Optimal. Leaf size=311 \[ -\frac{(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m-1} \left (-a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (4 d e-c f (2-m))+b^2 \left (-\left (c^2 f^2 \left (m^2-5 m+6\right )-8 c d e f (2-m)+12 d^2 e^2\right )\right )\right ) \, _2F_1\left (3,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{12 (m+1) (b e-a f)^5 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{2-m} (b (5 d e-c f (3-m))-a d f (m+2))}{12 (e+f x)^3 (b e-a f)^2 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{2-m}}{4 (e+f x)^4 (b e-a f) (d e-c f)} \]
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Rubi [A] time = 0.321836, antiderivative size = 310, normalized size of antiderivative = 1., 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} \[ -\frac{(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m-1} \left (-a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (4 d e-c f (2-m))+b^2 \left (-\left (c^2 f^2 \left (m^2-5 m+6\right )-8 c d e f (2-m)+12 d^2 e^2\right )\right )\right ) \, _2F_1\left (3,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{12 (m+1) (b e-a f)^5 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{2-m} (-a d f (m+2)-b c f (3-m)+5 b d e)}{12 (e+f x)^3 (b e-a f)^2 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{2-m}}{4 (e+f x)^4 (b e-a f) (d e-c f)} \]
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)^{1-m}}{(e+f x)^5} \, dx &=-\frac{f (a+b x)^{1+m} (c+d x)^{2-m}}{4 (b e-a f) (d e-c f) (e+f x)^4}-\frac{\int \frac{(a+b x)^m (c+d x)^{1-m} (-b (4 d e-c f (3-m))+a d f (2+m)+b d f x)}{(e+f x)^4} \, dx}{4 (b e-a f) (d e-c f)}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{2-m}}{4 (b e-a f) (d e-c f) (e+f x)^4}-\frac{f (5 b d e-b c f (3-m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{2-m}}{12 (b e-a f)^2 (d e-c f)^2 (e+f x)^3}+\frac{\int \frac{\left (-2 a b d f (4 d e-c f (2-m)) (1+m)+a^2 d^2 f^2 \left (2+3 m+m^2\right )+b^2 \left (12 d^2 e^2-8 c d e f (2-m)+c^2 f^2 \left (6-5 m+m^2\right )\right )\right ) (a+b x)^m (c+d x)^{1-m}}{(e+f x)^3} \, dx}{12 (b e-a f)^2 (d e-c f)^2}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{2-m}}{4 (b e-a f) (d e-c f) (e+f x)^4}-\frac{f (5 b d e-b c f (3-m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{2-m}}{12 (b e-a f)^2 (d e-c f)^2 (e+f x)^3}-\frac{\left (2 a b d f (4 d e-c f (2-m)) (1+m)-a^2 d^2 f^2 \left (2+3 m+m^2\right )-b^2 \left (12 d^2 e^2-8 c d e f (2-m)+c^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)^3} \, dx}{12 (b e-a f)^2 (d e-c f)^2}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{2-m}}{4 (b e-a f) (d e-c f) (e+f x)^4}-\frac{f (5 b d e-b c f (3-m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{2-m}}{12 (b e-a f)^2 (d e-c f)^2 (e+f x)^3}-\frac{(b c-a d)^2 \left (2 a b d f (4 d e-c f (2-m)) (1+m)-a^2 d^2 f^2 \left (2+3 m+m^2\right )-b^2 \left (12 d^2 e^2-8 c d e f (2-m)+c^2 f^2 \left (6-5 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m} \, _2F_1\left (3,1+m;2+m;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{12 (b e-a f)^5 (d e-c f)^2 (1+m)}\\ \end{align*}
Mathematica [A] time = 0.607688, size = 271, normalized size = 0.87 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-1} \left (\frac{(b c-a d)^2 \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )-2 a b d f (m+1) (c f (m-2)+4 d e)+b^2 \left (c^2 f^2 \left (m^2-5 m+6\right )+8 c d e f (m-2)+12 d^2 e^2\right )\right ) \, _2F_1\left (3,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (b e-a f)^4 (d e-c f)}-\frac{f (c+d x)^3 (-a d f (m+2)+b c f (m-3)+5 b d e)}{(e+f x)^3 (b e-a f) (d e-c f)}-\frac{3 f (c+d x)^3}{(e+f x)^4}\right )}{12 (b e-a f) (d e-c f)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.211, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{1-m}}{ \left ( fx+e \right ) ^{5}}}\, 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 + 1}}{{\left (f x + e\right )}^{5}}\,{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 + 1}}{f^{5} x^{5} + 5 \, e f^{4} x^{4} + 10 \, e^{2} f^{3} x^{3} + 10 \, e^{3} f^{2} x^{2} + 5 \, e^{4} f x + e^{5}}, 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 + 1}}{{\left (f x + e\right )}^{5}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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