My cousin who is in high school asked me if it is possible to integrate
$$ \int \frac{1}{(1 + x^5)(1 + x^7)} \, dx $$
I checked the list of integrals of rational functions on Wikipedia link and it doesn't seem to be here.
Is this not possible to do using elementary functions? Wolfram Alpha can do this but the answer is extremely complicated.
On
Using the substitution by Thomas Andrews:
$$x = -y$$
$$\int\dfrac{dx}{(x^5+1)(x^7+1)}dx = -\int\dfrac{dy}{(y^5-1)(y^7-1)}$$
Looking for expansion in the form:
$$\dfrac1{(y^5-1)(y^7-1)} = \dfrac A{(y-1)^2} + \dfrac B{y-1} $$
$$+ \sum_{k=1}^2\left(\dfrac{C_k}{y-\exp\left(\dfrac{2\pi ki}{5}\right)}+\dfrac{C_{-k}}{y-\exp\left(\dfrac{-2\pi ki}{5}\right)}\right)$$
$$+ \sum_{k=1}^3\left(\dfrac{D_k}{y-\exp\left(\dfrac{2\pi ki}{7}\right)}+\dfrac{D_{-k}}{y-\exp\left(\dfrac{-2\pi ki}{7}\right)}\right).$$
Notice, that: $$y^5-1 = (y-1)P_4(y),\quad y^7-1 = (y-1)P_6(y),$$ where $$P_4(y) = y^4 + y^3 + y^2 + y + 1,$$ $$P_6(y) = y^6 + y^5 + y^4 + y^3 + y^2 + y + 1,$$ $$P_4'(y) = 4y^3 + 3y^2 + 2y + 1,$$ $$P_6'(y) = 6y^5 + 5y^4 + 4y^3 + 3y^2 + 2y + 1.$$
Finding the coefficients. $$A = \lim_{y\to1}\dfrac{(y-1)^2}{(y^5-1)(y^7-1)} = \dfrac1{P_4(1)P_6(1))} = \dfrac1{35}.$$ $$B = \lim_{y\to1}(y-1)\left(\dfrac1{(y^5-1)(y^7-1)}-\dfrac A{(y-1)^2}\right) = \lim_{y\to-1}\dfrac{1-A\cdot P_4(y)P_6(y)}{(y-1)P_4(y)P_6(y)}.$$ Using the L'Hospital's Rule: $$B=-A\dfrac{P_4'(1)P_6(1) + P_4(1)P_6'(1)}{P_4(1)P_6(1)} = -\dfrac1{35}\dfrac{10\cdot7 + 5\cdot21}{5\cdot7} = -\dfrac17.$$ $$C_k = \lim_{y\to \exp\dfrac{2\pi ki}5}\dfrac{y -\exp\dfrac{2\pi ki}5}{(y^5-1)(y^7-1)} = \lim_{y\to \exp\dfrac{2\pi ki}5}\dfrac{y -\exp\dfrac{2\pi ki}5}{y^5-1}\lim_{y\to \exp\dfrac{2\pi ki}5}\dfrac1{y^7-1}.$$ Using the L'Hospital's Rule: $$C_k = \lim_{y\to \exp\dfrac{2\pi ki}5}\dfrac{1}{5y^4(y^7-1)} = \dfrac{1}{5\exp\dfrac{8\pi ki}5\left(\exp\dfrac{14\pi ki}5-1\right)} = \dfrac{1}{10i\exp\dfrac{15\pi ki}5\sin\dfrac{7\pi k}5},$$ $$C_k = \dfrac{(-1)^k}{10i\sin\dfrac{7\pi k}5}.$$ $$D_k = \lim_{y\to\exp\dfrac{2\pi ki}7}\dfrac{y -\exp\dfrac{2\pi ki}7}{(y^5-1)(y^7-1)} = \lim_{y\to \exp\dfrac{2\pi ki}7}\dfrac{y -\exp\dfrac{2\pi ki}7}{y^7-1}\lim_{y\to \exp\dfrac{2\pi ki}7}\dfrac1{y^5-1}.$$ Using the L'Hospital's Rule: $$D_k = \lim_{y\to \exp\dfrac{2\pi ki}7}\dfrac{1}{7y^6(y^5-1)} = \dfrac{1}{7\exp\dfrac{12\pi ki}7\left(\exp\dfrac{10\pi ki}7-1\right)} = \dfrac{1}{14i\exp\dfrac{17\pi ki}7\sin\dfrac{5\pi k}7},$$ $$D_k = \dfrac{\exp\left(-\dfrac{3\pi ki}7\right)}{14i\sin\dfrac{5\pi k}7}.$$ Then, $$\dfrac{C_k}{y-\exp\left(\dfrac{2\pi ki}{5}\right)}+\dfrac{C_{-k}}{y-\exp\left(\dfrac{-2\pi ki}{5}\right)} $$ $$= \dfrac{(-1)^k}{10i\sin{\dfrac{7\pi k}5}}\cdot\left(\dfrac1{y -\exp\left(\dfrac{2\pi ki}{5}\right)} - \dfrac1{y-\exp\left(-\dfrac{2\pi ki}{5}\right)}\right) $$ $$ = \dfrac{(-1)^k}{10i\sin{\dfrac{7\pi k}5}}\cdot\dfrac{\exp\left(\dfrac{2\pi ki}{5}\right)- \exp\left(-\dfrac{2\pi ki}{5}\right)}{\left(y - \cos{\dfrac{2\pi k}5}\right)^2 + \sin^2{\dfrac{2\pi k}5}}$$ $$ = \dfrac{(-1)^k}{5\sin{\dfrac{7\pi k}5}}\cdot\dfrac{\sin{\dfrac{2\pi k}5}}{\left(y - \cos{\dfrac{2\pi k}5}\right)^2 + \sin^2{\dfrac{2\pi k}5}},$$ $$\dfrac{D_k}{y-\exp\left(\dfrac{2\pi ki}7\right)}+\dfrac{D_{-k}}{y-\exp\left(-\dfrac{2\pi ki}7\right)} = $$ $$ = \dfrac{\cos\dfrac{3\pi k}7}{14i\sin\dfrac{5\pi k}7}\left(\dfrac1{x-\exp\left(\dfrac{2\pi ki}7\right)}-\dfrac1{y-\exp\left(-\dfrac{2\pi ki}7\right)}\right)$$ $$ -\, \dfrac{i\sin\dfrac{3\pi k}7}{14i\sin\dfrac{5\pi k}7}\left(\dfrac1{y-\exp\left(\dfrac{2\pi ki}7\right)}+\dfrac1{y-\exp\left(-\dfrac{2\pi ki}7\right)}\right)$$ $$ = \dfrac{\cos\dfrac{3\pi k}7}{14i\sin\dfrac{5\pi k}7}\cdot\dfrac{\exp\left(\dfrac{2\pi ki}7\right)-\exp\left(-\dfrac{2\pi ki}7\right)}{\left(y - \cos{\dfrac{2\pi k}7}\right)^2 + \sin^2{\dfrac{2\pi k}7}}$$ $$ -\, \dfrac{\sin\dfrac{3\pi k}7}{14\sin\dfrac{5\pi k}7}\cdot\dfrac{2\left(y-\cos{\dfrac{2\pi k}7}\right)}{y^2 - 2y\cos{\dfrac{2\pi k}7} + 1}$$ $$ = \dfrac{\cos\dfrac{3\pi k}7}{7\sin\dfrac{5\pi k}7}\cdot\dfrac{\sin\dfrac{2\pi k}7}{\left(y - \cos{\dfrac{2\pi k}7}\right)^2 + \sin^2{\dfrac{2\pi k}7}}$$ $$+ \dfrac{\sin\dfrac{3\pi k}7}{14\sin\dfrac{5\pi k}7}\cdot\dfrac{2\left(y-\cos{\dfrac{2\pi k}7}\right)}{y^2 - 2y\cos{\dfrac{2\pi k}7} + 1}.$$
So, $$\int\dfrac{dy}{(y^5-1)(y^7-1)} = \dfrac1{35}\int\dfrac{dy}{(y-1)^2} - \dfrac17\int\dfrac{dy}{y-1} $$$$ + \sum_{k=1}^2\dfrac{(-1)^k}{5\sin{\dfrac{7\pi k}5}}\int\dfrac{\sin{\dfrac{2\pi k}5}}{\left(y - \cos{\dfrac{2\pi k}5}\right)^2 + \sin^2{\dfrac{2\pi k}5}}\,dy$$ $$+\sum_{k=1}^3\dfrac{\cos\dfrac{3\pi k}7}{7\sin\dfrac{5\pi k}7}\int\dfrac{\sin\dfrac{2\pi k}7}{\left(y - \cos{\dfrac{2\pi k}7}\right)^2 + \sin^2{\dfrac{2\pi k}7}}\,dy$$ $${+ \sum_{k=1}^3\dfrac{\sin\dfrac{3\pi k}7}{14\sin\dfrac{5\pi k}7}\int\dfrac{2\left(y-\cos{\dfrac{2\pi k}7}\right)}{y^2 - 2y\cos{\dfrac{2\pi k}7} + 1}\,dy}$$ $$= -\dfrac1{35}\dfrac1{y-1} - \dfrac17\ln|y-1| + \sum_{k=1}^2\dfrac{(-1)^k}{5\sin\dfrac{7\pi k}5}\arctan\dfrac{y - \cos\dfrac{2\pi k}5}{\sin\dfrac{2\pi k}5}$$ $$+\sum_{k=1}^3\dfrac{\cos\dfrac{3\pi k}7}{7\sin\dfrac{5\pi k}7}\arctan\dfrac{y - \cos\dfrac{2\pi k}7}{\sin\dfrac{2\pi k}7}$$ $$+ \sum_{k=1}^3\dfrac{\sin\dfrac{3\pi k}7}{14\sin\dfrac{5\pi k}7}\ln\left(y^2 - 2y\cos{\dfrac{2\pi k}7} + 1\right) + const.$$ Thus $$\boxed{\int\dfrac{dx}{(x^5+1)(x^7+1)} = \dfrac1{35}\dfrac1{x+1} + \dfrac17\ln|x+1| + \sum_{k=1}^2\dfrac{(-1)^k}{5\sin\dfrac{7\pi k}5}\arctan\dfrac{x+ \cos\dfrac{2\pi k}5}{\sin\dfrac{2\pi k}5}+\sum_{k=1}^3\dfrac{\cos\dfrac{3\pi k}7}{7\sin\dfrac{5\pi k}7}\arctan\dfrac{x + \cos\dfrac{2\pi k}7}{\sin\dfrac{2\pi k}7} - \sum_{k=1}^3\dfrac{\sin\dfrac{3\pi k}7}{14\sin\dfrac{5\pi k}7}\ln\left(x^2 + 2x\cos{\dfrac{2\pi k}7} + 1\right) + const}.$$
Solve:
$$\begin{align}\frac{1}{(x^5+1)(x^7+1)} & =\frac{a_0}{(x+1)^2}+\frac{a_1}{x+1}\\ &\quad+\frac{a_2x+b_2}{x^2-2\cos(\pi/5)x +1}\\&+\frac{a_3x+b_3}{x^2-2\cos(3\pi/5)x+1}\\ &\quad+\frac{a_4x+b_4}{x^2-2\cos(\pi/7)x +1}\\ &\quad+\frac{a_5x+b_5}{x^2-2\cos(3\pi/7)x +1}\\ &\quad+\frac{a_6x+b_6}{x^2-2\cos(5\pi/7)x +1}\\ \end{align}$$
Then you can use basic integrals to solve each:
$$\int \frac{ax+b}{x^2-2cx+1}\,dx$$
You'll get lots of logarithms and arctans, and it will be just horrible. And that doesn't even count how horrible the $a_i$ and $b_i$ are going to be.