Integrand size = 26, antiderivative size = 362 \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=\frac {(b e-a f) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}+\frac {(a d f (2-m)-b (3 d e-c f (1+m))) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (d e-c f) (e+f x)}-\frac {\left (2 a b d f (2-m) (d e-c f m)-b^2 \left (2 d^2 e^2-2 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (2-3 m+m^2\right )\right ) (a+b x)^{-1+m} (c+d x)^{1-m} \operatorname {Hypergeometric2F1}\left (1,-1+m,m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{2 f^3 (b e-a f) (d e-c f) (1-m)}-\frac {d (b c-a d) (a+b x)^{-1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (-1+m,-1+m,m,-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)} \]
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Time = 0.49 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {132, 72, 71, 1627, 156, 12, 133} \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=-\frac {(a+b x)^{m-1} (c+d x)^{1-m} \left (-a^2 d^2 f^2 \left (m^2-3 m+2\right )+2 a b d f (2-m) (d e-c f m)-\left (b^2 \left (-c^2 f^2 (1-m) m-2 c d e f m+2 d^2 e^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m-1,m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{2 f^3 (1-m) (b e-a f) (d e-c f)}-\frac {(a+b x)^{m-1} (c+d x)^{2-m} (-a d f (2-m)-b c f (m+1)+3 b d e)}{2 f^2 (e+f x) (d e-c f)}+\frac {(b e-a f) (a+b x)^{m-1} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {d (b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)} \]
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Rule 12
Rule 71
Rule 72
Rule 132
Rule 133
Rule 156
Rule 1627
Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^2 d\right ) \int (a+b x)^{-2+m} (c+d x)^{1-m} \, dx}{f^3}+\int \frac {(a+b x)^{-2+m} (c+d x)^{1-m} \left (a^2 c-\frac {b^2 d e^3}{f^3}+\left (2 a b c+a^2 d-\frac {3 b^2 d e^2}{f^2}\right ) x+b \left (2 a d+b \left (c-\frac {3 d e}{f}\right )\right ) x^2\right )}{(e+f x)^3} \, dx \\ & = \frac {(b e-a f) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {\int \frac {(a+b x)^{-2+m} (c+d x)^{1-m} \left (\frac {(b e-a f) (d e-c f) \left (b^2 e (2 d e-c f (1-m))-a^2 d f^2 (2-m)+a b f (d e (2-m)-c f (1+m))\right )}{f^3}+\frac {b (b e-a f) (d e-c f) (5 b d e-2 b c f-3 a d f) x}{f^2}\right )}{(e+f x)^2} \, dx}{2 (b e-a f) (d e-c f)}+\frac {\left (b d (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)^{-2+m} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{1-m} \, dx}{f^3} \\ & = \frac {(b e-a f) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {(3 b d e-a d f (2-m)-b c f (1+m)) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (d e-c f) (e+f x)}-\frac {d (b c-a d) (a+b x)^{-1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,-1+m;m;-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)}+\frac {\int \frac {(b e-a f)^2 (d e-c f) \left (2 a b d f (2-m) (d e-c f m)-b^2 \left (2 d^2 e^2-2 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (2-3 m+m^2\right )\right ) (a+b x)^{-2+m} (c+d x)^{1-m}}{f^3 (e+f x)} \, dx}{2 (b e-a f)^2 (d e-c f)^2} \\ & = \frac {(b e-a f) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {(3 b d e-a d f (2-m)-b c f (1+m)) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (d e-c f) (e+f x)}-\frac {d (b c-a d) (a+b x)^{-1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,-1+m;m;-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)}+\frac {\left (2 a b d f (2-m) (d e-c f m)-b^2 \left (2 d^2 e^2-2 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (2-3 m+m^2\right )\right ) \int \frac {(a+b x)^{-2+m} (c+d x)^{1-m}}{e+f x} \, dx}{2 f^3 (d e-c f)} \\ & = \frac {(b e-a f) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {(3 b d e-a d f (2-m)-b c f (1+m)) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (d e-c f) (e+f x)}-\frac {\left (2 a b d f (2-m) (d e-c f m)-b^2 \left (2 d^2 e^2-2 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (2-3 m+m^2\right )\right ) (a+b x)^{-1+m} (c+d x)^{1-m} \, _2F_1\left (1,-1+m;m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{2 f^3 (b e-a f) (d e-c f) (1-m)}-\frac {d (b c-a d) (a+b x)^{-1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,-1+m;m;-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.25 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.30 \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=\frac {(b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {AppellF1}\left (1+m,-2+m,3,2+m,\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )}{(b e-a f)^3 (1+m)} \]
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\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{2-m}}{\left (f x +e \right )^{3}}d x\]
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\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (f x + e\right )}^{3}} \,d x } \]
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Exception generated. \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (f x + e\right )}^{3}} \,d x } \]
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\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (f x + e\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=\int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{2-m}}{{\left (e+f\,x\right )}^3} \,d x \]
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