Integrand size = 39, antiderivative size = 29 \[ \int \left (12-16 x+6 x^2+e^{3/4} \left (4-8 x+3 x^2\right )+\left (4-8 x+3 x^2\right ) \log (2)\right ) \, dx=x \left (x-(2-x)^2 \left (-2-e^{3/4}+\frac {1}{x}-\log (2)\right )\right ) \]
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Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.72, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6} \[ \int \left (12-16 x+6 x^2+e^{3/4} \left (4-8 x+3 x^2\right )+\left (4-8 x+3 x^2\right ) \log (2)\right ) \, dx=2 x^3+x^3 \left (e^{3/4}+\log (2)\right )-8 x^2-4 x^2 \left (e^{3/4}+\log (2)\right )+12 x+4 x \left (e^{3/4}+\log (2)\right ) \]
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Rule 6
Rubi steps \begin{align*} \text {integral}& = \int \left (12-16 x+6 x^2+\left (4-8 x+3 x^2\right ) \left (e^{3/4}+\log (2)\right )\right ) \, dx \\ & = 12 x-8 x^2+2 x^3+\left (e^{3/4}+\log (2)\right ) \int \left (4-8 x+3 x^2\right ) \, dx \\ & = 12 x-8 x^2+2 x^3+4 x \left (e^{3/4}+\log (2)\right )-4 x^2 \left (e^{3/4}+\log (2)\right )+x^3 \left (e^{3/4}+\log (2)\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \left (12-16 x+6 x^2+e^{3/4} \left (4-8 x+3 x^2\right )+\left (4-8 x+3 x^2\right ) \log (2)\right ) \, dx=\frac {1}{3} x \left (36+3 e^{3/4} (-2+x)^2-12 x (2+\log (2))+x^2 (6+\log (8))+\log (4096)\right ) \]
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Time = 0.11 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31
method | result | size |
norman | \(\left ({\mathrm e}^{\frac {3}{4}}+\ln \left (2\right )+2\right ) x^{3}+\left (-4 \ln \left (2\right )-4 \,{\mathrm e}^{\frac {3}{4}}-8\right ) x^{2}+\left (4 \ln \left (2\right )+4 \,{\mathrm e}^{\frac {3}{4}}+12\right ) x\) | \(38\) |
gosper | \(x \left (x^{2} \ln \left (2\right )+x^{2} {\mathrm e}^{\frac {3}{4}}-4 x \ln \left (2\right )-4 x \,{\mathrm e}^{\frac {3}{4}}+2 x^{2}+4 \ln \left (2\right )+4 \,{\mathrm e}^{\frac {3}{4}}-8 x +12\right )\) | \(43\) |
default | \(x^{3} \ln \left (2\right )+x^{3} {\mathrm e}^{\frac {3}{4}}-4 x^{2} \ln \left (2\right )-4 x^{2} {\mathrm e}^{\frac {3}{4}}+2 x^{3}+4 x \ln \left (2\right )+4 x \,{\mathrm e}^{\frac {3}{4}}-8 x^{2}+12 x\) | \(51\) |
risch | \(x^{3} \ln \left (2\right )+x^{3} {\mathrm e}^{\frac {3}{4}}-4 x^{2} \ln \left (2\right )-4 x^{2} {\mathrm e}^{\frac {3}{4}}+2 x^{3}+4 x \ln \left (2\right )+4 x \,{\mathrm e}^{\frac {3}{4}}-8 x^{2}+12 x\) | \(51\) |
parallelrisch | \(x^{3} \ln \left (2\right )+x^{3} {\mathrm e}^{\frac {3}{4}}-4 x^{2} \ln \left (2\right )-4 x^{2} {\mathrm e}^{\frac {3}{4}}+2 x^{3}+4 x \ln \left (2\right )+4 x \,{\mathrm e}^{\frac {3}{4}}-8 x^{2}+12 x\) | \(51\) |
parts | \(x^{3} \ln \left (2\right )+x^{3} {\mathrm e}^{\frac {3}{4}}-4 x^{2} \ln \left (2\right )-4 x^{2} {\mathrm e}^{\frac {3}{4}}+2 x^{3}+4 x \ln \left (2\right )+4 x \,{\mathrm e}^{\frac {3}{4}}-8 x^{2}+12 x\) | \(51\) |
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Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \left (12-16 x+6 x^2+e^{3/4} \left (4-8 x+3 x^2\right )+\left (4-8 x+3 x^2\right ) \log (2)\right ) \, dx=2 \, x^{3} - 8 \, x^{2} + {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} e^{\frac {3}{4}} + {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} \log \left (2\right ) + 12 \, x \]
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Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (20) = 40\).
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \left (12-16 x+6 x^2+e^{3/4} \left (4-8 x+3 x^2\right )+\left (4-8 x+3 x^2\right ) \log (2)\right ) \, dx=x^{3} \left (\log {\left (2 \right )} + 2 + e^{\frac {3}{4}}\right ) + x^{2} \left (- 4 e^{\frac {3}{4}} - 8 - 4 \log {\left (2 \right )}\right ) + x \left (4 \log {\left (2 \right )} + 4 e^{\frac {3}{4}} + 12\right ) \]
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Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \left (12-16 x+6 x^2+e^{3/4} \left (4-8 x+3 x^2\right )+\left (4-8 x+3 x^2\right ) \log (2)\right ) \, dx=2 \, x^{3} - 8 \, x^{2} + {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} e^{\frac {3}{4}} + {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} \log \left (2\right ) + 12 \, x \]
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Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \left (12-16 x+6 x^2+e^{3/4} \left (4-8 x+3 x^2\right )+\left (4-8 x+3 x^2\right ) \log (2)\right ) \, dx=2 \, x^{3} - 8 \, x^{2} + {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} e^{\frac {3}{4}} + {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} \log \left (2\right ) + 12 \, x \]
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Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \left (12-16 x+6 x^2+e^{3/4} \left (4-8 x+3 x^2\right )+\left (4-8 x+3 x^2\right ) \log (2)\right ) \, dx=\left ({\mathrm {e}}^{3/4}+\ln \left (2\right )+2\right )\,x^3+\left (-4\,{\mathrm {e}}^{3/4}-\ln \left (16\right )-8\right )\,x^2+\left (4\,{\mathrm {e}}^{3/4}+\ln \left (16\right )+12\right )\,x \]
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