Integrand size = 66, antiderivative size = 24 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=x^2+\frac {25 e^8 \left (1+x^2\right )}{4 x^2 (3+x)^2} \]
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Time = 0.14 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {12, 6820, 1634} \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=x^2+\frac {25 e^8}{36 x^2}+\frac {25 e^8}{54 (x+3)}+\frac {125 e^8}{18 (x+3)^2}-\frac {25 e^8}{54 x} \]
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Rule 12
Rule 1634
Rule 6820
Rubi steps \begin{align*} \text {integral}& = e^8 \int \frac {-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx \\ & = e^8 \int \left (\frac {2 x}{e^8}-\frac {25 \left (3+2 x+x^3\right )}{2 x^3 (3+x)^3}\right ) \, dx \\ & = x^2-\frac {1}{2} \left (25 e^8\right ) \int \frac {3+2 x+x^3}{x^3 (3+x)^3} \, dx \\ & = x^2-\frac {1}{2} \left (25 e^8\right ) \int \left (\frac {1}{9 x^3}-\frac {1}{27 x^2}+\frac {10}{9 (3+x)^3}+\frac {1}{27 (3+x)^2}\right ) \, dx \\ & = \frac {25 e^8}{36 x^2}-\frac {25 e^8}{54 x}+x^2+\frac {125 e^8}{18 (3+x)^2}+\frac {25 e^8}{54 (3+x)} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=x^2+\frac {25 e^8 \left (1+x^2\right )}{4 x^2 (3+x)^2} \]
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Time = 0.35 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17
method | result | size |
risch | \(x^{2}+\frac {{\mathrm e}^{8} \left (\frac {25 x^{2}}{4}+\frac {25}{4}\right )}{x^{2} \left (x^{2}+6 x +9\right )}\) | \(28\) |
default | \(x^{2}+\frac {25 \left (3 x^{4}+3 x^{2}\right ) {\mathrm e}^{8}}{12 x^{4} \left (x^{2}+6 x +9\right )}\) | \(38\) |
parts | \(x^{2}+\frac {25 \left (3 x^{4}+3 x^{2}\right ) {\mathrm e}^{8}}{12 x^{4} \left (x^{2}+6 x +9\right )}\) | \(38\) |
norman | \(\frac {\left (x^{6} {\mathrm e}^{4}-54 x^{3} {\mathrm e}^{4}+\frac {{\mathrm e}^{4} \left (25 \,{\mathrm e}^{8}-324\right ) x^{2}}{4}+\frac {25 \,{\mathrm e}^{12}}{4}+6 x^{5} {\mathrm e}^{4}\right ) {\mathrm e}^{-4}}{x^{2} \left (3+x \right )^{2}}\) | \(56\) |
parallelrisch | \(-\frac {\left (-36 x^{8} {\mathrm e}^{-8}-216 x^{7} {\mathrm e}^{-8}-324 x^{6} {\mathrm e}^{-8}-225 x^{4}-225 x^{2}\right ) {\mathrm e}^{8}}{36 x^{4} \left (x^{2}+6 x +9\right )}\) | \(70\) |
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Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=\frac {4 \, x^{6} + 24 \, x^{5} + 36 \, x^{4} + 25 \, {\left (x^{2} + 1\right )} e^{8}}{4 \, {\left (x^{4} + 6 \, x^{3} + 9 \, x^{2}\right )}} \]
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Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=x^{2} + \frac {25 x^{2} e^{8} + 25 e^{8}}{4 x^{4} + 24 x^{3} + 36 x^{2}} \]
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Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=\frac {1}{4} \, {\left (4 \, x^{2} e^{\left (-8\right )} + \frac {25 \, {\left (x^{2} + 1\right )}}{x^{4} + 6 \, x^{3} + 9 \, x^{2}}\right )} e^{8} \]
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=x^{2} + \frac {25 \, {\left (x^{2} e^{8} + e^{8}\right )}}{4 \, {\left (x^{2} + 3 \, x\right )}^{2}} \]
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Time = 10.67 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^8 \left (-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}\right )}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx=x^2+\frac {\frac {25\,{\mathrm {e}}^8\,x^2}{4}+\frac {25\,{\mathrm {e}}^8}{4}}{x^2\,{\left (x+3\right )}^2} \]
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