Integrand size = 26, antiderivative size = 158 \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\frac {d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)}+\frac {(a d f (1+m)-b (d e+c f m)) (a+b x)^{1+m} (c+d x)^{-1-m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(b e-a f)^2 (d e-c f) m (1+m)} \]
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Time = 0.05 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {98, 133} \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\frac {(a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+1)-b (c f m+d e)) \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{m (m+1) (b e-a f)^2 (d e-c f)}+\frac {d (a+b x)^{m+1} (c+d x)^{-m}}{m (e+f x) (b c-a d) (d e-c f)} \]
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Rule 98
Rule 133
Rubi steps \begin{align*} \text {integral}& = \frac {d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)}+\frac {(a d f (1+m)-b (d e+c f m)) \int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^2} \, dx}{(b c-a d) (d e-c f) m} \\ & = \frac {d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)}+\frac {(a d f (1+m)-b (d e+c f m)) (a+b x)^{1+m} (c+d x)^{-1-m} \, _2F_1\left (2,1+m;2+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(b e-a f)^2 (d e-c f) m (1+m)} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (-\frac {d}{e+f x}+\frac {(b c-a d) (-a d f (1+m)+b (d e+c f m)) \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(b e-a f)^2 (1+m) (c+d x)}\right )}{(b c-a d) (-d e+c f) m} \]
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\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-1-m}}{\left (f x +e \right )^{2}}d x\]
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\[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\text {Timed out} \]
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\[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{2}} \,d x } \]
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\[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\int \frac {{\left (a+b\,x\right )}^m}{{\left (e+f\,x\right )}^2\,{\left (c+d\,x\right )}^{m+1}} \,d x \]
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