Optimal. Leaf size=237 \[ \frac {2 \sqrt {e+f x} (d+e x)^m \left (-\frac {f (d+e x)}{e^2-d f}\right )^{-m} \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};\frac {e (e+f x)}{e^2-d f}\right ) \left (c \left (d^2 f^2+4 d e^2 f (m+1)-4 e^4 \left (m^2+3 m+2\right )\right )-e f (2 m+3) \left (a e f (2 m+1)+b \left (d f-2 e^2 (m+1)\right )\right )\right )}{e f^3 (2 m+3) \left (e^2-d f\right )}+\frac {2 (d+e x)^{m+1} \left (a+\frac {e (c e-b f)}{f^2}\right )}{\left (e^2-d f\right ) \sqrt {e+f x}}+\frac {2 c \sqrt {e+f x} (d+e x)^{m+1}}{e f^2 (2 m+3)} \]
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Rubi [A] time = 0.35, antiderivative size = 230, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {949, 80, 70, 69} \[ -\frac {2 \sqrt {e+f x} (d+e x)^m \left (-\frac {f (d+e x)}{e^2-d f}\right )^{-m} \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};\frac {e (e+f x)}{e^2-d f}\right ) \left (f \left (a e f (2 m+1)+b d f-2 b e^2 (m+1)\right )-\frac {c \left (d^2 f^2+4 d e^2 f (m+1)-4 e^4 \left (m^2+3 m+2\right )\right )}{e (2 m+3)}\right )}{f^3 \left (e^2-d f\right )}+\frac {2 (d+e x)^{m+1} \left (a+\frac {e (c e-b f)}{f^2}\right )}{\left (e^2-d f\right ) \sqrt {e+f x}}+\frac {2 c \sqrt {e+f x} (d+e x)^{m+1}}{e f^2 (2 m+3)} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 80
Rule 949
Rubi steps
\begin {align*} \int \frac {(d+e x)^m \left (a+b x+c x^2\right )}{(e+f x)^{3/2}} \, dx &=\frac {2 \left (a+\frac {e (c e-b f)}{f^2}\right ) (d+e x)^{1+m}}{\left (e^2-d f\right ) \sqrt {e+f x}}+\frac {2 \int \frac {(d+e x)^m \left (\frac {c \left (d e f-2 e^3 (1+m)\right )-f \left (b d f-2 b e^2 (1+m)+a e f (1+2 m)\right )}{2 f^2}-\frac {1}{2} c \left (d-\frac {e^2}{f}\right ) x\right )}{\sqrt {e+f x}} \, dx}{e^2-d f}\\ &=\frac {2 \left (a+\frac {e (c e-b f)}{f^2}\right ) (d+e x)^{1+m}}{\left (e^2-d f\right ) \sqrt {e+f x}}+\frac {2 c (d+e x)^{1+m} \sqrt {e+f x}}{e f^2 (3+2 m)}-\frac {\left (f \left (b d f-2 b e^2 (1+m)+a e f (1+2 m)\right )-\frac {c \left (d^2 f^2+4 d e^2 f (1+m)-4 e^4 \left (2+3 m+m^2\right )\right )}{e (3+2 m)}\right ) \int \frac {(d+e x)^m}{\sqrt {e+f x}} \, dx}{f^2 \left (e^2-d f\right )}\\ &=\frac {2 \left (a+\frac {e (c e-b f)}{f^2}\right ) (d+e x)^{1+m}}{\left (e^2-d f\right ) \sqrt {e+f x}}+\frac {2 c (d+e x)^{1+m} \sqrt {e+f x}}{e f^2 (3+2 m)}-\frac {\left (\left (f \left (b d f-2 b e^2 (1+m)+a e f (1+2 m)\right )-\frac {c \left (d^2 f^2+4 d e^2 f (1+m)-4 e^4 \left (2+3 m+m^2\right )\right )}{e (3+2 m)}\right ) (d+e x)^m \left (\frac {f (d+e x)}{-e^2+d f}\right )^{-m}\right ) \int \frac {\left (-\frac {d f}{e^2-d f}-\frac {e f x}{e^2-d f}\right )^m}{\sqrt {e+f x}} \, dx}{f^2 \left (e^2-d f\right )}\\ &=\frac {2 \left (a+\frac {e (c e-b f)}{f^2}\right ) (d+e x)^{1+m}}{\left (e^2-d f\right ) \sqrt {e+f x}}+\frac {2 c (d+e x)^{1+m} \sqrt {e+f x}}{e f^2 (3+2 m)}-\frac {2 \left (f \left (b d f-2 b e^2 (1+m)+a e f (1+2 m)\right )-\frac {c \left (d^2 f^2+4 d e^2 f (1+m)-4 e^4 \left (2+3 m+m^2\right )\right )}{e (3+2 m)}\right ) (d+e x)^m \left (-\frac {f (d+e x)}{e^2-d f}\right )^{-m} \sqrt {e+f x} \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};\frac {e (e+f x)}{e^2-d f}\right )}{f^3 \left (e^2-d f\right )}\\ \end {align*}
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Mathematica [A] time = 0.29, size = 171, normalized size = 0.72 \[ \frac {2 (d+e x)^m \left (\frac {f (d+e x)}{d f-e^2}\right )^{-m} \left (-3 \left (f (a f-b e)+c e^2\right ) \, _2F_1\left (-\frac {1}{2},-m;\frac {1}{2};\frac {e (e+f x)}{e^2-d f}\right )-(e+f x) \left ((6 c e-3 b f) \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};\frac {e (e+f x)}{e^2-d f}\right )-c (e+f x) \, _2F_1\left (\frac {3}{2},-m;\frac {5}{2};\frac {e (e+f x)}{e^2-d f}\right )\right )\right )}{3 f^3 \sqrt {e+f x}} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.12, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{2} + b x + a\right )} \sqrt {f x + e} {\left (e x + d\right )}^{m}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \]
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 \[ \int \frac {{\left (c x^{2} + b x + a\right )} {\left (e x + d\right )}^{m}}{{\left (f x + e\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \,x^{2}+b x +a \right ) \left (e x +d \right )^{m}}{\left (f x +e \right )^{\frac {3}{2}}}\, dx \]
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 \[ \int \frac {{\left (c x^{2} + b x + a\right )} {\left (e x + d\right )}^{m}}{{\left (f x + e\right )}^{\frac {3}{2}}}\,{d x} \]
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 \[ \int \frac {{\left (d+e\,x\right )}^m\,\left (c\,x^2+b\,x+a\right )}{{\left (e+f\,x\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]
Verification of antiderivative is not currently implemented for this CAS.
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