Integrand size = 34, antiderivative size = 558 \[ \int (a+b x)^m (A+B x) (c+d x)^n (e+f x)^{-4-m-n} \, dx=\frac {(B e-A f) (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}+\frac {(a f (A d f (2+m)+B (d e (1+n)-c f (3+m+n)))+b (B e (d e+c f (1+m))+A f (c f (2+n)-d e (4+m+n)))) (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-2-m-n}}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}+\frac {((2+m+n) (a b c d f (B e-A f)+b d e ((a B c f+A (b d e-b c f-a d f)) (3+m+n)-(B e-A f) (b c (1+m)+a d (1+n)))-(b c+a d) f ((a B c f+A (b d e-b c f-a d f)) (3+m+n)-(B e-A f) (b c (1+m)+a d (1+n))))-(b c (1+m)+a d (1+n)) (a f (A d f (2+m)+B (d e (1+n)-c f (3+m+n)))+b (B e (d e+c f (1+m))+A f (c f (2+n)-d e (4+m+n))))) (a+b x)^{1+m} (c+d x)^n \left (\frac {(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{-n} (e+f x)^{-1-m-n} \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,-\frac {(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(b e-a f)^3 (d e-c f)^2 (1+m) (2+m+n) (3+m+n)} \]
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Time = 0.64 (sec) , antiderivative size = 558, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {160, 12, 134} \[ \int (a+b x)^m (A+B x) (c+d x)^n (e+f x)^{-4-m-n} \, dx=\frac {(a+b x)^{m+1} (c+d x)^n (e+f x)^{-m-n-1} \left (\frac {(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{-n} ((m+n+2) (-b d e (a (A d f (m+2)-B c f (m+n+3)+B d e (n+1))+b (A c f (n+2)-A d e (m+n+3)+B c e (m+1)))+f (a d+b c) (a (A d f (m+2)-B c f (m+n+3)+B d e (n+1))+b (A c f (n+2)-A d e (m+n+3)+B c e (m+1)))+a b c d f (B e-A f))-(a d (n+1)+b c (m+1)) (a f (A d f (m+2)-B c f (m+n+3)+B d e (n+1))+b (A f (c f (n+2)-d e (m+n+4))+B e (c f (m+1)+d e)))) \operatorname {Hypergeometric2F1}\left (m+1,-n,m+2,-\frac {(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (m+n+2) (m+n+3) (b e-a f)^3 (d e-c f)^2}+\frac {(a+b x)^{m+1} (B e-A f) (c+d x)^{n+1} (e+f x)^{-m-n-3}}{(m+n+3) (b e-a f) (d e-c f)}+\frac {(a+b x)^{m+1} (c+d x)^{n+1} (e+f x)^{-m-n-2} (a f (A d f (m+2)-B c f (m+n+3)+B d e (n+1))+b (A f (c f (n+2)-d e (m+n+4))+B e (c f (m+1)+d e)))}{(m+n+2) (m+n+3) (b e-a f)^2 (d e-c f)^2} \]
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Rule 12
Rule 134
Rule 160
Rubi steps \begin{align*} \text {integral}& = \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*}
Time = 1.09 (sec) , antiderivative size = 508, normalized size of antiderivative = 0.91 \[ \int (a+b x)^m (A+B x) (c+d x)^n (e+f x)^{-4-m-n} \, dx=-\frac {(a+b x)^{1+m} (c+d x)^n (e+f x)^{-3-m-n} \left (-((B e-A f) (c+d x))-\frac {(a f (A d f (2+m)+B d e (1+n)-B c f (3+m+n))+b (B e (d e+c f (1+m))+A f (c f (2+n)-d e (4+m+n)))) (c+d x) (e+f x)}{(b e-a f) (d e-c f) (2+m+n)}-\frac {((2+m+n) (a b c d f (B e-A f)-b d e (b (B c e (1+m)+A c f (2+n)-A d e (3+m+n))+a (A d f (2+m)+B d e (1+n)-B c f (3+m+n)))+(b c+a d) f (b (B c e (1+m)+A c f (2+n)-A d e (3+m+n))+a (A d f (2+m)+B d e (1+n)-B c f (3+m+n))))-(b c (1+m)+a d (1+n)) (a f (A d f (2+m)+B d e (1+n)-B c f (3+m+n))+b (B e (d e+c f (1+m))+A f (c f (2+n)-d e (4+m+n))))) \left (\frac {(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{-n} (e+f x)^2 \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,\frac {(-d e+c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(b e-a f)^2 (d e-c f) (1+m) (2+m+n)}\right )}{(b e-a f) (d e-c f) (3+m+n)} \]
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\[\int \left (b x +a \right )^{m} \left (B x +A \right ) \left (d x +c \right )^{n} \left (f x +e \right )^{-4-m -n}d x\]
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\[ \int (a+b x)^m (A+B x) (c+d x)^n (e+f x)^{-4-m-n} \, dx=\int { {\left (B x + A\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} {\left (f x + e\right )}^{-m - n - 4} \,d x } \]
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Timed out. \[ \int (a+b x)^m (A+B x) (c+d x)^n (e+f x)^{-4-m-n} \, dx=\text {Timed out} \]
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\[ \int (a+b x)^m (A+B x) (c+d x)^n (e+f x)^{-4-m-n} \, dx=\int { {\left (B x + A\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} {\left (f x + e\right )}^{-m - n - 4} \,d x } \]
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\[ \int (a+b x)^m (A+B x) (c+d x)^n (e+f x)^{-4-m-n} \, dx=\int { {\left (B x + A\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} {\left (f x + e\right )}^{-m - n - 4} \,d x } \]
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Timed out. \[ \int (a+b x)^m (A+B x) (c+d x)^n (e+f x)^{-4-m-n} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^n}{{\left (e+f\,x\right )}^{m+n+4}} \,d x \]
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