Optimal. Leaf size=402 \[ \frac{(a+b x)^{m+1} (c+d x)^n (e+f x)^{-m-n-1} \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (c f (n+1)-d e (m+n+3))+b^2 \left (-\left (-c^2 f^2 \left (n^2+3 n+2\right )+2 c d e f (n+1) (m+n+3)-d^2 e^2 \left (m^2+m (2 n+5)+n^2+5 n+6\right )\right )\right )\right ) \left (\frac{(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{-n} \, _2F_1\left (m+1,-n;m+2;-\frac{(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (m+n+2) (m+n+3) (b e-a f)^3 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{n+1} (e+f x)^{-m-n-3}}{(m+n+3) (b e-a f) (d e-c f)}+\frac{f (a+b x)^{m+1} (c+d x)^{n+1} (e+f x)^{-m-n-2} (a d f (m+2)+b (c f (n+2)-d e (m+n+4)))}{(m+n+2) (m+n+3) (b e-a f)^2 (d e-c f)^2} \]
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Rubi [A] time = 0.538346, antiderivative size = 401, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {129, 155, 12, 132} \[ \frac{(a+b x)^{m+1} (c+d x)^n (e+f x)^{-m-n-1} \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (c f (n+1)-d e (m+n+3))+b^2 \left (-\left (-c^2 f^2 \left (n^2+3 n+2\right )+2 c d e f (n+1) (m+n+3)-d^2 e^2 \left (m^2+m (2 n+5)+n^2+5 n+6\right )\right )\right )\right ) \left (\frac{(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{-n} \, _2F_1\left (m+1,-n;m+2;-\frac{(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (m+n+2) (m+n+3) (b e-a f)^3 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{n+1} (e+f x)^{-m-n-3}}{(m+n+3) (b e-a f) (d e-c f)}+\frac{f (a+b x)^{m+1} (c+d x)^{n+1} (e+f x)^{-m-n-2} (a d f (m+2)+b c f (n+2)-b d e (m+n+4))}{(m+n+2) (m+n+3) (b e-a f)^2 (d e-c f)^2} \]
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)^n (e+f x)^{-4-m-n} \, dx &=-\frac{f (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}-\frac{\int (a+b x)^m (c+d x)^n (e+f x)^{-3-m-n} (a d f (2+m)+b c f (2+n)-b d e (3+m+n)+b d f x) \, dx}{(b e-a f) (d e-c f) (3+m+n)}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}+\frac{f (a d f (2+m)+b c f (2+n)-b d e (4+m+n)) (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-2-m-n}}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}+\frac{\int \left (-f (b c (1+m)+a d (1+n)) (a d f (2+m)+b c f (2+n)-b d e (4+m+n))-(2+m+n) \left (a b c d f^2+b d e (a d f (2+m)+b c f (2+n)-b d e (3+m+n))-(b c+a d) f (a d f (2+m)+b c f (2+n)-b d e (3+m+n))\right )\right ) (a+b x)^m (c+d x)^n (e+f x)^{-2-m-n} \, dx}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}+\frac{f (a d f (2+m)+b c f (2+n)-b d e (4+m+n)) (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-2-m-n}}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}-\frac{\left (f (b c (1+m)+a d (1+n)) (a d f (2+m)+b c f (2+n)-b d e (4+m+n))+(2+m+n) \left (a b c d f^2+b d e (a d f (2+m)+b c f (2+n)-b d e (3+m+n))-(b c+a d) f (a d f (2+m)+b c f (2+n)-b d e (3+m+n))\right )\right ) \int (a+b x)^m (c+d x)^n (e+f x)^{-2-m-n} \, dx}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}+\frac{f (a d f (2+m)+b c f (2+n)-b d e (4+m+n)) (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-2-m-n}}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}-\frac{\left (f (b c (1+m)+a d (1+n)) (a d f (2+m)+b c f (2+n)-b d e (4+m+n))+(2+m+n) \left (a b c d f^2+b d e (a d f (2+m)+b c f (2+n)-b d e (3+m+n))-(b c+a d) f (a d f (2+m)+b c f (2+n)-b d e (3+m+n))\right )\right ) (a+b x)^{1+m} (c+d x)^n \left (\frac{(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{-n} (e+f x)^{-1-m-n} \, _2F_1\left (1+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+m+n) (3+m+n)}\\ \end{align*}
Mathematica [A] time = 0.967656, size = 346, normalized size = 0.86 \[ -\frac{(a+b x)^{m+1} (c+d x)^n (e+f x)^{-m-n-3} \left (-\frac{(e+f x)^2 \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )-2 a b d f (m+1) (d e (m+n+3)-c f (n+1))+b^2 \left (c^2 f^2 \left (n^2+3 n+2\right )-2 c d e f (n+1) (m+n+3)+d^2 e^2 \left (m^2+m (2 n+5)+n^2+5 n+6\right )\right )\right ) \left (\frac{(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{-n} \, _2F_1\left (m+1,-n;m+2;\frac{(c f-d e) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (m+n+2) (b e-a f)^2 (d e-c f)}+\frac{f (c+d x) (e+f x) (-a d f (m+2)-b c f (n+2)+b d e (m+n+4))}{(m+n+2) (b e-a f) (d e-c f)}+f (c+d x)\right )}{(m+n+3) (b e-a f) (d e-c f)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.148, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{n} \left ( fx+e \right ) ^{-4-m-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 )}^{n}{\left (f x + e\right )}^{-m - 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 )}^{n}{\left (f x + e\right )}^{-m - 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 )}^{n}{\left (f x + e\right )}^{-m - n - 4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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