Integrand size = 17, antiderivative size = 110 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {\left (c d^2+a e^2\right )^2}{5 e^5 (d+e x)^5}+\frac {c d \left (c d^2+a e^2\right )}{e^5 (d+e x)^4}-\frac {2 c \left (3 c d^2+a e^2\right )}{3 e^5 (d+e x)^3}+\frac {2 c^2 d}{e^5 (d+e x)^2}-\frac {c^2}{e^5 (d+e x)} \]
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Time = 0.05 (sec) , antiderivative size = 110, 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} \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {2 c \left (a e^2+3 c d^2\right )}{3 e^5 (d+e x)^3}+\frac {c d \left (a e^2+c d^2\right )}{e^5 (d+e x)^4}-\frac {\left (a e^2+c d^2\right )^2}{5 e^5 (d+e x)^5}-\frac {c^2}{e^5 (d+e x)}+\frac {2 c^2 d}{e^5 (d+e x)^2} \]
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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)^6}-\frac {4 c d \left (c d^2+a e^2\right )}{e^4 (d+e x)^5}+\frac {2 c \left (3 c d^2+a e^2\right )}{e^4 (d+e x)^4}-\frac {4 c^2 d}{e^4 (d+e x)^3}+\frac {c^2}{e^4 (d+e x)^2}\right ) \, dx \\ & = -\frac {\left (c d^2+a e^2\right )^2}{5 e^5 (d+e x)^5}+\frac {c d \left (c d^2+a e^2\right )}{e^5 (d+e x)^4}-\frac {2 c \left (3 c d^2+a e^2\right )}{3 e^5 (d+e x)^3}+\frac {2 c^2 d}{e^5 (d+e x)^2}-\frac {c^2}{e^5 (d+e x)} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {3 a^2 e^4+a c e^2 \left (d^2+5 d e x+10 e^2 x^2\right )+3 c^2 \left (d^4+5 d^3 e x+10 d^2 e^2 x^2+10 d e^3 x^3+5 e^4 x^4\right )}{15 e^5 (d+e x)^5} \]
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Time = 2.13 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95
method | result | size |
gosper | \(-\frac {15 c^{2} x^{4} e^{4}+30 x^{3} c^{2} d \,e^{3}+10 x^{2} a c \,e^{4}+30 x^{2} c^{2} d^{2} e^{2}+5 x a c d \,e^{3}+15 x \,c^{2} d^{3} e +3 a^{2} e^{4}+a c \,d^{2} e^{2}+3 c^{2} d^{4}}{15 \left (e x +d \right )^{5} e^{5}}\) | \(105\) |
risch | \(\frac {-\frac {c^{2} x^{4}}{e}-\frac {2 c^{2} d \,x^{3}}{e^{2}}-\frac {2 c \left (e^{2} a +3 c \,d^{2}\right ) x^{2}}{3 e^{3}}-\frac {d c \left (e^{2} a +3 c \,d^{2}\right ) x}{3 e^{4}}-\frac {3 a^{2} e^{4}+a c \,d^{2} e^{2}+3 c^{2} d^{4}}{15 e^{5}}}{\left (e x +d \right )^{5}}\) | \(105\) |
parallelrisch | \(\frac {-15 c^{2} x^{4} e^{4}-30 x^{3} c^{2} d \,e^{3}-10 x^{2} a c \,e^{4}-30 x^{2} c^{2} d^{2} e^{2}-5 x a c d \,e^{3}-15 x \,c^{2} d^{3} e -3 a^{2} e^{4}-a c \,d^{2} e^{2}-3 c^{2} d^{4}}{15 e^{5} \left (e x +d \right )^{5}}\) | \(106\) |
norman | \(\frac {-\frac {c^{2} x^{4}}{e}-\frac {2 c^{2} d \,x^{3}}{e^{2}}-\frac {2 \left (a c \,e^{2}+3 c^{2} d^{2}\right ) x^{2}}{3 e^{3}}-\frac {d \left (a c \,e^{2}+3 c^{2} d^{2}\right ) x}{3 e^{4}}-\frac {3 a^{2} e^{4}+a c \,d^{2} e^{2}+3 c^{2} d^{4}}{15 e^{5}}}{\left (e x +d \right )^{5}}\) | \(109\) |
default | \(-\frac {c^{2}}{e^{5} \left (e x +d \right )}-\frac {2 c \left (e^{2} a +3 c \,d^{2}\right )}{3 e^{5} \left (e x +d \right )^{3}}+\frac {c d \left (e^{2} a +c \,d^{2}\right )}{e^{5} \left (e x +d \right )^{4}}+\frac {2 c^{2} d}{e^{5} \left (e x +d \right )^{2}}-\frac {a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}}{5 e^{5} \left (e x +d \right )^{5}}\) | \(119\) |
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Time = 0.27 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.37 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {15 \, c^{2} e^{4} x^{4} + 30 \, c^{2} d e^{3} x^{3} + 3 \, c^{2} d^{4} + a c d^{2} e^{2} + 3 \, a^{2} e^{4} + 10 \, {\left (3 \, c^{2} d^{2} e^{2} + a c e^{4}\right )} x^{2} + 5 \, {\left (3 \, c^{2} d^{3} e + a c d e^{3}\right )} x}{15 \, {\left (e^{10} x^{5} + 5 \, d e^{9} x^{4} + 10 \, d^{2} e^{8} x^{3} + 10 \, d^{3} e^{7} x^{2} + 5 \, d^{4} e^{6} x + d^{5} e^{5}\right )}} \]
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Time = 0.99 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {- 3 a^{2} e^{4} - a c d^{2} e^{2} - 3 c^{2} d^{4} - 30 c^{2} d e^{3} x^{3} - 15 c^{2} e^{4} x^{4} + x^{2} \left (- 10 a c e^{4} - 30 c^{2} d^{2} e^{2}\right ) + x \left (- 5 a c d e^{3} - 15 c^{2} d^{3} e\right )}{15 d^{5} e^{5} + 75 d^{4} e^{6} x + 150 d^{3} e^{7} x^{2} + 150 d^{2} e^{8} x^{3} + 75 d e^{9} x^{4} + 15 e^{10} x^{5}} \]
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Time = 0.19 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.37 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {15 \, c^{2} e^{4} x^{4} + 30 \, c^{2} d e^{3} x^{3} + 3 \, c^{2} d^{4} + a c d^{2} e^{2} + 3 \, a^{2} e^{4} + 10 \, {\left (3 \, c^{2} d^{2} e^{2} + a c e^{4}\right )} x^{2} + 5 \, {\left (3 \, c^{2} d^{3} e + a c d e^{3}\right )} x}{15 \, {\left (e^{10} x^{5} + 5 \, d e^{9} x^{4} + 10 \, d^{2} e^{8} x^{3} + 10 \, d^{3} e^{7} x^{2} + 5 \, d^{4} e^{6} x + d^{5} e^{5}\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {15 \, c^{2} e^{4} x^{4} + 30 \, c^{2} d e^{3} x^{3} + 30 \, c^{2} d^{2} e^{2} x^{2} + 10 \, a c e^{4} x^{2} + 15 \, c^{2} d^{3} e x + 5 \, a c d e^{3} x + 3 \, c^{2} d^{4} + a c d^{2} e^{2} + 3 \, a^{2} e^{4}}{15 \, {\left (e x + d\right )}^{5} e^{5}} \]
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Time = 0.07 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {\frac {3\,a^2\,e^4+a\,c\,d^2\,e^2+3\,c^2\,d^4}{15\,e^5}+\frac {c^2\,x^4}{e}+\frac {2\,c^2\,d\,x^3}{e^2}+\frac {2\,c\,x^2\,\left (3\,c\,d^2+a\,e^2\right )}{3\,e^3}+\frac {c\,d\,x\,\left (3\,c\,d^2+a\,e^2\right )}{3\,e^4}}{d^5+5\,d^4\,e\,x+10\,d^3\,e^2\,x^2+10\,d^2\,e^3\,x^3+5\,d\,e^4\,x^4+e^5\,x^5} \]
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