Integrand size = 15, antiderivative size = 54 \[ \int \frac {(a+b x)^m}{(c+d x)^3} \, dx=\frac {b^2 (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d)^3 (1+m)} \]
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Time = 0.01 (sec) , antiderivative size = 54, 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} \[ \int \frac {(a+b x)^m}{(c+d x)^3} \, dx=\frac {b^2 (a+b x)^{m+1} \operatorname {Hypergeometric2F1}\left (3,m+1,m+2,-\frac {d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d)^3} \]
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Rule 70
Rubi steps \begin{align*} \text {integral}& = \frac {b^2 (a+b x)^{1+m} \, _2F_1\left (3,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d)^3 (1+m)} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^m}{(c+d x)^3} \, dx=\frac {b^2 (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d)^3 (1+m)} \]
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\[\int \frac {\left (b x +a \right )^{m}}{\left (d x +c \right )^{3}}d x\]
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\[ \int \frac {(a+b x)^m}{(c+d x)^3} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{3}} \,d x } \]
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\[ \int \frac {(a+b x)^m}{(c+d x)^3} \, dx=\int \frac {\left (a + b x\right )^{m}}{\left (c + d x\right )^{3}}\, dx \]
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\[ \int \frac {(a+b x)^m}{(c+d x)^3} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{3}} \,d x } \]
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\[ \int \frac {(a+b x)^m}{(c+d x)^3} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^m}{(c+d x)^3} \, dx=\int \frac {{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^3} \,d x \]
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