Integrand size = 68, antiderivative size = 24 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=e^x \left (-\log (x)+e^3 \left (4 x-\log \left (\log \left (x^2\right )\right )\right )\right ) \]
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\[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=\int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (4 e^{3+x}-\frac {e^x}{x}+4 e^{3+x} x-e^x \log (x)-\frac {2 e^{3+x}}{x \log \left (x^2\right )}-e^{3+x} \log \left (\log \left (x^2\right )\right )\right ) \, dx \\ & = -\left (2 \int \frac {e^{3+x}}{x \log \left (x^2\right )} \, dx\right )+4 \int e^{3+x} \, dx+4 \int e^{3+x} x \, dx-\int \frac {e^x}{x} \, dx-\int e^x \log (x) \, dx-\int e^{3+x} \log \left (\log \left (x^2\right )\right ) \, dx \\ & = 4 e^{3+x}+4 e^{3+x} x-\text {Ei}(x)-e^x \log (x)-2 \int \frac {e^{3+x}}{x \log \left (x^2\right )} \, dx-4 \int e^{3+x} \, dx+\int \frac {e^x}{x} \, dx-\int e^{3+x} \log \left (\log \left (x^2\right )\right ) \, dx \\ & = 4 e^{3+x} x-e^x \log (x)-2 \int \frac {e^{3+x}}{x \log \left (x^2\right )} \, dx-\int e^{3+x} \log \left (\log \left (x^2\right )\right ) \, dx \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=e^x \left (-\log (x)+e^3 \left (4 x-\log \left (\log \left (x^2\right )\right )\right )\right ) \]
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Time = 7.47 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(4 x \,{\mathrm e}^{3} {\mathrm e}^{x}-{\mathrm e}^{3} {\mathrm e}^{x} \ln \left (\ln \left (x^{2}\right )\right )-{\mathrm e}^{x} \ln \left (x \right )\) | \(26\) |
risch | \(4 \,{\mathrm e}^{3+x} x -\ln \left (2 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right ) {\mathrm e}^{3+x}-{\mathrm e}^{x} \ln \left (x \right )\) | \(55\) |
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx={\left (4 \, x e^{\left (x + 6\right )} - e^{\left (x + 3\right )} \log \left (x\right ) - e^{\left (x + 6\right )} \log \left (2 \, \log \left (x\right )\right )\right )} e^{\left (-3\right )} \]
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Time = 7.62 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=\left (4 x e^{3} - \log {\left (x \right )} - e^{3} \log {\left (2 \log {\left (x \right )} \right )}\right ) e^{x} \]
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Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=4 \, {\left (x e^{3} - e^{3}\right )} e^{x} - e^{\left (x + 3\right )} \log \left (2\right ) - e^{x} \log \left (x\right ) - e^{\left (x + 3\right )} \log \left (\log \left (x\right )\right ) + 4 \, e^{\left (x + 3\right )} \]
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Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=4 \, x e^{\left (x + 3\right )} - e^{x} \log \left (x\right ) - e^{\left (x + 3\right )} \log \left (\log \left (x^{2}\right )\right ) \]
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Time = 11.85 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=4\,x\,{\mathrm {e}}^3\,{\mathrm {e}}^x-{\mathrm {e}}^x\,\ln \left (x\right )-{\mathrm {e}}^3\,{\mathrm {e}}^x\,\ln \left (\ln \left (x^2\right )\right ) \]
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