Integrand size = 15, antiderivative size = 52 \[ \int \frac {(a+b x)^m}{(e+f x)^2} \, dx=\frac {b (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f)^2 (1+m)} \]
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Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {70} \[ \int \frac {(a+b x)^m}{(e+f x)^2} \, dx=\frac {b (a+b x)^{m+1} \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2} \]
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Rule 70
Rubi steps \begin{align*} \text {integral}& = \frac {b (a+b x)^{1+m} \, _2F_1\left (2,1+m;2+m;-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f)^2 (1+m)} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^m}{(e+f x)^2} \, dx=\frac {b (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f)^2 (1+m)} \]
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\[\int \frac {\left (b x +a \right )^{m}}{\left (f x +e \right )^{2}}d x\]
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\[ \int \frac {(a+b x)^m}{(e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (f x + e\right )}^{2}} \,d x } \]
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\[ \int \frac {(a+b x)^m}{(e+f x)^2} \, dx=\int \frac {\left (a + b x\right )^{m}}{\left (e + f x\right )^{2}}\, dx \]
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\[ \int \frac {(a+b x)^m}{(e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (f x + e\right )}^{2}} \,d x } \]
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\[ \int \frac {(a+b x)^m}{(e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (f x + e\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^m}{(e+f x)^2} \, dx=\int \frac {{\left (a+b\,x\right )}^m}{{\left (e+f\,x\right )}^2} \,d x \]
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