Optimal. Leaf size=196 \[ -\frac{f^2 (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,-m;1-m;\frac{(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{m (d e-c f)^3}+\frac{d (a+b x)^{m+1} (c+d x)^{-m-2}}{(m+2) (b c-a d) (d e-c f)}+\frac{d (a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+2)+b (d e-c f (m+3)))}{(m+1) (m+2) (b c-a d)^2 (d e-c f)^2} \]
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Rubi [A] time = 0.182844, antiderivative size = 208, normalized size of antiderivative = 1.06, 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, 155, 12, 131} \[ \frac{f^2 (a+b x)^{m+1} (c+d x)^{-m-1} \, _2F_1\left (1,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (b e-a f) (d e-c f)^2}+\frac{d (a+b x)^{m+1} (c+d x)^{-m-2}}{(m+2) (b c-a d) (d e-c f)}+\frac{d (a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+2)-b c f (m+3)+b d e)}{(m+1) (m+2) (b c-a d)^2 (d e-c f)^2} \]
Antiderivative was successfully verified.
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Rule 129
Rule 155
Rule 12
Rule 131
Rubi steps
\begin{align*} \int \frac{(a+b x)^m (c+d x)^{-3-m}}{e+f x} \, dx &=\frac{d (a+b x)^{1+m} (c+d x)^{-2-m}}{(b c-a d) (d e-c f) (2+m)}+\frac{\int \frac{(a+b x)^m (c+d x)^{-2-m} (b d e-b c f (2+m)+a d f (2+m)+b d f x)}{e+f x} \, dx}{(b c-a d) (d e-c f) (2+m)}\\ &=\frac{d (a+b x)^{1+m} (c+d x)^{-2-m}}{(b c-a d) (d e-c f) (2+m)}+\frac{d (b d e+a d f (2+m)-b c f (3+m)) (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d)^2 (d e-c f)^2 (1+m) (2+m)}+\frac{\int \frac{(b c-a d)^2 f^2 (1+m) (2+m) (a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{(b c-a d)^2 (d e-c f)^2 (1+m) (2+m)}\\ &=\frac{d (a+b x)^{1+m} (c+d x)^{-2-m}}{(b c-a d) (d e-c f) (2+m)}+\frac{d (b d e+a d f (2+m)-b c f (3+m)) (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d)^2 (d e-c f)^2 (1+m) (2+m)}+\frac{f^2 \int \frac{(a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{(d e-c f)^2}\\ &=\frac{d (a+b x)^{1+m} (c+d x)^{-2-m}}{(b c-a d) (d e-c f) (2+m)}+\frac{d (b d e+a d f (2+m)-b c f (3+m)) (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d)^2 (d e-c f)^2 (1+m) (2+m)}+\frac{f^2 (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) (d e-c f)^2 (1+m)}\\ \end{align*}
Mathematica [A] time = 0.228978, size = 186, normalized size = 0.95 \[ -\frac{(a+b x)^{m+1} (c+d x)^{-m-2} \left (\frac{f^2 (m+2) (c+d x) (b c-a d) \, _2F_1\left (1,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (b e-a f) (d e-c f)}-\frac{d (c+d x) (a d f (m+2)-b c f (m+3)+b d e)}{(m+1) (b c-a d) (c f-d e)}+d\right )}{(m+2) (b c-a d) (c f-d e)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.064, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-3-m}}{fx+e}}\, 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 - 3}}{f x + e}\,{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 - 3}}{f x + e}, 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 - 3}}{f x + e}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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