Optimal. Leaf size=442 \[ \frac {\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) (d+e x)^{-2 p-1} \left (a+b x+c x^2\right )^p \left (-2 c e (a e+b d (2 p+3))+b^2 e^2 (p+2)+2 c^2 d^2 (2 p+3)\right ) \left (\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}{\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\right )^{-p} \, _2F_1\left (-2 p-1,-p;-2 p;-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt {b^2-4 a c}\right )}\right )}{2 (2 p+1) (2 p+3) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )^2}-\frac {e (d+e x)^{-2 p-3} \left (a+b x+c x^2\right )^{p+1}}{(2 p+3) \left (a e^2-b d e+c d^2\right )}-\frac {e (p+2) (2 c d-b e) (d+e x)^{-2 (p+1)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+1) (2 p+3) \left (a e^2-b d e+c d^2\right )^2} \]
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Rubi [A] time = 0.35, antiderivative size = 442, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {744, 806, 726} \[ \frac {\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) (d+e x)^{-2 p-1} \left (a+b x+c x^2\right )^p \left (-2 c e (a e+b d (2 p+3))+b^2 e^2 (p+2)+2 c^2 d^2 (2 p+3)\right ) \left (\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}{\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\right )^{-p} \, _2F_1\left (-2 p-1,-p;-2 p;-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt {b^2-4 a c}\right )}\right )}{2 (2 p+1) (2 p+3) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )^2}-\frac {e (d+e x)^{-2 p-3} \left (a+b x+c x^2\right )^{p+1}}{(2 p+3) \left (a e^2-b d e+c d^2\right )}-\frac {e (p+2) (2 c d-b e) (d+e x)^{-2 (p+1)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+1) (2 p+3) \left (a e^2-b d e+c d^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 726
Rule 744
Rule 806
Rubi steps
\begin {align*} \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 {\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*}
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Mathematica [A] time = 1.34, size = 399, normalized size = 0.90 \[ -\frac {(d+e x)^{-2 p-3} (a+x (b+c x))^p \left (\frac {(d+e x)^2 \left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (-2 c e (a e+b d (2 p+3))+b^2 e^2 (p+2)+2 c^2 d^2 (2 p+3)\right ) \left (\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (e \left (\sqrt {b^2-4 a c}-b\right )+2 c d\right )}{\left (\sqrt {b^2-4 a c}-b-2 c x\right ) \left (e \left (\sqrt {b^2-4 a c}+b\right )-2 c d\right )}\right )^{-p-1} \, _2F_1\left (-2 p-1,-p;-2 p;-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (\left (b+\sqrt {b^2-4 a c}\right ) e-2 c d\right ) \left (-b-2 c x+\sqrt {b^2-4 a c}\right )}\right )}{(2 p+1) \left (e \left (\sqrt {b^2-4 a c}+b\right )-2 c d\right ) \left (e (a e-b d)+c d^2\right )}+\frac {e (p+2) (d+e x) (a+x (b+c x)) (2 c d-b e)}{(p+1) \left (e (a e-b d)+c d^2\right )}+2 e (a+x (b+c x))\right )}{2 (2 p+3) \left (e (a e-b d)+c d^2\right )} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 4}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.52, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{-2 p -4} \left (c \,x^{2}+b x +a \right )^{p}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{{\left (d+e\,x\right )}^{2\,p+4}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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