Integrand size = 22, antiderivative size = 195 \[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\frac {(d+e x)^{1-2 p} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1-2 p)} \]
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Time = 0.08 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {773, 138} \[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\frac {(d+e x)^{1-2 p} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1-2 p)} \]
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Rule 138
Rule 773
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\left (a+b x+c x^2\right )^p \left (1-\frac {d+e x}{d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{-p} \left (1-\frac {d+e x}{d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{-p}\right ) \text {Subst}\left (\int x^{-2 p} \left (1-\frac {2 c x}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^p \left (1-\frac {2 c x}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^p \, dx,x,d+e x\right )}{e} \\ & = \frac {(d+e x)^{1-2 p} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} F_1\left (1-2 p;-p,-p;2-2 p;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1-2 p)} \\ \end{align*}
Time = 0.31 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.10 \[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=-\frac {\left (\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} (d+e x)^{1-2 p} (a+x (b+c x))^p \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (-1+2 p)} \]
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\[\int \left (c \,x^{2}+b x +a \right )^{p} \left (e x +d \right )^{-2 p}d x\]
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\[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (e x + d\right )}^{2 \, p}} \,d x } \]
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Timed out. \[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \]
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\[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (e x + d\right )}^{2 \, p}} \,d x } \]
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\[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (e x + d\right )}^{2 \, p}} \,d x } \]
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Timed out. \[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{{\left (d+e\,x\right )}^{2\,p}} \,d x \]
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