Integrand size = 17, antiderivative size = 109 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^5} \, dx=-\frac {\left (c d^2+a e^2\right )^2}{4 e^5 (d+e x)^4}+\frac {4 c d \left (c d^2+a e^2\right )}{3 e^5 (d+e x)^3}-\frac {c \left (3 c d^2+a e^2\right )}{e^5 (d+e x)^2}+\frac {4 c^2 d}{e^5 (d+e x)}+\frac {c^2 \log (d+e x)}{e^5} \]
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Time = 0.05 (sec) , antiderivative size = 109, 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} \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^5} \, dx=-\frac {c \left (a e^2+3 c d^2\right )}{e^5 (d+e x)^2}+\frac {4 c d \left (a e^2+c d^2\right )}{3 e^5 (d+e x)^3}-\frac {\left (a e^2+c d^2\right )^2}{4 e^5 (d+e x)^4}+\frac {4 c^2 d}{e^5 (d+e x)}+\frac {c^2 \log (d+e x)}{e^5} \]
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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)^5}-\frac {4 c d \left (c d^2+a e^2\right )}{e^4 (d+e x)^4}+\frac {2 c \left (3 c d^2+a e^2\right )}{e^4 (d+e x)^3}-\frac {4 c^2 d}{e^4 (d+e x)^2}+\frac {c^2}{e^4 (d+e x)}\right ) \, dx \\ & = -\frac {\left (c d^2+a e^2\right )^2}{4 e^5 (d+e x)^4}+\frac {4 c d \left (c d^2+a e^2\right )}{3 e^5 (d+e x)^3}-\frac {c \left (3 c d^2+a e^2\right )}{e^5 (d+e x)^2}+\frac {4 c^2 d}{e^5 (d+e x)}+\frac {c^2 \log (d+e x)}{e^5} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {-3 a^2 e^4-2 a c e^2 \left (d^2+4 d e x+6 e^2 x^2\right )+c^2 d \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )+12 c^2 (d+e x)^4 \log (d+e x)}{12 e^5 (d+e x)^4} \]
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Time = 2.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\frac {4 c^{2} d \,x^{3}}{e^{2}}-\frac {c \left (e^{2} a -9 c \,d^{2}\right ) x^{2}}{e^{3}}-\frac {2 c d \left (e^{2} a -11 c \,d^{2}\right ) x}{3 e^{4}}-\frac {3 a^{2} e^{4}+2 a c \,d^{2} e^{2}-25 c^{2} d^{4}}{12 e^{5}}}{\left (e x +d \right )^{4}}+\frac {c^{2} \ln \left (e x +d \right )}{e^{5}}\) | \(109\) |
norman | \(\frac {-\frac {3 a^{2} e^{4}+2 a c \,d^{2} e^{2}-25 c^{2} d^{4}}{12 e^{5}}-\frac {\left (a c \,e^{2}-9 c^{2} d^{2}\right ) x^{2}}{e^{3}}+\frac {4 c^{2} d \,x^{3}}{e^{2}}-\frac {2 d \left (a c \,e^{2}-11 c^{2} d^{2}\right ) x}{3 e^{4}}}{\left (e x +d \right )^{4}}+\frac {c^{2} \ln \left (e x +d \right )}{e^{5}}\) | \(113\) |
default | \(\frac {4 c^{2} d}{e^{5} \left (e x +d \right )}+\frac {4 c d \left (e^{2} a +c \,d^{2}\right )}{3 e^{5} \left (e x +d \right )^{3}}-\frac {a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}}{4 e^{5} \left (e x +d \right )^{4}}-\frac {c \left (e^{2} a +3 c \,d^{2}\right )}{e^{5} \left (e x +d \right )^{2}}+\frac {c^{2} \ln \left (e x +d \right )}{e^{5}}\) | \(118\) |
parallelrisch | \(\frac {12 \ln \left (e x +d \right ) x^{4} c^{2} e^{4}+48 \ln \left (e x +d \right ) x^{3} c^{2} d \,e^{3}+72 \ln \left (e x +d \right ) x^{2} c^{2} d^{2} e^{2}+48 x^{3} c^{2} d \,e^{3}+48 \ln \left (e x +d \right ) x \,c^{2} d^{3} e -12 x^{2} a c \,e^{4}+108 x^{2} c^{2} d^{2} e^{2}+12 \ln \left (e x +d \right ) c^{2} d^{4}-8 x a c d \,e^{3}+88 x \,c^{2} d^{3} e -3 a^{2} e^{4}-2 a c \,d^{2} e^{2}+25 c^{2} d^{4}}{12 e^{5} \left (e x +d \right )^{4}}\) | \(180\) |
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Time = 0.25 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.78 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {48 \, c^{2} d e^{3} x^{3} + 25 \, c^{2} d^{4} - 2 \, a c d^{2} e^{2} - 3 \, a^{2} e^{4} + 12 \, {\left (9 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} x^{2} + 8 \, {\left (11 \, c^{2} d^{3} e - a c d e^{3}\right )} x + 12 \, {\left (c^{2} e^{4} x^{4} + 4 \, c^{2} d e^{3} x^{3} + 6 \, c^{2} d^{2} e^{2} x^{2} + 4 \, c^{2} d^{3} e x + c^{2} d^{4}\right )} \log \left (e x + d\right )}{12 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} \]
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Time = 0.74 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.38 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^{2} \log {\left (d + e x \right )}}{e^{5}} + \frac {- 3 a^{2} e^{4} - 2 a c d^{2} e^{2} + 25 c^{2} d^{4} + 48 c^{2} d e^{3} x^{3} + x^{2} \left (- 12 a c e^{4} + 108 c^{2} d^{2} e^{2}\right ) + x \left (- 8 a c d e^{3} + 88 c^{2} d^{3} e\right )}{12 d^{4} e^{5} + 48 d^{3} e^{6} x + 72 d^{2} e^{7} x^{2} + 48 d e^{8} x^{3} + 12 e^{9} x^{4}} \]
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Time = 0.22 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.34 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {48 \, c^{2} d e^{3} x^{3} + 25 \, c^{2} d^{4} - 2 \, a c d^{2} e^{2} - 3 \, a^{2} e^{4} + 12 \, {\left (9 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} x^{2} + 8 \, {\left (11 \, c^{2} d^{3} e - a c d e^{3}\right )} x}{12 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} + \frac {c^{2} \log \left (e x + d\right )}{e^{5}} \]
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Time = 0.27 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.51 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^5} \, dx=-\frac {c^{2} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{5}} + \frac {\frac {48 \, c^{2} d e^{15}}{e x + d} - \frac {36 \, c^{2} d^{2} e^{15}}{{\left (e x + d\right )}^{2}} + \frac {16 \, c^{2} d^{3} e^{15}}{{\left (e x + d\right )}^{3}} - \frac {3 \, c^{2} d^{4} e^{15}}{{\left (e x + d\right )}^{4}} - \frac {12 \, a c e^{17}}{{\left (e x + d\right )}^{2}} + \frac {16 \, a c d e^{17}}{{\left (e x + d\right )}^{3}} - \frac {6 \, a c d^{2} e^{17}}{{\left (e x + d\right )}^{4}} - \frac {3 \, a^{2} e^{19}}{{\left (e x + d\right )}^{4}}}{12 \, e^{20}} \]
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Time = 9.48 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.32 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^2\,\ln \left (d+e\,x\right )}{e^5}-\frac {\frac {3\,a^2\,e^4+2\,a\,c\,d^2\,e^2-25\,c^2\,d^4}{12\,e^5}-\frac {2\,x\,\left (11\,c^2\,d^3-a\,c\,d\,e^2\right )}{3\,e^4}-\frac {4\,c^2\,d\,x^3}{e^2}+\frac {c\,x^2\,\left (a\,e^2-9\,c\,d^2\right )}{e^3}}{d^4+4\,d^3\,e\,x+6\,d^2\,e^2\,x^2+4\,d\,e^3\,x^3+e^4\,x^4} \]
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