For \(m\) integer \(\neq 1\) which can be negative or positive, the ode is\begin {equation} \frac {dy}{dx}=\left ( a+bx+cy\right ) ^{m} \tag {1} \end {equation} Let \(z=a+bx+cy\) then\begin {align*} \frac {dz}{dx} & =b+c\frac {dy}{dx}\\ \frac {dy}{dx} & =\left ( \frac {dz}{dx}-b\right ) \frac {1}{c} \end {align*}
Hence (1) becomes\begin {align} \left ( \frac {dz}{dx}-b\right ) \frac {1}{c} & =z^{m}\nonumber \\ \frac {dz}{dx} & =cz^{m}+b\nonumber \\ \int \frac {dz}{cz^{m}+b} & =\int dx \tag {2} \end {align}
If the left side is integrable, then the solution to (1) can be found. For \(m\) integer it is possible to find antiderivative. For example for \(n=2\) then (2) becomes\[ \frac {1}{\sqrt {bc}}\arctan \left ( \sqrt {\frac {c}{b}}z\right ) =x+C_{1}\] Replacing back \(z=a+bx+cy\) the above becomes\begin {equation} \frac {1}{\sqrt {bc}}\arctan \left ( \sqrt {\frac {c}{b}}\left ( a+bx+cy\right ) \right ) =x+C_{1} \tag {3} \end {equation} Which is the implicit solution to (1).
for \(m=2\). For an example, for \(a=1,b=1,c=1\) Eq. (1) becomes\[ \frac {dy}{dx}=\left ( 1+x+y\right ) ^{2}\] And (3) becomes\begin {align} \arctan \left ( 1+x+y\right ) & =x+C_{1}\nonumber \\ 1+x+y & =\tan \left ( x+C_{1}\right ) \nonumber \\ y & =\tan \left ( x+C_{1}\right ) -1-x \tag {4} \end {align}
And for \(m=3\) Eq. (2) becomes\[ \frac {-1}{6b^{\frac {2}{3}}c^{\frac {1}{3}}}\left ( 2\sqrt {3}\arctan \left ( \frac {1-2\left ( \frac {c}{b}\right ) ^{\frac {1}{3}}z}{\sqrt {3}}\right ) -2\ln \left ( b^{\frac {1}{3}}+c^{\frac {1}{3}}z\right ) +\ln \left ( b^{\frac {2}{3}}-b^{\frac {1}{3}}c^{\frac {1}{3}}+c^{\frac {2}{3}}z^{2}\right ) \right ) =x+C_{1}\] Replacing back \(z=a+bx+cy\) the above becomes\begin {equation} \frac {-1}{6b^{\frac {2}{3}}c^{\frac {1}{3}}}\left ( 2\sqrt {3}\arctan \left ( \frac {1-2\left ( \frac {c}{b}\right ) ^{\frac {1}{3}}\left ( a+bx+cy\right ) }{\sqrt {3}}\right ) -2\ln \left ( b^{\frac {1}{3}}+c^{\frac {1}{3}}\left ( a+bx+cy\right ) \right ) +\ln \left ( b^{\frac {2}{3}}-b^{\frac {1}{3}}c^{\frac {1}{3}}+c^{\frac {2}{3}}\left ( a+bx+cy\right ) ^{2}\right ) \right ) =x+C_{1} \tag {5} \end {equation} Which is the implicit solution to (1) for \(m=3\). Using \(a=1,b=1,c=1\) then (1) becomes\[ \frac {dy}{dx}=\left ( 1+x+y\right ) ^{3}\] And its solution (5) now simplifies to\[ \frac {-1}{6}\left ( 2\sqrt {3}\arctan \left ( \frac {1-2\left ( 1+x+y\right ) }{\sqrt {3}}\right ) -2\ln \left ( 2+x+y\right ) +\ln \left ( \left ( 1+x+y\right ) ^{2}\right ) \right ) =x+C_{1}\] And so on for higher values of \(m\), but solution get complicated very quickly. This method also works for negative \(m\).
For example, for \(m=-2\) then (1) becomes\[ \frac {dy}{dx}=\left ( a+bx+cy\right ) ^{-2}\] And the integral equation (2) now becomes\[ \int \frac {dz}{cz^{-2}+b}=\int dx \] Which gives\[ \frac {z}{b}-\frac {\sqrt {c}\arctan \left ( \sqrt {\frac {b}{c}}z\right ) }{b^{\frac {3}{2}}}=x+C_{1}\] Replacing back \(z=a+bx+cy\) the above becomes\[ \frac {a+bx+cy}{b}-\frac {\sqrt {c}\arctan \left ( \sqrt {\frac {b}{c}}\left ( a+bx+cy\right ) \right ) }{b^{\frac {3}{2}}}=x+C_{1}\] For \(a=1,b=1,c=1\) the above becomes\begin {align*} \left ( 1+x+y\right ) -\arctan \left ( 1+x+y\right ) & =x+C_{1}\\ \arctan \left ( 1+x+y\right ) & =\left ( 1+x+y\right ) -x-C_{1}\\ \arctan \left ( 1+x+y\right ) & =1+y-C_{1}\\ \arctan \left ( 1+x+y\right ) & =y+C_{2} \end {align*}
And and so on for \(=-3,-4,\cdots \) as all of these are integrable but become complicated very quickly and the computer is needed to find the antiderivatives in these cases.