Evaluate $\int \:\frac{3x^5+13x^4+32x^3+8x^2-40x-75}{x^2\left(x^2+3x+5\right)^2}\:dx$

As per the WA result from the comment section by @Alexey Burdin , your partial fraction work is found to be correct so I shall continue the work from where you got stuck. $$2\ln x +\frac{3}{x}+\underbrace{\int\frac{x+1}{x^2+3x+5}}_{I_1}+\underbrace{\int\frac{4x}{(x^2+3x+5)^2}}_{I_2}$$ Since $$I_1 =2^{-1}\int\left(\frac{2x+3}{x^2+3x+5}-\frac{1}{(\left(x+\frac{3}{2}\right)^2+\frac{11}{4}}\right)\\=\frac{1}{2}\ln\left(x^2+3x+5\right)-\frac{1}{\sqrt{11}}\tan^{-1}\left(\frac{2x+3}{\sqrt{11}}\right)$$ And $$I_2=\int\frac{4x}{(x^2+3x+5)^2}=2\int\left(\frac{2x+3}{(x^2+3x+5)^2}-\frac{3}{(x^2+3x+5)^2}\right)\\=-\frac{2}{x^2+3x+5}-\underbrace{\int\frac{6}{(x^2+3x+5)^2}}_{I_3}$$ Notice that $$(x^2+3x+5)^2= \frac{(2x+3)^2+11}{16}$$ To evaluate the integral $I_3$ recall the reduction formula of $$\int\frac{1}{(pu^2+q)^n}=\frac{2n-3}{2q(n-1)}\int\frac{1}{(pu^2+q)^{n-1}}+\frac{u}{2q(n-1)(pu^2+q)^{n-1}}$$ with our case $p=1,q=11, u=2x+3$ and $n=2$ and hence we have $$I_3=\int\frac{-96}{((2x+3)^2+11)^2}=\frac{-96}{2}\left(\frac{1}{22}\int\frac{1}{(2x+3)^2+11}+\frac{(2x+3)}{22((2x+3)^2+11)}\right)=-\frac{24}{11\sqrt{11}}\tan^{-1}\left(\frac{2x+3}{\sqrt {11}}\right)-\frac{24(2x+3)}{11((2x+3)^2+11)}$$ and we obtained $I_2$ as $$ -\frac{24}{11\sqrt{11}}\tan^{-1}\left(\frac{2x+3}{\sqrt{11}}\right)-\frac{24(2x+3)}{11{(2x+3)^2+11)}}-\frac{2}{x^2+3x+5}$$ and hence we have $$2\ln x+\frac{3}{x}+\frac{1}{2}\ln(x^2+3x+5)-\frac{1}{\sqrt{11}}\tan^{-}\left(\frac{2x+3}{\sqrt{11}}\right)-\frac{2}{x^2+3x+5}-\frac{24}{11\sqrt{11}}\tan^{-1}\left(\frac{2x+3}{\sqrt{11}}\right)-\frac{24(2x+3)}{11(2x+3)^2+11)}$$ on further simplification we yield the indefinite integral as $$\ln x^2+\frac{3}{x}+\frac{1}{2}\ln(x^2+3x+5)-\frac{35}{11\sqrt{11}}\tan^{-1}\left(\frac{2x+3}{\sqrt{11}}\right)-\frac{4(3x+10)}{11(x^2+3x+5)}$$