Optimal. Leaf size=442 \[ -\frac {e (d+e x)^{-3-2 p} \left (a+b x+c x^2\right )^{1+p}}{\left (c d^2-b d e+a e^2\right ) (3+2 p)}-\frac {e (2 c d-b e) (2+p) (d+e x)^{-2 (1+p)} \left (a+b x+c x^2\right )^{1+p}}{2 \left (c d^2-b d e+a e^2\right )^2 (1+p) (3+2 p)}+\frac {\left (b^2 e^2 (2+p)+2 c^2 d^2 (3+2 p)-2 c e (a e+b d (3+2 p))\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )^{-p} (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \, _2F_1\left (-1-2 p,-p;-2 p;-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )}{2 \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right )^2 (1+2 p) (3+2 p)} \]
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Rubi [A]
time = 0.24, 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 = {758, 820, 740}
\begin {gather*} \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} \end {gather*}
Antiderivative was successfully verified.
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Rule 740
Rule 758
Rule 820
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 = 1.01, size = 399, normalized size = 0.90 \begin {gather*} -\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 \, _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 ) \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)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.56, size = 0, normalized size = 0.00 \[\int \left (e x +d \right )^{-4-2 p} \left (c \,x^{2}+b x +a \right )^{p}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{{\left (d+e\,x\right )}^{2\,p+4}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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