Integrand size = 27, antiderivative size = 177 \[ \int \frac {(A+B x) (c+d x)^n (e+f x)^p}{a+b x} \, dx=-\frac {(A b-a B) (c+d x)^{1+n} (e+f x)^p \left (\frac {d (e+f x)}{d e-c f}\right )^{-p} \operatorname {AppellF1}\left (1+n,1,-p,2+n,\frac {b (c+d x)}{b c-a d},-\frac {f (c+d x)}{d e-c f}\right )}{b (b c-a d) (1+n)}-\frac {B (c+d x)^{1+n} (e+f x)^{1+p} \operatorname {Hypergeometric2F1}\left (1,2+n+p,2+p,\frac {d (e+f x)}{d e-c f}\right )}{b (d e-c f) (1+p)} \]
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Time = 0.08 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {163, 72, 71, 142, 141} \[ \int \frac {(A+B x) (c+d x)^n (e+f x)^p}{a+b x} \, dx=\frac {B (c+d x)^{n+1} (e+f x)^p \left (\frac {d (e+f x)}{d e-c f}\right )^{-p} \operatorname {Hypergeometric2F1}\left (n+1,-p,n+2,-\frac {f (c+d x)}{d e-c f}\right )}{b d (n+1)}-\frac {(A b-a B) (c+d x)^{n+1} (e+f x)^p \left (\frac {d (e+f x)}{d e-c f}\right )^{-p} \operatorname {AppellF1}\left (n+1,-p,1,n+2,-\frac {f (c+d x)}{d e-c f},\frac {b (c+d x)}{b c-a d}\right )}{b (n+1) (b c-a d)} \]
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Rule 71
Rule 72
Rule 141
Rule 142
Rule 163
Rubi steps \begin{align*} \text {integral}& = \frac {B \int (c+d x)^n (e+f x)^p \, dx}{b}+\frac {(A b-a B) \int \frac {(c+d x)^n (e+f x)^p}{a+b x} \, dx}{b} \\ & = \frac {\left (B (e+f x)^p \left (\frac {d (e+f x)}{d e-c f}\right )^{-p}\right ) \int (c+d x)^n \left (\frac {d e}{d e-c f}+\frac {d f x}{d e-c f}\right )^p \, dx}{b}+\frac {\left ((A b-a B) (e+f x)^p \left (\frac {d (e+f x)}{d e-c f}\right )^{-p}\right ) \int \frac {(c+d x)^n \left (\frac {d e}{d e-c f}+\frac {d f x}{d e-c f}\right )^p}{a+b x} \, dx}{b} \\ & = -\frac {(A b-a B) (c+d x)^{1+n} (e+f x)^p \left (\frac {d (e+f x)}{d e-c f}\right )^{-p} F_1\left (1+n;-p,1;2+n;-\frac {f (c+d x)}{d e-c f},\frac {b (c+d x)}{b c-a d}\right )}{b (b c-a d) (1+n)}+\frac {B (c+d x)^{1+n} (e+f x)^p \left (\frac {d (e+f x)}{d e-c f}\right )^{-p} \, _2F_1\left (1+n,-p;2+n;-\frac {f (c+d x)}{d e-c f}\right )}{b d (1+n)} \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) (c+d x)^n (e+f x)^p}{a+b x} \, dx=\frac {(c+d x)^n (e+f x)^p \left (\frac {(A b-a B) \left (\frac {b (c+d x)}{d (a+b x)}\right )^{-n} \left (\frac {b (e+f x)}{f (a+b x)}\right )^{-p} \operatorname {AppellF1}\left (-n-p,-n,-p,1-n-p,\frac {-b c+a d}{d (a+b x)},\frac {-b e+a f}{f (a+b x)}\right )}{n+p}+\frac {b B \left (\frac {f (c+d x)}{-d e+c f}\right )^{-n} (e+f x) \operatorname {Hypergeometric2F1}\left (-n,1+p,2+p,\frac {d (e+f x)}{d e-c f}\right )}{f (1+p)}\right )}{b^2} \]
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\[\int \frac {\left (B x +A \right ) \left (d x +c \right )^{n} \left (f x +e \right )^{p}}{b x +a}d x\]
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\[ \int \frac {(A+B x) (c+d x)^n (e+f x)^p}{a+b x} \, dx=\int { \frac {{\left (B x + A\right )} {\left (d x + c\right )}^{n} {\left (f x + e\right )}^{p}}{b x + a} \,d x } \]
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Timed out. \[ \int \frac {(A+B x) (c+d x)^n (e+f x)^p}{a+b x} \, dx=\text {Timed out} \]
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\[ \int \frac {(A+B x) (c+d x)^n (e+f x)^p}{a+b x} \, dx=\int { \frac {{\left (B x + A\right )} {\left (d x + c\right )}^{n} {\left (f x + e\right )}^{p}}{b x + a} \,d x } \]
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\[ \int \frac {(A+B x) (c+d x)^n (e+f x)^p}{a+b x} \, dx=\int { \frac {{\left (B x + A\right )} {\left (d x + c\right )}^{n} {\left (f x + e\right )}^{p}}{b x + a} \,d x } \]
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Timed out. \[ \int \frac {(A+B x) (c+d x)^n (e+f x)^p}{a+b x} \, dx=\int \frac {{\left (e+f\,x\right )}^p\,\left (A+B\,x\right )\,{\left (c+d\,x\right )}^n}{a+b\,x} \,d x \]
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