Integrand size = 29, antiderivative size = 203 \[ \int (a+b x)^m (c+d x)^{-2-m} (e+f x) (g+h x) \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-1-m} \left (b d^2 e g+b c^2 f h (2+m)-c d (b (f g+e h)+a f h (1+m))+d (b c-a d) f h (1+m) x\right )}{b d^2 (b c-a d) (1+m)}-\frac {(a d f h m+b (d (f g+e h)-c f h (2+m))) (a+b x)^m \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{b d^3 m} \]
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Time = 0.07 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {148, 72, 71} \[ \int (a+b x)^m (c+d x)^{-2-m} (e+f x) (g+h x) \, dx=-\frac {(a+b x)^{m+1} (c+d x)^{-m-1} \left (-d f h (m+1) x (b c-a d)+a c d f h (m+1)-b \left (c^2 f h (m+2)-c d (e h+f g)+d^2 e g\right )\right )}{b d^2 (m+1) (b c-a d)}-\frac {(a+b x)^m (c+d x)^{-m} \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right ) (a d f h m-b c f h (m+2)+b d (e h+f g))}{b d^3 m} \]
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Rule 71
Rule 72
Rule 148
Rubi steps \begin{align*} \text {integral}& = -\frac {(a+b x)^{1+m} (c+d x)^{-1-m} \left (a c d f h (1+m)-b \left (d^2 e g-c d (f g+e h)+c^2 f h (2+m)\right )-d (b c-a d) f h (1+m) x\right )}{b d^2 (b c-a d) (1+m)}+\frac {(b d (f g+e h)+a d f h m-b c f h (2+m)) \int (a+b x)^m (c+d x)^{-1-m} \, dx}{b d^2} \\ & = -\frac {(a+b x)^{1+m} (c+d x)^{-1-m} \left (a c d f h (1+m)-b \left (d^2 e g-c d (f g+e h)+c^2 f h (2+m)\right )-d (b c-a d) f h (1+m) x\right )}{b d^2 (b c-a d) (1+m)}+\frac {\left ((b d (f g+e h)+a d f h m-b c f h (2+m)) (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}{b d^2} \\ & = -\frac {(a+b x)^{1+m} (c+d x)^{-1-m} \left (a c d f h (1+m)-b \left (d^2 e g-c d (f g+e h)+c^2 f h (2+m)\right )-d (b c-a d) f h (1+m) x\right )}{b d^2 (b c-a d) (1+m)}-\frac {(b d (f g+e h)+a d f h m-b c f h (2+m)) (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 )}{b d^3 m} \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.98 \[ \int (a+b x)^m (c+d x)^{-2-m} (e+f x) (g+h x) \, dx=\frac {(a+b x)^m (c+d x)^{-m} \left (-\frac {d (a+b x) \left (a d f h (1+m) (c+d x)-b \left (d^2 e g+c^2 f h (2+m)+c d (-f g-e h+f h (1+m) x)\right )\right )}{c+d x}+\frac {(b c-a d) (1+m) (-b d (f g+e h)-a d f h m+b c f h (2+m)) \left (\frac {d (a+b x)}{-b c+a d}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{m}\right )}{b d^3 (b c-a d) (1+m)} \]
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\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{-2-m} \left (f x +e \right ) \left (h x +g \right )d x\]
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\[ \int (a+b x)^m (c+d x)^{-2-m} (e+f x) (g+h x) \, dx=\int { {\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 2} \,d x } \]
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Exception generated. \[ \int (a+b x)^m (c+d x)^{-2-m} (e+f x) (g+h x) \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int (a+b x)^m (c+d x)^{-2-m} (e+f x) (g+h x) \, dx=\int { {\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 2} \,d x } \]
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\[ \int (a+b x)^m (c+d x)^{-2-m} (e+f x) (g+h x) \, dx=\int { {\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 2} \,d x } \]
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Timed out. \[ \int (a+b x)^m (c+d x)^{-2-m} (e+f x) (g+h x) \, dx=\int \frac {\left (e+f\,x\right )\,\left (g+h\,x\right )\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^{m+2}} \,d x \]
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