Integrand size = 17, antiderivative size = 33 \[ \int \frac {(d x)^m}{\left (b x+c x^2\right )^2} \, dx=-\frac {d (d x)^{-1+m} \operatorname {Hypergeometric2F1}\left (2,-1+m,m,-\frac {c x}{b}\right )}{b^2 (1-m)} \]
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Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {661, 66} \[ \int \frac {(d x)^m}{\left (b x+c x^2\right )^2} \, dx=-\frac {d (d x)^{m-1} \operatorname {Hypergeometric2F1}\left (2,m-1,m,-\frac {c x}{b}\right )}{b^2 (1-m)} \]
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Rule 66
Rule 661
Rubi steps \begin{align*} \text {integral}& = d^2 \int \frac {(d x)^{-2+m}}{(b+c x)^2} \, dx \\ & = -\frac {d (d x)^{-1+m} \, _2F_1\left (2,-1+m;m;-\frac {c x}{b}\right )}{b^2 (1-m)} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {(d x)^m}{\left (b x+c x^2\right )^2} \, dx=\frac {(d x)^m \operatorname {Hypergeometric2F1}\left (2,-1+m,m,-\frac {c x}{b}\right )}{b^2 (-1+m) x} \]
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\[\int \frac {\left (d x \right )^{m}}{\left (c \,x^{2}+b x \right )^{2}}d x\]
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\[ \int \frac {(d x)^m}{\left (b x+c x^2\right )^2} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{2} + b x\right )}^{2}} \,d x } \]
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\[ \int \frac {(d x)^m}{\left (b x+c x^2\right )^2} \, dx=\int \frac {\left (d x\right )^{m}}{x^{2} \left (b + c x\right )^{2}}\, dx \]
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\[ \int \frac {(d x)^m}{\left (b x+c x^2\right )^2} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{2} + b x\right )}^{2}} \,d x } \]
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\[ \int \frac {(d x)^m}{\left (b x+c x^2\right )^2} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{2} + b x\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(d x)^m}{\left (b x+c x^2\right )^2} \, dx=\int \frac {{\left (d\,x\right )}^m}{{\left (c\,x^2+b\,x\right )}^2} \,d x \]
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