Integrand size = 36, antiderivative size = 25 \[ \int \frac {-4 e^{4/x}-x^2-50 e^{2 e^x+x} x^2}{5 x^2} \, dx=\frac {1}{5} \left (3-25 e^{2 e^x}+e^{4/x}-x\right ) \]
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Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {12, 14, 2320, 2225, 2240} \[ \int \frac {-4 e^{4/x}-x^2-50 e^{2 e^x+x} x^2}{5 x^2} \, dx=-\frac {x}{5}-5 e^{2 e^x}+\frac {e^{4/x}}{5} \]
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Rule 12
Rule 14
Rule 2225
Rule 2240
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {-4 e^{4/x}-x^2-50 e^{2 e^x+x} x^2}{x^2} \, dx \\ & = \frac {1}{5} \int \left (-50 e^{2 e^x+x}-\frac {4 e^{4/x}+x^2}{x^2}\right ) \, dx \\ & = -\left (\frac {1}{5} \int \frac {4 e^{4/x}+x^2}{x^2} \, dx\right )-10 \int e^{2 e^x+x} \, dx \\ & = -\left (\frac {1}{5} \int \left (1+\frac {4 e^{4/x}}{x^2}\right ) \, dx\right )-10 \text {Subst}\left (\int e^{2 x} \, dx,x,e^x\right ) \\ & = -5 e^{2 e^x}-\frac {x}{5}-\frac {4}{5} \int \frac {e^{4/x}}{x^2} \, dx \\ & = -5 e^{2 e^x}+\frac {e^{4/x}}{5}-\frac {x}{5} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-4 e^{4/x}-x^2-50 e^{2 e^x+x} x^2}{5 x^2} \, dx=-5 e^{2 e^x}+\frac {e^{4/x}}{5}-\frac {x}{5} \]
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Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {{\mathrm e}^{\frac {4}{x}}}{5}-5 \,{\mathrm e}^{2 \,{\mathrm e}^{x}}-\frac {x}{5}\) | \(20\) |
risch | \(\frac {{\mathrm e}^{\frac {4}{x}}}{5}-5 \,{\mathrm e}^{2 \,{\mathrm e}^{x}}-\frac {x}{5}\) | \(20\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {4}{x}}}{5}-5 \,{\mathrm e}^{2 \,{\mathrm e}^{x}}-\frac {x}{5}\) | \(22\) |
parts | \(\frac {{\mathrm e}^{\frac {4}{x}}}{5}-5 \,{\mathrm e}^{2 \,{\mathrm e}^{x}}-\frac {x}{5}\) | \(22\) |
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Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-4 e^{4/x}-x^2-50 e^{2 e^x+x} x^2}{5 x^2} \, dx=-\frac {1}{5} \, {\left ({\left (x - e^{\frac {4}{x}}\right )} e^{x} + 25 \, e^{\left (x + 2 \, e^{x}\right )}\right )} e^{\left (-x\right )} \]
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Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {-4 e^{4/x}-x^2-50 e^{2 e^x+x} x^2}{5 x^2} \, dx=- \frac {x}{5} + \frac {e^{\frac {4}{x}}}{5} - 5 e^{2 e^{x}} \]
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Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {-4 e^{4/x}-x^2-50 e^{2 e^x+x} x^2}{5 x^2} \, dx=-\frac {1}{5} \, x - 5 \, e^{\left (2 \, e^{x}\right )} + \frac {1}{5} \, e^{\frac {4}{x}} \]
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-4 e^{4/x}-x^2-50 e^{2 e^x+x} x^2}{5 x^2} \, dx=-\frac {1}{5} \, {\left (x e^{x} + 25 \, e^{\left (x + 2 \, e^{x}\right )}\right )} e^{\left (-x\right )} + \frac {1}{5} \, e^{\frac {4}{x}} \]
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Time = 9.59 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {-4 e^{4/x}-x^2-50 e^{2 e^x+x} x^2}{5 x^2} \, dx=\frac {{\mathrm {e}}^{4/x}}{5}-\frac {x}{5}-5\,{\mathrm {e}}^{2\,{\mathrm {e}}^x} \]
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