Integrand size = 90, antiderivative size = 22 \[ \int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{-3 x-x^2+e^{-2+2^x x^{2+x}} (3+x)+(-3-x) \log (3+x)} \, dx=\log \left (-e^{-2+2^x x^{2+x}}+x+\log (3+x)\right ) \]
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\[ \int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{-3 x-x^2+e^{-2+2^x x^{2+x}} (3+x)+(-3-x) \log (3+x)} \, dx=\int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{-3 x-x^2+e^{-2+2^x x^{2+x}} (3+x)+(-3-x) \log (3+x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^2 \left (-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )\right )}{(3+x) \left (e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)\right )} \, dx \\ & = e^2 \int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{(3+x) \left (e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)\right )} \, dx \\ & = e^2 \int \left (\frac {2^x e^{-2+2^x x^{2+x}} x^{1+x} (2+x+x \log (2 x))}{e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)}+\frac {4+x}{(3+x) \left (-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)\right )}\right ) \, dx \\ & = e^2 \int \frac {2^x e^{-2+2^x x^{2+x}} x^{1+x} (2+x+x \log (2 x))}{e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)} \, dx+e^2 \int \frac {4+x}{(3+x) \left (-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)\right )} \, dx \\ & = e^2 \int \left (-\frac {1}{e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)}+\frac {1}{(3+x) \left (-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)\right )}\right ) \, dx+e^2 \int \left (\frac {2^{1+x} e^{-2+2^x x^{2+x}} x^{1+x}}{e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)}-\frac {2^x e^{-2+2^x x^{2+x}} x^{2+x}}{-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)}-\frac {2^x e^{-2+2^x x^{2+x}} x^{2+x} \log (2 x)}{-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)}\right ) \, dx \\ & = -\left (e^2 \int \frac {1}{e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)} \, dx\right )+e^2 \int \frac {2^{1+x} e^{-2+2^x x^{2+x}} x^{1+x}}{e^{2^x x^{2+x}}-e^2 x-e^2 \log (3+x)} \, dx-e^2 \int \frac {2^x e^{-2+2^x x^{2+x}} x^{2+x}}{-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)} \, dx+e^2 \int \frac {1}{(3+x) \left (-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)\right )} \, dx-e^2 \int \frac {2^x e^{-2+2^x x^{2+x}} x^{2+x} \log (2 x)}{-e^{2^x x^{2+x}}+e^2 x+e^2 \log (3+x)} \, dx \\ \end{align*}
\[ \int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{-3 x-x^2+e^{-2+2^x x^{2+x}} (3+x)+(-3-x) \log (3+x)} \, dx=\int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{-3 x-x^2+e^{-2+2^x x^{2+x}} (3+x)+(-3-x) \log (3+x)} \, dx \]
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Time = 62.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09
method | result | size |
parallelrisch | \(\ln \left (x -{\mathrm e}^{x^{2} {\mathrm e}^{x \ln \left (2 x \right )}-2}+\ln \left (3+x \right )\right )\) | \(24\) |
risch | \(2+\ln \left (-x +{\mathrm e}^{x^{2} \left (2 x \right )^{x}-2}-\ln \left (3+x \right )\right )\) | \(26\) |
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Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{-3 x-x^2+e^{-2+2^x x^{2+x}} (3+x)+(-3-x) \log (3+x)} \, dx=\log \left (-x + e^{\left (\left (2 \, x\right )^{x} x^{2} - 2\right )} - \log \left (x + 3\right )\right ) \]
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Time = 0.74 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{-3 x-x^2+e^{-2+2^x x^{2+x}} (3+x)+(-3-x) \log (3+x)} \, dx=\log {\left (- x + e^{x^{2} e^{x \log {\left (2 x \right )}} - 2} - \log {\left (x + 3 \right )} \right )} \]
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Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{-3 x-x^2+e^{-2+2^x x^{2+x}} (3+x)+(-3-x) \log (3+x)} \, dx=\log \left (-x e^{2} - e^{2} \log \left (x + 3\right ) + e^{\left (x^{2} e^{\left (x \log \left (2\right ) + x \log \left (x\right )\right )}\right )}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (21) = 42\).
Time = 3.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.32 \[ \int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{-3 x-x^2+e^{-2+2^x x^{2+x}} (3+x)+(-3-x) \log (3+x)} \, dx=-\left (2 \, x\right )^{x} x^{3} \log \left (2 \, x\right ) - 2 \, \left (2 \, x\right )^{x} x^{2} \log \left (2 \, x\right ) + \left (2 \, x\right )^{x} x^{2} + \log \left (-2 \, x e^{2} + 2 \, e^{2} \log \left (2\right ) - 2 \, e^{2} \log \left (2 \, x + 6\right ) + 2 \, e^{\left (\left (2 \, x\right )^{x} x^{2}\right )}\right ) \]
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Time = 9.93 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-4-x+2^x e^{-2+2^x x^{2+x}} x^x \left (6 x+5 x^2+x^3+\left (3 x^2+x^3\right ) \log (2 x)\right )}{-3 x-x^2+e^{-2+2^x x^{2+x}} (3+x)+(-3-x) \log (3+x)} \, dx=\ln \left (x+\ln \left (x+3\right )-{\mathrm {e}}^{2^x\,x^{x+2}-2}\right ) \]
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