Integrand size = 33, antiderivative size = 144 \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=-\frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (2,m,1+m,-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d m}+\frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (-1+m,m,1+m,-\frac {d (a+b x)}{b c-a d}\right )}{4 b^3 d m} \]
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Time = 0.06 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {130, 72, 71, 133} \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=\frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m,m+1,-\frac {d (a+b x)}{b c-a d}\right )}{4 b^3 d m}-\frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (2,m,m+1,-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d m} \]
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Rule 71
Rule 72
Rule 130
Rule 133
Rubi steps \begin{align*} \text {integral}& = \frac {\int (a+b x)^{-1+m} (c+d x)^{1-m} \, dx}{4 b d}-\frac {(b c-a d)^2 \int \frac {(a+b x)^{-1+m} (c+d x)^{1-m}}{(b c+a d+2 b d x)^2} \, dx}{4 b d} \\ & = -\frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \, _2F_1\left (2,m;1+m;-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d m}+\frac {\left ((b c-a d) (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^{-1+m} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{1-m} \, dx}{4 b^2 d} \\ & = -\frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \, _2F_1\left (2,m;1+m;-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d m}+\frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,m;1+m;-\frac {d (a+b x)}{b c-a d}\right )}{4 b^3 d m} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 1.83 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.14 \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=\frac {(a+b x)^m (c+d x)^{-m} \left (-\frac {(b c-a d)^2 \left (\frac {d (a+b x)}{a d+b (c+2 d x)}\right )^{1-m} \left (\frac {b (c+d x)}{a d+b (c+2 d x)}\right )^m \operatorname {AppellF1}\left (1,m,-m,2,\frac {-b c+a d}{a d+b (c+2 d x)},\frac {b c-a d}{b c+a d+2 b d x}\right )}{d^2 (a+b x)}+\frac {\left (\frac {d (a+b x)}{-b c+a d}\right )^{-m} \left (4 b (1+m) (c+d x) \operatorname {AppellF1}\left (1-m,-m,1,2-m,\frac {b (c+d x)}{b c-a d},\frac {2 b (c+d x)}{b c-a d}\right )+2 d (-1+m) (a+b x) \left (-\frac {b d (a+b x) (c+d x)}{(b c-a d)^2}\right )^m \operatorname {Hypergeometric2F1}\left (m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )\right )}{d (-1+m) (1+m)}\right )}{8 b^3} \]
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\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{2-m}}{\left (2 b d x +a d +b c \right )^{2}}d x\]
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\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (2 \, b d x + b c + a d\right )}^{2}} \,d x } \]
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Exception generated. \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (2 \, b d x + b c + a d\right )}^{2}} \,d x } \]
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\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (2 \, b d x + b c + a d\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx=\int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{2-m}}{{\left (a\,d+b\,c+2\,b\,d\,x\right )}^2} \,d x \]
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