Integrand size = 18, antiderivative size = 69 \[ \int \frac {a+b x+c x^2}{(d+e x)^5} \, dx=-\frac {c d^2-b d e+a e^2}{4 e^3 (d+e x)^4}+\frac {2 c d-b e}{3 e^3 (d+e x)^3}-\frac {c}{2 e^3 (d+e x)^2} \]
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Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {712} \[ \int \frac {a+b x+c x^2}{(d+e x)^5} \, dx=-\frac {a e^2-b d e+c d^2}{4 e^3 (d+e x)^4}+\frac {2 c d-b e}{3 e^3 (d+e x)^3}-\frac {c}{2 e^3 (d+e x)^2} \]
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Rule 712
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {c d^2-b d e+a e^2}{e^2 (d+e x)^5}+\frac {-2 c d+b e}{e^2 (d+e x)^4}+\frac {c}{e^2 (d+e x)^3}\right ) \, dx \\ & = -\frac {c d^2-b d e+a e^2}{4 e^3 (d+e x)^4}+\frac {2 c d-b e}{3 e^3 (d+e x)^3}-\frac {c}{2 e^3 (d+e x)^2} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.71 \[ \int \frac {a+b x+c x^2}{(d+e x)^5} \, dx=-\frac {c \left (d^2+4 d e x+6 e^2 x^2\right )+e (3 a e+b (d+4 e x))}{12 e^3 (d+e x)^4} \]
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Time = 2.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.74
method | result | size |
gosper | \(-\frac {6 c \,x^{2} e^{2}+4 x b \,e^{2}+4 x c d e +3 e^{2} a +b d e +c \,d^{2}}{12 e^{3} \left (e x +d \right )^{4}}\) | \(51\) |
risch | \(\frac {-\frac {c \,x^{2}}{2 e}-\frac {\left (b e +c d \right ) x}{3 e^{2}}-\frac {3 e^{2} a +b d e +c \,d^{2}}{12 e^{3}}}{\left (e x +d \right )^{4}}\) | \(53\) |
parallelrisch | \(\frac {-6 c \,e^{3} x^{2}-4 b \,e^{3} x -4 c d \,e^{2} x -3 a \,e^{3}-b d \,e^{2}-d^{2} e c}{12 e^{4} \left (e x +d \right )^{4}}\) | \(58\) |
norman | \(\frac {-\frac {c \,x^{2}}{2 e}-\frac {\left (e^{2} b +c d e \right ) x}{3 e^{3}}-\frac {3 a \,e^{3}+b d \,e^{2}+d^{2} e c}{12 e^{4}}}{\left (e x +d \right )^{4}}\) | \(59\) |
default | \(-\frac {b e -2 c d}{3 e^{3} \left (e x +d \right )^{3}}-\frac {e^{2} a -b d e +c \,d^{2}}{4 e^{3} \left (e x +d \right )^{4}}-\frac {c}{2 e^{3} \left (e x +d \right )^{2}}\) | \(63\) |
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Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25 \[ \int \frac {a+b x+c x^2}{(d+e x)^5} \, dx=-\frac {6 \, c e^{2} x^{2} + c d^{2} + b d e + 3 \, a e^{2} + 4 \, {\left (c d e + b e^{2}\right )} x}{12 \, {\left (e^{7} x^{4} + 4 \, d e^{6} x^{3} + 6 \, d^{2} e^{5} x^{2} + 4 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]
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Time = 0.61 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.33 \[ \int \frac {a+b x+c x^2}{(d+e x)^5} \, dx=\frac {- 3 a e^{2} - b d e - c d^{2} - 6 c e^{2} x^{2} + x \left (- 4 b e^{2} - 4 c d e\right )}{12 d^{4} e^{3} + 48 d^{3} e^{4} x + 72 d^{2} e^{5} x^{2} + 48 d e^{6} x^{3} + 12 e^{7} x^{4}} \]
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Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25 \[ \int \frac {a+b x+c x^2}{(d+e x)^5} \, dx=-\frac {6 \, c e^{2} x^{2} + c d^{2} + b d e + 3 \, a e^{2} + 4 \, {\left (c d e + b e^{2}\right )} x}{12 \, {\left (e^{7} x^{4} + 4 \, d e^{6} x^{3} + 6 \, d^{2} e^{5} x^{2} + 4 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25 \[ \int \frac {a+b x+c x^2}{(d+e x)^5} \, dx=-\frac {\frac {3 \, a}{{\left (e x + d\right )}^{4}} + \frac {6 \, c}{{\left (e x + d\right )}^{2} e^{2}} - \frac {8 \, c d}{{\left (e x + d\right )}^{3} e^{2}} + \frac {3 \, c d^{2}}{{\left (e x + d\right )}^{4} e^{2}} + \frac {4 \, b}{{\left (e x + d\right )}^{3} e} - \frac {3 \, b d}{{\left (e x + d\right )}^{4} e}}{12 \, e} \]
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Time = 9.79 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25 \[ \int \frac {a+b x+c x^2}{(d+e x)^5} \, dx=-\frac {\frac {c\,d^2+b\,d\,e+3\,a\,e^2}{12\,e^3}+\frac {x\,\left (b\,e+c\,d\right )}{3\,e^2}+\frac {c\,x^2}{2\,e}}{d^4+4\,d^3\,e\,x+6\,d^2\,e^2\,x^2+4\,d\,e^3\,x^3+e^4\,x^4} \]
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