Integrand size = 111, antiderivative size = 25 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5}{16} x^2 \left (1+x-e^x (3-\log (2 x))\right )^2 \]
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\[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \int \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx \\ & = \frac {5 x^2}{16}+\frac {5 x^3}{8}+\frac {5 x^4}{16}+\frac {1}{8} \int e^{2 x} \left (30 x+45 x^2\right ) \, dx+\frac {1}{8} \int e^x \left (-25 x-55 x^2-15 x^3\right ) \, dx+\frac {1}{8} \int \left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x) \, dx+\frac {1}{8} \int e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x) \, dx \\ & = \frac {5 x^2}{16}+\frac {5 x^3}{8}+\frac {5 x^4}{16}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {1}{8} \int e^{2 x} x (30+45 x) \, dx+\frac {1}{8} \int e^x x \left (-25-55 x-15 x^2\right ) \, dx-\frac {1}{8} \int e^x \left (e^x \left (\frac {5}{2}-\frac {5}{4 x}-15 x\right )+5 x (1+x)\right ) \, dx+\frac {1}{8} \int e^{2 x} x (5+5 x) \log ^2(2 x) \, dx \\ & = \frac {5 x^2}{16}+\frac {5 x^3}{8}+\frac {5 x^4}{16}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {1}{8} \int \left (30 e^{2 x} x+45 e^{2 x} x^2\right ) \, dx+\frac {1}{8} \int \left (-25 e^x x-55 e^x x^2-15 e^x x^3\right ) \, dx-\frac {1}{8} \int \left (5 e^x x (1+x)-\frac {5 e^{2 x} \left (1-2 x+12 x^2\right )}{4 x}\right ) \, dx+\frac {1}{8} \int \left (5 e^{2 x} x \log ^2(2 x)+5 e^{2 x} x^2 \log ^2(2 x)\right ) \, dx \\ & = \frac {5 x^2}{16}+\frac {5 x^3}{8}+\frac {5 x^4}{16}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {5}{32} \int \frac {e^{2 x} \left (1-2 x+12 x^2\right )}{x} \, dx-\frac {5}{8} \int e^x x (1+x) \, dx+\frac {5}{8} \int e^{2 x} x \log ^2(2 x) \, dx+\frac {5}{8} \int e^{2 x} x^2 \log ^2(2 x) \, dx-\frac {15}{8} \int e^x x^3 \, dx-\frac {25}{8} \int e^x x \, dx+\frac {15}{4} \int e^{2 x} x \, dx+\frac {45}{8} \int e^{2 x} x^2 \, dx-\frac {55}{8} \int e^x x^2 \, dx \\ & = -\frac {25 e^x x}{8}+\frac {15}{8} e^{2 x} x+\frac {5 x^2}{16}-\frac {55 e^x x^2}{8}+\frac {45}{16} e^{2 x} x^2+\frac {5 x^3}{8}-\frac {15 e^x x^3}{8}+\frac {5 x^4}{16}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {5}{32} \int \left (-2 e^{2 x}+\frac {e^{2 x}}{x}+12 e^{2 x} x\right ) \, dx-\frac {5}{8} \int \left (e^x x+e^x x^2\right ) \, dx+\frac {5}{8} \int e^{2 x} x \log ^2(2 x) \, dx+\frac {5}{8} \int e^{2 x} x^2 \log ^2(2 x) \, dx-\frac {15}{8} \int e^{2 x} \, dx+\frac {25 \int e^x \, dx}{8}-\frac {45}{8} \int e^{2 x} x \, dx+\frac {45}{8} \int e^x x^2 \, dx+\frac {55}{4} \int e^x x \, dx \\ & = \frac {25 e^x}{8}-\frac {15 e^{2 x}}{16}+\frac {85 e^x x}{8}-\frac {15}{16} e^{2 x} x+\frac {5 x^2}{16}-\frac {5 e^x x^2}{4}+\frac {45}{16} e^{2 x} x^2+\frac {5 x^3}{8}-\frac {15 e^x x^3}{8}+\frac {5 x^4}{16}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {5}{32} \int \frac {e^{2 x}}{x} \, dx-\frac {5}{16} \int e^{2 x} \, dx-\frac {5}{8} \int e^x x \, dx-\frac {5}{8} \int e^x x^2 \, dx+\frac {5}{8} \int e^{2 x} x \log ^2(2 x) \, dx+\frac {5}{8} \int e^{2 x} x^2 \log ^2(2 x) \, dx+\frac {15}{8} \int e^{2 x} x \, dx+\frac {45}{16} \int e^{2 x} \, dx-\frac {45}{4} \int e^x x \, dx-\frac {55 \int e^x \, dx}{4} \\ & = -\frac {85 e^x}{8}+\frac {5 e^{2 x}}{16}-\frac {5 e^x x}{4}+\frac {5 x^2}{16}-\frac {15 e^x x^2}{8}+\frac {45}{16} e^{2 x} x^2+\frac {5 x^3}{8}-\frac {15 e^x x^3}{8}+\frac {5 x^4}{16}+\frac {5 \text {Ei}(2 x)}{32}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {5 \int e^x \, dx}{8}+\frac {5}{8} \int e^{2 x} x \log ^2(2 x) \, dx+\frac {5}{8} \int e^{2 x} x^2 \log ^2(2 x) \, dx-\frac {15}{16} \int e^{2 x} \, dx+\frac {5}{4} \int e^x x \, dx+\frac {45 \int e^x \, dx}{4} \\ & = \frac {5 e^x}{4}-\frac {5 e^{2 x}}{32}+\frac {5 x^2}{16}-\frac {15 e^x x^2}{8}+\frac {45}{16} e^{2 x} x^2+\frac {5 x^3}{8}-\frac {15 e^x x^3}{8}+\frac {5 x^4}{16}+\frac {5 \text {Ei}(2 x)}{32}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {5}{8} \int e^{2 x} x \log ^2(2 x) \, dx+\frac {5}{8} \int e^{2 x} x^2 \log ^2(2 x) \, dx-\frac {5 \int e^x \, dx}{4} \\ & = -\frac {5 e^{2 x}}{32}+\frac {5 x^2}{16}-\frac {15 e^x x^2}{8}+\frac {45}{16} e^{2 x} x^2+\frac {5 x^3}{8}-\frac {15 e^x x^3}{8}+\frac {5 x^4}{16}+\frac {5 \text {Ei}(2 x)}{32}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {5}{8} \int e^{2 x} x \log ^2(2 x) \, dx+\frac {5}{8} \int e^{2 x} x^2 \log ^2(2 x) \, dx \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5}{16} x^2 \left (1-3 e^x+x+e^x \log (2 x)\right )^2 \]
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Leaf count of result is larger than twice the leaf count of optimal. \(89\) vs. \(2(22)=44\).
Time = 0.25 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.60
method | result | size |
default | \(\frac {45 \,{\mathrm e}^{2 x} x^{2}}{16}-\frac {15 \,{\mathrm e}^{2 x} \ln \left (2 x \right ) x^{2}}{8}+\frac {5 \,{\mathrm e}^{2 x} \ln \left (2 x \right )^{2} x^{2}}{16}-\frac {15 \,{\mathrm e}^{x} x^{2}}{8}-\frac {15 \,{\mathrm e}^{x} x^{3}}{8}+\frac {5 \ln \left (2 x \right ) {\mathrm e}^{x} x^{2}}{8}+\frac {5 \ln \left (2 x \right ) {\mathrm e}^{x} x^{3}}{8}+\frac {5 x^{2}}{16}+\frac {5 x^{3}}{8}+\frac {5 x^{4}}{16}\) | \(90\) |
risch | \(\frac {45 \,{\mathrm e}^{2 x} x^{2}}{16}-\frac {15 \,{\mathrm e}^{2 x} \ln \left (2 x \right ) x^{2}}{8}+\frac {5 \,{\mathrm e}^{2 x} \ln \left (2 x \right )^{2} x^{2}}{16}-\frac {15 \,{\mathrm e}^{x} x^{2}}{8}-\frac {15 \,{\mathrm e}^{x} x^{3}}{8}+\frac {5 \ln \left (2 x \right ) {\mathrm e}^{x} x^{2}}{8}+\frac {5 \ln \left (2 x \right ) {\mathrm e}^{x} x^{3}}{8}+\frac {5 x^{2}}{16}+\frac {5 x^{3}}{8}+\frac {5 x^{4}}{16}\) | \(90\) |
parallelrisch | \(\frac {45 \,{\mathrm e}^{2 x} x^{2}}{16}-\frac {15 \,{\mathrm e}^{2 x} \ln \left (2 x \right ) x^{2}}{8}+\frac {5 \,{\mathrm e}^{2 x} \ln \left (2 x \right )^{2} x^{2}}{16}-\frac {15 \,{\mathrm e}^{x} x^{2}}{8}-\frac {15 \,{\mathrm e}^{x} x^{3}}{8}+\frac {5 \ln \left (2 x \right ) {\mathrm e}^{x} x^{2}}{8}+\frac {5 \ln \left (2 x \right ) {\mathrm e}^{x} x^{3}}{8}+\frac {5 x^{2}}{16}+\frac {5 x^{3}}{8}+\frac {5 x^{4}}{16}\) | \(90\) |
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Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (19) = 38\).
Time = 0.26 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.12 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5}{16} \, x^{2} e^{\left (2 \, x\right )} \log \left (2 \, x\right )^{2} + \frac {5}{16} \, x^{4} + \frac {5}{8} \, x^{3} + \frac {45}{16} \, x^{2} e^{\left (2 \, x\right )} + \frac {5}{16} \, x^{2} - \frac {15}{8} \, {\left (x^{3} + x^{2}\right )} e^{x} - \frac {5}{8} \, {\left (3 \, x^{2} e^{\left (2 \, x\right )} - {\left (x^{3} + x^{2}\right )} e^{x}\right )} \log \left (2 \, x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (20) = 40\).
Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.52 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5 x^{4}}{16} + \frac {5 x^{3}}{8} + \frac {5 x^{2}}{16} + \frac {\left (40 x^{2} \log {\left (2 x \right )}^{2} - 240 x^{2} \log {\left (2 x \right )} + 360 x^{2}\right ) e^{2 x}}{128} + \frac {\left (80 x^{3} \log {\left (2 x \right )} - 240 x^{3} + 80 x^{2} \log {\left (2 x \right )} - 240 x^{2}\right ) e^{x}}{128} \]
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Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (19) = 38\).
Time = 0.31 (sec) , antiderivative size = 125, normalized size of antiderivative = 5.00 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5}{16} \, x^{4} + \frac {5}{8} \, x^{3} + \frac {5}{16} \, x^{2} + \frac {5}{32} \, {\left (4 \, x^{2} {\left (\log \left (2\right ) - 3\right )} \log \left (x\right ) + 2 \, x^{2} \log \left (x\right )^{2} + 2 \, {\left (\log \left (2\right )^{2} - 6 \, \log \left (2\right )\right )} x^{2} + 6 \, x - 3\right )} e^{\left (2 \, x\right )} + \frac {15}{32} \, {\left (6 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + \frac {5}{8} \, {\left (x^{3} \log \left (2\right ) + x^{2} {\left (\log \left (2\right ) - 1\right )} + {\left (x^{3} + x^{2}\right )} \log \left (x\right ) + x - 1\right )} e^{x} - \frac {5}{8} \, {\left (3 \, x^{3} + 2 \, x^{2} + x - 1\right )} e^{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (19) = 38\).
Time = 0.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 5.64 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5}{16} \, x^{2} e^{\left (2 \, x\right )} \log \left (2 \, x\right )^{2} + \frac {5}{16} \, x^{4} + \frac {5}{8} \, x^{3} - \frac {5}{8} \, x^{2} e^{x} - \frac {5}{32} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \log \left (2 \, x\right ) + \frac {5}{16} \, x^{2} + \frac {15}{32} \, {\left (6 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + \frac {15}{16} \, x e^{\left (2 \, x\right )} - \frac {5}{8} \, {\left (3 \, x^{3} + 2 \, x^{2} + x - 1\right )} e^{x} + \frac {5}{8} \, x e^{x} - \frac {5}{32} \, {\left ({\left (12 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} - 4 \, {\left (x^{3} + x^{2}\right )} e^{x}\right )} \log \left (2 \, x\right ) - \frac {15}{32} \, e^{\left (2 \, x\right )} - \frac {5}{8} \, e^{x} \]
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Time = 12.76 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5\,x^2\,{\left (x-3\,{\mathrm {e}}^x+\ln \left (2\,x\right )\,{\mathrm {e}}^x+1\right )}^2}{16} \]
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