Optimal. Leaf size=204 \[ -\frac {f (a+b x)^m (c+d x)^{-m} \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} (a d f m+b (2 d e-c f (m+2))) \, _2F_1\left (-m,-m;1-m;\frac {b (c+d x)}{b c-a d}\right )}{b d^3 m}+\frac {(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-1} (a d f (m+1)+b (d e-c f (m+2)))}{b d^2 (m+1) (b c-a d)}+\frac {f (e+f x) (a+b x)^{m+1} (c+d x)^{-m-1}}{b d} \]
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Rubi [A] time = 0.19, antiderivative size = 202, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {90, 79, 70, 69} \[ -\frac {f (a+b x)^m (c+d x)^{-m} \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} (a d f m-b c f (m+2)+2 b d e) \, _2F_1\left (-m,-m;1-m;\frac {b (c+d x)}{b c-a d}\right )}{b d^3 m}+\frac {(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-1} (a d f (m+1)-b c f (m+2)+b d e)}{b d^2 (m+1) (b c-a d)}+\frac {f (e+f x) (a+b x)^{m+1} (c+d x)^{-m-1}}{b d} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 79
Rule 90
Rubi steps
\begin {align*} \int (a+b x)^m (c+d x)^{-2-m} (e+f x)^2 \, dx &=\frac {f (a+b x)^{1+m} (c+d x)^{-1-m} (e+f x)}{b d}+\frac {\int (a+b x)^m (c+d x)^{-2-m} (-a f (c f-d e (1+m))+b e (d e-c f (1+m))+f (2 b d e+a d f m-b c f (2+m)) x) \, dx}{b d}\\ &=\frac {(d e-c f) (b d e+a d f (1+m)-b c f (2+m)) (a+b x)^{1+m} (c+d x)^{-1-m}}{b d^2 (b c-a d) (1+m)}+\frac {f (a+b x)^{1+m} (c+d x)^{-1-m} (e+f x)}{b d}+\frac {(f (2 b d e+a d f m-b c f (2+m))) \int (a+b x)^m (c+d x)^{-1-m} \, dx}{b d^2}\\ &=\frac {(d e-c f) (b d e+a d f (1+m)-b c f (2+m)) (a+b x)^{1+m} (c+d x)^{-1-m}}{b d^2 (b c-a d) (1+m)}+\frac {f (a+b x)^{1+m} (c+d x)^{-1-m} (e+f x)}{b d}+\frac {\left (f (2 b d e+a d f m-b c f (2+m)) (a+b x)^m \left (\frac {d (a+b x)}{-b c+a d}\right )^{-m}\right ) \int (c+d x)^{-1-m} \left (-\frac {a d}{b c-a d}-\frac {b d x}{b c-a d}\right )^m \, dx}{b d^2}\\ &=\frac {(d e-c f) (b d e+a d f (1+m)-b c f (2+m)) (a+b x)^{1+m} (c+d x)^{-1-m}}{b d^2 (b c-a d) (1+m)}+\frac {f (a+b x)^{1+m} (c+d x)^{-1-m} (e+f x)}{b d}-\frac {f (2 b d e+a d f m-b c f (2+m)) (a+b x)^m \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} (c+d x)^{-m} \, _2F_1\left (-m,-m;1-m;\frac {b (c+d x)}{b c-a d}\right )}{b d^3 m}\\ \end {align*}
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Mathematica [A] time = 0.42, size = 179, normalized size = 0.88 \[ \frac {(a+b x)^m (c+d x)^{-m} \left (\frac {f \left (\frac {d (a+b x)}{a d-b c}\right )^{-m} (-a d f m+b c f (m+2)-2 b d e) \, _2F_1\left (-m,-m;1-m;\frac {b (c+d x)}{b c-a d}\right )}{d^2 m}+\frac {(a+b x) (d e-c f) (a d f (m+1)-b c f (m+2)+b d e)}{d (m+1) (c+d x) (b c-a d)}+\frac {f (a+b x) (e+f x)}{c+d x}\right )}{b d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (f^{2} x^{2} + 2 \, e f x + e^{2}\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 2}, x\right ) \]
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 \[ \int {\left (f x + e\right )}^{2} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.23, size = 0, normalized size = 0.00 \[ \int \left (f x +e \right )^{2} \left (b x +a \right )^{m} \left (d x +c \right )^{-m -2}\, dx \]
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 \[ \int {\left (f x + e\right )}^{2} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 2}\,{d x} \]
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 \[ \int \frac {{\left (e+f\,x\right )}^2\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^{m+2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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