$\int \frac{6\sin x \cos^2 x+\sin 2x-23\sin x} {(\cos x - 1)^2(5-\sin^2x)} \mathrm{dx}$ I got this question $\int \frac{6\sin x \cos^2 x+\sin 2x-23\sin x} {(\cos x - 1)^2(5-\sin^2x)} \mathrm{dx}$ and I solved it as thus:
$$\int \frac{6\sin x \cos^2 x+\sin 2x-23\sin x} {(\cos x - 1)^2(5-\sin^2x)} \mathrm{dx}$$
$$=\int \frac{6\sin x \cos^2 x+2 \sin x\cos x-23\sin x} {(\cos x - 1)^2(5-\sin^2x)} \mathrm{dx}$$
$$=\int \frac{\sin x(6\cos^2x+2\cos x-23)} {(\cos x - 1)^2(5-(1-\cos^2x))}\mathrm{dx}$$
$$=\int \frac{\sin x(6\cos^2x+2\cos x-23)} {(\cos x - 1)^2(4+\cos^2x))}\mathrm{dx}\\$$
$$\mathrm{put\;} \color{red}{u=\cos x}\quad \mathrm{\frac{du} {dx}}=-\sin x\quad \mathrm{dx} =\frac{\mathrm{du} } {-\sin x} \\
\require{cancel}
\int \frac{\sin x(6u^2+2u-23)} {(u- 1)^2(u^2+4)}\frac{\mathrm{du}} {-\sin x} =$$
$$=\int \frac{\cancel{\sin x}(6u^2+2u-23)} {(u- 1)^2(u^2+4)}\frac{\mathrm{du}} {-\cancel{\sin x}}\\$$
$$=\int \frac{23-2u-6u^2 } {(u-1)^2(u^2+4)}\mathrm{du}$$
$$\color{red} {\mathrm{Resolving\;}} \frac{23-2u-6u^2 } {(u-1)^2(u^2+4)}\color{red} {\mathrm{\; into\; partial\; fractions\;we \;have:}} \\$$
$$\frac{4u-5} {u^2+4} - \frac{4} {u-1}+ \frac{3} {(u-1)^2} =\frac{4u} {u^2+4} - \frac{5 } {u^2+ 4} - \frac{4} {u-1}+ \frac{3} {(u-1)^2} \\$$
$$\int\ \frac{23-2u-6u^2 } {(u-1)^2(u^2+4)}=\int \left(\frac{4u} {u^2+4} - \frac{5 } {u^2+ 4} - \frac{4} {u-1}+ \frac{3} {(u-1)^2}\right) \mathrm{du}$$
$$2\ln \{u^2 +4\}-\frac{5}{2 }\tan^{-1} \left\{\frac{u}{2} \right\}-4\ln\{u-1\}-\frac{3}{u-1}
\\$$
$$\mathrm{But} \;u=\cos x\\$$
Edit
$$\underline{2\ln \left|\cos^2x +4\right|-\frac{5}{2 }\tan^{-1} \left(\frac{\cos x}{2} \right) -4\ln\left|\cos x-1\right|-\frac{3}{\cos x-1}+\mathrm{C}}$$
Is there anything wrong with these steps? 
Wolfram alpha made me extremely skeptical. 
Are there alternative steps? 

Mathematica gives exactly the same answer for you $u$-integral as
$-(3/(-1 + u)) - 5/2 ArcTan[u/2] - 4 Log[-1 + u] + 2 Log[4 + u^2], u=\cos x$
There is indeed no discrepancy with Mathematica and your answer. Cheer up!
