Integrand size = 85, antiderivative size = 23 \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=e^{-1+e^{x^2}} \left (4-2 x+\log \left (\left (5+e^x\right )^2\right )\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(74\) vs. \(2(23)=46\).
Time = 0.38 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.22, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6820, 12, 2326} \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=\frac {e^{-x^2+e^{x^2}-1} \left (10 e^{x^2} (2-x) x+2 e^{x^2+x} (2-x) x+e^{x^2} \left (e^x+5\right ) x \log \left (\left (e^x+5\right )^2\right )\right )}{\left (e^x+5\right ) x} \]
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Rule 12
Rule 2326
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {2 e^{-1+e^{x^2}} \left (-5-10 e^{x^2} (-2+x) x-2 e^{x+x^2} (-2+x) x+e^{x^2} \left (5+e^x\right ) x \log \left (\left (5+e^x\right )^2\right )\right )}{5+e^x} \, dx \\ & = 2 \int \frac {e^{-1+e^{x^2}} \left (-5-10 e^{x^2} (-2+x) x-2 e^{x+x^2} (-2+x) x+e^{x^2} \left (5+e^x\right ) x \log \left (\left (5+e^x\right )^2\right )\right )}{5+e^x} \, dx \\ & = \frac {e^{-1+e^{x^2}-x^2} \left (10 e^{x^2} (2-x) x+2 e^{x+x^2} (2-x) x+e^{x^2} \left (5+e^x\right ) x \log \left (\left (5+e^x\right )^2\right )\right )}{\left (5+e^x\right ) x} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=e^{-1+e^{x^2}} \left (4-2 x+\log \left (\left (5+e^x\right )^2\right )\right ) \]
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Time = 0.36 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74
method | result | size |
parallelrisch | \(-2 \,{\mathrm e}^{{\mathrm e}^{x^{2}}-1} x +\ln \left ({\mathrm e}^{2 x}+10 \,{\mathrm e}^{x}+25\right ) {\mathrm e}^{{\mathrm e}^{x^{2}}-1}+4 \,{\mathrm e}^{{\mathrm e}^{x^{2}}-1}\) | \(40\) |
risch | \(2 \,{\mathrm e}^{{\mathrm e}^{x^{2}}-1} \ln \left ({\mathrm e}^{x}+5\right )-\frac {\left (i \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5\right )\right )}^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5\right )^{2}\right )-2 i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5\right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5\right )^{2}\right )}^{2}+i \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5\right )^{2}\right )}^{3}+4 x -8\right ) {\mathrm e}^{{\mathrm e}^{x^{2}}-1}}{2}\) | \(94\) |
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Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04 \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=-{\left (2 \, {\left (x - 2\right )} e^{\left (x^{2} + e^{\left (x^{2}\right )} - 1\right )} - e^{\left (x^{2} + e^{\left (x^{2}\right )} - 1\right )} \log \left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right )\right )} e^{\left (-x^{2}\right )} \]
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Timed out. \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=\text {Timed out} \]
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Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=-2 \, {\left (x - \log \left (e^{x} + 5\right ) - 2\right )} e^{\left (e^{\left (x^{2}\right )} - 1\right )} \]
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\[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=\int { \frac {2 \, {\left ({\left (x e^{x} + 5 \, x\right )} e^{\left (x^{2} + e^{\left (x^{2}\right )} - 1\right )} \log \left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right ) - {\left (2 \, {\left (5 \, x^{2} + {\left (x^{2} - 2 \, x\right )} e^{x} - 10 \, x\right )} e^{\left (x^{2}\right )} + 5\right )} e^{\left (e^{\left (x^{2}\right )} - 1\right )}\right )}}{e^{x} + 5} \,d x } \]
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Time = 8.42 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx={\mathrm {e}}^{{\mathrm {e}}^{x^2}-1}\,\left (\ln \left ({\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^x+25\right )-2\,x+4\right ) \]
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