Integrand size = 17, antiderivative size = 114 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^8} \, dx=-\frac {\left (c d^2+a e^2\right )^2}{7 e^5 (d+e x)^7}+\frac {2 c d \left (c d^2+a e^2\right )}{3 e^5 (d+e x)^6}-\frac {2 c \left (3 c d^2+a e^2\right )}{5 e^5 (d+e x)^5}+\frac {c^2 d}{e^5 (d+e x)^4}-\frac {c^2}{3 e^5 (d+e x)^3} \]
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Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {711} \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^8} \, dx=-\frac {2 c \left (a e^2+3 c d^2\right )}{5 e^5 (d+e x)^5}+\frac {2 c d \left (a e^2+c d^2\right )}{3 e^5 (d+e x)^6}-\frac {\left (a e^2+c d^2\right )^2}{7 e^5 (d+e x)^7}-\frac {c^2}{3 e^5 (d+e x)^3}+\frac {c^2 d}{e^5 (d+e x)^4} \]
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Rule 711
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (c d^2+a e^2\right )^2}{e^4 (d+e x)^8}-\frac {4 c d \left (c d^2+a e^2\right )}{e^4 (d+e x)^7}+\frac {2 c \left (3 c d^2+a e^2\right )}{e^4 (d+e x)^6}-\frac {4 c^2 d}{e^4 (d+e x)^5}+\frac {c^2}{e^4 (d+e x)^4}\right ) \, dx \\ & = -\frac {\left (c d^2+a e^2\right )^2}{7 e^5 (d+e x)^7}+\frac {2 c d \left (c d^2+a e^2\right )}{3 e^5 (d+e x)^6}-\frac {2 c \left (3 c d^2+a e^2\right )}{5 e^5 (d+e x)^5}+\frac {c^2 d}{e^5 (d+e x)^4}-\frac {c^2}{3 e^5 (d+e x)^3} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^8} \, dx=-\frac {15 a^2 e^4+2 a c e^2 \left (d^2+7 d e x+21 e^2 x^2\right )+c^2 \left (d^4+7 d^3 e x+21 d^2 e^2 x^2+35 d e^3 x^3+35 e^4 x^4\right )}{105 e^5 (d+e x)^7} \]
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Time = 2.16 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92
| method | result | size |
| gosper | \(-\frac {35 c^{2} x^{4} e^{4}+35 x^{3} c^{2} d \,e^{3}+42 x^{2} a c \,e^{4}+21 x^{2} c^{2} d^{2} e^{2}+14 x a c d \,e^{3}+7 x \,c^{2} d^{3} e +15 a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}}{105 e^{5} \left (e x +d \right )^{7}}\) | \(105\) |
| risch | \(\frac {-\frac {c^{2} x^{4}}{3 e}-\frac {c^{2} d \,x^{3}}{3 e^{2}}-\frac {c \left (2 e^{2} a +c \,d^{2}\right ) x^{2}}{5 e^{3}}-\frac {c d \left (2 e^{2} a +c \,d^{2}\right ) x}{15 e^{4}}-\frac {15 a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}}{105 e^{5}}}{\left (e x +d \right )^{7}}\) | \(105\) |
| parallelrisch | \(\frac {-35 x^{4} c^{2} e^{6}-35 c^{2} d \,x^{3} e^{5}-42 a c \,e^{6} x^{2}-21 c^{2} d^{2} e^{4} x^{2}-14 a c d \,e^{5} x -7 c^{2} d^{3} e^{3} x -15 a^{2} e^{6}-2 a c \,d^{2} e^{4}-c^{2} d^{4} e^{2}}{105 e^{7} \left (e x +d \right )^{7}}\) | \(111\) |
| norman | \(\frac {-\frac {c^{2} x^{4}}{3 e}-\frac {c^{2} d \,x^{3}}{3 e^{2}}-\frac {\left (2 a c \,e^{4}+d^{2} e^{2} c^{2}\right ) x^{2}}{5 e^{5}}-\frac {d \left (2 a c \,e^{4}+d^{2} e^{2} c^{2}\right ) x}{15 e^{6}}-\frac {15 a^{2} e^{6}+2 a c \,d^{2} e^{4}+c^{2} d^{4} e^{2}}{105 e^{7}}}{\left (e x +d \right )^{7}}\) | \(118\) |
| default | \(-\frac {a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}}{7 e^{5} \left (e x +d \right )^{7}}+\frac {2 c d \left (e^{2} a +c \,d^{2}\right )}{3 e^{5} \left (e x +d \right )^{6}}-\frac {c^{2}}{3 e^{5} \left (e x +d \right )^{3}}+\frac {c^{2} d}{e^{5} \left (e x +d \right )^{4}}-\frac {2 c \left (e^{2} a +3 c \,d^{2}\right )}{5 e^{5} \left (e x +d \right )^{5}}\) | \(119\) |
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Time = 0.24 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^8} \, dx=-\frac {35 \, c^{2} e^{4} x^{4} + 35 \, c^{2} d e^{3} x^{3} + c^{2} d^{4} + 2 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} + 21 \, {\left (c^{2} d^{2} e^{2} + 2 \, a c e^{4}\right )} x^{2} + 7 \, {\left (c^{2} d^{3} e + 2 \, a c d e^{3}\right )} x}{105 \, {\left (e^{12} x^{7} + 7 \, d e^{11} x^{6} + 21 \, d^{2} e^{10} x^{5} + 35 \, d^{3} e^{9} x^{4} + 35 \, d^{4} e^{8} x^{3} + 21 \, d^{5} e^{7} x^{2} + 7 \, d^{6} e^{6} x + d^{7} e^{5}\right )}} \]
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Time = 1.72 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.62 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^8} \, dx=\frac {- 15 a^{2} e^{4} - 2 a c d^{2} e^{2} - c^{2} d^{4} - 35 c^{2} d e^{3} x^{3} - 35 c^{2} e^{4} x^{4} + x^{2} \left (- 42 a c e^{4} - 21 c^{2} d^{2} e^{2}\right ) + x \left (- 14 a c d e^{3} - 7 c^{2} d^{3} e\right )}{105 d^{7} e^{5} + 735 d^{6} e^{6} x + 2205 d^{5} e^{7} x^{2} + 3675 d^{4} e^{8} x^{3} + 3675 d^{3} e^{9} x^{4} + 2205 d^{2} e^{10} x^{5} + 735 d e^{11} x^{6} + 105 e^{12} x^{7}} \]
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Time = 0.21 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^8} \, dx=-\frac {35 \, c^{2} e^{4} x^{4} + 35 \, c^{2} d e^{3} x^{3} + c^{2} d^{4} + 2 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} + 21 \, {\left (c^{2} d^{2} e^{2} + 2 \, a c e^{4}\right )} x^{2} + 7 \, {\left (c^{2} d^{3} e + 2 \, a c d e^{3}\right )} x}{105 \, {\left (e^{12} x^{7} + 7 \, d e^{11} x^{6} + 21 \, d^{2} e^{10} x^{5} + 35 \, d^{3} e^{9} x^{4} + 35 \, d^{4} e^{8} x^{3} + 21 \, d^{5} e^{7} x^{2} + 7 \, d^{6} e^{6} x + d^{7} e^{5}\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^8} \, dx=-\frac {35 \, c^{2} e^{4} x^{4} + 35 \, c^{2} d e^{3} x^{3} + 21 \, c^{2} d^{2} e^{2} x^{2} + 42 \, a c e^{4} x^{2} + 7 \, c^{2} d^{3} e x + 14 \, a c d e^{3} x + c^{2} d^{4} + 2 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4}}{105 \, {\left (e x + d\right )}^{7} e^{5}} \]
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Time = 0.07 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^8} \, dx=-\frac {\frac {15\,a^2\,e^4+2\,a\,c\,d^2\,e^2+c^2\,d^4}{105\,e^5}+\frac {c^2\,x^4}{3\,e}+\frac {c^2\,d\,x^3}{3\,e^2}+\frac {c\,x^2\,\left (c\,d^2+2\,a\,e^2\right )}{5\,e^3}+\frac {c\,d\,x\,\left (c\,d^2+2\,a\,e^2\right )}{15\,e^4}}{d^7+7\,d^6\,e\,x+21\,d^5\,e^2\,x^2+35\,d^4\,e^3\,x^3+35\,d^3\,e^4\,x^4+21\,d^2\,e^5\,x^5+7\,d\,e^6\,x^6+e^7\,x^7} \]
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