Optimal. Leaf size=442 \[ -\frac {2^{p-1} \left (a+b x+c x^2\right )^{p+1} \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \left (12 a^2 B c^2+6 a A b c^2 (2 p+5)-12 a b^2 B c (p+3)-A b^3 c \left (2 p^2+11 p+15\right )+b^4 B \left (p^2+7 p+12\right )\right ) \, _2F_1\left (-p,p+1;p+2;\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{c^4 (p+1) (2 p+3) (2 p+5) \sqrt {b^2-4 a c}}+\frac {\left (a+b x+c x^2\right )^{p+1} \left (-2 c (p+1) x \left (6 a B c (p+2)+A b c \left (2 p^2+11 p+15\right )+b^2 (-B) \left (p^2+7 p+12\right )\right )+b (p+2) \left (6 a B c (p+2)+A b c \left (2 p^2+11 p+15\right )+b^2 (-B) \left (p^2+7 p+12\right )\right )+2 a c (2 p+3) (b B (p+4)-A c (2 p+5))\right )}{4 c^4 (p+1) (p+2) (2 p+3) (2 p+5)}-\frac {x^2 \left (a+b x+c x^2\right )^{p+1} (b B (p+4)-A c (2 p+5))}{2 c^2 (p+2) (2 p+5)}+\frac {B x^3 \left (a+b x+c x^2\right )^{p+1}}{c (2 p+5)} \]
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Rubi [A] time = 0.74, antiderivative size = 442, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {832, 779, 624} \[ -\frac {2^{p-1} \left (a+b x+c x^2\right )^{p+1} \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \left (12 a^2 B c^2+6 a A b c^2 (2 p+5)-12 a b^2 B c (p+3)-A b^3 c \left (2 p^2+11 p+15\right )+b^4 B \left (p^2+7 p+12\right )\right ) \, _2F_1\left (-p,p+1;p+2;\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{c^4 (p+1) (2 p+3) (2 p+5) \sqrt {b^2-4 a c}}+\frac {\left (a+b x+c x^2\right )^{p+1} \left (-2 c (p+1) x \left (6 a B c (p+2)+A b c \left (2 p^2+11 p+15\right )+b^2 (-B) \left (p^2+7 p+12\right )\right )+b (p+2) \left (6 a B c (p+2)+A b c \left (2 p^2+11 p+15\right )+b^2 (-B) \left (p^2+7 p+12\right )\right )+2 a c (2 p+3) (b B (p+4)-A c (2 p+5))\right )}{4 c^4 (p+1) (p+2) (2 p+3) (2 p+5)}-\frac {x^2 \left (a+b x+c x^2\right )^{p+1} (b B (p+4)-A c (2 p+5))}{2 c^2 (p+2) (2 p+5)}+\frac {B x^3 \left (a+b x+c x^2\right )^{p+1}}{c (2 p+5)} \]
Antiderivative was successfully verified.
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Rule 624
Rule 779
Rule 832
Rubi steps
\begin {align*} \int x^3 (A+B x) \left (a+b x+c x^2\right )^p \, dx &=\frac {B x^3 \left (a+b x+c x^2\right )^{1+p}}{c (5+2 p)}+\frac {\int x^2 (-3 a B-(b B (4+p)-A c (5+2 p)) x) \left (a+b x+c x^2\right )^p \, dx}{c (5+2 p)}\\ &=-\frac {(b B (4+p)-A c (5+2 p)) x^2 \left (a+b x+c x^2\right )^{1+p}}{2 c^2 (2+p) (5+2 p)}+\frac {B x^3 \left (a+b x+c x^2\right )^{1+p}}{c (5+2 p)}+\frac {\int x \left (2 a (b B (4+p)-A c (5+2 p))-\left (6 a B c (2+p)-b^2 B \left (12+7 p+p^2\right )+A b c \left (15+11 p+2 p^2\right )\right ) x\right ) \left (a+b x+c x^2\right )^p \, dx}{2 c^2 (2+p) (5+2 p)}\\ &=-\frac {(b B (4+p)-A c (5+2 p)) x^2 \left (a+b x+c x^2\right )^{1+p}}{2 c^2 (2+p) (5+2 p)}+\frac {B x^3 \left (a+b x+c x^2\right )^{1+p}}{c (5+2 p)}+\frac {\left (2 a c (3+2 p) (b B (4+p)-A c (5+2 p))+b (2+p) \left (6 a B c (2+p)-b^2 B \left (12+7 p+p^2\right )+A b c \left (15+11 p+2 p^2\right )\right )-2 c (1+p) \left (6 a B c (2+p)-b^2 B \left (12+7 p+p^2\right )+A b c \left (15+11 p+2 p^2\right )\right ) x\right ) \left (a+b x+c x^2\right )^{1+p}}{4 c^4 (1+p) (2+p) (3+2 p) (5+2 p)}+\frac {\left (12 a^2 B c^2-12 a b^2 B c (3+p)+6 a A b c^2 (5+2 p)+b^4 B \left (12+7 p+p^2\right )-A b^3 c \left (15+11 p+2 p^2\right )\right ) \int \left (a+b x+c x^2\right )^p \, dx}{4 c^4 (3+2 p) (5+2 p)}\\ &=-\frac {(b B (4+p)-A c (5+2 p)) x^2 \left (a+b x+c x^2\right )^{1+p}}{2 c^2 (2+p) (5+2 p)}+\frac {B x^3 \left (a+b x+c x^2\right )^{1+p}}{c (5+2 p)}+\frac {\left (2 a c (3+2 p) (b B (4+p)-A c (5+2 p))+b (2+p) \left (6 a B c (2+p)-b^2 B \left (12+7 p+p^2\right )+A b c \left (15+11 p+2 p^2\right )\right )-2 c (1+p) \left (6 a B c (2+p)-b^2 B \left (12+7 p+p^2\right )+A b c \left (15+11 p+2 p^2\right )\right ) x\right ) \left (a+b x+c x^2\right )^{1+p}}{4 c^4 (1+p) (2+p) (3+2 p) (5+2 p)}-\frac {2^{-1+p} \left (12 a^2 B c^2-12 a b^2 B c (3+p)+6 a A b c^2 (5+2 p)+b^4 B \left (12+7 p+p^2\right )-A b^3 c \left (15+11 p+2 p^2\right )\right ) \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x+c x^2\right )^{1+p} \, _2F_1\left (-p,1+p;2+p;\frac {b+\sqrt {b^2-4 a c}+2 c x}{2 \sqrt {b^2-4 a c}}\right )}{c^4 \sqrt {b^2-4 a c} (1+p) (3+2 p) (5+2 p)}\\ \end {align*}
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Mathematica [C] time = 0.40, size = 210, normalized size = 0.48 \[ \frac {1}{20} x^4 \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}+b}\right )^{-p} (a+x (b+c x))^p \left (5 A F_1\left (4;-p,-p;5;-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{\sqrt {b^2-4 a c}-b}\right )+4 B x F_1\left (5;-p,-p;6;-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{\sqrt {b^2-4 a c}-b}\right )\right ) \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B x^{4} + A x^{3}\right )} {\left (c x^{2} + b x + a\right )}^{p}, x\right ) \]
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 \[ \int {\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p} x^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.57, size = 0, normalized size = 0.00 \[ \int \left (B x +A \right ) x^{3} \left (c \,x^{2}+b x +a \right )^{p}\, dx \]
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 \[ \int {\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p} x^{3}\,{d x} \]
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 \[ \int x^3\,\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \]
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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