Integrand size = 23, antiderivative size = 134 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^6} \, dx=\frac {(B d-A e) \left (c d^2-b d e+a e^2\right )}{5 e^4 (d+e x)^5}+\frac {A e (2 c d-b e)-B \left (3 c d^2-e (2 b d-a e)\right )}{4 e^4 (d+e x)^4}+\frac {3 B c d-b B e-A c e}{3 e^4 (d+e x)^3}-\frac {B c}{2 e^4 (d+e x)^2} \]
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Time = 0.07 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {785} \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^6} \, dx=-\frac {-B e (2 b d-a e)-A e (2 c d-b e)+3 B c d^2}{4 e^4 (d+e x)^4}+\frac {(B d-A e) \left (a e^2-b d e+c d^2\right )}{5 e^4 (d+e x)^5}+\frac {-A c e-b B e+3 B c d}{3 e^4 (d+e x)^3}-\frac {B c}{2 e^4 (d+e x)^2} \]
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Rule 785
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(-B d+A e) \left (c d^2-b d e+a e^2\right )}{e^3 (d+e x)^6}+\frac {3 B c d^2-B e (2 b d-a e)-A e (2 c d-b e)}{e^3 (d+e x)^5}+\frac {-3 B c d+b B e+A c e}{e^3 (d+e x)^4}+\frac {B c}{e^3 (d+e x)^3}\right ) \, dx \\ & = \frac {(B d-A e) \left (c d^2-b d e+a e^2\right )}{5 e^4 (d+e x)^5}-\frac {3 B c d^2-B e (2 b d-a e)-A e (2 c d-b e)}{4 e^4 (d+e x)^4}+\frac {3 B c d-b B e-A c e}{3 e^4 (d+e x)^3}-\frac {B c}{2 e^4 (d+e x)^2} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^6} \, dx=-\frac {A e \left (3 e (b d+4 a e+5 b e x)+2 c \left (d^2+5 d e x+10 e^2 x^2\right )\right )+B \left (3 c \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )+e \left (3 a e (d+5 e x)+2 b \left (d^2+5 d e x+10 e^2 x^2\right )\right )\right )}{60 e^4 (d+e x)^5} \]
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Time = 0.29 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01
method | result | size |
risch | \(\frac {-\frac {B c \,x^{3}}{2 e}-\frac {\left (2 A c e +2 B b e +3 B c d \right ) x^{2}}{6 e^{2}}-\frac {\left (3 A b \,e^{2}+2 A c d e +3 B a \,e^{2}+2 B b d e +3 B c \,d^{2}\right ) x}{12 e^{3}}-\frac {12 A a \,e^{3}+3 A b d \,e^{2}+2 A c \,d^{2} e +3 B a d \,e^{2}+2 B b \,d^{2} e +3 B c \,d^{3}}{60 e^{4}}}{\left (e x +d \right )^{5}}\) | \(136\) |
default | \(-\frac {A a \,e^{3}-A b d \,e^{2}+A c \,d^{2} e -B a d \,e^{2}+B b \,d^{2} e -B c \,d^{3}}{5 e^{4} \left (e x +d \right )^{5}}-\frac {A c e +B b e -3 B c d}{3 e^{4} \left (e x +d \right )^{3}}-\frac {A b \,e^{2}-2 A c d e +B a \,e^{2}-2 B b d e +3 B c \,d^{2}}{4 e^{4} \left (e x +d \right )^{4}}-\frac {B c}{2 e^{4} \left (e x +d \right )^{2}}\) | \(142\) |
gosper | \(-\frac {30 B c \,x^{3} e^{3}+20 A c \,e^{3} x^{2}+20 B \,x^{2} b \,e^{3}+30 B \,x^{2} c d \,e^{2}+15 A b \,e^{3} x +10 A c d \,e^{2} x +15 B x a \,e^{3}+10 B x b d \,e^{2}+15 B c \,d^{2} e x +12 A a \,e^{3}+3 A b d \,e^{2}+2 A c \,d^{2} e +3 B a d \,e^{2}+2 B b \,d^{2} e +3 B c \,d^{3}}{60 e^{4} \left (e x +d \right )^{5}}\) | \(144\) |
norman | \(\frac {-\frac {B c \,x^{3}}{2 e}-\frac {\left (2 A c \,e^{2}+2 B \,e^{2} b +3 B c d e \right ) x^{2}}{6 e^{3}}-\frac {\left (3 A b \,e^{3}+2 A c d \,e^{2}+3 B \,e^{3} a +2 B b d \,e^{2}+3 B c \,d^{2} e \right ) x}{12 e^{4}}-\frac {12 A a \,e^{4}+3 A b d \,e^{3}+2 A c \,d^{2} e^{2}+3 B a d \,e^{3}+2 B b \,d^{2} e^{2}+3 B c \,d^{3} e}{60 e^{5}}}{\left (e x +d \right )^{5}}\) | \(151\) |
parallelrisch | \(-\frac {30 B c \,x^{3} e^{4}+20 A c \,e^{4} x^{2}+20 B b \,e^{4} x^{2}+30 B c d \,e^{3} x^{2}+15 A b \,e^{4} x +10 d x \,e^{3} c A +15 B a \,e^{4} x +10 B b d \,e^{3} x +15 d^{2} e^{2} x c B +12 A a \,e^{4}+3 A b d \,e^{3}+2 A c \,d^{2} e^{2}+3 B a d \,e^{3}+2 B b \,d^{2} e^{2}+3 B c \,d^{3} e}{60 e^{5} \left (e x +d \right )^{5}}\) | \(151\) |
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Time = 0.33 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.29 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^6} \, dx=-\frac {30 \, B c e^{3} x^{3} + 3 \, B c d^{3} + 12 \, A a e^{3} + 2 \, {\left (B b + A c\right )} d^{2} e + 3 \, {\left (B a + A b\right )} d e^{2} + 10 \, {\left (3 \, B c d e^{2} + 2 \, {\left (B b + A c\right )} e^{3}\right )} x^{2} + 5 \, {\left (3 \, B c d^{2} e + 2 \, {\left (B b + A c\right )} d e^{2} + 3 \, {\left (B a + A b\right )} e^{3}\right )} x}{60 \, {\left (e^{9} x^{5} + 5 \, d e^{8} x^{4} + 10 \, d^{2} e^{7} x^{3} + 10 \, d^{3} e^{6} x^{2} + 5 \, d^{4} e^{5} x + d^{5} e^{4}\right )}} \]
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Time = 80.69 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.58 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^6} \, dx=\frac {- 12 A a e^{3} - 3 A b d e^{2} - 2 A c d^{2} e - 3 B a d e^{2} - 2 B b d^{2} e - 3 B c d^{3} - 30 B c e^{3} x^{3} + x^{2} \left (- 20 A c e^{3} - 20 B b e^{3} - 30 B c d e^{2}\right ) + x \left (- 15 A b e^{3} - 10 A c d e^{2} - 15 B a e^{3} - 10 B b d e^{2} - 15 B c d^{2} e\right )}{60 d^{5} e^{4} + 300 d^{4} e^{5} x + 600 d^{3} e^{6} x^{2} + 600 d^{2} e^{7} x^{3} + 300 d e^{8} x^{4} + 60 e^{9} x^{5}} \]
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Time = 0.20 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.29 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^6} \, dx=-\frac {30 \, B c e^{3} x^{3} + 3 \, B c d^{3} + 12 \, A a e^{3} + 2 \, {\left (B b + A c\right )} d^{2} e + 3 \, {\left (B a + A b\right )} d e^{2} + 10 \, {\left (3 \, B c d e^{2} + 2 \, {\left (B b + A c\right )} e^{3}\right )} x^{2} + 5 \, {\left (3 \, B c d^{2} e + 2 \, {\left (B b + A c\right )} d e^{2} + 3 \, {\left (B a + A b\right )} e^{3}\right )} x}{60 \, {\left (e^{9} x^{5} + 5 \, d e^{8} x^{4} + 10 \, d^{2} e^{7} x^{3} + 10 \, d^{3} e^{6} x^{2} + 5 \, d^{4} e^{5} x + d^{5} e^{4}\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^6} \, dx=-\frac {30 \, B c e^{3} x^{3} + 30 \, B c d e^{2} x^{2} + 20 \, B b e^{3} x^{2} + 20 \, A c e^{3} x^{2} + 15 \, B c d^{2} e x + 10 \, B b d e^{2} x + 10 \, A c d e^{2} x + 15 \, B a e^{3} x + 15 \, A b e^{3} x + 3 \, B c d^{3} + 2 \, B b d^{2} e + 2 \, A c d^{2} e + 3 \, B a d e^{2} + 3 \, A b d e^{2} + 12 \, A a e^{3}}{60 \, {\left (e x + d\right )}^{5} e^{4}} \]
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Time = 0.09 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.34 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^6} \, dx=-\frac {\frac {12\,A\,a\,e^3+3\,B\,c\,d^3+3\,A\,b\,d\,e^2+3\,B\,a\,d\,e^2+2\,A\,c\,d^2\,e+2\,B\,b\,d^2\,e}{60\,e^4}+\frac {x^2\,\left (2\,A\,c\,e+2\,B\,b\,e+3\,B\,c\,d\right )}{6\,e^2}+\frac {x\,\left (3\,A\,b\,e^2+3\,B\,a\,e^2+3\,B\,c\,d^2+2\,A\,c\,d\,e+2\,B\,b\,d\,e\right )}{12\,e^3}+\frac {B\,c\,x^3}{2\,e}}{d^5+5\,d^4\,e\,x+10\,d^3\,e^2\,x^2+10\,d^2\,e^3\,x^3+5\,d\,e^4\,x^4+e^5\,x^5} \]
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