Optimal. Leaf size=558 \[ \frac{(a+b x)^{m+1} (c+d x)^n (e+f x)^{-m-n-1} \left (\frac{(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{-n} ((m+n+2) (b d e ((m+n+3) (A (-a d f-b c f+b d e)+a B c f)-(B e-A f) (a d (n+1)+b c (m+1)))-f (a d+b c) ((m+n+3) (A (-a d f-b c f+b d e)+a B c f)-(B e-A f) (a d (n+1)+b c (m+1)))+a b c d f (B e-A f))-(a d (n+1)+b c (m+1)) (a f (A d f (m+2)+B (d e (n+1)-c f (m+n+3)))+b (A f (c f (n+2)-d e (m+n+4))+B e (c f (m+1)+d e)))) \, _2F_1\left (m+1,-n;m+2;-\frac{(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (m+n+2) (m+n+3) (b e-a f)^3 (d e-c f)^2}+\frac{(a+b x)^{m+1} (B e-A f) (c+d x)^{n+1} (e+f x)^{-m-n-3}}{(m+n+3) (b e-a f) (d e-c f)}+\frac{(a+b x)^{m+1} (c+d x)^{n+1} (e+f x)^{-m-n-2} (a f (A d f (m+2)+B (d e (n+1)-c f (m+n+3)))+b (A f (c f (n+2)-d e (m+n+4))+B e (c f (m+1)+d e)))}{(m+n+2) (m+n+3) (b e-a f)^2 (d e-c f)^2} \]
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Rubi [A] time = 0.9808, antiderivative size = 558, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {155, 12, 132} \[ \frac{(a+b x)^{m+1} (c+d x)^n (e+f x)^{-m-n-1} \left (\frac{(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{-n} ((m+n+2) (-b d e (a (A d f (m+2)-B c f (m+n+3)+B d e (n+1))+b (A c f (n+2)-A d e (m+n+3)+B c e (m+1)))+f (a d+b c) (a (A d f (m+2)-B c f (m+n+3)+B d e (n+1))+b (A c f (n+2)-A d e (m+n+3)+B c e (m+1)))+a b c d f (B e-A f))-(a d (n+1)+b c (m+1)) (a f (A d f (m+2)-B c f (m+n+3)+B d e (n+1))+b (A f (c f (n+2)-d e (m+n+4))+B e (c f (m+1)+d e)))) \, _2F_1\left (m+1,-n;m+2;-\frac{(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (m+n+2) (m+n+3) (b e-a f)^3 (d e-c f)^2}+\frac{(a+b x)^{m+1} (B e-A f) (c+d x)^{n+1} (e+f x)^{-m-n-3}}{(m+n+3) (b e-a f) (d e-c f)}+\frac{(a+b x)^{m+1} (c+d x)^{n+1} (e+f x)^{-m-n-2} (a f (A d f (m+2)-B c f (m+n+3)+B d e (n+1))+b (A f (c f (n+2)-d e (m+n+4))+B e (c f (m+1)+d e)))}{(m+n+2) (m+n+3) (b e-a f)^2 (d e-c f)^2} \]
Antiderivative was successfully verified.
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Rule 155
Rule 12
Rule 132
Rubi steps
\begin{align*} \int (a+b x)^m (A+B x) (c+d x)^n (e+f x)^{-4-m-n} \, dx &=\frac{(B e-A 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} (b (B c e (1+m)+A c f (2+n)-A d e (3+m+n))+a (A d f (2+m)+B d e (1+n)-B c f (3+m+n))-b d (B e-A f) x) \, dx}{(b e-a f) (d e-c f) (3+m+n)}\\ &=\frac{(B e-A 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{(a f (A d f (2+m)+B d e (1+n)-B c f (3+m+n))+b (B e (d e+c f (1+m))+A f (c f (2+n)-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 ((2+m+n) (a b c d f (B e-A f)-b d e (b (B c e (1+m)+A c f (2+n)-A d e (3+m+n))+a (A d f (2+m)+B d e (1+n)-B c f (3+m+n)))+(b c+a d) f (b (B c e (1+m)+A c f (2+n)-A d e (3+m+n))+a (A d f (2+m)+B d e (1+n)-B c f (3+m+n))))-(b c (1+m)+a d (1+n)) (a f (A d f (2+m)+B d e (1+n)-B c f (3+m+n))+b (B e (d e+c f (1+m))+A f (c f (2+n)-d e (4+m+n))))) (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{(B e-A 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{(a f (A d f (2+m)+B d e (1+n)-B c f (3+m+n))+b (B e (d e+c f (1+m))+A f (c f (2+n)-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{((2+m+n) (a b c d f (B e-A f)-b d e (b (B c e (1+m)+A c f (2+n)-A d e (3+m+n))+a (A d f (2+m)+B d e (1+n)-B c f (3+m+n)))+(b c+a d) f (b (B c e (1+m)+A c f (2+n)-A d e (3+m+n))+a (A d f (2+m)+B d e (1+n)-B c f (3+m+n))))-(b c (1+m)+a d (1+n)) (a f (A d f (2+m)+B d e (1+n)-B c f (3+m+n))+b (B e (d e+c f (1+m))+A f (c f (2+n)-d e (4+m+n))))) \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{(B e-A 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{(a f (A d f (2+m)+B d e (1+n)-B c f (3+m+n))+b (B e (d e+c f (1+m))+A f (c f (2+n)-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{((2+m+n) (a b c d f (B e-A f)-b d e (b (B c e (1+m)+A c f (2+n)-A d e (3+m+n))+a (A d f (2+m)+B d e (1+n)-B c f (3+m+n)))+(b c+a d) f (b (B c e (1+m)+A c f (2+n)-A d e (3+m+n))+a (A d f (2+m)+B d e (1+n)-B c f (3+m+n))))-(b c (1+m)+a d (1+n)) (a f (A d f (2+m)+B d e (1+n)-B c f (3+m+n))+b (B e (d e+c f (1+m))+A f (c f (2+n)-d e (4+m+n))))) (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 = 1.87166, size = 508, normalized size = 0.91 \[ -\frac{(a+b x)^{m+1} (c+d x)^n (e+f x)^{-m-n-3} \left (-\frac{(e+f x)^2 \left (\frac{(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{-n} ((m+n+2) (-b d e (a (A d f (m+2)-B c f (m+n+3)+B d e (n+1))+b (A c f (n+2)-A d e (m+n+3)+B c e (m+1)))+f (a d+b c) (a (A d f (m+2)-B c f (m+n+3)+B d e (n+1))+b (A c f (n+2)-A d e (m+n+3)+B c e (m+1)))+a b c d f (B e-A f))-(a d (n+1)+b c (m+1)) (a f (A d f (m+2)-B c f (m+n+3)+B d e (n+1))+b (A f (c f (n+2)-d e (m+n+4))+B e (c f (m+1)+d e)))) \, _2F_1\left (m+1,-n;m+2;\frac{(c f-d e) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (m+n+2) (b e-a f)^2 (d e-c f)}-\frac{(c+d x) (e+f x) (a f (A d f (m+2)-B c f (m+n+3)+B d e (n+1))+b (A f (c f (n+2)-d e (m+n+4))+B e (c f (m+1)+d e)))}{(m+n+2) (b e-a f) (d e-c f)}-(c+d x) (B e-A f)\right )}{(m+n+3) (b e-a f) (d e-c f)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.152, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( Bx+A \right ) \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 )}{\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 )}{\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 )}{\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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