Integrand size = 26, antiderivative size = 406 \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\frac {(d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-3-m}}{d^3 (b c-a d) (3+m)}+\frac {3 f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d) (2+m)}+\frac {2 b (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d)^2 (2+m) (3+m)}+\frac {3 f^2 (d e-c f) (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d) (1+m)}+\frac {3 b f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d)^2 (1+m) (2+m)}+\frac {2 b^2 (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d)^3 (1+m) (2+m) (3+m)}-\frac {f^3 (a+b x)^m \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{d^4 m} \]
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Time = 0.20 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {128, 47, 37, 72, 71} \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\frac {2 b^2 (a+b x)^{m+1} (d e-c f)^3 (c+d x)^{-m-1}}{d^3 (m+1) (m+2) (m+3) (b c-a d)^3}-\frac {f^3 (a+b x)^m (c+d x)^{-m} \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{d^4 m}+\frac {3 f^2 (a+b x)^{m+1} (d e-c f) (c+d x)^{-m-1}}{d^3 (m+1) (b c-a d)}+\frac {(a+b x)^{m+1} (d e-c f)^3 (c+d x)^{-m-3}}{d^3 (m+3) (b c-a d)}+\frac {3 f (a+b x)^{m+1} (d e-c f)^2 (c+d x)^{-m-2}}{d^3 (m+2) (b c-a d)}+\frac {2 b (a+b x)^{m+1} (d e-c f)^3 (c+d x)^{-m-2}}{d^3 (m+2) (m+3) (b c-a d)^2}+\frac {3 b f (a+b x)^{m+1} (d e-c f)^2 (c+d x)^{-m-1}}{d^3 (m+1) (m+2) (b c-a d)^2} \]
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Rule 37
Rule 47
Rule 71
Rule 72
Rule 128
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(d e-c f)^3 (a+b x)^m (c+d x)^{-4-m}}{d^3}+\frac {3 f (d e-c f)^2 (a+b x)^m (c+d x)^{-3-m}}{d^3}+\frac {3 f^2 (d e-c f) (a+b x)^m (c+d x)^{-2-m}}{d^3}+\frac {f^3 (a+b x)^m (c+d x)^{-1-m}}{d^3}\right ) \, dx \\ & = \frac {f^3 \int (a+b x)^m (c+d x)^{-1-m} \, dx}{d^3}+\frac {\left (3 f^2 (d e-c f)\right ) \int (a+b x)^m (c+d x)^{-2-m} \, dx}{d^3}+\frac {\left (3 f (d e-c f)^2\right ) \int (a+b x)^m (c+d x)^{-3-m} \, dx}{d^3}+\frac {(d e-c f)^3 \int (a+b x)^m (c+d x)^{-4-m} \, dx}{d^3} \\ & = \frac {(d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-3-m}}{d^3 (b c-a d) (3+m)}+\frac {3 f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d) (2+m)}+\frac {3 f^2 (d e-c f) (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d) (1+m)}+\frac {\left (3 b f (d e-c f)^2\right ) \int (a+b x)^m (c+d x)^{-2-m} \, dx}{d^3 (b c-a d) (2+m)}+\frac {\left (2 b (d e-c f)^3\right ) \int (a+b x)^m (c+d x)^{-3-m} \, dx}{d^3 (b c-a d) (3+m)}+\frac {\left (f^3 (a+b x)^m \left (\frac {d (a+b x)}{-b c+a d}\right )^{-m}\right ) \int (c+d x)^{-1-m} \left (-\frac {a d}{b c-a d}-\frac {b d x}{b c-a d}\right )^m \, dx}{d^3} \\ & = \frac {(d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-3-m}}{d^3 (b c-a d) (3+m)}+\frac {3 f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d) (2+m)}+\frac {2 b (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d)^2 (2+m) (3+m)}+\frac {3 f^2 (d e-c f) (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d) (1+m)}+\frac {3 b f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d)^2 (1+m) (2+m)}-\frac {f^3 (a+b x)^m \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} (c+d x)^{-m} \, _2F_1\left (-m,-m;1-m;\frac {b (c+d x)}{b c-a d}\right )}{d^4 m}+\frac {\left (2 b^2 (d e-c f)^3\right ) \int (a+b x)^m (c+d x)^{-2-m} \, dx}{d^3 (b c-a d)^2 (2+m) (3+m)} \\ & = \frac {(d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-3-m}}{d^3 (b c-a d) (3+m)}+\frac {3 f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d) (2+m)}+\frac {2 b (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d)^2 (2+m) (3+m)}+\frac {3 f^2 (d e-c f) (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d) (1+m)}+\frac {3 b f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d)^2 (1+m) (2+m)}+\frac {2 b^2 (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d)^3 (1+m) (2+m) (3+m)}-\frac {f^3 (a+b x)^m \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} (c+d x)^{-m} \, _2F_1\left (-m,-m;1-m;\frac {b (c+d x)}{b c-a d}\right )}{d^4 m} \\ \end{align*}
Time = 12.60 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.69 \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\frac {(a+b x)^m (c+d x)^{-m} \left (\frac {6 f^2 (a+b x)^3 (e+f x)}{(b c-a d)^3 (c+d x)}-\frac {3 f (a+b x)^2 (-a d (1+m)+b c (3+m)+2 b d x) (e+f x)^2}{(b c-a d)^3 (c+d x)^2}+\frac {(a+b x) (e+f x)^3 \left (a^2 d^2 \left (2+3 m+m^2\right )-2 a b d (1+m) (c (3+m)+d x)+b^2 \left (c^2 \left (6+5 m+m^2\right )+2 c d (3+m) x+2 d^2 x^2\right )\right )}{(b c-a d)^3 (c+d x)^3}-\frac {6 f^3 \left (\frac {d (a+b x)}{-b c+a d}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-3-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{d^4 m}\right )}{(1+m) (2+m) (3+m)} \]
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\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{-4-m} \left (f x +e \right )^{3}d x\]
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\[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\int { {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 4} \,d x } \]
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Exception generated. \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\int { {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 4} \,d x } \]
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\[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\int { {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 4} \,d x } \]
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Timed out. \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\int \frac {{\left (e+f\,x\right )}^3\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^{m+4}} \,d x \]
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