Integrand size = 26, antiderivative size = 498 \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=-\frac {f (a+b x)^{1+m} (c+d x)^{-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {f (b (5 d e-c f (2-m))-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}-\frac {f \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f \left (d e (15+8 m)-c f \left (3-2 m-2 m^2\right )\right )+b^2 \left (11 d^2 e^2-c d e f (7-8 m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m}}{6 (b e-a f)^3 (d e-c f)^3 (e+f x)}+\frac {\left (3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )+a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) (a+b x)^m (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (1,-m,1-m,\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{6 (b e-a f)^3 (d e-c f)^4 m} \]
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Time = 0.45 (sec) , antiderivative size = 497, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {105, 156, 12, 133} \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=-\frac {f (a+b x)^{m+1} (c+d x)^{-m} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (d e (8 m+15)-c f \left (-2 m^2-2 m+3\right )\right )+b^2 \left (c^2 f^2 \left (m^2-3 m+2\right )-c d e f (7-8 m)+11 d^2 e^2\right )\right )}{6 (e+f x) (b e-a f)^3 (d e-c f)^3}+\frac {(a+b x)^m (c+d x)^{-m} \left (a^3 d^3 f^3 \left (m^3+6 m^2+11 m+6\right )-3 a^2 b d^2 f^2 \left (m^2+3 m+2\right ) (c f m+3 d e)+3 a b^2 d f (m+1) \left (-c^2 f^2 (1-m) m+6 c d e f m+6 d^2 e^2\right )-\left (b^3 \left (c^3 f^3 m \left (m^2-3 m+2\right )-9 c^2 d e f^2 (1-m) m+18 c d^2 e^2 f m+6 d^3 e^3\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,-m,1-m,\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{6 m (b e-a f)^3 (d e-c f)^4}-\frac {f (a+b x)^{m+1} (c+d x)^{-m} (-a d f (m+3)-b c f (2-m)+5 b d e)}{6 (e+f x)^2 (b e-a f)^2 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)} \]
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Rule 12
Rule 105
Rule 133
Rule 156
Rubi steps \begin{align*} \text {integral}& = -\frac {f (a+b x)^{1+m} (c+d x)^{-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {\int \frac {(a+b x)^m (c+d x)^{-1-m} (-b (3 d e-c f (2-m))+a d f (3+m)+2 b d f x)}{(e+f x)^3} \, dx}{3 (b e-a f) (d e-c f)} \\ & = -\frac {f (a+b x)^{1+m} (c+d x)^{-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {f (5 b d e-b c f (2-m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}+\frac {\int \frac {(a+b x)^m (c+d x)^{-1-m} \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f \left (d e (12+7 m)-c f \left (3-2 m-2 m^2\right )\right )+b^2 \left (6 d^2 e^2-c d e f (5-7 m)+c^2 f^2 \left (2-3 m+m^2\right )\right )-b d f (5 b d e-b c f (2-m)-a d f (3+m)) x\right )}{(e+f x)^2} \, dx}{6 (b e-a f)^2 (d e-c f)^2} \\ & = -\frac {f (a+b x)^{1+m} (c+d x)^{-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {f (5 b d e-b c f (2-m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}-\frac {f \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f \left (d e (15+8 m)-c f \left (3-2 m-2 m^2\right )\right )+b^2 \left (11 d^2 e^2-c d e f (7-8 m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m}}{6 (b e-a f)^3 (d e-c f)^3 (e+f x)}-\frac {\int \frac {\left (3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )+a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) (a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{6 (b e-a f)^3 (d e-c f)^3} \\ & = -\frac {f (a+b x)^{1+m} (c+d x)^{-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {f (5 b d e-b c f (2-m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}-\frac {f \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f \left (d e (15+8 m)-c f \left (3-2 m-2 m^2\right )\right )+b^2 \left (11 d^2 e^2-c d e f (7-8 m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m}}{6 (b e-a f)^3 (d e-c f)^3 (e+f x)}-\frac {\left (3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )+a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) \int \frac {(a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{6 (b e-a f)^3 (d e-c f)^3} \\ & = -\frac {f (a+b x)^{1+m} (c+d x)^{-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {f (5 b d e-b c f (2-m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}-\frac {f \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f \left (d e (15+8 m)-c f \left (3-2 m-2 m^2\right )\right )+b^2 \left (11 d^2 e^2-c d e f (7-8 m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m}}{6 (b e-a f)^3 (d e-c f)^3 (e+f x)}+\frac {\left (3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )+a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,-m;1-m;\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{6 (b e-a f)^3 (d e-c f)^4 m} \\ \end{align*}
Time = 1.03 (sec) , antiderivative size = 466, normalized size of antiderivative = 0.94 \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=-\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (6 d (b e-a f)^4 (d e-c f)^2 (1+m)+2 f (-b e+a f)^3 (-d e+c f) (1+m) (-a d f (3+m)+b (3 d e+c f m)) (c+d x)-\frac {(e+f x) \left (f (b e-a f)^2 (1+m) \left (-b^2 \left (6 d^2 e^2+7 c d e f m+c^2 f^2 (-2+m) m\right )-a^2 d^2 f^2 \left (6+5 m+m^2\right )+a b d f (c f m (3+2 m)+d e (12+7 m))\right ) (c+d x)^2+(b c-a d) \left (-3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m+c^2 f^2 (-1+m) m\right )+3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )-a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )+b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m+9 c^2 d e f^2 (-1+m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) (e+f x)^2 \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )\right )}{c+d x}\right )}{6 (b c-a d) (b e-a f)^4 (d e-c f)^2 (-d e+c f) m (1+m) (e+f x)^3} \]
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\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-1-m}}{\left (f x +e \right )^{4}}d x\]
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\[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{4}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=\text {Timed out} \]
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\[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{4}} \,d x } \]
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\[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{4}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=\int \frac {{\left (a+b\,x\right )}^m}{{\left (e+f\,x\right )}^4\,{\left (c+d\,x\right )}^{m+1}} \,d x \]
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