Optimal. Leaf size=298 \[ \frac{(d+e x)^{m+1} \left (2 c e g (a e g-b (d g+2 e f))+b^2 e^2 g^2+c^2 \left (d^2 g^2+2 d e f g+3 e^2 f^2\right )\right )}{e^3 g^4 (m+1)}+\frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right ) \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right ) (c f (4 d g-e f (m+4))-g (a e g m+b (2 d g-e f (m+2))))}{g^4 (m+1) (e f-d g)^2}+\frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2}{g^4 (f+g x) (e f-d g)}-\frac{2 c (d+e x)^{m+2} (-b e g+c d g+c e f)}{e^3 g^3 (m+2)}+\frac{c^2 (d+e x)^{m+3}}{e^3 g^2 (m+3)} \]
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Rubi [A] time = 1.14088, antiderivative size = 298, 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 = {949, 1620, 68} \[ \frac{(d+e x)^{m+1} \left (2 c e g (a e g-b (d g+2 e f))+b^2 e^2 g^2+c^2 \left (d^2 g^2+2 d e f g+3 e^2 f^2\right )\right )}{e^3 g^4 (m+1)}-\frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right ) \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right ) (g (a e g m+2 b d g-b e f (m+2))-c f (4 d g-e f (m+4)))}{g^4 (m+1) (e f-d g)^2}+\frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2}{g^4 (f+g x) (e f-d g)}-\frac{2 c (d+e x)^{m+2} (-b e g+c d g+c e f)}{e^3 g^3 (m+2)}+\frac{c^2 (d+e x)^{m+3}}{e^3 g^2 (m+3)} \]
Antiderivative was successfully verified.
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Rule 949
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)^2} \, dx &=\frac{\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{g^4 (e f-d g) (f+g x)}+\frac{\int \frac{(d+e x)^m \left (\frac{c^2 f^3 (d g-e f (1+m))-2 c f g (b f-a g) (d g-e f (1+m))-g^2 \left (a^2 e g^2 m-b^2 f (d g-e f (1+m))+2 a b g (d g-e f (1+m))\right )}{g^4}+\frac{(e f-d g) \left (c^2 f^2+b^2 g^2-2 c g (b f-a g)\right ) x}{g^3}-\frac{c (c f-2 b g) (e f-d g) x^2}{g^2}-c^2 \left (d-\frac{e f}{g}\right ) x^3\right )}{f+g x} \, dx}{e f-d g}\\ &=\frac{\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{g^4 (e f-d g) (f+g x)}+\frac{\int \left (\frac{(e f-d g) \left (b^2 e^2 g^2+c^2 \left (3 e^2 f^2+2 d e f g+d^2 g^2\right )+2 c e g (a e g-b (2 e f+d g))\right ) (d+e x)^m}{e^2 g^4}-\frac{2 c (e f-d g) (c e f+c d g-b e g) (d+e x)^{1+m}}{e^2 g^3}+\frac{c^2 (e f-d g) (d+e x)^{2+m}}{e^2 g^2}+\frac{\left (c f^2-b f g+a g^2\right ) (-g (2 b d g+a e g m-b e f (2+m))+c f (4 d g-e f (4+m))) (d+e x)^m}{g^4 (f+g x)}\right ) \, dx}{e f-d g}\\ &=\frac{\left (b^2 e^2 g^2+c^2 \left (3 e^2 f^2+2 d e f g+d^2 g^2\right )+2 c e g (a e g-b (2 e f+d g))\right ) (d+e x)^{1+m}}{e^3 g^4 (1+m)}-\frac{2 c (c e f+c d g-b e g) (d+e x)^{2+m}}{e^3 g^3 (2+m)}+\frac{c^2 (d+e x)^{3+m}}{e^3 g^2 (3+m)}+\frac{\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{g^4 (e f-d g) (f+g x)}-\frac{\left (\left (c f^2-b f g+a g^2\right ) (g (2 b d g+a e g m-b e f (2+m))-c f (4 d g-e f (4+m)))\right ) \int \frac{(d+e x)^m}{f+g x} \, dx}{g^4 (e f-d g)}\\ &=\frac{\left (b^2 e^2 g^2+c^2 \left (3 e^2 f^2+2 d e f g+d^2 g^2\right )+2 c e g (a e g-b (2 e f+d g))\right ) (d+e x)^{1+m}}{e^3 g^4 (1+m)}-\frac{2 c (c e f+c d g-b e g) (d+e x)^{2+m}}{e^3 g^3 (2+m)}+\frac{c^2 (d+e x)^{3+m}}{e^3 g^2 (3+m)}+\frac{\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{g^4 (e f-d g) (f+g x)}-\frac{\left (c f^2-b f g+a g^2\right ) (g (2 b d g+a e g m-b e f (2+m))-c f (4 d g-e f (4+m))) (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)^2 (1+m)}\\ \end{align*}
Mathematica [A] time = 0.394679, size = 261, normalized size = 0.88 \[ \frac{(d+e x)^{m+1} \left (\frac{2 c e g (a e g-b (d g+2 e f))+b^2 e^2 g^2+c^2 \left (d^2 g^2+2 d e f g+3 e^2 f^2\right )}{e^3 (m+1)}+\frac{e \left (g (a g-b f)+c f^2\right )^2 \, _2F_1\left (2,m+1;m+2;\frac{g (d+e x)}{d g-e f}\right )}{(m+1) (e f-d g)^2}-\frac{2 (2 c f-b g) \left (g (a g-b f)+c f^2\right ) \, _2F_1\left (1,m+1;m+2;\frac{g (d+e x)}{d g-e f}\right )}{(m+1) (e f-d g)}-\frac{2 c g (d+e x) (-b e g+c d g+c e f)}{e^3 (m+2)}+\frac{c^2 g^2 (d+e x)^2}{e^3 (m+3)}\right )}{g^4} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.626, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+bx+a \right ) ^{2} \left ( ex+d \right ) ^{m}}{ \left ( gx+f \right ) ^{2}}}\, 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}}{{\left (g x + f\right )}^{2}}\,{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^{2} x^{2} + 2 \, f g x + f^{2}}, 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}}{\left (f + g x\right )^{2}}\, 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}}{{\left (g x + f\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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