Optimal. Leaf size=432 \[ -\frac{f (a+b x)^m (c+d x)^{-m} \left (-a^2 d^2 f^2 \left (m^2+5 m+6\right )+2 a b d f (m+2) (c f m+3 d e)+b^2 \left (-\left (-c^2 f^2 (1-m) m+6 c d e f m+6 d^2 e^2\right )\right )\right ) \, _2F_1\left (1,-m;1-m;\frac{(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{2 m (b e-a f)^2 (d e-c f)^4}+\frac{d (a+b x)^{m+1} (c+d x)^{-m-1} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (c f \left (2 m^2+5 m+3\right )+d e (5 m+9)\right )+b^2 \left (-c^2 f^2 \left (1-m^2\right )+5 c d e f (m+1)+2 d^2 e^2\right )\right )}{2 (m+1) (b c-a d) (b e-a f)^2 (d e-c f)^3}-\frac{f (a+b x)^{m+1} (c+d x)^{-m-1} (b (4 d e-c f (1-m))-a d f (m+3))}{2 (e+f x) (b e-a f)^2 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{-m-1}}{2 (e+f x)^2 (b e-a f) (d e-c f)} \]
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Rubi [A] time = 0.565419, antiderivative size = 452, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {129, 151, 12, 131} \[ \frac{f (a+b x)^{m+1} (c+d x)^{-m-1} \left (-a^2 d^2 f^2 \left (m^2+5 m+6\right )+2 a b d f (m+2) (c f m+3 d e)+b^2 \left (-\left (-c^2 f^2 (1-m) m+6 c d e f m+6 d^2 e^2\right )\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{2 (m+1) (b e-a f)^3 (d e-c f)^3}+\frac{f (a+b x)^{m+1} (c+d x)^{-m} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (c f \left (2 m^2+5 m+3\right )+d e (5 m+9)\right )+b^2 \left (-c^2 f^2 \left (1-m^2\right )+5 c d e f (m+1)+2 d^2 e^2\right )\right )}{2 (m+1) (e+f x) (b c-a d) (b e-a f)^2 (d e-c f)^3}+\frac{d (a+b x)^{m+1} (c+d x)^{-m-1}}{(m+1) (e+f x)^2 (b c-a d) (d e-c f)}+\frac{f (a+b x)^{m+1} (c+d x)^{-m} (-a d f (m+3)+b c f (m+1)+2 b d e)}{2 (m+1) (e+f x)^2 (b c-a d) (b e-a f) (d e-c f)^2} \]
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)^3} \, 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)^2}+\frac{\int \frac{(a+b x)^m (c+d x)^{-1-m} (-f (b c (1+m)-a d (3+m))+2 b d f x)}{(e+f x)^3} \, 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)^2}+\frac{f (2 b d e+b c f (1+m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)^2}-\frac{\int \frac{(a+b x)^m (c+d x)^{-1-m} \left (f \left (b^2 c (4 d e-c f (1-m)) (1+m)+a^2 d^2 f \left (6+5 m+m^2\right )-a b d (3+2 m) (2 d e+c f (1+m))\right )-b d f (2 b d e+b c f (1+m)-a d f (3+m)) x\right )}{(e+f x)^2} \, dx}{2 (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)^2}+\frac{f (2 b d e+b c f (1+m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)^2}+\frac{f \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )+b^2 \left (2 d^2 e^2+5 c d e f (1+m)-c^2 f^2 \left (1-m^2\right )\right )-a b d f \left (d e (9+5 m)+c f \left (3+5 m+2 m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f)^2 (d e-c f)^3 (1+m) (e+f x)}+\frac{\int \frac{(b c-a d) f (1+m) \left (2 a b d f (2+m) (3 d e+c f m)-b^2 \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (6+5 m+m^2\right )\right ) (a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{2 (b c-a d) (b e-a f)^2 (d e-c f)^3 (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)^2}+\frac{f (2 b d e+b c f (1+m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)^2}+\frac{f \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )+b^2 \left (2 d^2 e^2+5 c d e f (1+m)-c^2 f^2 \left (1-m^2\right )\right )-a b d f \left (d e (9+5 m)+c f \left (3+5 m+2 m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f)^2 (d e-c f)^3 (1+m) (e+f x)}+\frac{\left (f \left (2 a b d f (2+m) (3 d e+c f m)-b^2 \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (6+5 m+m^2\right )\right )\right ) \int \frac{(a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{2 (b e-a f)^2 (d e-c f)^3}\\ &=\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)^2}+\frac{f (2 b d e+b c f (1+m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)^2}+\frac{f \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )+b^2 \left (2 d^2 e^2+5 c d e f (1+m)-c^2 f^2 \left (1-m^2\right )\right )-a b d f \left (d e (9+5 m)+c f \left (3+5 m+2 m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f)^2 (d e-c f)^3 (1+m) (e+f x)}+\frac{f \left (2 a b d f (2+m) (3 d e+c f m)-b^2 \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (6+5 m+m^2\right )\right ) (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 )}{2 (b e-a f)^3 (d e-c f)^3 (1+m)}\\ \end{align*}
Mathematica [A] time = 0.869794, size = 374, normalized size = 0.87 \[ -\frac{(a+b x)^{m+1} (c+d x)^{-m} \left ((e+f x) \left (f (b e-a f) \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (c f \left (2 m^2+5 m+3\right )+d e (5 m+9)\right )+b^2 \left (c^2 f^2 \left (m^2-1\right )+5 c d e f (m+1)+2 d^2 e^2\right )\right )-\frac{f (e+f x) (b c-a d) \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-2 a b d f (m+2) (c f m+3 d e)+b^2 \left (c^2 f^2 (m-1) m+6 c d e f m+6 d^2 e^2\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{c+d x}\right )+f (b e-a f)^2 (d e-c f) (-a d f (m+3)+b c f (m+1)+2 b d e)+\frac{2 d (b e-a f)^3 (d e-c f)^2}{c+d x}\right )}{2 (m+1) (e+f x)^2 (b c-a d) (b e-a f)^3 (d e-c f)^2 (c f-d e)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.092, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-2-m}}{ \left ( fx+e \right ) ^{3}}}\, 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 )}^{3}}\,{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^{3} x^{3} + 3 \, e f^{2} x^{2} + 3 \, e^{2} f x + e^{3}}, 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 )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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