Integrand size = 20, antiderivative size = 156 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {\left (c d^2-b d e+a e^2\right )^2}{6 e^5 (d+e x)^6}+\frac {2 (2 c d-b e) \left (c d^2-b d e+a e^2\right )}{5 e^5 (d+e x)^5}-\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{4 e^5 (d+e x)^4}+\frac {2 c (2 c d-b e)}{3 e^5 (d+e x)^3}-\frac {c^2}{2 e^5 (d+e x)^2} \]
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Time = 0.07 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {712} \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2}{4 e^5 (d+e x)^4}+\frac {2 (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{5 e^5 (d+e x)^5}-\frac {\left (a e^2-b d e+c d^2\right )^2}{6 e^5 (d+e x)^6}+\frac {2 c (2 c d-b e)}{3 e^5 (d+e x)^3}-\frac {c^2}{2 e^5 (d+e x)^2} \]
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Rule 712
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (c d^2-b d e+a e^2\right )^2}{e^4 (d+e x)^7}+\frac {2 (-2 c d+b e) \left (c d^2-b d e+a e^2\right )}{e^4 (d+e x)^6}+\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{e^4 (d+e x)^5}-\frac {2 c (2 c d-b e)}{e^4 (d+e x)^4}+\frac {c^2}{e^4 (d+e x)^3}\right ) \, dx \\ & = -\frac {\left (c d^2-b d e+a e^2\right )^2}{6 e^5 (d+e x)^6}+\frac {2 (2 c d-b e) \left (c d^2-b d e+a e^2\right )}{5 e^5 (d+e x)^5}-\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{4 e^5 (d+e x)^4}+\frac {2 c (2 c d-b e)}{3 e^5 (d+e x)^3}-\frac {c^2}{2 e^5 (d+e x)^2} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {2 c^2 \left (d^4+6 d^3 e x+15 d^2 e^2 x^2+20 d e^3 x^3+15 e^4 x^4\right )+e^2 \left (10 a^2 e^2+4 a b e (d+6 e x)+b^2 \left (d^2+6 d e x+15 e^2 x^2\right )\right )+2 c e \left (a e \left (d^2+6 d e x+15 e^2 x^2\right )+b \left (d^3+6 d^2 e x+15 d e^2 x^2+20 e^3 x^3\right )\right )}{60 e^5 (d+e x)^6} \]
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Time = 2.93 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.14
method | result | size |
risch | \(\frac {-\frac {c^{2} x^{4}}{2 e}-\frac {2 c \left (b e +c d \right ) x^{3}}{3 e^{2}}-\frac {\left (2 a c \,e^{2}+b^{2} e^{2}+2 b c d e +2 c^{2} d^{2}\right ) x^{2}}{4 e^{3}}-\frac {\left (4 a b \,e^{3}+2 d \,e^{2} a c +b^{2} d \,e^{2}+2 b c e \,d^{2}+2 c^{2} d^{3}\right ) x}{10 e^{4}}-\frac {10 a^{2} e^{4}+4 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}+2 d^{3} e b c +2 c^{2} d^{4}}{60 e^{5}}}{\left (e x +d \right )^{6}}\) | \(178\) |
norman | \(\frac {-\frac {c^{2} x^{4}}{2 e}-\frac {2 \left (b c \,e^{2}+d e \,c^{2}\right ) x^{3}}{3 e^{3}}-\frac {\left (2 a c \,e^{3}+b^{2} e^{3}+2 b c d \,e^{2}+2 d^{2} e \,c^{2}\right ) x^{2}}{4 e^{4}}-\frac {\left (4 a b \,e^{4}+2 a c d \,e^{3}+b^{2} d \,e^{3}+2 b c \,d^{2} e^{2}+2 d^{3} e \,c^{2}\right ) x}{10 e^{5}}-\frac {10 a^{2} e^{5}+4 a b d \,e^{4}+2 a \,d^{2} e^{3} c +b^{2} d^{2} e^{3}+2 b c \,d^{3} e^{2}+2 d^{4} e \,c^{2}}{60 e^{6}}}{\left (e x +d \right )^{6}}\) | \(192\) |
gosper | \(-\frac {30 c^{2} x^{4} e^{4}+40 x^{3} b c \,e^{4}+40 x^{3} c^{2} d \,e^{3}+30 x^{2} a c \,e^{4}+15 x^{2} b^{2} e^{4}+30 x^{2} b c d \,e^{3}+30 x^{2} c^{2} d^{2} e^{2}+24 x a b \,e^{4}+12 x a c d \,e^{3}+6 x \,b^{2} d \,e^{3}+12 x b c \,d^{2} e^{2}+12 x \,c^{2} d^{3} e +10 a^{2} e^{4}+4 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}+2 d^{3} e b c +2 c^{2} d^{4}}{60 e^{5} \left (e x +d \right )^{6}}\) | \(193\) |
default | \(-\frac {2 a b \,e^{3}-4 d \,e^{2} a c -2 b^{2} d \,e^{2}+6 b c e \,d^{2}-4 c^{2} d^{3}}{5 e^{5} \left (e x +d \right )^{5}}-\frac {2 c \left (b e -2 c d \right )}{3 e^{5} \left (e x +d \right )^{3}}-\frac {2 a c \,e^{2}+b^{2} e^{2}-6 b c d e +6 c^{2} d^{2}}{4 e^{5} \left (e x +d \right )^{4}}-\frac {c^{2}}{2 e^{5} \left (e x +d \right )^{2}}-\frac {a^{2} e^{4}-2 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}-2 d^{3} e b c +c^{2} d^{4}}{6 e^{5} \left (e x +d \right )^{6}}\) | \(195\) |
parallelrisch | \(\frac {-30 c^{2} e^{5} x^{4}-40 b c \,e^{5} x^{3}-40 c^{2} d \,e^{4} x^{3}-30 a c \,e^{5} x^{2}-15 b^{2} e^{5} x^{2}-30 b c d \,e^{4} x^{2}-30 c^{2} d^{2} e^{3} x^{2}-24 a b \,e^{5} x -12 a c d \,e^{4} x -6 b^{2} d \,e^{4} x -12 b c \,d^{2} e^{3} x -12 c^{2} d^{3} e^{2} x -10 a^{2} e^{5}-4 a b d \,e^{4}-2 a \,d^{2} e^{3} c -b^{2} d^{2} e^{3}-2 b c \,d^{3} e^{2}-2 d^{4} e \,c^{2}}{60 e^{6} \left (e x +d \right )^{6}}\) | \(199\) |
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Time = 0.64 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {30 \, c^{2} e^{4} x^{4} + 2 \, c^{2} d^{4} + 2 \, b c d^{3} e + 4 \, a b d e^{3} + 10 \, a^{2} e^{4} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + 40 \, {\left (c^{2} d e^{3} + b c e^{4}\right )} x^{3} + 15 \, {\left (2 \, c^{2} d^{2} e^{2} + 2 \, b c d e^{3} + {\left (b^{2} + 2 \, a c\right )} e^{4}\right )} x^{2} + 6 \, {\left (2 \, c^{2} d^{3} e + 2 \, b c d^{2} e^{2} + 4 \, a b e^{4} + {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x}{60 \, {\left (e^{11} x^{6} + 6 \, d e^{10} x^{5} + 15 \, d^{2} e^{9} x^{4} + 20 \, d^{3} e^{8} x^{3} + 15 \, d^{4} e^{7} x^{2} + 6 \, d^{5} e^{6} x + d^{6} e^{5}\right )}} \]
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Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=\text {Timed out} \]
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Time = 0.21 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {30 \, c^{2} e^{4} x^{4} + 2 \, c^{2} d^{4} + 2 \, b c d^{3} e + 4 \, a b d e^{3} + 10 \, a^{2} e^{4} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + 40 \, {\left (c^{2} d e^{3} + b c e^{4}\right )} x^{3} + 15 \, {\left (2 \, c^{2} d^{2} e^{2} + 2 \, b c d e^{3} + {\left (b^{2} + 2 \, a c\right )} e^{4}\right )} x^{2} + 6 \, {\left (2 \, c^{2} d^{3} e + 2 \, b c d^{2} e^{2} + 4 \, a b e^{4} + {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x}{60 \, {\left (e^{11} x^{6} + 6 \, d e^{10} x^{5} + 15 \, d^{2} e^{9} x^{4} + 20 \, d^{3} e^{8} x^{3} + 15 \, d^{4} e^{7} x^{2} + 6 \, d^{5} e^{6} x + d^{6} e^{5}\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {30 \, c^{2} e^{4} x^{4} + 40 \, c^{2} d e^{3} x^{3} + 40 \, b c e^{4} x^{3} + 30 \, c^{2} d^{2} e^{2} x^{2} + 30 \, b c d e^{3} x^{2} + 15 \, b^{2} e^{4} x^{2} + 30 \, a c e^{4} x^{2} + 12 \, c^{2} d^{3} e x + 12 \, b c d^{2} e^{2} x + 6 \, b^{2} d e^{3} x + 12 \, a c d e^{3} x + 24 \, a b e^{4} x + 2 \, c^{2} d^{4} + 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} + 4 \, a b d e^{3} + 10 \, a^{2} e^{4}}{60 \, {\left (e x + d\right )}^{6} e^{5}} \]
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Time = 9.90 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.49 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {\frac {10\,a^2\,e^4+4\,a\,b\,d\,e^3+2\,a\,c\,d^2\,e^2+b^2\,d^2\,e^2+2\,b\,c\,d^3\,e+2\,c^2\,d^4}{60\,e^5}+\frac {x\,\left (b^2\,d\,e^2+2\,b\,c\,d^2\,e+4\,a\,b\,e^3+2\,c^2\,d^3+2\,a\,c\,d\,e^2\right )}{10\,e^4}+\frac {c^2\,x^4}{2\,e}+\frac {x^2\,\left (b^2\,e^2+2\,b\,c\,d\,e+2\,c^2\,d^2+2\,a\,c\,e^2\right )}{4\,e^3}+\frac {2\,c\,x^3\,\left (b\,e+c\,d\right )}{3\,e^2}}{d^6+6\,d^5\,e\,x+15\,d^4\,e^2\,x^2+20\,d^3\,e^3\,x^3+15\,d^2\,e^4\,x^4+6\,d\,e^5\,x^5+e^6\,x^6} \]
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