Integrand size = 29, antiderivative size = 502 \[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {g^2 (d+e x)^{1+m} \sqrt {a+b x+c x^2}}{c e (2+m)}+\frac {\left (e (b d-a e) g^2 (1+m)+c \left (d^2 g^2+e^2 f^2 (2+m)-2 d e f g (2+m)\right )\right ) (d+e x)^{1+m} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},\frac {1}{2},2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c e^3 (1+m) (2+m) \sqrt {a+b x+c x^2}}-\frac {g (b e g (3+2 m)+c (2 d g-4 e f (2+m))) (d+e x)^{2+m} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \operatorname {AppellF1}\left (2+m,\frac {1}{2},\frac {1}{2},3+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c e^3 (2+m)^2 \sqrt {a+b x+c x^2}} \]
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Time = 0.38 (sec) , antiderivative size = 500, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1667, 857, 773, 138} \[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+b x+c x^2}} \, dx=\frac {(d+e x)^{m+1} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} \operatorname {AppellF1}\left (m+1,\frac {1}{2},\frac {1}{2},m+2,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right ) \left (g^2 (b d-a e)+\frac {c \left (d^2 g^2-2 d e f g (m+2)+e^2 f^2 (m+2)\right )}{e (m+1)}\right )}{c e^2 (m+2) \sqrt {a+b x+c x^2}}-\frac {g (d+e x)^{m+2} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} (b e g (2 m+3)+2 c d g-4 c e f (m+2)) \operatorname {AppellF1}\left (m+2,\frac {1}{2},\frac {1}{2},m+3,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c e^3 (m+2)^2 \sqrt {a+b x+c x^2}}+\frac {g^2 \sqrt {a+b x+c x^2} (d+e x)^{m+1}}{c e (m+2)} \]
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Rule 138
Rule 773
Rule 857
Rule 1667
Rubi steps \begin{align*} \text {integral}& = \frac {g^2 (d+e x)^{1+m} \sqrt {a+b x+c x^2}}{c e (2+m)}+\frac {\int \frac {(d+e x)^m \left (\frac {1}{2} e \left (2 c e f^2 (2+m)-g^2 (b d+2 a e (1+m))\right )-\frac {1}{2} e g (2 c d g-4 c e f (2+m)+b e g (3+2 m)) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{c e^2 (2+m)} \\ & = \frac {g^2 (d+e x)^{1+m} \sqrt {a+b x+c x^2}}{c e (2+m)}-\frac {(g (2 c d g-4 c e f (2+m)+b e g (3+2 m))) \int \frac {(d+e x)^{1+m}}{\sqrt {a+b x+c x^2}} \, dx}{2 c e^2 (2+m)}+\frac {\left (e (b d-a e) g^2 (1+m)+c \left (d^2 g^2+e^2 f^2 (2+m)-2 d e f g (2+m)\right )\right ) \int \frac {(d+e x)^m}{\sqrt {a+b x+c x^2}} \, dx}{c e^2 (2+m)} \\ & = \frac {g^2 (d+e x)^{1+m} \sqrt {a+b x+c x^2}}{c e (2+m)}-\frac {\left (g (2 c d g-4 c e f (2+m)+b e g (3+2 m)) \sqrt {1-\frac {d+e x}{d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c}}} \sqrt {1-\frac {d+e x}{d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c}}}\right ) \text {Subst}\left (\int \frac {x^{1+m}}{\sqrt {1-\frac {2 c x}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c x}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}} \, dx,x,d+e x\right )}{2 c e^3 (2+m) \sqrt {a+b x+c x^2}}+\frac {\left (\left (e (b d-a e) g^2 (1+m)+c \left (d^2 g^2+e^2 f^2 (2+m)-2 d e f g (2+m)\right )\right ) \sqrt {1-\frac {d+e x}{d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c}}} \sqrt {1-\frac {d+e x}{d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c}}}\right ) \text {Subst}\left (\int \frac {x^m}{\sqrt {1-\frac {2 c x}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c x}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}} \, dx,x,d+e x\right )}{c e^3 (2+m) \sqrt {a+b x+c x^2}} \\ & = \frac {g^2 (d+e x)^{1+m} \sqrt {a+b x+c x^2}}{c e (2+m)}+\frac {\left (e (b d-a e) g^2 (1+m)+c \left (d^2 g^2+e^2 f^2 (2+m)-2 d e f g (2+m)\right )\right ) (d+e x)^{1+m} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} F_1\left (1+m;\frac {1}{2},\frac {1}{2};2+m;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c e^3 (1+m) (2+m) \sqrt {a+b x+c x^2}}-\frac {g (2 c d g-4 c e f (2+m)+b e g (3+2 m)) (d+e x)^{2+m} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} F_1\left (2+m;\frac {1}{2},\frac {1}{2};3+m;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c e^3 (2+m)^2 \sqrt {a+b x+c x^2}} \\ \end{align*}
\[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+b x+c x^2}} \, dx \]
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\[\int \frac {\left (e x +d \right )^{m} \left (g x +f \right )^{2}}{\sqrt {c \,x^{2}+b x +a}}d x\]
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\[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+b x+c x^2}} \, dx=\int { \frac {{\left (g x + f\right )}^{2} {\left (e x + d\right )}^{m}}{\sqrt {c x^{2} + b x + a}} \,d x } \]
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\[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {\left (d + e x\right )^{m} \left (f + g x\right )^{2}}{\sqrt {a + b x + c x^{2}}}\, dx \]
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\[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+b x+c x^2}} \, dx=\int { \frac {{\left (g x + f\right )}^{2} {\left (e x + d\right )}^{m}}{\sqrt {c x^{2} + b x + a}} \,d x } \]
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\[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+b x+c x^2}} \, dx=\int { \frac {{\left (g x + f\right )}^{2} {\left (e x + d\right )}^{m}}{\sqrt {c x^{2} + b x + a}} \,d x } \]
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Timed out. \[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {{\left (f+g\,x\right )}^2\,{\left (d+e\,x\right )}^m}{\sqrt {c\,x^2+b\,x+a}} \,d x \]
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