Optimal. Leaf size=376 \[ \frac{4^{p-1} \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \left (A \left (2 a c-b^2 (1-p)\right )+2 a b B\right ) F_1\left (-2 p;-p,-p;1-2 p;-\frac{b-\sqrt{b^2-4 a c}}{2 c x},-\frac{b+\sqrt{b^2-4 a c}}{2 c x}\right )}{a^2}-\frac{c 2^p (2 p+1) (2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1} \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p-1} \, _2F_1\left (-p,p+1;p+2;\frac{b+2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{a^2 (p+1) \sqrt{b^2-4 a c}}-\frac{(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1}}{2 a^2 x}-\frac{A \left (a+b x+c x^2\right )^{p+1}}{2 a x^2} \]
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Rubi [A] time = 0.355723, antiderivative size = 375, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {834, 843, 624, 758, 133} \[ \frac{4^{p-1} \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \left (2 a A c+2 a b B-A b^2 (1-p)\right ) F_1\left (-2 p;-p,-p;1-2 p;-\frac{b-\sqrt{b^2-4 a c}}{2 c x},-\frac{b+\sqrt{b^2-4 a c}}{2 c x}\right )}{a^2}-\frac{c 2^p (2 p+1) (2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1} \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p-1} \, _2F_1\left (-p,p+1;p+2;\frac{b+2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{a^2 (p+1) \sqrt{b^2-4 a c}}-\frac{(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1}}{2 a^2 x}-\frac{A \left (a+b x+c x^2\right )^{p+1}}{2 a x^2} \]
Antiderivative was successfully verified.
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Rule 834
Rule 843
Rule 624
Rule 758
Rule 133
Rubi steps
\begin{align*} \int \frac{(A+B x) \left (a+b x+c x^2\right )^p}{x^3} \, dx &=-\frac{A \left (a+b x+c x^2\right )^{1+p}}{2 a x^2}-\frac{\int \frac{(-2 a B+A (b-b p)-2 A c p x) \left (a+b x+c x^2\right )^p}{x^2} \, dx}{2 a}\\ &=-\frac{A \left (a+b x+c x^2\right )^{1+p}}{2 a x^2}-\frac{(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{1+p}}{2 a^2 x}+\frac{\int \frac{\left (\left (2 a b B+2 a A c-A b^2 (1-p)\right ) p+c (1+2 p) (2 a B-A (b-b p)) x\right ) \left (a+b x+c x^2\right )^p}{x} \, dx}{2 a^2}\\ &=-\frac{A \left (a+b x+c x^2\right )^{1+p}}{2 a x^2}-\frac{(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{1+p}}{2 a^2 x}+\frac{\left (\left (2 a b B+2 a A c-A b^2 (1-p)\right ) p\right ) \int \frac{\left (a+b x+c x^2\right )^p}{x} \, dx}{2 a^2}+\frac{(c (2 a B-A b (1-p)) (1+2 p)) \int \left (a+b x+c x^2\right )^p \, dx}{2 a^2}\\ &=-\frac{A \left (a+b x+c x^2\right )^{1+p}}{2 a x^2}-\frac{(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{1+p}}{2 a^2 x}-\frac{2^p c (2 a B-A b (1-p)) (1+2 p) \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 )}{a^2 \sqrt{b^2-4 a c} (1+p)}-\frac{\left (2^{-1+2 p} \left (2 a b B+2 a A c-A b^2 (1-p)\right ) p \left (\frac{1}{x}\right )^{2 p} \left (\frac{b-\sqrt{b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (\frac{b+\sqrt{b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p\right ) \operatorname{Subst}\left (\int x^{1-2 (1+p)} \left (1+\frac{\left (b-\sqrt{b^2-4 a c}\right ) x}{2 c}\right )^p \left (1+\frac{\left (b+\sqrt{b^2-4 a c}\right ) x}{2 c}\right )^p \, dx,x,\frac{1}{x}\right )}{a^2}\\ &=-\frac{A \left (a+b x+c x^2\right )^{1+p}}{2 a x^2}-\frac{(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{1+p}}{2 a^2 x}+\frac{4^{-1+p} \left (2 a b B+2 a A c-A b^2 (1-p)\right ) \left (\frac{b-\sqrt{b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (\frac{b+\sqrt{b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (-2 p;-p,-p;1-2 p;-\frac{b-\sqrt{b^2-4 a c}}{2 c x},-\frac{b+\sqrt{b^2-4 a c}}{2 c x}\right )}{a^2}-\frac{2^p c (2 a B-A b (1-p)) (1+2 p) \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 )}{a^2 \sqrt{b^2-4 a c} (1+p)}\\ \end{align*}
Mathematica [A] time = 0.316734, size = 295, normalized size = 0.78 \[ \frac{\left (\frac{b-\sqrt{b^2-4 a c}}{2 c x}+1\right )^{-p} \left (\frac{b-\sqrt{b^2-4 a c}}{2 c}+x\right )^{-p} \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{c}\right )^p \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{c x}\right )^{-p} (a+x (b+c x))^p \left (A (2 p-1) F_1\left (2-2 p;-p,-p;3-2 p;-\frac{b+\sqrt{b^2-4 a c}}{2 c x},\frac{\sqrt{b^2-4 a c}-b}{2 c x}\right )+2 B (p-1) x F_1\left (1-2 p;-p,-p;2-2 p;-\frac{b+\sqrt{b^2-4 a c}}{2 c x},\frac{\sqrt{b^2-4 a c}-b}{2 c x}\right )\right )}{2 (p-1) (2 p-1) x^2} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.086, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( Bx+A \right ) \left ( c{x}^{2}+bx+a \right ) ^{p}}{{x}^{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (c x^{2} + b x + a\right )}^{p}}{x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x + A\right )}{\left (c x^{2} + b x + a\right )}^{p}}{x^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (c x^{2} + b x + a\right )}^{p}}{x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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