\[ -a \cos (y(x))+b+y'(x)=0 \] ✓ Mathematica : cpu = 0.204094 (sec), leaf count = 116
\[\left \{\left \{y(x)\to 2 \tan ^{-1}\left (\frac {a \tanh \left (\frac {1}{2} \left (x \sqrt {(a-b) (a+b)}-c_1 \sqrt {(a-b) (a+b)}\right )\right )}{\sqrt {(a-b) (a+b)}}-\frac {b \tanh \left (\frac {1}{2} \left (x \sqrt {(a-b) (a+b)}-c_1 \sqrt {(a-b) (a+b)}\right )\right )}{\sqrt {(a-b) (a+b)}}\right )\right \}\right \}\] ✓ Maple : cpu = 0.054 (sec), leaf count = 41
\[y \relax (x ) = 2 \arctan \left (\frac {\tanh \left (\frac {\sqrt {\left (a -b \right ) \left (a +b \right )}\, \left (x +c_{1}\right )}{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}{a +b}\right )\]
Hand solution
\[ y^{\prime }=a\cos y+b \]
This is separable.
\begin {align} \frac {dy}{a\cos y+b} & =dx\nonumber \\ \int \frac {dy}{a\cos y+b} & =x+C\tag {1} \end {align}
Using standard Tangent half-angle substitution, let \(t=\tan \frac {y}{2},\cos y=\frac {1-t^{2}}{1+t^{2}},dy=\frac {2}{1+t^{2}}dt\), then the integral becomes
\begin {align*} \int \frac {dy}{a\cos y+b} & =\int \frac {2}{1+t^{2}}\frac {1}{\left ( a\frac {1-t^{2}}{1+t^{2}}+b\right ) }dt\\ & =2\int \frac {1+t^{2}}{\left (1+t^{2}\right ) \left (a\left (1-t^{2}\right ) +b\left (1+t^{2}\right ) \right ) }dt\\ & =2\int \frac {dt}{a-at^{2}+b+bt^{2}}\\ & =2\int \frac {dt}{\left (a+b\right ) +t^{2}\left (b-a\right ) }\\ & =2\int \frac {dt}{\left (a+b\right ) \left (1+\frac {t^{2}\left (b-a\right ) }{\left (a+b\right ) }\right ) }\\ & =\frac {2}{a+b}\int \frac {dt}{\left (1+\frac {t^{2}\left (b-a\right ) }{\left (a+b\right ) }\right ) } \end {align*}
Let \(z^{2}=\frac {t^{2}\left (b-a\right ) }{\left (a+b\right ) }\), or \(z=\frac {t\sqrt {b-a}}{\sqrt {a+b}}\), then \(\frac {dz}{dt}=\frac {\sqrt {b-a}}{\sqrt {a+b}}\) and the above integral becomes
\begin {align*} \frac {2}{a+b}\int \frac {dt}{\left (1+\frac {t^{2}\left (b-a\right ) }{\left ( a+b\right ) }\right ) } & =\frac {2}{a+b}\int \frac {\sqrt {a+b}}{\sqrt {b-a}}\frac {dz}{\left (1+z^{2}\right ) }\\ & =\frac {2}{a+b}\frac {\sqrt {a+b}}{\sqrt {b-a}}\int \frac {dz}{\left ( 1+z^{2}\right ) }\\ & =\frac {2}{\sqrt {a+b}}\frac {1}{\sqrt {b-a}}\int \frac {dz}{\left ( 1+z^{2}\right ) }\\ & =\frac {2}{\sqrt {\left (a+b\right ) \left (b-a\right ) }}\int \frac {dz}{\left (1+z^{2}\right ) }\\ & =\frac {2}{\sqrt {b^{2}-a^{2}}}\int \frac {dz}{\left (1+z^{2}\right ) } \end {align*}
Now, \(\int \frac {dz}{\left (1+z^{2}\right ) }=\arctan \relax (z) \), hence
\begin {align*} \frac {2}{\sqrt {b^{2}-a^{2}}}\int \frac {dz}{\left (1+z^{2}\right ) } & =\frac {2}{\sqrt {b^{2}-a^{2}}}\arctan \relax (z) \\ & =\frac {2}{\sqrt {b^{2}-a^{2}}}\arctan \left (\frac {t\sqrt {b-a}}{\sqrt {a+b}}\right ) \end {align*}
But \(t=\tan \frac {y}{2}\) therefore
\[ \frac {2}{\sqrt {b^{2}-a^{2}}}\arctan \left (\frac {t\sqrt {b-a}}{\sqrt {a+b}}\right ) =\frac {2}{\sqrt {b^{2}-a^{2}}}\arctan \left (\frac {\tan \left ( \frac {y}{2}\right ) \sqrt {b-a}}{\sqrt {a+b}}\right ) \]
Going back to (1)
\begin {align*} \int \frac {dy}{a\cos y+b} & =x+C\\ \frac {2}{\sqrt {b^{2}-a^{2}}}\arctan \left (\frac {\tan \left (\frac {y}{2}\right ) \sqrt {b-a}}{\sqrt {a+b}}\right ) & =x+C\\ \arctan \left (\frac {\tan \left (\frac {y}{2}\right ) \sqrt {b-a}}{\sqrt {a+b}}\right ) & =\frac {1}{2}\sqrt {b^{2}-a^{2}}\left (x+C\right ) \\ \frac {\tan \left (\frac {y}{2}\right ) \sqrt {b-a}}{\sqrt {a+b}} & =\tan \left ( \frac {1}{2}\sqrt {b^{2}-a^{2}}\left (x+C\right ) \right ) \\ \tan \left (\frac {y}{2}\right ) & =\frac {\sqrt {a+b}}{\sqrt {b-a}}\tan \left ( \frac {1}{2}\sqrt {b^{2}-a^{2}}\left (x+C\right ) \right ) \\ \frac {y}{2} & =\arctan \left (\frac {\left (a+b\right ) }{\sqrt {\left ( a+b\right ) \left (b-a\right ) }}\tan \left (\frac {1}{2}\sqrt {b^{2}-a^{2}}\left (x+C\right ) \right ) \right ) \\ & =\arctan \left (\frac {\left (a+b\right ) }{\sqrt {b^{2}-a^{2}}}\tan \left ( \frac {1}{2}\sqrt {b^{2}-a^{2}}\left (x+C\right ) \right ) \right ) \\ y & =2\arctan \left (\frac {a+b}{\sqrt {b^{2}-a^{2}}}\tan \left (\frac {1}{2}\sqrt {b^{2}-a^{2}}\left (x+C\right ) \right ) \right ) \end {align*}
Verification