Integrand size = 29, antiderivative size = 261 \[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (2 b d^2 e g+b c^2 f h (2+m)-c d (2 b (f g+e h)+a f h m)+d (b c-a d) f h m x\right )}{2 b d^2 (b c-a d) m}-\frac {\left (b^2 c^2 f h (1+m) (2+m)-2 b c d (1+m) (b f g+b e h+a f h m)+d^2 \left (2 b^2 e g+2 a b (f g+e h) m-a^2 f h (1-m) m\right )\right ) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,1+m,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{2 b^2 d^2 (b c-a d) m (1+m)} \]
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Time = 0.11 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {151, 72, 71} \[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\frac {(a+b x)^{m+1} (c+d x)^{-m} \left (-c d (a f h m+2 b (e h+f g))+d f h m x (b c-a d)+b c^2 f h (m+2)+2 b d^2 e g\right )}{2 b d^2 m (b c-a d)}-\frac {(a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,m+1,m+2,-\frac {d (a+b x)}{b c-a d}\right ) \left (d^2 \left (a^2 (-f) h (1-m) m+2 a b m (e h+f g)+2 b^2 e g\right )-2 b c d (m+1) (a f h m+b e h+b f g)+b^2 c^2 f h (m+1) (m+2)\right )}{2 b^2 d^2 m (m+1) (b c-a d)} \]
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Rule 71
Rule 72
Rule 151
Rubi steps \begin{align*} \text {integral}& = \frac {(a+b x)^{1+m} (c+d x)^{-m} \left (2 b d^2 e g+b c^2 f h (2+m)-c d (2 b (f g+e h)+a f h m)+d (b c-a d) f h m x\right )}{2 b d^2 (b c-a d) m}-\frac {\left (b^2 c^2 f h (1+m) (2+m)-2 b c d (1+m) (b f g+b e h+a f h m)+d^2 \left (2 b^2 e g+2 a b (f g+e h) m-a^2 f h (1-m) m\right )\right ) \int (a+b x)^m (c+d x)^{-m} \, dx}{2 b d^2 (b c-a d) m} \\ & = \frac {(a+b x)^{1+m} (c+d x)^{-m} \left (2 b d^2 e g+b c^2 f h (2+m)-c d (2 b (f g+e h)+a f h m)+d (b c-a d) f h m x\right )}{2 b d^2 (b c-a d) m}-\frac {\left (\left (b^2 c^2 f h (1+m) (2+m)-2 b c d (1+m) (b f g+b e h+a f h m)+d^2 \left (2 b^2 e g+2 a b (f g+e h) m-a^2 f h (1-m) m\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}{2 b d^2 (b c-a d) m} \\ & = \frac {(a+b x)^{1+m} (c+d x)^{-m} \left (2 b d^2 e g+b c^2 f h (2+m)-c d (2 b (f g+e h)+a f h m)+d (b c-a d) f h m x\right )}{2 b d^2 (b c-a d) m}-\frac {\left (b^2 c^2 f h (1+m) (2+m)-2 b c d (1+m) (b f g+b e h+a f h m)+d^2 \left (2 b^2 e g+2 a b (f g+e h) m-a^2 f h (1-m) m\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 )}{2 b^2 d^2 (b c-a d) m (1+m)} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.85 \[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (b \left (a d f h m (c+d x)-b \left (2 d^2 e g+c^2 f h (2+m)+c d (-2 f g-2 e h+f h m x)\right )\right )+\frac {\left (a^2 d^2 f h (-1+m) m+2 a b d m (d (f g+e h)-c f h (1+m))+b^2 \left (2 d^2 e g-2 c d (f g+e h) (1+m)+c^2 f h \left (2+3 m+m^2\right )\right )\right ) \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )}{1+m}\right )}{2 b^2 d^2 (-b c+a d) m} \]
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\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{-1-m} \left (f x +e \right ) \left (h x +g \right )d x\]
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\[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\int { {\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1} \,d x } \]
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Exception generated. \[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\int { {\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1} \,d x } \]
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\[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\int { {\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1} \,d x } \]
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Timed out. \[ \int (a+b x)^m (c+d x)^{-1-m} (e+f x) (g+h x) \, dx=\int \frac {\left (e+f\,x\right )\,\left (g+h\,x\right )\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^{m+1}} \,d x \]
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