Integrand size = 27, antiderivative size = 461 \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=-\frac {c (3 c e f+c d g-2 b e g) (d+e x)^{1+m}}{e^2 g^4 (1+m)}+\frac {c^2 (d+e x)^{2+m}}{e^2 g^3 (2+m)}+\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}+\frac {\left (c f^2-b f g+a g^2\right ) (c f (8 d g-e f (7+m))+g (a e g (1-m)-b (4 d g-e f (3+m)))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac {\left (c^2 f^2 \left (12 d^2 g^2-8 d e f g (3+m)+e^2 f^2 \left (12+7 m+m^2\right )\right )-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (2+m)+e^2 f^2 \left (6+5 m+m^2\right )\right )\right )\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {g (d+e x)}{e f-d g}\right )}{2 g^4 (e f-d g)^3 (1+m)} \]
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Time = 0.89 (sec) , antiderivative size = 461, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {963, 1635, 965, 81, 70} \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=\frac {(d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {g (d+e x)}{e f-d g}\right ) \left (-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (m+1))-\left (b^2 \left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (m+2)+e^2 f^2 \left (m^2+5 m+6\right )\right )\right )+c^2 f^2 \left (12 d^2 g^2-8 d e f g (m+3)+e^2 f^2 \left (m^2+7 m+12\right )\right )\right )}{2 g^4 (m+1) (e f-d g)^3}-\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right ) (g (-a e g (1-m)+4 b d g-b e f (m+3))-c f (8 d g-e f (m+7)))}{2 g^4 (f+g x) (e f-d g)^2}+\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2}{2 g^4 (f+g x)^2 (e f-d g)}-\frac {c (d+e x)^{m+1} (-2 b e g+c d g+3 c e f)}{e^2 g^4 (m+1)}+\frac {c^2 (d+e x)^{m+2}}{e^2 g^3 (m+2)} \]
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Rule 70
Rule 81
Rule 963
Rule 965
Rule 1635
Rubi steps \begin{align*} \text {integral}& = \frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}+\frac {\int \frac {(d+e x)^m \left (\frac {c^2 f^3 (2 d g-e f (1+m))-2 c f g (b f-a g) (2 d g-e f (1+m))+g^2 \left (a^2 e g^2 (1-m)+b^2 f (2 d g-e f (1+m))-2 a b g (2 d g-e f (1+m))\right )}{g^4}+\frac {2 (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 {2 c (c f-2 b g) (e f-d g) x^2}{g^2}-2 c^2 \left (d-\frac {e f}{g}\right ) x^3\right )}{(f+g x)^2} \, dx}{2 (e f-d g)} \\ & = \frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}-\frac {\left (c f^2-b f g+a g^2\right ) (g (4 b d g-a e g (1-m)-b e f (3+m))-c f (8 d g-e f (7+m))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac {\int \frac {(d+e x)^m \left (\frac {c^2 f^2 \left (6 d^2 g^2-4 d e f g (3+2 m)+e^2 f^2 \left (6+7 m+m^2\right )\right )-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (4 d^2 g^2-2 d e f g (4+3 m)+e^2 f^2 \left (4+5 m+m^2\right )\right )\right )}{g^4}-\frac {4 c (c f-b g) (e f-d g)^2 x}{g^3}+\frac {2 c^2 (e f-d g)^2 x^2}{g^2}\right )}{f+g x} \, dx}{2 (e f-d g)^2} \\ & = \frac {c^2 (d+e x)^{2+m}}{e^2 g^3 (2+m)}+\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}-\frac {\left (c f^2-b f g+a g^2\right ) (g (4 b d g-a e g (1-m)-b e f (3+m))-c f (8 d g-e f (7+m))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac {\int \frac {(d+e x)^m \left (\frac {e (2+m) \left (c^2 f \left (10 d^2 e f g^2-2 d^3 g^3-2 d e^2 f^2 g (7+4 m)+e^3 f^3 \left (6+7 m+m^2\right )\right )-e g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c e g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (4 d^2 g^2-2 d e f g (4+3 m)+e^2 f^2 \left (4+5 m+m^2\right )\right )\right )\right )}{g^3}-\frac {2 c e (e f-d g)^2 (3 c e f+c d g-2 b e g) (2+m) x}{g^2}\right )}{f+g x} \, dx}{2 e^2 g (e f-d g)^2 (2+m)} \\ & = -\frac {c (3 c e f+c d g-2 b e g) (d+e x)^{1+m}}{e^2 g^4 (1+m)}+\frac {c^2 (d+e x)^{2+m}}{e^2 g^3 (2+m)}+\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}-\frac {\left (c f^2-b f g+a g^2\right ) (g (4 b d g-a e g (1-m)-b e f (3+m))-c f (8 d g-e f (7+m))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac {\left (c^2 f^2 \left (12 d^2 g^2-8 d e f g (3+m)+e^2 f^2 \left (12+7 m+m^2\right )\right )-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (2+m)+e^2 f^2 \left (6+5 m+m^2\right )\right )\right )\right ) \int \frac {(d+e x)^m}{f+g x} \, dx}{2 g^4 (e f-d g)^2} \\ & = -\frac {c (3 c e f+c d g-2 b e g) (d+e x)^{1+m}}{e^2 g^4 (1+m)}+\frac {c^2 (d+e x)^{2+m}}{e^2 g^3 (2+m)}+\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}-\frac {\left (c f^2-b f g+a g^2\right ) (g (4 b d g-a e g (1-m)-b e f (3+m))-c f (8 d g-e f (7+m))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac {\left (c^2 f^2 \left (12 d^2 g^2-8 d e f g (3+m)+e^2 f^2 \left (12+7 m+m^2\right )\right )-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (2+m)+e^2 f^2 \left (6+5 m+m^2\right )\right )\right )\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {g (d+e x)}{e f-d g}\right )}{2 g^4 (e f-d g)^3 (1+m)} \\ \end{align*}
Time = 1.18 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.56 \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=\frac {(d+e x)^{1+m} \left (-\frac {c (3 c e f+c d g-2 b e g)}{e^2 (1+m)}+\frac {c^2 g (d+e x)}{e^2 (2+m)}+\frac {\left (6 c^2 f^2+b^2 g^2+2 c g (-3 b f+a g)\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {g (d+e x)}{-e f+d g}\right )}{(e f-d g) (1+m)}-\frac {2 e (2 c f-b g) \left (c f^2+g (-b f+a g)\right ) \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {g (d+e x)}{-e f+d g}\right )}{(e f-d g)^2 (1+m)}+\frac {e^2 \left (c f^2+g (-b f+a g)\right )^2 \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,\frac {g (d+e x)}{-e f+d g}\right )}{(e f-d g)^3 (1+m)}\right )}{g^4} \]
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\[\int \frac {\left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )^{2}}{\left (g x +f \right )^{3}}d x\]
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\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{2} {\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{3}} \,d x } \]
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\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=\int \frac {\left (d + e x\right )^{m} \left (a + b x + c x^{2}\right )^{2}}{\left (f + g x\right )^{3}}\, dx \]
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\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{2} {\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{3}} \,d x } \]
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\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{2} {\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=\int \frac {{\left (d+e\,x\right )}^m\,{\left (c\,x^2+b\,x+a\right )}^2}{{\left (f+g\,x\right )}^3} \,d x \]
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