Optimal. Leaf size=502 \[ \frac {(d+e x)^{m+1} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F_1\left (m+1;\frac {1}{2},\frac {1}{2};m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right ) \left (e g^2 (m+1) (b d-a e)+c \left (d^2 g^2-2 d e f g (m+2)+e^2 f^2 (m+2)\right )\right )}{c e^3 (m+1) (m+2) \sqrt {a+b x+c x^2}}-\frac {g (d+e x)^{m+2} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} (b e g (2 m+3)+c (2 d g-4 e f (m+2))) F_1\left (m+2;\frac {1}{2},\frac {1}{2};m+3;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c e^3 (m+2)^2 \sqrt {a+b x+c x^2}}+\frac {g^2 \sqrt {a+b x+c x^2} (d+e x)^{m+1}}{c e (m+2)} \]
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Rubi [A] time = 0.68, antiderivative size = 500, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1653, 843, 759, 133} \[ \frac {(d+e x)^{m+1} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F_1\left (m+1;\frac {1}{2},\frac {1}{2};m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right ) \left (g^2 (b d-a e)+\frac {c \left (d^2 g^2-2 d e f g (m+2)+e^2 f^2 (m+2)\right )}{e (m+1)}\right )}{c e^2 (m+2) \sqrt {a+b x+c x^2}}-\frac {g (d+e x)^{m+2} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} (b e g (2 m+3)+2 c d g-4 c e f (m+2)) F_1\left (m+2;\frac {1}{2},\frac {1}{2};m+3;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c e^3 (m+2)^2 \sqrt {a+b x+c x^2}}+\frac {g^2 \sqrt {a+b x+c x^2} (d+e x)^{m+1}}{c e (m+2)} \]
Antiderivative was successfully verified.
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Rule 133
Rule 759
Rule 843
Rule 1653
Rubi steps
\begin {align*} \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 {\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 ) \operatorname {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 ) \operatorname {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*}
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Mathematica [F] time = 1.30, size = 0, normalized size = 0.00 \[ \int \frac {(d+e x)^m (f+g x)^2}{\sqrt {a+b x+c x^2}} \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 1.30, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (g^{2} x^{2} + 2 \, f g x + f^{2}\right )} {\left (e x + d\right )}^{m}}{\sqrt {c x^{2} + b x + a}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (g x + f\right )}^{2} {\left (e x + d\right )}^{m}}{\sqrt {c x^{2} + b x + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.13, size = 0, normalized size = 0.00 \[ \int \frac {\left (g x +f \right )^{2} \left (e x +d \right )^{m}}{\sqrt {c \,x^{2}+b x +a}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (g x + f\right )}^{2} {\left (e x + d\right )}^{m}}{\sqrt {c x^{2} + b x + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (f+g\,x\right )}^2\,{\left (d+e\,x\right )}^m}{\sqrt {c\,x^2+b\,x+a}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{m} \left (f + g x\right )^{2}}{\sqrt {a + b x + c x^{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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