1. **State the problem:** Evaluate the integral $$\int \frac{\left(\frac{5}{3}\right)^5 \sec^5 \theta}{9 \cdot \left(\frac{5}{3}\right)^2 \sec^2 \theta - 25}^{3/2} \cdot \frac{5}{3} \sec \theta \tan \theta \, d\theta$$. 2. **Rewrite the integral:** Let $a = \frac{5}{3}$. Then the integral becomes: $$\int \frac{a^5 \sec^5 \theta}{9 a^2 \sec^2 \theta - 25}^{3/2} \cdot a \sec \theta \tan \theta \, d\theta = \int \frac{a^6 \sec^6 \theta \tan \theta}{\left(9 a^2 \sec^2 \theta - 25\right)^{3/2}} \, d\theta$$. 3. **Simplify the denominator:** Calculate $9 a^2$: $$9 \times \left(\frac{5}{3}\right)^2 = 9 \times \frac{25}{9} = 25$$. So denominator inside the power is: $$25 \sec^2 \theta - 25 = 25 (\sec^2 \theta - 1) = 25 \tan^2 \theta$$. 4. **Rewrite the integral using this:** $$\int \frac{a^6 \sec^6 \theta \tan \theta}{(25 \tan^2 \theta)^{3/2}} \, d\theta = \int \frac{a^6 \sec^6 \theta \tan \theta}{25^{3/2} \tan^3 \theta} \, d\theta$$. 5. **Simplify powers:** $$25^{3/2} = (5^2)^{3/2} = 5^3 = 125$$. So integral is: $$\int \frac{a^6 \sec^6 \theta \tan \theta}{125 \tan^3 \theta} \, d\theta = \int \frac{a^6 \sec^6 \theta}{125 \tan^2 \theta} \, d\theta$$. 6. **Rewrite $\tan^2 \theta$ as $\sec^2 \theta - 1$:** $$\int \frac{a^6 \sec^6 \theta}{125 (\sec^2 \theta - 1)} \, d\theta$$. 7. **Substitute $u = \sec \theta$:** Then $du = \sec \theta \tan \theta \, d\theta$. But we need to express $d\theta$ in terms of $du$ and $\theta$. Alternatively, rewrite the integral in terms of $u$: Since $\sec \theta = u$, $\sec^2 \theta = u^2$, and $\tan^2 \theta = u^2 - 1$. Rewrite integral: $$\int \frac{a^6 u^6}{125 (u^2 - 1)} \, d\theta$$. We need $d\theta$ in terms of $du$: $$du = \sec \theta \tan \theta \, d\theta = u \tan \theta \, d\theta \Rightarrow d\theta = \frac{du}{u \tan \theta}$$. But $\tan \theta = \sqrt{u^2 - 1}$, so: $$d\theta = \frac{du}{u \sqrt{u^2 - 1}}$$. 8. **Substitute $d\theta$ back into the integral:** $$\int \frac{a^6 u^6}{125 (u^2 - 1)} \cdot \frac{du}{u \sqrt{u^2 - 1}} = \int \frac{a^6 u^5}{125 (u^2 - 1)^{3/2}} \, du$$. 9. **Recall $a = \frac{5}{3}$, so $a^6 = \left(\frac{5}{3}\right)^6 = \frac{5^6}{3^6} = \frac{15625}{729}$:** Integral becomes: $$\int \frac{\frac{15625}{729} u^5}{125 (u^2 - 1)^{3/2}} \, du = \frac{15625}{729 \times 125} \int \frac{u^5}{(u^2 - 1)^{3/2}} \, du$$. Simplify constant: $$\frac{15625}{729 \times 125} = \frac{15625}{91125} = \frac{125}{729}$$. So integral is: $$\frac{125}{729} \int \frac{u^5}{(u^2 - 1)^{3/2}} \, du$$. 10. **Evaluate $\int \frac{u^5}{(u^2 - 1)^{3/2}} \, du$:** Rewrite numerator: $$u^5 = u^4 \cdot u = (u^2)^2 \cdot u$$. Try substitution $w = u^2 - 1$, so $dw = 2u \, du$, or $u \, du = \frac{dw}{2}$. Rewrite integral: $$\int \frac{u^5}{w^{3/2}} \, du = \int \frac{u^4 \cdot u}{w^{3/2}} \, du = \int \frac{(w+1)^2}{w^{3/2}} \cdot u \, du$$. Substitute $u \, du = \frac{dw}{2}$: $$= \int \frac{(w+1)^2}{w^{3/2}} \cdot \frac{dw}{2} = \frac{1}{2} \int \frac{(w+1)^2}{w^{3/2}} \, dw$$. 11. **Expand numerator:** $$(w+1)^2 = w^2 + 2w + 1$$. Integral becomes: $$\frac{1}{2} \int \frac{w^2 + 2w + 1}{w^{3/2}} \, dw = \frac{1}{2} \int \left(w^{2 - \frac{3}{2}} + 2 w^{1 - \frac{3}{2}} + w^{0 - \frac{3}{2}}\right) \, dw$$ $$= \frac{1}{2} \int \left(w^{\frac{1}{2}} + 2 w^{-\frac{1}{2}} + w^{-\frac{3}{2}}\right) \, dw$$. 12. **Integrate term by term:** $$\int w^{1/2} \, dw = \frac{2}{3} w^{3/2}$$ $$\int w^{-1/2} \, dw = 2 w^{1/2}$$ $$\int w^{-3/2} \, dw = -2 w^{-1/2}$$ So the integral is: $$\frac{1}{2} \left( \frac{2}{3} w^{3/2} + 2 \times 2 w^{1/2} - 2 w^{-1/2} \right) + C = \frac{1}{2} \left( \frac{2}{3} w^{3/2} + 4 w^{1/2} - 2 w^{-1/2} \right) + C$$ Simplify: $$= \frac{1}{3} w^{3/2} + 2 w^{1/2} - w^{-1/2} + C$$. 13. **Substitute back $w = u^2 - 1$:** $$= \frac{1}{3} (u^2 - 1)^{3/2} + 2 (u^2 - 1)^{1/2} - \frac{1}{(u^2 - 1)^{1/2}} + C$$. 14. **Recall the constant factor $\frac{125}{729}$:** Final integral: $$\frac{125}{729} \left[ \frac{1}{3} (u^2 - 1)^{3/2} + 2 (u^2 - 1)^{1/2} - \frac{1}{(u^2 - 1)^{1/2}} \right] + C$$ 15. **Substitute back $u = \sec \theta$:** $$= \frac{125}{729} \left[ \frac{1}{3} (\sec^2 \theta - 1)^{3/2} + 2 (\sec^2 \theta - 1)^{1/2} - \frac{1}{(\sec^2 \theta - 1)^{1/2}} \right] + C$$ 16. **Simplify $\sec^2 \theta - 1 = \tan^2 \theta$:** $$= \frac{125}{729} \left[ \frac{1}{3} \tan^3 \theta + 2 \tan \theta - \frac{1}{\tan \theta} \right] + C$$. **Final answer:** $$\boxed{\frac{125}{729} \left( \frac{1}{3} \tan^3 \theta + 2 \tan \theta - \frac{1}{\tan \theta} \right) + C}$$