\partial \,\mathrm{dc}}{\partial u} = {k^{\prime}}^2\mathrm{sn}\,\mathrm{nc}^2

Then we multiply both sides by $$\mathrm{sc}\,\mathrm{nc}$$, integrate by parts, and then make substitutions for squares of Jacobi elliptic functions.
\begin{align}
{k^{\prime}}^{2} \int \mathrm{nc}^2 \mathrm{sc}^2 &= \int \left( \frac{\partial \,\mathrm{dc}}{\partial u} \right) \mathrm{sc}\,\mathrm{nc} \\
\label{eq:1}
\tag{1}
&= \mathrm{dc}\,\mathrm{sc}\,\mathrm{nc} \,-\, \int \mathrm{dc}^2 \mathrm{sc}^2 \,-\, \int \mathrm{dc}^2 \mathrm{nc}^2 \\
&= \mathrm{dc}\,\mathrm{sc}\,\mathrm{nc} \,-\, \int ({k^{\prime}}^{2}\mathrm{nc}^2 + k^{2}) \mathrm{sc}^2 \,-\, \int ({k^{\prime}}^{2}\mathrm{sc}^2 + 1) \mathrm{nc}^2 \\
\end{align}
Collecting terms, we have

3{k^{\prime}}^{4} \int \mathrm{nc}^2 \mathrm{sc}^2 = \mathrm{sc}\,\mathrm{dn} ({k^{\prime}}^{2}\mathrm{nc}^2 \,-\, 1 \,-\, k^{2}) \,-\, {k^{\prime}}^{2}u + (k^{2} + 1)E

## BF 361.15

\begin{align}
\int \mathrm{dc}^2 \mathrm{sc}^2 &= \int (k^{2} + {k^{\prime}}^{2} \mathrm{nc}^2)\mathrm{sc}^2 = k^{2} \int \mathrm{sc}^2 + {k^{\prime}}^{2} \int \mathrm{nc}^2 \mathrm{sc}^2 \\
&= \frac{1}{3{k^{\prime}}^{2}} [(1 \,-\, 2k^{2})E \,-\, {k^{\prime}}^{2}u + \mathrm{dn}\,\mathrm{sc}(2k^{2} \,-\, 1 + {k^{\prime}}^{2} \mathrm{nc}^2)]
\end{align}
We used the result from BF 361.07 above. The final expression is simpler than that of BF.

## BF 361.13

We use equation \eqref{eq:1} above, as well as results from BF 361.15 and BF 361.07.
\begin{align}
\int \mathrm{nc}^2 \mathrm{dc}^2 &= \mathrm{dc}\,\mathrm{sc}\,\mathrm{nc} \,-\, \int \mathrm{dc}^2 \mathrm{sc}^2 \,-\, {k^{\prime}}^{2} \int \mathrm{nc}^2 \mathrm{sc}^2 \\
&= \frac{1}{3{k^{\prime}}^{2}} [2{k^{\prime}}^{2}u + (k^{2} \,-\, 2)E + \mathrm{dn}\,\mathrm{sc}(2 \,-\, k^{2} + {k^{\prime}}^{2} \mathrm{nc}^2)]
\end{align}

## BF 361.21

We begin with

\frac{\partial \,\mathrm{ns}}{\partial u} = -\mathrm{cs}\,\mathrm{ds}

square it, integrate by parts, and make substitutions for squares of Jacobi elliptic functions.

\label{eq:2}
\tag{2}
\int \mathrm{cs}^{2} \mathrm{ds}^{2} = \int \frac{\partial \,\mathrm{ns}}{\partial u} \frac{\partial \,\mathrm{ns}}{\partial u} = -\mathrm{cs}\,\mathrm{ds}\,\mathrm{ns} \,-\, \int \mathrm{cs}^{2} \mathrm{ns}^{2} \,-\, \int \mathrm{ds}^{2} \mathrm{ns}^{2}

We will use the equation above for the next integral also. Continuing, we have

\int \mathrm{cs}^{2} (\mathrm{ns}^{2} \,-\, k^{2}) = -\mathrm{cs}\,\mathrm{ds}\,\mathrm{ns} \,-\, \int \mathrm{cs}^{2} \mathrm{ns}^{2} \,-\, \int (\mathrm{cs}^{2} + {k^{\prime}}^{2}) \mathrm{ns}^{2}

Rearranging yields
\begin{align}
3\int \mathrm{cs}^{2} \mathrm{ns}^{2} &= -\mathrm{cs}\,\mathrm{ds}\,\mathrm{ns} + k^{2} \int \mathrm{cs}^{2} \,-\, {k^{\prime}}^{2} \int \mathrm{ns}^{2} \\
&= (1 \,-\, 2k^{2})E \,-\, {k^{\prime}}^{2}u \,-\, \mathrm{dn}\,\mathrm{cs}(2k^{2} \,-\, 1 + \mathrm{ns}^2)
\end{align}

## BF 361.22

From equation \eqref{eq:2} we have

\int \mathrm{cs}^{2} \mathrm{ds}^{2} = -\mathrm{cs}\,\mathrm{ds}\,\mathrm{ns} \,-\, \int \mathrm{cs}^{2} \mathrm{ns}^{2} \,-\, \int \mathrm{ds}^{2} \mathrm{ns}^{2}

Substitutions yield

\int (\mathrm{ns}^{2} \,-\, 1) \mathrm{ds}^{2} = -\mathrm{cs}\,\mathrm{ds}\,\mathrm{ns} \,-\, \int (\mathrm{ds}^{2} \,-\, {k^{\prime}}^{2}) \mathrm{ns}^{2} \,-\, \int \mathrm{ds}^{2} \mathrm{ns}^{2}

Rearranging yields
\begin{align}
3\int \mathrm{ds}^{2} \mathrm{ns}^{2} &= -\mathrm{cs}\,\mathrm{ds}\,\mathrm{ns} + {k^{\prime}}^{2} \int \mathrm{ns}^{2} + \int \mathrm{ds}^{2} \\
&= 2{k^{\prime}}^{2}u + (k^{2} \,-\, 2)E \,-\, \mathrm{dn}\,\mathrm{cs}(2 \,-\, k^{2} + \mathrm{ns}^2)
\end{align}

## BF 361.27

We begin with

\frac{\partial \,\mathrm{nd}}{\partial u} = k^{2}\mathrm{cd}\,\mathrm{sd}

Multiplying by $$\mathrm{cd}\,\mathrm{sd}$$, rearranging, integrating by parts, and simplifying, yields
\begin{align}
k^{2}\int \mathrm{cd}^{2} \mathrm{sd}^{2} &= \int \mathrm{sd}\,\mathrm{cd} \frac{\partial \,\mathrm{nd}}{\partial u} \\
&= \mathrm{cd}\,\mathrm{nd}\,\mathrm{sd} \,-\, \int \mathrm{cd}\,\mathrm{cn}\,\mathrm{nd}^{3} + {k^{\prime}}^{2} \int \mathrm{sd}\,\mathrm{sn}\,\mathrm{nd}^{3} \\
\label{eq:3}
\tag{3}
&= \mathrm{cd}\,\mathrm{nd}\,\mathrm{sd} \,-\, \int \mathrm{cd}^{2} \mathrm{nd}^{2} + {k^{\prime}}^{2} \int \mathrm{sd}^{2} \mathrm{nd}^{2} \\
&= \mathrm{cd}\,\mathrm{nd}\,\mathrm{sd} \,-\, \int \mathrm{cd}^{2} (k^{2}\mathrm{sd}^{2} + 1) + {k^{\prime}}^{2} \int \mathrm{sd}^{2} (1 \,-\, k^{2}\mathrm{cd}^{2})
\end{align}
Collecting like terms, we have
\begin{align}
\int \mathrm{cd}^{2} \mathrm{sd}^{2} &= \frac{1}{3k^{2}} \left[\mathrm{cd}\,\mathrm{nd}\,\mathrm{sd} \,-\, \int \mathrm{cd}^{2} + \int \mathrm{sd}^{2} \right] \\
&= \frac{1}{3k^{4} {k^{\prime}}^{2}} \left[(1 + {k^{\prime}}^{2})E \,-\, 2{k^{\prime}}^{2}u + k^{2}\,\mathrm{sn}\,\mathrm{cd}\,({k^{\prime}}^{2} \mathrm{nd}^{2} \,-\, 1 \,-\, {k^{\prime}}^{2}) \right]
\end{align}

## BF 361.16

\begin{align}
\int \mathrm{cd}^{2} \mathrm{nd}^{2} &= \int \mathrm{cd}^{2} (k^{2}\mathrm{sd}^{2} + 1) = \int \mathrm{cd}^{2} + k^{2}\int \mathrm{cd}^{2} \mathrm{sd}^{2} \\
&= \frac{1}{3k^{2} {k^{\prime}}^{2}} \left[{k^{\prime}}^{2}u + (2k^{2} \,-\, 1)E + \mathrm{sn}\,\mathrm{cd}\,(k^{2} \,-\, 2 k^{4} + {k^{\prime}}^{2} k^{2} \mathrm{nd}^{2}) \right]
\end{align}
We used the result from BF 361.27.

## BF 361.19

From equation \eqref{eq:3} we have

\int \mathrm{sd}^{2} (1 \,-\, {k^{\prime}}^{2}\mathrm{nd}^{2}) = \mathrm{cd}\,\mathrm{nd}\,\mathrm{sd} \,-\, \int \mathrm{nd}^{2} (1 \,-\, {k^{\prime}}^{2}\mathrm{sd}^{2}) + {k^{\prime}}^{2}\int \mathrm{nd}^{2} \mathrm{sd}^{2}

Rearranging, we have
\begin{align}
\int \mathrm{nd}^{2} \mathrm{sd}^{2} &= \frac{1}{3 {k^{\prime}}^{2}} \left[-\mathrm{cd}\,\mathrm{nd}\,\mathrm{sd} + \int \mathrm{sd}^{2} + \int \mathrm{nd}^{2} \right] \\
&= \frac{1}{3k^{2} {k^{\prime}}^{4}} \left[(1 + k^{2})E \,-\, {k^{\prime}}^{2}u + k^{2}\,\mathrm{sn}\,\mathrm{cd}\,({k^{\prime}}^{2} \,-\, {k^{\prime}}^{2} \mathrm{nd}^{2} \,-\, 2) \right]
\end{align}