Integrand size = 26, antiderivative size = 133 \[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\frac {(b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (1+m,-2+m,-p,2+m,-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (1+m)} \]
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Time = 0.07 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {145, 144, 143} \[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\frac {(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m} (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (m+1,m-2,-p,m+2,-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (m+1)} \]
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Rule 143
Rule 144
Rule 145
Rubi steps \begin{align*} \text {integral}& = \frac {\left ((b c-a d)^2 (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 )^{2-m} (e+f x)^p \, dx}{b^2} \\ & = \frac {\left ((b c-a d)^2 (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p}\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{2-m} \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^p \, dx}{b^2} \\ & = \frac {(b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (1+m;-2+m,-p;2+m;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (1+m)} \\ \end{align*}
Time = 0.29 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98 \[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\frac {(b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (1+m,-2+m,-p,2+m,\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )}{b^3 (1+m)} \]
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\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{2-m} \left (f x +e \right )^{p}d x\]
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\[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2} {\left (f x + e\right )}^{p} \,d x } \]
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Timed out. \[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\text {Timed out} \]
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\[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2} {\left (f x + e\right )}^{p} \,d x } \]
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\[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2} {\left (f x + e\right )}^{p} \,d x } \]
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Timed out. \[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\int {\left (e+f\,x\right )}^p\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{2-m} \,d x \]
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