Integrand size = 34, antiderivative size = 25 \[ \int \left (8 x-139 x^2+e^{e^x} \left (-2 x-e^x x^2\right )+51 x^2 \log (x)\right ) \, dx=x^2 \left (4-e^{e^x}-x-17 x (3-\log (x))\right ) \]
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Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2326, 2341} \[ \int \left (8 x-139 x^2+e^{e^x} \left (-2 x-e^x x^2\right )+51 x^2 \log (x)\right ) \, dx=-52 x^3+17 x^3 \log (x)-e^{e^x} x^2+4 x^2 \]
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Rule 2326
Rule 2341
Rubi steps \begin{align*} \text {integral}& = 4 x^2-\frac {139 x^3}{3}+51 \int x^2 \log (x) \, dx+\int e^{e^x} \left (-2 x-e^x x^2\right ) \, dx \\ & = 4 x^2-e^{e^x} x^2-52 x^3+17 x^3 \log (x) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \left (8 x-139 x^2+e^{e^x} \left (-2 x-e^x x^2\right )+51 x^2 \log (x)\right ) \, dx=x^2 \left (4-e^{e^x}-52 x+17 x \log (x)\right ) \]
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Time = 0.57 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08
method | result | size |
default | \(-{\mathrm e}^{{\mathrm e}^{x}} x^{2}+4 x^{2}-52 x^{3}+17 x^{3} \ln \left (x \right )\) | \(27\) |
risch | \(-{\mathrm e}^{{\mathrm e}^{x}} x^{2}+4 x^{2}-52 x^{3}+17 x^{3} \ln \left (x \right )\) | \(27\) |
parallelrisch | \(-{\mathrm e}^{{\mathrm e}^{x}} x^{2}+4 x^{2}-52 x^{3}+17 x^{3} \ln \left (x \right )\) | \(27\) |
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \left (8 x-139 x^2+e^{e^x} \left (-2 x-e^x x^2\right )+51 x^2 \log (x)\right ) \, dx=17 \, x^{3} \log \left (x\right ) - 52 \, x^{3} - x^{2} e^{\left (e^{x}\right )} + 4 \, x^{2} \]
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Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \left (8 x-139 x^2+e^{e^x} \left (-2 x-e^x x^2\right )+51 x^2 \log (x)\right ) \, dx=17 x^{3} \log {\left (x \right )} - 52 x^{3} - x^{2} e^{e^{x}} + 4 x^{2} \]
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Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \left (8 x-139 x^2+e^{e^x} \left (-2 x-e^x x^2\right )+51 x^2 \log (x)\right ) \, dx=17 \, x^{3} \log \left (x\right ) - 52 \, x^{3} - x^{2} e^{\left (e^{x}\right )} + 4 \, x^{2} \]
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \left (8 x-139 x^2+e^{e^x} \left (-2 x-e^x x^2\right )+51 x^2 \log (x)\right ) \, dx=17 \, x^{3} \log \left (x\right ) - 52 \, x^{3} - x^{2} e^{\left (e^{x}\right )} + 4 \, x^{2} \]
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Time = 10.44 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \left (8 x-139 x^2+e^{e^x} \left (-2 x-e^x x^2\right )+51 x^2 \log (x)\right ) \, dx=-x^2\,\left (52\,x+{\mathrm {e}}^{{\mathrm {e}}^x}-17\,x\,\ln \left (x\right )-4\right ) \]
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