Integrand size = 24, antiderivative size = 442 \[ \int (d+e x)^{-4-2 p} \left (a+b x+c x^2\right )^p \, dx=-\frac {e (d+e x)^{-3-2 p} \left (a+b x+c x^2\right )^{1+p}}{\left (c d^2-b d e+a e^2\right ) (3+2 p)}-\frac {e (2 c d-b e) (2+p) (d+e x)^{-2 (1+p)} \left (a+b x+c x^2\right )^{1+p}}{2 \left (c d^2-b d e+a e^2\right )^2 (1+p) (3+2 p)}+\frac {\left (b^2 e^2 (2+p)+2 c^2 d^2 (3+2 p)-2 c e (a e+b d (3+2 p))\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )^{-p} (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-1-2 p,-p,-2 p,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )}{2 \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right )^2 (1+2 p) (3+2 p)} \]
[Out]
Time = 0.25 (sec) , antiderivative size = 442, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {758, 820, 740} \[ \int (d+e x)^{-4-2 p} \left (a+b x+c x^2\right )^p \, dx=\frac {\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) (d+e x)^{-2 p-1} \left (a+b x+c x^2\right )^p \left (-2 c e (a e+b d (2 p+3))+b^2 e^2 (p+2)+2 c^2 d^2 (2 p+3)\right ) \left (\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}{\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-2 p-1,-p,-2 p,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt {b^2-4 a c}\right )}\right )}{2 (2 p+1) (2 p+3) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )^2}-\frac {e (d+e x)^{-2 p-3} \left (a+b x+c x^2\right )^{p+1}}{(2 p+3) \left (a e^2-b d e+c d^2\right )}-\frac {e (p+2) (2 c d-b e) (d+e x)^{-2 (p+1)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+1) (2 p+3) \left (a e^2-b d e+c d^2\right )^2} \]
[In]
[Out]
Rule 740
Rule 758
Rule 820
Rubi steps \begin{align*} \text {integral}& = -\frac {e (d+e x)^{-3-2 p} \left (a+b x+c x^2\right )^{1+p}}{\left (c d^2-b d e+a e^2\right ) (3+2 p)}-\frac {\int (d+e x)^{-3-2 p} (b e (2+p)-c d (3+2 p)+c e x) \left (a+b x+c x^2\right )^p \, dx}{\left (c d^2-b d e+a e^2\right ) (3+2 p)} \\ & = -\frac {e (d+e x)^{-3-2 p} \left (a+b x+c x^2\right )^{1+p}}{\left (c d^2-b d e+a e^2\right ) (3+2 p)}-\frac {e (2 c d-b e) (2+p) (d+e x)^{-2 (1+p)} \left (a+b x+c x^2\right )^{1+p}}{2 \left (c d^2-b d e+a e^2\right )^2 (1+p) (3+2 p)}+\frac {\left (b^2 e^2 (2+p)+2 c^2 d^2 (3+2 p)-2 c e (a e+b d (3+2 p))\right ) \int (d+e x)^{-2-2 p} \left (a+b x+c x^2\right )^p \, dx}{2 \left (c d^2-b d e+a e^2\right )^2 (3+2 p)} \\ & = -\frac {e (d+e x)^{-3-2 p} \left (a+b x+c x^2\right )^{1+p}}{\left (c d^2-b d e+a e^2\right ) (3+2 p)}-\frac {e (2 c d-b e) (2+p) (d+e x)^{-2 (1+p)} \left (a+b x+c x^2\right )^{1+p}}{2 \left (c d^2-b d e+a e^2\right )^2 (1+p) (3+2 p)}+\frac {\left (b^2 e^2 (2+p)+2 c^2 d^2 (3+2 p)-2 c e (a e+b d (3+2 p))\right ) \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 )}{2 \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right )^2 (1+2 p) (3+2 p)} \\ \end{align*}
Time = 0.94 (sec) , antiderivative size = 399, normalized size of antiderivative = 0.90 \[ \int (d+e x)^{-4-2 p} \left (a+b x+c x^2\right )^p \, dx=-\frac {(d+e x)^{-3-2 p} (a+x (b+c x))^p \left (2 e (a+x (b+c x))+\frac {e (2 c d-b e) (2+p) (d+e x) (a+x (b+c x))}{\left (c d^2+e (-b d+a e)\right ) (1+p)}+\frac {\left (b^2 e^2 (2+p)+2 c^2 d^2 (3+2 p)-2 c e (a e+b d (3+2 p))\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )^{-1-p} (d+e x)^2 \operatorname {Hypergeometric2F1}\left (-1-2 p,-p,-2 p,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2+e (-b d+a e)\right ) (1+2 p)}\right )}{2 \left (c d^2+e (-b d+a e)\right ) (3+2 p)} \]
[In]
[Out]
\[\int \left (e x +d \right )^{-4-2 p} \left (c \,x^{2}+b x +a \right )^{p}d x\]
[In]
[Out]
\[ \int (d+e x)^{-4-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 4} \,d x } \]
[In]
[Out]
Timed out. \[ \int (d+e x)^{-4-2 p} \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int (d+e x)^{-4-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 4} \,d x } \]
[In]
[Out]
\[ \int (d+e x)^{-4-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 4} \,d x } \]
[In]
[Out]
Timed out. \[ \int (d+e x)^{-4-2 p} \left (a+b x+c x^2\right )^p \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{{\left (d+e\,x\right )}^{2\,p+4}} \,d x \]
[In]
[Out]