Optimal. Leaf size=287 \[ \frac{(d+e x)^{m+2} \left (2 c e g (a e g-b (2 d g+e f))+b^2 e^2 g^2+c^2 \left (3 d^2 g^2+2 d e f g+e^2 f^2\right )\right )}{e^4 g^3 (m+2)}+\frac{(d+e x)^{m+1} (b e g-c (d g+e f)) \left (e g (2 a e g-b (d g+e f))+c \left (d^2 g^2+e^2 f^2\right )\right )}{e^4 g^4 (m+1)}+\frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2 \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right )}{g^4 (m+1) (e f-d g)}-\frac{c (d+e x)^{m+3} (-2 b e g+3 c d g+c e f)}{e^4 g^2 (m+3)}+\frac{c^2 (d+e x)^{m+4}}{e^4 g (m+4)} \]
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Rubi [A] time = 0.865186, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {951, 1620, 68} \[ \frac{(d+e x)^{m+2} \left (2 c e g (a e g-b (2 d g+e f))+b^2 e^2 g^2+c^2 \left (3 d^2 g^2+2 d e f g+e^2 f^2\right )\right )}{e^4 g^3 (m+2)}+\frac{(d+e x)^{m+1} (b e g-c (d g+e f)) \left (e g (2 a e g-b (d g+e f))+c \left (d^2 g^2+e^2 f^2\right )\right )}{e^4 g^4 (m+1)}+\frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2 \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right )}{g^4 (m+1) (e f-d g)}-\frac{c (d+e x)^{m+3} (-2 b e g+3 c d g+c e f)}{e^4 g^2 (m+3)}+\frac{c^2 (d+e x)^{m+4}}{e^4 g (m+4)} \]
Antiderivative was successfully verified.
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Rule 951
Rule 1620
Rule 68
Rubi steps
\begin{align*} \int \frac{(d+e x)^m \left (a+b x+c x^2\right )^2}{f+g x} \, dx &=\frac{c^2 (d+e x)^{4+m}}{e^4 g (4+m)}+\frac{\int \frac{(d+e x)^m \left (-e \left (c^2 d^3 f-a^2 e^3 g\right ) (4+m)+e \left (2 a b e^3 g-c^2 d^2 (3 e f+d g)\right ) (4+m) x+e^2 \left (b^2 e^2 g+2 a c e^2 g-3 c^2 d (e f+d g)\right ) (4+m) x^2-c e^3 (c e f+3 c d g-2 b e g) (4+m) x^3\right )}{f+g x} \, dx}{e^4 g (4+m)}\\ &=\frac{c^2 (d+e x)^{4+m}}{e^4 g (4+m)}+\frac{\int \left (\frac{e (b e g-c (e f+d g)) \left (c \left (e^2 f^2+d^2 g^2\right )+e g (2 a e g-b (e f+d g))\right ) (4+m) (d+e x)^m}{g^3}+\frac{e \left (b^2 e^2 g^2+c^2 \left (e^2 f^2+2 d e f g+3 d^2 g^2\right )+2 c e g (a e g-b (e f+2 d g))\right ) (4+m) (d+e x)^{1+m}}{g^2}-\frac{c e (c e f+3 c d g-2 b e g) (4+m) (d+e x)^{2+m}}{g}+\frac{e^4 \left (c f^2-b f g+a g^2\right )^2 (4+m) (d+e x)^m}{g^3 (f+g x)}\right ) \, dx}{e^4 g (4+m)}\\ &=\frac{(b e g-c (e f+d g)) \left (c \left (e^2 f^2+d^2 g^2\right )+e g (2 a e g-b (e f+d g))\right ) (d+e x)^{1+m}}{e^4 g^4 (1+m)}+\frac{\left (b^2 e^2 g^2+c^2 \left (e^2 f^2+2 d e f g+3 d^2 g^2\right )+2 c e g (a e g-b (e f+2 d g))\right ) (d+e x)^{2+m}}{e^4 g^3 (2+m)}-\frac{c (c e f+3 c d g-2 b e g) (d+e x)^{3+m}}{e^4 g^2 (3+m)}+\frac{c^2 (d+e x)^{4+m}}{e^4 g (4+m)}+\frac{\left (c f^2-b f g+a g^2\right )^2 \int \frac{(d+e x)^m}{f+g x} \, dx}{g^4}\\ &=\frac{(b e g-c (e f+d g)) \left (c \left (e^2 f^2+d^2 g^2\right )+e g (2 a e g-b (e f+d g))\right ) (d+e x)^{1+m}}{e^4 g^4 (1+m)}+\frac{\left (b^2 e^2 g^2+c^2 \left (e^2 f^2+2 d e f g+3 d^2 g^2\right )+2 c e g (a e g-b (e f+2 d g))\right ) (d+e x)^{2+m}}{e^4 g^3 (2+m)}-\frac{c (c e f+3 c d g-2 b e g) (d+e x)^{3+m}}{e^4 g^2 (3+m)}+\frac{c^2 (d+e x)^{4+m}}{e^4 g (4+m)}+\frac{\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{g (d+e x)}{e f-d g}\right )}{g^4 (e f-d g) (1+m)}\\ \end{align*}
Mathematica [A] time = 0.447522, size = 265, normalized size = 0.92 \[ \frac{(d+e x)^{m+1} \left (\frac{g (d+e x) \left (2 c e g (a e g-b (2 d g+e f))+b^2 e^2 g^2+c^2 \left (3 d^2 g^2+2 d e f g+e^2 f^2\right )\right )}{e^4 (m+2)}-\frac{(-b e g+c d g+c e f) \left (e g (2 a e g-b (d g+e f))+c \left (d^2 g^2+e^2 f^2\right )\right )}{e^4 (m+1)}+\frac{\left (g (a g-b f)+c f^2\right )^2 \, _2F_1\left (1,m+1;m+2;\frac{g (d+e x)}{d g-e f}\right )}{(m+1) (e f-d g)}-\frac{c g^2 (d+e x)^2 (-2 b e g+3 c d g+c e f)}{e^4 (m+3)}+\frac{c^2 g^3 (d+e x)^3}{e^4 (m+4)}\right )}{g^4} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.641, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+bx+a \right ) ^{2} \left ( ex+d \right ) ^{m}}{gx+f}}\, 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 (c x^{2} + b x + a\right )}^{2}{\left (e x + d\right )}^{m}}{g x + f}\,{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 (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}{\left (e x + d\right )}^{m}}{g x + f}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{m} \left (a + b x + c x^{2}\right )^{2}}{f + g x}\, dx \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 (c x^{2} + b x + a\right )}^{2}{\left (e x + d\right )}^{m}}{g x + f}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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