Integrand size = 65, antiderivative size = 21 \[ \int \frac {e^{1+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} \left (6-10 e^{\frac {3 x^3}{2}}+2 x+\left (2 x-45 e^{\frac {3 x^3}{2}} x^3\right ) \log (x)\right )}{2 x} \, dx=e^{1+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} \]
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\[ \int \frac {e^{1+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} \left (6-10 e^{\frac {3 x^3}{2}}+2 x+\left (2 x-45 e^{\frac {3 x^3}{2}} x^3\right ) \log (x)\right )}{2 x} \, dx=\int \frac {e^{1+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} \left (6-10 e^{\frac {3 x^3}{2}}+2 x+\left (2 x-45 e^{\frac {3 x^3}{2}} x^3\right ) \log (x)\right )}{2 x} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {e^{1+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} \left (6-10 e^{\frac {3 x^3}{2}}+2 x+\left (2 x-45 e^{\frac {3 x^3}{2}} x^3\right ) \log (x)\right )}{x} \, dx \\ & = \frac {1}{2} \int \left (\frac {2 e^{1+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} (3+x+x \log (x))}{x}-\frac {5 e^{1+\frac {3 x^3}{2}+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} \left (2+9 x^3 \log (x)\right )}{x}\right ) \, dx \\ & = -\left (\frac {5}{2} \int \frac {e^{1+\frac {3 x^3}{2}+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} \left (2+9 x^3 \log (x)\right )}{x} \, dx\right )+\int \frac {e^{1+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} (3+x+x \log (x))}{x} \, dx \\ & = -\left (\frac {5}{2} \int \left (\frac {2 e^{1+\frac {3 x^3}{2}+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)}}{x}+9 e^{1+\frac {3 x^3}{2}+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} x^2 \log (x)\right ) \, dx\right )+\int e x^{2-5 e^{\frac {3 x^3}{2}}+x} (3+x+x \log (x)) \, dx \\ & = -\left (5 \int \frac {e^{1+\frac {3 x^3}{2}+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)}}{x} \, dx\right )-\frac {45}{2} \int e^{1+\frac {3 x^3}{2}+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} x^2 \log (x) \, dx+e \int x^{2-5 e^{\frac {3 x^3}{2}}+x} (3+x+x \log (x)) \, dx \\ & = -\left (5 \int \frac {e^{1+\frac {3 x^3}{2}+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)}}{x} \, dx\right )-\frac {45}{2} \int e^{1+\frac {3 x^3}{2}+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} x^2 \log (x) \, dx+e \int \left (3 x^{2-5 e^{\frac {3 x^3}{2}}+x}+x^{3-5 e^{\frac {3 x^3}{2}}+x}+x^{3-5 e^{\frac {3 x^3}{2}}+x} \log (x)\right ) \, dx \\ & = -\left (5 \int \frac {e^{1+\frac {3 x^3}{2}+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)}}{x} \, dx\right )-\frac {45}{2} \int e^{1+\frac {3 x^3}{2}+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} x^2 \log (x) \, dx+e \int x^{3-5 e^{\frac {3 x^3}{2}}+x} \, dx+e \int x^{3-5 e^{\frac {3 x^3}{2}}+x} \log (x) \, dx+(3 e) \int x^{2-5 e^{\frac {3 x^3}{2}}+x} \, dx \\ & = -\left (5 \int \frac {e^{1+\frac {3 x^3}{2}+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)}}{x} \, dx\right )-\frac {45}{2} \int e^{1+\frac {3 x^3}{2}+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} x^2 \log (x) \, dx+e \int x^{3-5 e^{\frac {3 x^3}{2}}+x} \, dx-e \int \frac {\int x^{3-5 e^{\frac {3 x^3}{2}}+x} \, dx}{x} \, dx+(3 e) \int x^{2-5 e^{\frac {3 x^3}{2}}+x} \, dx+(e \log (x)) \int x^{3-5 e^{\frac {3 x^3}{2}}+x} \, dx \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {e^{1+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} \left (6-10 e^{\frac {3 x^3}{2}}+2 x+\left (2 x-45 e^{\frac {3 x^3}{2}} x^3\right ) \log (x)\right )}{2 x} \, dx=e x^{3-5 e^{\frac {3 x^3}{2}}+x} \]
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Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81
method | result | size |
risch | \(x^{-5 \,{\mathrm e}^{\frac {3 x^{3}}{2}}+3+x} {\mathrm e}\) | \(17\) |
parallelrisch | \({\mathrm e}^{\left (-5 \,{\mathrm e}^{\frac {3 x^{3}}{2}}+3+x \right ) \ln \left (x \right )+1}\) | \(18\) |
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Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {e^{1+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} \left (6-10 e^{\frac {3 x^3}{2}}+2 x+\left (2 x-45 e^{\frac {3 x^3}{2}} x^3\right ) \log (x)\right )}{2 x} \, dx=e^{\left ({\left (x - 5 \, e^{\left (\frac {3}{2} \, x^{3}\right )} + 3\right )} \log \left (x\right ) + 1\right )} \]
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Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{1+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} \left (6-10 e^{\frac {3 x^3}{2}}+2 x+\left (2 x-45 e^{\frac {3 x^3}{2}} x^3\right ) \log (x)\right )}{2 x} \, dx=e^{\left (x - 5 e^{\frac {3 x^{3}}{2}} + 3\right ) \log {\left (x \right )} + 1} \]
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Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} \left (6-10 e^{\frac {3 x^3}{2}}+2 x+\left (2 x-45 e^{\frac {3 x^3}{2}} x^3\right ) \log (x)\right )}{2 x} \, dx=x^{3} e^{\left (x \log \left (x\right ) - 5 \, e^{\left (\frac {3}{2} \, x^{3}\right )} \log \left (x\right ) + 1\right )} \]
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Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} \left (6-10 e^{\frac {3 x^3}{2}}+2 x+\left (2 x-45 e^{\frac {3 x^3}{2}} x^3\right ) \log (x)\right )}{2 x} \, dx=e^{\left (x \log \left (x\right ) - 5 \, e^{\left (\frac {3}{2} \, x^{3}\right )} \log \left (x\right ) + 3 \, \log \left (x\right ) + 1\right )} \]
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Time = 13.41 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+\left (3-5 e^{\frac {3 x^3}{2}}+x\right ) \log (x)} \left (6-10 e^{\frac {3 x^3}{2}}+2 x+\left (2 x-45 e^{\frac {3 x^3}{2}} x^3\right ) \log (x)\right )}{2 x} \, dx=\frac {x^x\,x^3\,\mathrm {e}}{x^{5\,{\mathrm {e}}^{\frac {3\,x^3}{2}}}} \]
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