Optimal. Leaf size=309 \[ -\frac{(b c-a d) (a+b x)^{m+1} (c+d x)^{-m-1} \left (-a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (3 d e-c f (1-m))+b^2 \left (-\left (c^2 f^2 \left (m^2-3 m+2\right )-6 c d e f (1-m)+6 d^2 e^2\right )\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{6 (m+1) (b e-a f)^4 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{1-m} (b (4 d e-c f (2-m))-a d f (m+2))}{6 (e+f x)^2 (b e-a f)^2 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{1-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)} \]
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Rubi [A] time = 0.315863, antiderivative size = 308, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {129, 151, 12, 131} \[ -\frac{(b c-a d) (a+b x)^{m+1} (c+d x)^{-m-1} \left (-a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (3 d e-c f (1-m))+b^2 \left (-\left (c^2 f^2 \left (m^2-3 m+2\right )-6 c d e f (1-m)+6 d^2 e^2\right )\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{6 (m+1) (b e-a f)^4 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{1-m} (-a d f (m+2)-b c f (2-m)+4 b d e)}{6 (e+f x)^2 (b e-a f)^2 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{1-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)} \]
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)^{-m}}{(e+f x)^4} \, dx &=-\frac{f (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac{\int \frac{(a+b x)^m (c+d x)^{-m} (-b (3 d e-c f (2-m))+a d f (2+m)+b d f x)}{(e+f x)^3} \, dx}{3 (b e-a f) (d e-c f)}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac{f (4 b d e-b c f (2-m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}+\frac{\int \frac{\left (-2 a b d f (3 d e-c f (1-m)) (1+m)+a^2 d^2 f^2 \left (2+3 m+m^2\right )+b^2 \left (6 d^2 e^2-6 c d e f (1-m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) (a+b x)^m (c+d x)^{-m}}{(e+f x)^2} \, dx}{6 (b e-a f)^2 (d e-c f)^2}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac{f (4 b d e-b c f (2-m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}-\frac{\left (2 a b d f (3 d e-c f (1-m)) (1+m)-a^2 d^2 f^2 \left (2+3 m+m^2\right )-b^2 \left (6 d^2 e^2-6 c d e f (1-m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) \int \frac{(a+b x)^m (c+d x)^{-m}}{(e+f x)^2} \, dx}{6 (b e-a f)^2 (d e-c f)^2}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac{f (4 b d e-b c f (2-m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}-\frac{(b c-a d) \left (2 a b d f (3 d e-c f (1-m)) (1+m)-a^2 d^2 f^2 \left (2+3 m+m^2\right )-b^2 \left (6 d^2 e^2-6 c d e f (1-m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m} \, _2F_1\left (2,1+m;2+m;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{6 (b e-a f)^4 (d e-c f)^2 (1+m)}\\ \end{align*}
Mathematica [A] time = 0.391852, size = 255, normalized size = 0.83 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-1} \left (\frac{(b c-a d) \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )-2 a b d f (m+1) (c f (m-1)+3 d e)+b^2 \left (c^2 f^2 \left (m^2-3 m+2\right )+6 c d e f (m-1)+6 d^2 e^2\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (b e-a f)^2}-\frac{f (c+d x)^2 (-a d f (m+2)+b c f (m-2)+4 b d e)}{(e+f x)^2}-\frac{2 f (c+d x)^2 (a f-b e) (c f-d e)}{(e+f x)^3}\right )}{6 (b e-a f)^2 (d e-c f)^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.115, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m}}{ \left ( dx+c \right ) ^{m} \left ( fx+e \right ) ^{4}}}\, 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 (f x + e\right )}^{4}{\left (d x + c\right )}^{m}}\,{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 (f^{4} x^{4} + 4 \, e f^{3} x^{3} + 6 \, e^{2} f^{2} x^{2} + 4 \, e^{3} f x + e^{4}\right )}{\left (d x + c\right )}^{m}}, 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 (f x + e\right )}^{4}{\left (d x + c\right )}^{m}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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