Integrand size = 24, antiderivative size = 248 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {d^2 (B d-A e) (c d-b e)^2}{5 e^6 (d+e x)^5}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{4 e^6 (d+e x)^4}-\frac {A e \left (6 c^2 d^2-6 b c d e+b^2 e^2\right )-B d \left (10 c^2 d^2-12 b c d e+3 b^2 e^2\right )}{3 e^6 (d+e x)^3}+\frac {2 A c e (2 c d-b e)-B \left (10 c^2 d^2-8 b c d e+b^2 e^2\right )}{2 e^6 (d+e x)^2}+\frac {c (5 B c d-2 b B e-A c e)}{e^6 (d+e x)}+\frac {B c^2 \log (d+e x)}{e^6} \]
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Time = 0.14 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {785} \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {2 A c e (2 c d-b e)-B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )}{2 e^6 (d+e x)^2}-\frac {A e \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )-B d \left (3 b^2 e^2-12 b c d e+10 c^2 d^2\right )}{3 e^6 (d+e x)^3}+\frac {d^2 (B d-A e) (c d-b e)^2}{5 e^6 (d+e x)^5}+\frac {c (-A c e-2 b B e+5 B c d)}{e^6 (d+e x)}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{4 e^6 (d+e x)^4}+\frac {B c^2 \log (d+e x)}{e^6} \]
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Rule 785
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {d^2 (B d-A e) (c d-b e)^2}{e^5 (d+e x)^6}+\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{e^5 (d+e x)^5}+\frac {A e \left (6 c^2 d^2-6 b c d e+b^2 e^2\right )-B d \left (10 c^2 d^2-12 b c d e+3 b^2 e^2\right )}{e^5 (d+e x)^4}+\frac {-2 A c e (2 c d-b e)+B \left (10 c^2 d^2-8 b c d e+b^2 e^2\right )}{e^5 (d+e x)^3}+\frac {c (-5 B c d+2 b B e+A c e)}{e^5 (d+e x)^2}+\frac {B c^2}{e^5 (d+e x)}\right ) \, dx \\ & = \frac {d^2 (B d-A e) (c d-b e)^2}{5 e^6 (d+e x)^5}-\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{4 e^6 (d+e x)^4}-\frac {A e \left (6 c^2 d^2-6 b c d e+b^2 e^2\right )-B d \left (10 c^2 d^2-12 b c d e+3 b^2 e^2\right )}{3 e^6 (d+e x)^3}+\frac {2 A c e (2 c d-b e)-B \left (10 c^2 d^2-8 b c d e+b^2 e^2\right )}{2 e^6 (d+e x)^2}+\frac {c (5 B c d-2 b B e-A c e)}{e^6 (d+e x)}+\frac {B c^2 \log (d+e x)}{e^6} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {-2 A e \left (b^2 e^2 \left (d^2+5 d e x+10 e^2 x^2\right )+3 b c e \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )+6 c^2 \left (d^4+5 d^3 e x+10 d^2 e^2 x^2+10 d e^3 x^3+5 e^4 x^4\right )\right )+B \left (-3 b^2 e^2 \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )-24 b c e \left (d^4+5 d^3 e x+10 d^2 e^2 x^2+10 d e^3 x^3+5 e^4 x^4\right )+c^2 d \left (137 d^4+625 d^3 e x+1100 d^2 e^2 x^2+900 d e^3 x^3+300 e^4 x^4\right )\right )+60 B c^2 (d+e x)^5 \log (d+e x)}{60 e^6 (d+e x)^5} \]
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Time = 0.23 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.18
method | result | size |
risch | \(\frac {-\frac {c \left (A c e +2 B b e -5 B c d \right ) x^{4}}{e^{2}}-\frac {\left (2 A b c \,e^{2}+4 A \,c^{2} d e +B \,b^{2} e^{2}+8 B b c d e -30 B \,c^{2} d^{2}\right ) x^{3}}{2 e^{3}}-\frac {\left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -110 B \,c^{2} d^{3}\right ) x^{2}}{6 e^{4}}-\frac {d \left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -125 B \,c^{2} d^{3}\right ) x}{12 e^{5}}-\frac {d^{2} \left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -137 B \,c^{2} d^{3}\right )}{60 e^{6}}}{\left (e x +d \right )^{5}}+\frac {B \,c^{2} \ln \left (e x +d \right )}{e^{6}}\) | \(292\) |
norman | \(\frac {-\frac {d^{2} \left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -137 B \,c^{2} d^{3}\right )}{60 e^{6}}-\frac {\left (A \,c^{2} e +2 B e b c -5 B \,c^{2} d \right ) x^{4}}{e^{2}}-\frac {\left (2 A b c \,e^{2}+4 A \,c^{2} d e +B \,b^{2} e^{2}+8 B b c d e -30 B \,c^{2} d^{2}\right ) x^{3}}{2 e^{3}}-\frac {\left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -110 B \,c^{2} d^{3}\right ) x^{2}}{6 e^{4}}-\frac {d \left (2 A \,b^{2} e^{3}+6 A b c d \,e^{2}+12 A \,c^{2} d^{2} e +3 B \,b^{2} d \,e^{2}+24 B b c \,d^{2} e -125 B \,c^{2} d^{3}\right ) x}{12 e^{5}}}{\left (e x +d \right )^{5}}+\frac {B \,c^{2} \ln \left (e x +d \right )}{e^{6}}\) | \(296\) |
default | \(-\frac {d^{2} \left (A \,b^{2} e^{3}-2 A b c d \,e^{2}+A \,c^{2} d^{2} e -B \,b^{2} d \,e^{2}+2 B b c \,d^{2} e -B \,c^{2} d^{3}\right )}{5 e^{6} \left (e x +d \right )^{5}}-\frac {A \,b^{2} e^{3}-6 A b c d \,e^{2}+6 A \,c^{2} d^{2} e -3 B \,b^{2} d \,e^{2}+12 B b c \,d^{2} e -10 B \,c^{2} d^{3}}{3 e^{6} \left (e x +d \right )^{3}}-\frac {c \left (A c e +2 B b e -5 B c d \right )}{e^{6} \left (e x +d \right )}+\frac {B \,c^{2} \ln \left (e x +d \right )}{e^{6}}+\frac {d \left (2 A \,b^{2} e^{3}-6 A b c d \,e^{2}+4 A \,c^{2} d^{2} e -3 B \,b^{2} d \,e^{2}+8 B b c \,d^{2} e -5 B \,c^{2} d^{3}\right )}{4 e^{6} \left (e x +d \right )^{4}}-\frac {2 A b c \,e^{2}-4 A \,c^{2} d e +B \,b^{2} e^{2}-8 B b c d e +10 B \,c^{2} d^{2}}{2 e^{6} \left (e x +d \right )^{2}}\) | \(305\) |
parallelrisch | \(-\frac {-137 B \,c^{2} d^{5}-300 B \ln \left (e x +d \right ) x \,c^{2} d^{4} e -600 B \ln \left (e x +d \right ) x^{3} c^{2} d^{2} e^{3}-300 B \ln \left (e x +d \right ) x^{4} c^{2} d \,e^{4}-600 B \ln \left (e x +d \right ) x^{2} c^{2} d^{3} e^{2}-60 B \ln \left (e x +d \right ) c^{2} d^{5}+60 A \,x^{4} c^{2} e^{5}+30 B \,x^{3} b^{2} e^{5}+20 A \,x^{2} b^{2} e^{5}+120 B x b c \,d^{3} e^{2}+30 A x b c \,d^{2} e^{3}+240 B \,x^{2} b c \,d^{2} e^{3}+60 A \,x^{2} b c d \,e^{4}+240 B \,x^{3} b c d \,e^{4}+2 A \,b^{2} d^{2} e^{3}+12 A \,c^{2} d^{4} e +3 B \,b^{2} d^{3} e^{2}+6 A b c \,d^{3} e^{2}+24 B b c \,d^{4} e -60 B \ln \left (e x +d \right ) x^{5} c^{2} e^{5}+60 A \,x^{3} b c \,e^{5}+120 A \,x^{3} c^{2} d \,e^{4}-900 B \,x^{3} c^{2} d^{2} e^{3}+120 A \,x^{2} c^{2} d^{2} e^{3}+30 B \,x^{2} b^{2} d \,e^{4}-1100 B \,x^{2} c^{2} d^{3} e^{2}+10 A x \,b^{2} d \,e^{4}+60 A x \,c^{2} d^{3} e^{2}+15 B x \,b^{2} d^{2} e^{3}-625 B x \,c^{2} d^{4} e +120 B \,x^{4} b c \,e^{5}-300 B \,x^{4} c^{2} d \,e^{4}}{60 e^{6} \left (e x +d \right )^{5}}\) | \(440\) |
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Time = 0.29 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.64 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {137 \, B c^{2} d^{5} - 2 \, A b^{2} d^{2} e^{3} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} + 60 \, {\left (5 \, B c^{2} d e^{4} - {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 30 \, {\left (30 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} - {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} + 10 \, {\left (110 \, B c^{2} d^{3} e^{2} - 2 \, A b^{2} e^{5} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} e - 2 \, A b^{2} d e^{4} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x + 60 \, {\left (B c^{2} e^{5} x^{5} + 5 \, B c^{2} d e^{4} x^{4} + 10 \, B c^{2} d^{2} e^{3} x^{3} + 10 \, B c^{2} d^{3} e^{2} x^{2} + 5 \, B c^{2} d^{4} e x + B c^{2} d^{5}\right )} \log \left (e x + d\right )}{60 \, {\left (e^{11} x^{5} + 5 \, d e^{10} x^{4} + 10 \, d^{2} e^{9} x^{3} + 10 \, d^{3} e^{8} x^{2} + 5 \, d^{4} e^{7} x + d^{5} e^{6}\right )}} \]
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Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\text {Timed out} \]
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Time = 0.20 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {137 \, B c^{2} d^{5} - 2 \, A b^{2} d^{2} e^{3} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} + 60 \, {\left (5 \, B c^{2} d e^{4} - {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 30 \, {\left (30 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} - {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} + 10 \, {\left (110 \, B c^{2} d^{3} e^{2} - 2 \, A b^{2} e^{5} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} e - 2 \, A b^{2} d e^{4} - 12 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x}{60 \, {\left (e^{11} x^{5} + 5 \, d e^{10} x^{4} + 10 \, d^{2} e^{9} x^{3} + 10 \, d^{3} e^{8} x^{2} + 5 \, d^{4} e^{7} x + d^{5} e^{6}\right )}} + \frac {B c^{2} \log \left (e x + d\right )}{e^{6}} \]
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Time = 0.26 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.27 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {B c^{2} \log \left ({\left | e x + d \right |}\right )}{e^{6}} + \frac {60 \, {\left (5 \, B c^{2} d e^{3} - 2 \, B b c e^{4} - A c^{2} e^{4}\right )} x^{4} + 30 \, {\left (30 \, B c^{2} d^{2} e^{2} - 8 \, B b c d e^{3} - 4 \, A c^{2} d e^{3} - B b^{2} e^{4} - 2 \, A b c e^{4}\right )} x^{3} + 10 \, {\left (110 \, B c^{2} d^{3} e - 24 \, B b c d^{2} e^{2} - 12 \, A c^{2} d^{2} e^{2} - 3 \, B b^{2} d e^{3} - 6 \, A b c d e^{3} - 2 \, A b^{2} e^{4}\right )} x^{2} + 5 \, {\left (125 \, B c^{2} d^{4} - 24 \, B b c d^{3} e - 12 \, A c^{2} d^{3} e - 3 \, B b^{2} d^{2} e^{2} - 6 \, A b c d^{2} e^{2} - 2 \, A b^{2} d e^{3}\right )} x + \frac {137 \, B c^{2} d^{5} - 24 \, B b c d^{4} e - 12 \, A c^{2} d^{4} e - 3 \, B b^{2} d^{3} e^{2} - 6 \, A b c d^{3} e^{2} - 2 \, A b^{2} d^{2} e^{3}}{e}}{60 \, {\left (e x + d\right )}^{5} e^{5}} \]
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Time = 10.87 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.38 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {B\,c^2\,\ln \left (d+e\,x\right )}{e^6}-\frac {\frac {3\,B\,b^2\,d^3\,e^2+2\,A\,b^2\,d^2\,e^3+24\,B\,b\,c\,d^4\,e+6\,A\,b\,c\,d^3\,e^2-137\,B\,c^2\,d^5+12\,A\,c^2\,d^4\,e}{60\,e^6}+\frac {x^3\,\left (B\,b^2\,e^2+8\,B\,b\,c\,d\,e+2\,A\,b\,c\,e^2-30\,B\,c^2\,d^2+4\,A\,c^2\,d\,e\right )}{2\,e^3}+\frac {x\,\left (3\,B\,b^2\,d^2\,e^2+2\,A\,b^2\,d\,e^3+24\,B\,b\,c\,d^3\,e+6\,A\,b\,c\,d^2\,e^2-125\,B\,c^2\,d^4+12\,A\,c^2\,d^3\,e\right )}{12\,e^5}+\frac {x^2\,\left (3\,B\,b^2\,d\,e^2+2\,A\,b^2\,e^3+24\,B\,b\,c\,d^2\,e+6\,A\,b\,c\,d\,e^2-110\,B\,c^2\,d^3+12\,A\,c^2\,d^2\,e\right )}{6\,e^4}+\frac {c\,x^4\,\left (A\,c\,e+2\,B\,b\,e-5\,B\,c\,d\right )}{e^2}}{d^5+5\,d^4\,e\,x+10\,d^3\,e^2\,x^2+10\,d^2\,e^3\,x^3+5\,d\,e^4\,x^4+e^5\,x^5} \]
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