Integrand size = 215, antiderivative size = 24 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx=e^{x^2}+\frac {5}{-5+\log \left (e^{e^{3+x}}+\frac {1}{x}\right )} \]
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Time = 1.88 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6820, 6874, 2240, 6818} \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx=e^{x^2}-\frac {5}{5-\log \left (e^{e^{x+3}}+\frac {1}{x}\right )} \]
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Rule 2240
Rule 6818
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {5+50 e^{x^2} x^2-5 e^{3+e^{3+x}+x} x^2+50 e^{e^{3+x}+x^2} x^3-20 e^{x^2} x^2 \left (1+e^{e^{3+x}} x\right ) \log \left (e^{e^{3+x}}+\frac {1}{x}\right )+2 e^{x^2} x^2 \left (1+e^{e^{3+x}} x\right ) \log ^2\left (e^{e^{3+x}}+\frac {1}{x}\right )}{x \left (1+e^{e^{3+x}} x\right ) \left (5-\log \left (e^{e^{3+x}}+\frac {1}{x}\right )\right )^2} \, dx \\ & = \int \left (2 e^{x^2} x-\frac {5 \left (-1+e^{3+e^{3+x}+x} x^2\right )}{x \left (1+e^{e^{3+x}} x\right ) \left (-5+\log \left (e^{e^{3+x}}+\frac {1}{x}\right )\right )^2}\right ) \, dx \\ & = 2 \int e^{x^2} x \, dx-5 \int \frac {-1+e^{3+e^{3+x}+x} x^2}{x \left (1+e^{e^{3+x}} x\right ) \left (-5+\log \left (e^{e^{3+x}}+\frac {1}{x}\right )\right )^2} \, dx \\ & = e^{x^2}-\frac {5}{5-\log \left (e^{e^{3+x}}+\frac {1}{x}\right )} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx=e^{x^2}+\frac {5}{-5+\log \left (e^{e^{3+x}}+\frac {1}{x}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(63\) vs. \(2(21)=42\).
Time = 127.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.67
method | result | size |
parallelrisch | \(-\frac {-10 \ln \left (\frac {x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1}{x}\right ) {\mathrm e}^{x^{2}}+50 \,{\mathrm e}^{x^{2}}-10 \ln \left (\frac {x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1}{x}\right )}{10 \left (\ln \left (\frac {x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1}{x}\right )-5\right )}\) | \(64\) |
risch | \({\mathrm e}^{x^{2}}-\frac {10 i}{\pi \,\operatorname {csgn}\left (i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )}{x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )\right ) \operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )-\pi {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )}{x}\right )}^{3}+\pi {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 i \ln \left (x \right )-2 i \ln \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )+10 i}\) | \(152\) |
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Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (21) = 42\).
Time = 0.42 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.00 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx=\frac {e^{\left (x^{2}\right )} \log \left (\frac {{\left (x e^{\left (x^{2} + e^{\left (x + 3\right )}\right )} + e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}}{x}\right ) - 5 \, e^{\left (x^{2}\right )} + 5}{\log \left (\frac {{\left (x e^{\left (x^{2} + e^{\left (x + 3\right )}\right )} + e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}}{x}\right ) - 5} \]
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Time = 0.54 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx=e^{x^{2}} + \frac {5}{\log {\left (\frac {x e^{e^{x + 3}} + 1}{x} \right )} - 5} \]
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Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (21) = 42\).
Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx=-\frac {{\left (\log \left (x\right ) + 5\right )} e^{\left (x^{2}\right )} - e^{\left (x^{2}\right )} \log \left (x e^{\left (e^{\left (x + 3\right )}\right )} + 1\right ) - 5}{\log \left (x e^{\left (e^{\left (x + 3\right )}\right )} + 1\right ) - \log \left (x\right ) - 5} \]
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Leaf count of result is larger than twice the leaf count of optimal. 932 vs. \(2 (21) = 42\).
Time = 0.53 (sec) , antiderivative size = 932, normalized size of antiderivative = 38.83 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx=\text {Too large to display} \]
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Time = 12.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx={\mathrm {e}}^{x^2}+\frac {5}{\ln \left (\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^3\,{\mathrm {e}}^x}+1}{x}\right )-5} \]
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