Integrand size = 25, antiderivative size = 22 \[ \int \frac {1}{16} \left (e^3 (-80+40 x)+e^3 (-37-80 x) \log (x)\right ) \, dx=e^3 \left (-\frac {37}{8}-5 x\right ) \left (1+\frac {1}{2} x (-1+\log (x))\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(55\) vs. \(2(22)=44\).
Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.50, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {12, 2350} \[ \int \frac {1}{16} \left (e^3 (-80+40 x)+e^3 (-37-80 x) \log (x)\right ) \, dx=\frac {5 e^3 x^2}{4}-\frac {5}{2} e^3 x^2 \log (x)+\frac {5}{4} e^3 (2-x)^2+\frac {37 e^3 x}{16}-\frac {37}{16} e^3 x \log (x) \]
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Rule 12
Rule 2350
Rubi steps \begin{align*} \text {integral}& = \frac {1}{16} \int \left (e^3 (-80+40 x)+e^3 (-37-80 x) \log (x)\right ) \, dx \\ & = \frac {5}{4} e^3 (2-x)^2+\frac {1}{16} e^3 \int (-37-80 x) \log (x) \, dx \\ & = \frac {5}{4} e^3 (2-x)^2-\frac {37}{16} e^3 x \log (x)-\frac {5}{2} e^3 x^2 \log (x)-\frac {1}{16} e^3 \int (-37-40 x) \, dx \\ & = \frac {5}{4} e^3 (2-x)^2+\frac {37 e^3 x}{16}+\frac {5 e^3 x^2}{4}-\frac {37}{16} e^3 x \log (x)-\frac {5}{2} e^3 x^2 \log (x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86 \[ \int \frac {1}{16} \left (e^3 (-80+40 x)+e^3 (-37-80 x) \log (x)\right ) \, dx=-\frac {43 e^3 x}{16}+\frac {5 e^3 x^2}{2}-\frac {37}{16} e^3 x \log (x)-\frac {5}{2} e^3 x^2 \log (x) \]
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Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32
method | result | size |
risch | \(\frac {{\mathrm e}^{3} \left (-40 x^{2}-37 x \right ) \ln \left (x \right )}{16}+\frac {5 x^{2} {\mathrm e}^{3}}{2}-\frac {43 x \,{\mathrm e}^{3}}{16}\) | \(29\) |
norman | \(-\frac {43 x \,{\mathrm e}^{3}}{16}+\frac {5 x^{2} {\mathrm e}^{3}}{2}-\frac {37 x \,{\mathrm e}^{3} \ln \left (x \right )}{16}-\frac {5 \,{\mathrm e}^{3} \ln \left (x \right ) x^{2}}{2}\) | \(30\) |
parallelrisch | \(-\frac {43 x \,{\mathrm e}^{3}}{16}+\frac {5 x^{2} {\mathrm e}^{3}}{2}-\frac {37 x \,{\mathrm e}^{3} \ln \left (x \right )}{16}-\frac {5 \,{\mathrm e}^{3} \ln \left (x \right ) x^{2}}{2}\) | \(30\) |
default | \(\frac {5 \,{\mathrm e}^{3} \left (-2 x +\frac {1}{2} x^{2}\right )}{2}+\frac {{\mathrm e}^{3} \left (-40 x^{2} \ln \left (x \right )+20 x^{2}-37 x \ln \left (x \right )+37 x \right )}{16}\) | \(40\) |
parts | \(\frac {5 \,{\mathrm e}^{3} \left (-2 x +\frac {1}{2} x^{2}\right )}{2}-\frac {{\mathrm e}^{3} \left (40 x^{2} \ln \left (x \right )-20 x^{2}+37 x \ln \left (x \right )-37 x \right )}{16}\) | \(40\) |
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none
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {1}{16} \left (e^3 (-80+40 x)+e^3 (-37-80 x) \log (x)\right ) \, dx=-\frac {1}{16} \, {\left (40 \, x^{2} + 37 \, x\right )} e^{3} \log \left (x\right ) + \frac {1}{16} \, {\left (40 \, x^{2} - 43 \, x\right )} e^{3} \]
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Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86 \[ \int \frac {1}{16} \left (e^3 (-80+40 x)+e^3 (-37-80 x) \log (x)\right ) \, dx=\frac {5 x^{2} e^{3}}{2} - \frac {43 x e^{3}}{16} + \left (- \frac {5 x^{2} e^{3}}{2} - \frac {37 x e^{3}}{16}\right ) \log {\left (x \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (17) = 34\).
Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73 \[ \int \frac {1}{16} \left (e^3 (-80+40 x)+e^3 (-37-80 x) \log (x)\right ) \, dx=\frac {1}{16} \, {\left (20 \, x^{2} - {\left (40 \, x^{2} + 37 \, x\right )} \log \left (x\right ) + 37 \, x\right )} e^{3} + \frac {5}{4} \, {\left (x^{2} - 4 \, x\right )} e^{3} \]
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Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (17) = 34\).
Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {1}{16} \left (e^3 (-80+40 x)+e^3 (-37-80 x) \log (x)\right ) \, dx=-\frac {1}{16} \, {\left (40 \, x^{2} \log \left (x\right ) - 20 \, x^{2} + 37 \, x \log \left (x\right ) - 37 \, x\right )} e^{3} + \frac {5}{4} \, {\left (x^{2} - 4 \, x\right )} e^{3} \]
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Time = 12.67 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1}{16} \left (e^3 (-80+40 x)+e^3 (-37-80 x) \log (x)\right ) \, dx=-\frac {x\,{\mathrm {e}}^3\,\left (37\,\ln \left (x\right )-40\,x+40\,x\,\ln \left (x\right )+43\right )}{16} \]
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