Integrand size = 96, antiderivative size = 22 \[ \int \frac {-40 e^x+8 e^{2 x}+e^{\frac {x^2}{8}} x+e^{\frac {x^2}{16}} \left (e^x (-8-x)+5 x\right )}{100-40 e^x+4 e^{2 x}+4 e^{\frac {x^2}{8}}+e^{\frac {x^2}{16}} \left (40-8 e^x\right )+4 \log (4)} \, dx=\log \left (\left (5-e^x+e^{\frac {x^2}{16}}\right )^2+\log (4)\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(54\) vs. \(2(22)=44\).
Time = 0.54 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.45, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {6873, 12, 6816} \[ \int \frac {-40 e^x+8 e^{2 x}+e^{\frac {x^2}{8}} x+e^{\frac {x^2}{16}} \left (e^x (-8-x)+5 x\right )}{100-40 e^x+4 e^{2 x}+4 e^{\frac {x^2}{8}}+e^{\frac {x^2}{16}} \left (40-8 e^x\right )+4 \log (4)} \, dx=\log \left (-10 e^{\frac {x^2}{16}}-e^{\frac {x^2}{8}}+2 e^{\frac {x^2}{16}+x}+10 e^x-e^{2 x}-25-\log (4)\right ) \]
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Rule 12
Rule 6816
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (5-e^x+e^{\frac {x^2}{16}}\right ) \left (8 e^x-e^{\frac {x^2}{16}} x\right )}{40 e^x-4 e^{2 x}-4 e^{\frac {x^2}{8}}-e^{\frac {x^2}{16}} \left (40-8 e^x\right )-100 \left (1+\frac {2 \log (2)}{25}\right )} \, dx \\ & = \int \frac {\left (5-e^x+e^{\frac {x^2}{16}}\right ) \left (8 e^x-e^{\frac {x^2}{16}} x\right )}{4 \left (10 e^x-e^{2 x}-10 e^{\frac {x^2}{16}}-e^{\frac {x^2}{8}}+2 e^{x+\frac {x^2}{16}}-25 \left (1+\frac {2 \log (2)}{25}\right )\right )} \, dx \\ & = \frac {1}{4} \int \frac {\left (5-e^x+e^{\frac {x^2}{16}}\right ) \left (8 e^x-e^{\frac {x^2}{16}} x\right )}{10 e^x-e^{2 x}-10 e^{\frac {x^2}{16}}-e^{\frac {x^2}{8}}+2 e^{x+\frac {x^2}{16}}-25 \left (1+\frac {2 \log (2)}{25}\right )} \, dx \\ & = \log \left (-25+10 e^x-e^{2 x}-10 e^{\frac {x^2}{16}}-e^{\frac {x^2}{8}}+2 e^{x+\frac {x^2}{16}}-\log (4)\right ) \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(48\) vs. \(2(22)=44\).
Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.18 \[ \int \frac {-40 e^x+8 e^{2 x}+e^{\frac {x^2}{8}} x+e^{\frac {x^2}{16}} \left (e^x (-8-x)+5 x\right )}{100-40 e^x+4 e^{2 x}+4 e^{\frac {x^2}{8}}+e^{\frac {x^2}{16}} \left (40-8 e^x\right )+4 \log (4)} \, dx=\log \left (25-10 e^x+e^{2 x}+10 e^{\frac {x^2}{16}}+e^{\frac {x^2}{8}}-2 e^{x+\frac {x^2}{16}}+\log (4)\right ) \]
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Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59
method | result | size |
risch | \(\ln \left ({\mathrm e}^{\frac {x^{2}}{8}}+\left (-2 \,{\mathrm e}^{x}+10\right ) {\mathrm e}^{\frac {x^{2}}{16}}+{\mathrm e}^{2 x}+2 \ln \left (2\right )-10 \,{\mathrm e}^{x}+25\right )\) | \(35\) |
parallelrisch | \(\ln \left (-2 \,{\mathrm e}^{x} {\mathrm e}^{\frac {x^{2}}{16}}-10 \,{\mathrm e}^{x}+2 \ln \left (2\right )+10 \,{\mathrm e}^{\frac {x^{2}}{16}}+{\mathrm e}^{\frac {x^{2}}{8}}+{\mathrm e}^{2 x}+25\right )\) | \(42\) |
norman | \(\ln \left (4 \,{\mathrm e}^{2 x}-8 \,{\mathrm e}^{x} {\mathrm e}^{\frac {x^{2}}{16}}+4 \,{\mathrm e}^{\frac {x^{2}}{8}}-40 \,{\mathrm e}^{x}+8 \ln \left (2\right )+40 \,{\mathrm e}^{\frac {x^{2}}{16}}+100\right )\) | \(46\) |
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Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {-40 e^x+8 e^{2 x}+e^{\frac {x^2}{8}} x+e^{\frac {x^2}{16}} \left (e^x (-8-x)+5 x\right )}{100-40 e^x+4 e^{2 x}+4 e^{\frac {x^2}{8}}+e^{\frac {x^2}{16}} \left (40-8 e^x\right )+4 \log (4)} \, dx=\log \left (-2 \, {\left (e^{x} - 5\right )} e^{\left (\frac {1}{16} \, x^{2}\right )} + e^{\left (\frac {1}{8} \, x^{2}\right )} + e^{\left (2 \, x\right )} - 10 \, e^{x} + 2 \, \log \left (2\right ) + 25\right ) \]
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Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-40 e^x+8 e^{2 x}+e^{\frac {x^2}{8}} x+e^{\frac {x^2}{16}} \left (e^x (-8-x)+5 x\right )}{100-40 e^x+4 e^{2 x}+4 e^{\frac {x^2}{8}}+e^{\frac {x^2}{16}} \left (40-8 e^x\right )+4 \log (4)} \, dx=\log {\left (\left (10 - 2 e^{x}\right ) e^{\frac {x^{2}}{16}} + e^{2 x} - 10 e^{x} + e^{\frac {x^{2}}{8}} + 2 \log {\left (2 \right )} + 25 \right )} \]
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Time = 0.33 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {-40 e^x+8 e^{2 x}+e^{\frac {x^2}{8}} x+e^{\frac {x^2}{16}} \left (e^x (-8-x)+5 x\right )}{100-40 e^x+4 e^{2 x}+4 e^{\frac {x^2}{8}}+e^{\frac {x^2}{16}} \left (40-8 e^x\right )+4 \log (4)} \, dx=\log \left (-2 \, {\left (e^{x} - 5\right )} e^{\left (\frac {1}{16} \, x^{2}\right )} + e^{\left (\frac {1}{8} \, x^{2}\right )} + e^{\left (2 \, x\right )} - 10 \, e^{x} + 2 \, \log \left (2\right ) + 25\right ) \]
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Time = 0.32 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {-40 e^x+8 e^{2 x}+e^{\frac {x^2}{8}} x+e^{\frac {x^2}{16}} \left (e^x (-8-x)+5 x\right )}{100-40 e^x+4 e^{2 x}+4 e^{\frac {x^2}{8}}+e^{\frac {x^2}{16}} \left (40-8 e^x\right )+4 \log (4)} \, dx=\log \left (e^{\left (\frac {1}{8} \, x^{2}\right )} + 10 \, e^{\left (\frac {1}{16} \, x^{2}\right )} - 2 \, e^{\left (\frac {1}{16} \, x^{2} + x\right )} + e^{\left (2 \, x\right )} - 10 \, e^{x} + 2 \, \log \left (2\right ) + 25\right ) \]
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Time = 9.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-40 e^x+8 e^{2 x}+e^{\frac {x^2}{8}} x+e^{\frac {x^2}{16}} \left (e^x (-8-x)+5 x\right )}{100-40 e^x+4 e^{2 x}+4 e^{\frac {x^2}{8}}+e^{\frac {x^2}{16}} \left (40-8 e^x\right )+4 \log (4)} \, dx=\ln \left ({\mathrm {e}}^{2\,x}+\ln \left (4\right )-2\,{\mathrm {e}}^{\frac {x^2}{16}+x}+{\mathrm {e}}^{\frac {x^2}{8}}+10\,{\mathrm {e}}^{\frac {x^2}{16}}-10\,{\mathrm {e}}^x+25\right ) \]
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