# Help to solve an integral of type: $\int\frac{1}{a+bx}\cos(cx)^2dx$

I am stuck with this integral: $$\int_{0}^{L}\frac{1}{a+bx}\cos(\frac{i\pi x}{L})^2dx$$

which is one term of the stiffness matrix of a conical shell. I thought about expanding $\cos(cx)^2/(a + bx)$ in Taylor series around zero, which will do the following:

$$\frac{\cos(cx)^2}{a+bx}=\frac{1}{a+\text{bx}}-\frac{c^2 x^2}{a+\text{bx}}+\frac{c^4 x^4}{3 (a+\text{bx})}-\frac{\left(2 c^6\right) x^6}{45 (a+\text{bx})}+\frac{c^8 x^8}{315 (a+\text{bx})}-\frac{\left(2 c^{10}\right) x^{10}}{14175 (a+\text{bx})}+O\left(x^{11}\right)$$

Making the integral solvable up to the desired precision.

The question is, is there a better way to perform this integration?

$$\begin{eqnarray} \int_{0}^{L}\frac{\cos(\frac{i\pi x}{L})^2}{a+bx}dx&=&\frac{1}{4}\int_{0}^{L}\frac{\left(e^{i\frac{i\pi x}{L}}+e^{-i\frac{i\pi x}{L}}\right)^2}{a+bx}dx=\\ &=&\frac{1}{4}\int_{0}^{L}\frac{e^{\frac{2\pi x}{L}}}{a+bx}dx+\frac{1}{4}\int_{0}^{L}\frac{e^{-\frac{2\pi x}{L}}}{a+bx}dx+\frac{1}{4}\int_{0}^{L}\frac{2}{a+bx}dx=\\ &=&\frac{e^{-\frac{2 \pi a}{b L}} \left(\text{Ei}\left(2 \left(\frac{a}{b L}+1\right) \pi \right)-\text{Ei}\left(\frac{2 a \pi }{b L}\right)\right)}{b}+\\ &+&\frac{e^{\frac{2 \pi a}{b L}} \left(\text{Ei}\left(-\frac{2 (a+b L) \pi }{b L}\right)-\text{Ei}\left(-\frac{2 a \pi }{b L}\right)\right)}{b}+\frac{\log(1+\frac{b L}{a})}{b} \end{eqnarray}$$ which is valid if $\left(a\neq 0\land b\neq 0\land L\neq 0\land \Re{(\frac{a}{b L})}\geq 0\right)\lor \Re{(\frac{a}{b L})}<-1\lor \frac{a}{b L}\notin \mathbb{R}$
• thank you... I will have look at this approach. The problem was the $Ei$ function created at the end... – Saullo G. P. Castro Sep 25 '13 at 8:02
• The problem is that after integrating I transfer the resulting expression to a Cython routine, where $Ei$ should be translated in terms of another known function, preferably a function which is available in C... – Saullo G. P. Castro Sep 25 '13 at 8:10