Integrand size = 24, antiderivative size = 248 \[ \int (d+e x)^{-2-2 p} \left (a+b x+c x^2\right )^p \, dx=\frac {\left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )^{-p} (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-1-2 p,-p,-2 p,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )}{\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1+2 p)} \]
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Time = 0.10 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {740} \[ \int (d+e x)^{-2-2 p} \left (a+b x+c x^2\right )^p \, dx=\frac {\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) (d+e x)^{-2 p-1} \left (a+b x+c x^2\right )^p \left (\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}{\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-2 p-1,-p,-2 p,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt {b^2-4 a c}\right )}\right )}{(2 p+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )} \]
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Rule 740
Rubi steps \begin{align*} \text {integral}& = \frac {\left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )^{-p} (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \, _2F_1\left (-1-2 p,-p;-2 p;-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )}{\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1+2 p)} \\ \end{align*}
Time = 0.51 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.00 \[ \int (d+e x)^{-2-2 p} \left (a+b x+c x^2\right )^p \, dx=-\frac {\left (\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )^{1-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} (d+e x)^{-1-2 p} (a+x (b+c x))^p \left (1-\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )^{2 p} \operatorname {Hypergeometric2F1}\left (-1-2 p,-p,-2 p,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )}{e+2 e p} \]
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\[\int \left (e x +d \right )^{-2-2 p} \left (c \,x^{2}+b x +a \right )^{p}d x\]
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\[ \int (d+e x)^{-2-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 2} \,d x } \]
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Timed out. \[ \int (d+e x)^{-2-2 p} \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \]
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\[ \int (d+e x)^{-2-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 2} \,d x } \]
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\[ \int (d+e x)^{-2-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 2} \,d x } \]
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Timed out. \[ \int (d+e x)^{-2-2 p} \left (a+b x+c x^2\right )^p \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{{\left (d+e\,x\right )}^{2\,p+2}} \,d x \]
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