Integrand size = 27, antiderivative size = 235 \[ \int (a+b x)^m (c+d x)^{-m} (e+f x) (g+h x) \, dx=\frac {(a+b x)^{1+m} (c+d x)^{1-m} (3 b d (f g+e h)-a d f h (2-m)-b c f h (2+m)+2 b d f h x)}{6 b^2 d^2}+\frac {\left (a^2 d^2 f h \left (2-3 m+m^2\right )-a b d (1-m) (3 d (f g+e h)-2 c f h (1+m))+b^2 \left (6 d^2 e g-3 c d (f g+e h) (1+m)+c^2 f h \left (2+3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,1+m,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (1+m)} \]
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Time = 0.09 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {152, 72, 71} \[ \int (a+b x)^m (c+d x)^{-m} (e+f x) (g+h x) \, dx=\frac {(a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,m+1,m+2,-\frac {d (a+b x)}{b c-a d}\right ) \left (a^2 d^2 f h \left (m^2-3 m+2\right )-a b d (1-m) (3 d (e h+f g)-2 c f h (m+1))+b^2 \left (c^2 f h \left (m^2+3 m+2\right )-3 c d (m+1) (e h+f g)+6 d^2 e g\right )\right )}{6 b^3 d^2 (m+1)}+\frac {(a+b x)^{m+1} (c+d x)^{1-m} (-a d f h (2-m)-b c f h (m+2)+3 b d (e h+f g)+2 b d f h x)}{6 b^2 d^2} \]
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Rule 71
Rule 72
Rule 152
Rubi steps \begin{align*} \text {integral}& = \frac {(a+b x)^{1+m} (c+d x)^{1-m} (3 b d (f g+e h)-a d f h (2-m)-b c f h (2+m)+2 b d f h x)}{6 b^2 d^2}+\frac {\left (a^2 d^2 f h \left (2-3 m+m^2\right )-a b d (1-m) (3 d (f g+e h)-2 c f h (1+m))+b^2 \left (6 d^2 e g-3 c d (f g+e h) (1+m)+c^2 f h \left (2+3 m+m^2\right )\right )\right ) \int (a+b x)^m (c+d x)^{-m} \, dx}{6 b^2 d^2} \\ & = \frac {(a+b x)^{1+m} (c+d x)^{1-m} (3 b d (f g+e h)-a d f h (2-m)-b c f h (2+m)+2 b d f h x)}{6 b^2 d^2}+\frac {\left (\left (a^2 d^2 f h \left (2-3 m+m^2\right )-a b d (1-m) (3 d (f g+e h)-2 c f h (1+m))+b^2 \left (6 d^2 e g-3 c d (f g+e h) (1+m)+c^2 f h \left (2+3 m+m^2\right )\right )\right ) (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{-m} \, dx}{6 b^2 d^2} \\ & = \frac {(a+b x)^{1+m} (c+d x)^{1-m} (3 b d (f g+e h)-a d f h (2-m)-b c f h (2+m)+2 b d f h x)}{6 b^2 d^2}+\frac {\left (a^2 d^2 f h \left (2-3 m+m^2\right )-a b d (1-m) (3 d (f g+e h)-2 c f h (1+m))+b^2 \left (6 d^2 e g-3 c d (f g+e h) (1+m)+c^2 f h \left (2+3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (1+m)} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.80 \[ \int (a+b x)^m (c+d x)^{-m} (e+f x) (g+h x) \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \left ((b c-a d)^2 f h \operatorname {Hypergeometric2F1}\left (-2+m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )+b \left (-\left ((b c-a d) (2 c f h-d (f g+e h)) \operatorname {Hypergeometric2F1}\left (-1+m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )\right )+b (d e-c f) (d g-c h) \operatorname {Hypergeometric2F1}\left (m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )\right )\right )}{b^3 d^2 (1+m)} \]
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\[\int \left (b x +a \right )^{m} \left (f x +e \right ) \left (h x +g \right ) \left (d x +c \right )^{-m}d x\]
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\[ \int (a+b x)^m (c+d x)^{-m} (e+f x) (g+h x) \, dx=\int { \frac {{\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{m}} \,d x } \]
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Exception generated. \[ \int (a+b x)^m (c+d x)^{-m} (e+f x) (g+h x) \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int (a+b x)^m (c+d x)^{-m} (e+f x) (g+h x) \, dx=\int { \frac {{\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{m}} \,d x } \]
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\[ \int (a+b x)^m (c+d x)^{-m} (e+f x) (g+h x) \, dx=\int { \frac {{\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{m}} \,d x } \]
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Timed out. \[ \int (a+b x)^m (c+d x)^{-m} (e+f x) (g+h x) \, dx=\int \frac {\left (e+f\,x\right )\,\left (g+h\,x\right )\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^m} \,d x \]
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