How to solve the given integral of the type: $\int_{0}^{\infty}\tfrac{1}{(a + b x)(c + d x) (1 + p x)}dx$

Question

Can anyone give a hint how to solve integral of the following type:

$$\int_{0}^{\infty}\dfrac{1}{(a + b x)(c + d x) (1 + p x)}dx$$

The problem is if we proceed with partial fraction, we get $$\dfrac{1}{(a + bx)(c + dx)(1 + px)} = \dfrac{b^2}{(bc - ad)(b - ap)(a + bx)} + \dfrac{d^2}{(bc - ad)(cp - d)(c + dx)}- \dfrac{p^2}{(b - ap)(cp-d)(1 + px)}$$

Now solving the integral for the first part,

$$\int_{0}^{\infty}\dfrac{b^2}{(bc - ad)(b - ap)(a + bx)} dx = \left.\dfrac{b \ln(a + bx)}{(bc - ad)(b - ap)}\right\vert_0^{\infty}$$

Hence a $\ln(\infty)$ term is encountered. Same is the case with all the three parts.

Any hints appreciated.

Edit: The final solution using Mathematica:

Answer 1

Let $P(x) = \prod\limits_{k=1}^n( a_k + b_k x)$ where $n > 1$, $a_k, b_k > 0$ and the roots $-\frac{a_k}{b_k}$ are distinct.
Consider the integral

$$\int_0^\infty \frac{dx}{P(x)} = \lim_{\Lambda\to\infty}\int_0^\Lambda \frac{dx}{P(x)}\tag{*1}$$

Since all the roots of $P(x)$ are distinct, we have following partial fraction decomposition

$$\frac{1}{P(x)} = \sum_{j=1}^n \frac{1}{a_j + b_j x}\prod_{k=1,\ne j}^n \frac{1}{a_k - b_k\frac{a_j}{b_j}} = \sum_{j=1}^n \frac{1}{a_j + b_j x}\prod_{k=1,\ne j}^n \frac{b_j}{a_k b_j - b_k a_j}\tag{*2}$$

Notice $$\int_0^\Lambda \frac{dx}{a_j + b_j x} = \frac{1}{b_j}\left(\log(a_j + b_j \Lambda) - \log(a_j)\right) = \frac{1}{b_j}\left[\log\Lambda + \log\left(\frac{b_j}{a_j} + \frac{1}{\Lambda}\right)\right]\tag{*3}$$ When you plug $(*2)$ into RHS of $(*1)$ and integrate, $(*3)$ tells us there is a term proportional to $\log\Lambda$. The coefficient of that term equals to $$ \sum_{j=1}^n \frac{1}{b_j}\prod_{k=1,\ne j}^n \frac{b_j}{a_k b_j - b_k a_j} = \lim_{x\to\infty} \sum_{j=1}^n \frac{x}{a_j+b_jx}\prod_{k=1,\ne j}^n \frac{b_j}{a_k b_j - b_k a_j} = \lim_{x\to\infty} \frac{x}{P(x)} = 0$$ Notice the remaining term in $(*3)$ converges to $\frac{1}{b_j}\log\left(\frac{b_j}{a_j}\right)$ as $\Lambda \to \infty$, we obtain:

$$\int_0^\infty \frac{dx}{P(x)} = \sum_{j=1}^n \log\left(\frac{b_j}{a_j}\right)b_j^{n-2} \prod_{k=1,\ne j}^n \frac{1}{a_k b_j - b_k a_j}$$

For $(a_1,b_1,a_2,b_2,a_3,b_3) = (a,b,c,d,1,p)$, the integral becomes $$\frac{p\,\mathrm{log}\left( p\right) }{\left( a\,p-b\right) \,\left( c\,p-d\right) }+\frac{d\,\mathrm{log}\left( \frac{d}{c}\right) }{\left( a\,d-b\,c\right) \,\left( d-c\,p\right) }+\frac{b\,\mathrm{log}\left( \frac{b}{a}\right) }{\left( b\,c-a\,d\right) \,\left( b-a\,p\right) }$$

With help of a CAS, I have verified this matches the answer you get from Mathematica.

Answer 2

For simplicity, write this as $$J = \dfrac{1}{bdp} \int_0^\infty \dfrac{dx}{(A+x)(B+x)(C+x)}$$ where $A = a/b$, and $B = c/d$, $C = 1/p$. We're assuming $a,b,c,d,p > 0$ and $A,B,C$ are all distinct. The partial fraction decomposition of the integrand is $$ F(x) = \frac{1}{(A-B)(A-C)(A+x)} + \frac{1}{(B-A)(B-C)(B+x)} + \frac{1}{(C-A)(C-B)(C+x)}$$ so that $$ J = \frac{1}{bdp} \lim_{R \to \infty} \left(\frac{\ln(A+R)-\ln(A)}{(A-B)(A-C)} + \frac{\ln(B+R)-\ln(B)}{(B-A)(B-C)} + \frac{\ln(C+R)-\ln(C)}{(C-A)(C- B)}\right) $$ Now write $\ln(A+R) = \ln(R) + \ln(1+A/R)$ etc, and notice that (as Achille Hui remarked) the coefficients of $\ln(R)$ cancel (as they must, because the improper integral is known to converge), while $\ln(1+A/R), \ln(1+B/R), \ln(1+C/R) \to 0$ as $R \to \infty$.