Optimal. Leaf size=428 \[ \frac{(a+b x)^{m+1} (e+f x)^{n-1} (c+d x)^{-m-n} \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )-2 a b d f (m+1) (d e (3-n)-c f (-m-n+1))+b^2 \left (-\left (-c^2 f^2 \left (m^2-m (3-2 n)+n^2-3 n+2\right )+2 c d e f (3-n) (-m-n+1)-d^2 e^2 \left (n^2-5 n+6\right )\right )\right )\right ) \left (\frac{(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{m+n} \, _2F_1\left (m+1,m+n;m+2;-\frac{(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (2-n) (3-n) (b e-a f)^3 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (e+f x)^{n-3} (c+d x)^{-m-n+1}}{(3-n) (b e-a f) (d e-c f)}+\frac{f (a+b x)^{m+1} (e+f x)^{n-2} (c+d x)^{-m-n+1} (a d f (m+2)-b (d e (4-n)-c f (-m-n+2)))}{(2-n) (3-n) (b e-a f)^2 (d e-c f)^2} \]
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Rubi [A] time = 0.526338, antiderivative size = 426, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {129, 155, 12, 132} \[ \frac{(a+b x)^{m+1} (e+f x)^{n-1} (c+d x)^{-m-n} \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )-2 a b d f (m+1) (d e (3-n)-c f (-m-n+1))+b^2 \left (-\left (-c^2 f^2 \left (m^2-m (3-2 n)+n^2-3 n+2\right )+2 c d e f (3-n) (-m-n+1)-d^2 e^2 \left (n^2-5 n+6\right )\right )\right )\right ) \left (\frac{(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{m+n} \, _2F_1\left (m+1,m+n;m+2;-\frac{(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (2-n) (3-n) (b e-a f)^3 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (e+f x)^{n-3} (c+d x)^{-m-n+1}}{(3-n) (b e-a f) (d e-c f)}+\frac{f (a+b x)^{m+1} (e+f x)^{n-2} (c+d x)^{-m-n+1} (a d f (m+2)+b c f (-m-n+2)-b d e (4-n))}{(2-n) (3-n) (b e-a f)^2 (d e-c f)^2} \]
Antiderivative was successfully verified.
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Rule 129
Rule 155
Rule 12
Rule 132
Rubi steps
\begin{align*} \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-4+n} \, dx &=-\frac{f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-3+n}}{(b e-a f) (d e-c f) (3-n)}-\frac{\int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-3+n} (a d f (2+m)-b d e (3-n)+b c f (2-m-n)+b d f x) \, dx}{(b e-a f) (d e-c f) (3-n)}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-3+n}}{(b e-a f) (d e-c f) (3-n)}+\frac{f (a d f (2+m)-b d e (4-n)+b c f (2-m-n)) (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-2+n}}{(b e-a f)^2 (d e-c f)^2 (2-n) (3-n)}+\frac{\int \left (-f (b c (1+m)+a d (1-m-n)) (a d f (2+m)-b d e (4-n)+b c f (2-m-n))-\left (a b c d f^2+b d e (a d f (2+m)-b d e (3-n)+b c f (2-m-n))-(b c+a d) f (a d f (2+m)-b d e (3-n)+b c f (2-m-n))\right ) (2-n)\right ) (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-2+n} \, dx}{(b e-a f)^2 (d e-c f)^2 (2-n) (3-n)}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-3+n}}{(b e-a f) (d e-c f) (3-n)}+\frac{f (a d f (2+m)-b d e (4-n)+b c f (2-m-n)) (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-2+n}}{(b e-a f)^2 (d e-c f)^2 (2-n) (3-n)}+\frac{\left (a^2 d^2 f^2 \left (2+3 m+m^2\right )-2 a b d f (1+m) (d e (3-n)-c f (1-m-n))-b^2 \left (2 c d e f (3-n) (1-m-n)-d^2 e^2 \left (6-5 n+n^2\right )-c^2 f^2 \left (2+m^2-m (3-2 n)-3 n+n^2\right )\right )\right ) \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-2+n} \, dx}{(b e-a f)^2 (d e-c f)^2 (2-n) (3-n)}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-3+n}}{(b e-a f) (d e-c f) (3-n)}+\frac{f (a d f (2+m)-b d e (4-n)+b c f (2-m-n)) (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-2+n}}{(b e-a f)^2 (d e-c f)^2 (2-n) (3-n)}+\frac{\left (a^2 d^2 f^2 \left (2+3 m+m^2\right )-2 a b d f (1+m) (d e (3-n)-c f (1-m-n))-b^2 \left (2 c d e f (3-n) (1-m-n)-d^2 e^2 \left (6-5 n+n^2\right )-c^2 f^2 \left (2+m^2-m (3-2 n)-3 n+n^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m-n} \left (\frac{(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{m+n} (e+f x)^{-1+n} \, _2F_1\left (1+m,m+n;2+m;-\frac{(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(b e-a f)^3 (d e-c f)^2 (1+m) (2-n) (3-n)}\\ \end{align*}
Mathematica [A] time = 1.05979, size = 341, normalized size = 0.8 \[ \frac{(a+b x)^{m+1} (e+f x)^{n-3} (c+d x)^{-m-n} \left (\frac{(e+f x)^2 \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (d e (n-3)-c f (m+n-1))+b^2 \left (c^2 f^2 \left (m^2+m (2 n-3)+n^2-3 n+2\right )-2 c d e f (n-3) (m+n-1)+d^2 e^2 \left (n^2-5 n+6\right )\right )\right ) \left (\frac{(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{m+n} \, _2F_1\left (m+1,m+n;m+2;\frac{(c f-d e) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (n-2) (b e-a f)^2 (d e-c f)}+\frac{f (c+d x) (e+f x) (a d f (m+2)-b c f (m+n-2)+b d e (n-4))}{(n-2) (b e-a f) (d e-c f)}+f (c+d x)\right )}{(n-3) (b e-a f) (d e-c f)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.151, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-n-m} \left ( fx+e \right ) ^{-4+n}\, 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{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - n}{\left (f x + e\right )}^{n - 4}\,{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 ({\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - n}{\left (f x + e\right )}^{n - 4}, 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{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - n}{\left (f x + e\right )}^{n - 4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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