Optimal. Leaf size=311 \[ -\frac {(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m-1} \left (-a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (4 d e-c f (2-m))-\left (b^2 \left (c^2 f^2 \left (m^2-5 m+6\right )-8 c d e f (2-m)+12 d^2 e^2\right )\right )\right ) \, _2F_1\left (3,m+1;m+2;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{12 (m+1) (b e-a f)^5 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} (b (5 d e-c f (3-m))-a d f (m+2))}{12 (e+f x)^3 (b e-a f)^2 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m}}{4 (e+f x)^4 (b e-a f) (d e-c f)} \]
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Rubi [A] time = 0.32, antiderivative size = 310, normalized size of antiderivative = 1.00, 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 12
Rule 129
Rule 131
Rule 151
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*}
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Mathematica [A] time = 0.90, 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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fricas [F] time = 1.15, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}}{f^{5} x^{5} + 5 \, e f^{4} x^{4} + 10 \, e^{2} f^{3} x^{3} + 10 \, e^{3} f^{2} x^{2} + 5 \, e^{4} f x + e^{5}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}}{{\left (f x + e\right )}^{5}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-m +1}}{\left (f x +e \right )^{5}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}}{{\left (f x + e\right )}^{5}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{1-m}}{{\left (e+f\,x\right )}^5} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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