Optimal. Leaf size=233 \[ -\frac{f (a+b x)^m (c+d x)^{-m} (a d f (m+2)-b (c f m+2 d e)) \, _2F_1\left (1,-m;1-m;\frac{(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{m (b e-a f) (d e-c f)^3}-\frac{d (a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+2)-b (c f (m+1)+d e))}{(m+1) (b c-a d) (b e-a f) (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{-m-1}}{(e+f x) (b e-a f) (d e-c f)} \]
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Rubi [A] time = 0.203663, antiderivative size = 243, normalized size of antiderivative = 1.04, 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 = {129, 151, 12, 131} \[ \frac{f (a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+2)-b (c f m+2 d e)) \, _2F_1\left (1,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (b e-a f)^2 (d e-c f)^2}+\frac{d (a+b x)^{m+1} (c+d x)^{-m-1}}{(m+1) (e+f x) (b c-a d) (d e-c f)}+\frac{f (a+b x)^{m+1} (c+d x)^{-m} (-a d f (m+2)+b c f (m+1)+b d e)}{(m+1) (e+f x) (b c-a d) (b e-a f) (d e-c f)^2} \]
Antiderivative was successfully verified.
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Rule 129
Rule 151
Rule 12
Rule 131
Rubi steps
\begin{align*} \int \frac{(a+b x)^m (c+d x)^{-2-m}}{(e+f x)^2} \, dx &=\frac{d (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (d e-c f) (1+m) (e+f x)}+\frac{\int \frac{(a+b x)^m (c+d x)^{-1-m} (-f (b c (1+m)-a d (2+m))+b d f x)}{(e+f x)^2} \, dx}{(b c-a d) (d e-c f) (1+m)}\\ &=\frac{d (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (d e-c f) (1+m) (e+f x)}+\frac{f (b d e+b c f (1+m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)}+\frac{\int \frac{(b c-a d) f (1+m) (a d f (2+m)-b (2 d e+c f m)) (a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{(b c-a d) (b e-a f) (d e-c f)^2 (1+m)}\\ &=\frac{d (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (d e-c f) (1+m) (e+f x)}+\frac{f (b d e+b c f (1+m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)}+\frac{(f (a d f (2+m)-b (2 d e+c f m))) \int \frac{(a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{(b e-a f) (d e-c f)^2}\\ &=\frac{d (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (d e-c f) (1+m) (e+f x)}+\frac{f (b d e+b c f (1+m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)}+\frac{f (a d f (2+m)-b (2 d e+c f m)) (a+b x)^{1+m} (c+d x)^{-1-m} \, _2F_1\left (1,1+m;2+m;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(b e-a f)^2 (d e-c f)^2 (1+m)}\\ \end{align*}
Mathematica [A] time = 0.250391, size = 193, normalized size = 0.83 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-1} \left (f (e+f x) (b c-a d) (a d f (m+2)-b (c f m+2 d e)) \, _2F_1\left (1,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )+f (c+d x) (b e-a f) (-a d f (m+2)+b c f (m+1)+b d e)+d (b e-a f)^2 (d e-c f)\right )}{(m+1) (e+f x) (b c-a d) (b e-a f)^2 (d e-c f)^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.07, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-2-m}}{ \left ( fx+e \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 2}}{{\left (f x + e\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 2}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 2}}{{\left (f x + e\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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