The indefinite integral `I=int(sec^(2)xtanx(secx+tanx)dx)/((sec^(5)x+sec^(2)xtan^(3)x-sec^(3)x tan^(2)x-tan^(5)x))` simplifies to `(1)/(3)ln |f(x)|+c`, where `f((pi)/(4))=2sqrt2+1` and c is the constant of integration. If the value of `f((pi)/(3))` is `a+sqrtb`, then the value of `b-3a` is equal to

Answer

To solve the given problem, we need to evaluate the indefinite integral and simplify it step by step. Let's break down the solution:

### Step 1: Write the Integral
We start with the integral:
\[ I = \int \frac{\sec^2 x \tan x (\sec x + \tan x)}{\sec^5 x + \sec^2 x \tan^3 x - \sec^3 x \tan^2 x - \tan^5 x} \, dx \]

### Step 2: Factor the Denominator
We can factor the denominator:
\[ \sec^5 x + \sec^2 x \tan^3 x - \sec^3 x \tan^2 x - \tan^5 x \]
By taking \(\sec^2 x\) common from the first two terms and \(\tan^2 x\) from the last two terms, we can rewrite it as:
\[ \sec^2 x (\sec^3 x + \tan^3 x) - \tan^2 x (\sec^3 x + \tan^3 x) = (\sec^3 x + \tan^3 x)(\sec^2 x - \tan^2 x) \]

### Step 3: Use the Identity
Using the identity \(\sec^2 x - \tan^2 x = 1\), we simplify the denominator:
\[ \sec^5 x + \sec^2 x \tan^3 x - \sec^3 x \tan^2 x - \tan^5 x = \sec^3 x + \tan^3 x \]

### Step 4: Substitute and Simplify
Now we can substitute this back into the integral:
\[ I = \int \frac{\sec^2 x \tan x (\sec x + \tan x)}{\sec^3 x + \tan^3 x} \, dx \]

### Step 5: Let \(T = \sec^3 x + \tan^3 x\)
Let \(T = \sec^3 x + \tan^3 x\). Then, differentiate \(T\):
\[ \frac{dT}{dx} = 3\sec^2 x \tan x + 3\tan^2 x \sec^2 x = 3\sec^2 x \tan x (\sec x + \tan x) \]

### Step 6: Change of Variables
Thus, we can express \(dx\) in terms of \(dT\):
\[ dx = \frac{dT}{3\sec^2 x \tan x (\sec x + \tan x)} \]

### Step 7: Substitute Back into the Integral
Substituting back into the integral gives:
\[ I = \int \frac{1}{3} \frac{dT}{T} = \frac{1}{3} \ln |T| + C \]

### Step 8: Substitute \(T\) Back
Substituting \(T\) back gives:
\[ I = \frac{1}{3} \ln |\sec^3 x + \tan^3 x| + C \]

### Step 9: Evaluate \(f\left(\frac{\pi}{4}\right)\)
Given that \(f\left(\frac{\pi}{4}\right) = 2\sqrt{2} + 1\), we can find \(f\left(\frac{\pi}{3}\right)\):
\[ f\left(\frac{\pi}{3}\right) = \sec^3\left(\frac{\pi}{3}\right) + \tan^3\left(\frac{\pi}{3}\right) \]
Calculating:
\[ \sec\left(\frac{\pi}{3}\right) = 2, \quad \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]
Thus,
\[ f\left(\frac{\pi}{3}\right) = 2^3 + (\sqrt{3})^3 = 8 + 3\sqrt{3} \]

### Step 10: Identify \(a\) and \(b\)
From \(f\left(\frac{\pi}{3}\right) = a + \sqrt{b}\), we have:
\[ a = 8, \quad b = 27 \]

### Step 11: Calculate \(b - 3a\)
Now, we calculate:
\[ b - 3a = 27 - 3 \cdot 8 = 27 - 24 = 3 \]

### Final Answer
Thus, the value of \(b - 3a\) is:
\[ \boxed{3} \]