Integrand size = 43, antiderivative size = 28 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=\frac {(-1+2 x)^2}{2 x}+\frac {-1-\frac {16 x^2}{25}}{e^4} \]
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Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {1607, 1600, 14} \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=-\frac {16 x^2}{25 e^4}+2 x+\frac {1}{2 x} \]
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Rule 14
Rule 1600
Rule 1607
Rubi steps \begin{align*} \text {integral}& = \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{x^2 \left (50+32 x^2\right )} \, dx \\ & = \int \frac {-\frac {1}{2}+2 x^2-\frac {32 x^3}{25 e^4}}{x^2} \, dx \\ & = \int \left (2-\frac {1}{2 x^2}-\frac {32 x}{25 e^4}\right ) \, dx \\ & = \frac {1}{2 x}+2 x-\frac {16 x^2}{25 e^4} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=\frac {1}{2 x}+2 x-\frac {16 x^2}{25 e^4} \]
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Time = 0.63 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61
method | result | size |
risch | \(-\frac {16 \,{\mathrm e}^{-4} x^{2}}{25}+2 x +\frac {1}{2 x}\) | \(17\) |
default | \(2 x +\frac {1}{2 x}+{\mathrm e}^{\ln \left (-\frac {16 x^{2}}{25}-1\right )-4}\) | \(21\) |
norman | \(\frac {\frac {1}{2}+2 x^{2}-\frac {16 \,{\mathrm e}^{-4} x^{3}}{25}}{x}\) | \(21\) |
parts | \(2 x +\frac {1}{2 x}+{\mathrm e}^{\ln \left (-\frac {16 x^{2}}{25}-1\right )-4}\) | \(21\) |
parallelrisch | \(-\frac {-4000000-6553600 x^{6}-3276800 \,{\mathrm e}^{\ln \left (-\frac {16 x^{2}}{25}-1\right )-4} x^{5}-22118400 x^{4}-21120000 x^{2}+8000000 \,{\mathrm e}^{\ln \left (-\frac {16 x^{2}}{25}-1\right )-4} x}{12800 \left (16 x^{2}+25\right )^{2} x}\) | \(62\) |
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Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=-\frac {{\left (32 \, x^{3} - 25 \, {\left (4 \, x^{2} + 1\right )} e^{4}\right )} e^{\left (-4\right )}}{50 \, x} \]
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Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=\frac {- 32 x^{2} + 100 x e^{4} + \frac {25 e^{4}}{x}}{50 e^{4}} \]
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Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=-\frac {2}{25} \, {\left (8 \, x^{2} - 25 \, x e^{4}\right )} e^{\left (-4\right )} + \frac {1}{2 \, x} \]
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Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=-\frac {2}{25} \, {\left (8 \, x^{2} e^{4} - 25 \, x e^{8}\right )} e^{\left (-8\right )} + \frac {1}{2 \, x} \]
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Time = 11.78 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=\frac {-32\,{\mathrm {e}}^{-4}\,x^3+100\,x^2+25}{50\,x} \]
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