Integrand size = 75, antiderivative size = 27 \[ \int \frac {1}{25} \left (10 x+16 x^3+e^{4 x} \left (6 x^5+4 x^6\right )+12 x^2 \log (2)+2 x \log ^2(2)+e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right )\right ) \, dx=\frac {1}{25} x^2 \left (5+\left (2 x-e^{2 x} x^2+\log (2)\right )^2\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(68\) vs. \(2(27)=54\).
Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.52, number of steps used = 43, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6, 12, 1607, 2227, 2207, 2225} \[ \int \frac {1}{25} \left (10 x+16 x^3+e^{4 x} \left (6 x^5+4 x^6\right )+12 x^2 \log (2)+2 x \log ^2(2)+e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right )\right ) \, dx=\frac {1}{25} e^{4 x} x^6-\frac {4}{25} e^{2 x} x^5+\frac {4 x^4}{25}-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {4}{25} x^3 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right ) \]
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Rule 6
Rule 12
Rule 1607
Rule 2207
Rule 2225
Rule 2227
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{25} \left (16 x^3+e^{4 x} \left (6 x^5+4 x^6\right )+12 x^2 \log (2)+e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right )+x \left (10+2 \log ^2(2)\right )\right ) \, dx \\ & = \frac {1}{25} \int \left (16 x^3+e^{4 x} \left (6 x^5+4 x^6\right )+12 x^2 \log (2)+e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right )+x \left (10+2 \log ^2(2)\right )\right ) \, dx \\ & = \frac {4 x^4}{25}+\frac {4}{25} x^3 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {1}{25} \int e^{4 x} \left (6 x^5+4 x^6\right ) \, dx+\frac {1}{25} \int e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right ) \, dx \\ & = \frac {4 x^4}{25}+\frac {4}{25} x^3 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {1}{25} \int e^{4 x} x^5 (6+4 x) \, dx+\frac {1}{25} \int \left (-20 e^{2 x} x^4-8 e^{2 x} x^5-4 e^{2 x} x^3 (2+x) \log (2)\right ) \, dx \\ & = \frac {4 x^4}{25}+\frac {4}{25} x^3 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {1}{25} \int \left (6 e^{4 x} x^5+4 e^{4 x} x^6\right ) \, dx-\frac {8}{25} \int e^{2 x} x^5 \, dx-\frac {4}{5} \int e^{2 x} x^4 \, dx-\frac {1}{25} (4 \log (2)) \int e^{2 x} x^3 (2+x) \, dx \\ & = \frac {4 x^4}{25}-\frac {2}{5} e^{2 x} x^4-\frac {4}{25} e^{2 x} x^5+\frac {4}{25} x^3 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {4}{25} \int e^{4 x} x^6 \, dx+\frac {6}{25} \int e^{4 x} x^5 \, dx+\frac {4}{5} \int e^{2 x} x^4 \, dx+\frac {8}{5} \int e^{2 x} x^3 \, dx-\frac {1}{25} (4 \log (2)) \int \left (2 e^{2 x} x^3+e^{2 x} x^4\right ) \, dx \\ & = \frac {4}{5} e^{2 x} x^3+\frac {4 x^4}{25}-\frac {4}{25} e^{2 x} x^5+\frac {3}{50} e^{4 x} x^5+\frac {1}{25} e^{4 x} x^6+\frac {4}{25} x^3 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )-\frac {6}{25} \int e^{4 x} x^5 \, dx-\frac {3}{10} \int e^{4 x} x^4 \, dx-\frac {8}{5} \int e^{2 x} x^3 \, dx-\frac {12}{5} \int e^{2 x} x^2 \, dx-\frac {1}{25} (4 \log (2)) \int e^{2 x} x^4 \, dx-\frac {1}{25} (8 \log (2)) \int e^{2 x} x^3 \, dx \\ & = -\frac {6}{5} e^{2 x} x^2+\frac {4 x^4}{25}-\frac {3}{40} e^{4 x} x^4-\frac {4}{25} e^{2 x} x^5+\frac {1}{25} e^{4 x} x^6+\frac {4}{25} x^3 \log (2)-\frac {4}{25} e^{2 x} x^3 \log (2)-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {3}{10} \int e^{4 x} x^3 \, dx+\frac {3}{10} \int e^{4 x} x^4 \, dx+\frac {12}{5} \int e^{2 x} x \, dx+\frac {12}{5} \int e^{2 x} x^2 \, dx+\frac {1}{25} (8 \log (2)) \int e^{2 x} x^3 \, dx+\frac {1}{25} (12 \log (2)) \int e^{2 x} x^2 \, dx \\ & = \frac {6}{5} e^{2 x} x+\frac {3}{40} e^{4 x} x^3+\frac {4 x^4}{25}-\frac {4}{25} e^{2 x} x^5+\frac {1}{25} e^{4 x} x^6+\frac {6}{25} e^{2 x} x^2 \log (2)+\frac {4}{25} x^3 \log (2)-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )-\frac {9}{40} \int e^{4 x} x^2 \, dx-\frac {3}{10} \int e^{4 x} x^3 \, dx-\frac {6}{5} \int e^{2 x} \, dx-\frac {12}{5} \int e^{2 x} x \, dx-\frac {1}{25} (12 \log (2)) \int e^{2 x} x \, dx-\frac {1}{25} (12 \log (2)) \int e^{2 x} x^2 \, dx \\ & = -\frac {3 e^{2 x}}{5}-\frac {9}{160} e^{4 x} x^2+\frac {4 x^4}{25}-\frac {4}{25} e^{2 x} x^5+\frac {1}{25} e^{4 x} x^6-\frac {6}{25} e^{2 x} x \log (2)+\frac {4}{25} x^3 \log (2)-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {9}{80} \int e^{4 x} x \, dx+\frac {9}{40} \int e^{4 x} x^2 \, dx+\frac {6}{5} \int e^{2 x} \, dx+\frac {1}{25} (6 \log (2)) \int e^{2 x} \, dx+\frac {1}{25} (12 \log (2)) \int e^{2 x} x \, dx \\ & = \frac {9}{320} e^{4 x} x+\frac {4 x^4}{25}-\frac {4}{25} e^{2 x} x^5+\frac {1}{25} e^{4 x} x^6+\frac {3}{25} e^{2 x} \log (2)+\frac {4}{25} x^3 \log (2)-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )-\frac {9}{320} \int e^{4 x} \, dx-\frac {9}{80} \int e^{4 x} x \, dx-\frac {1}{25} (6 \log (2)) \int e^{2 x} \, dx \\ & = -\frac {9 e^{4 x}}{1280}+\frac {4 x^4}{25}-\frac {4}{25} e^{2 x} x^5+\frac {1}{25} e^{4 x} x^6+\frac {4}{25} x^3 \log (2)-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {9}{320} \int e^{4 x} \, dx \\ & = \frac {4 x^4}{25}-\frac {4}{25} e^{2 x} x^5+\frac {1}{25} e^{4 x} x^6+\frac {4}{25} x^3 \log (2)-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right ) \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93 \[ \int \frac {1}{25} \left (10 x+16 x^3+e^{4 x} \left (6 x^5+4 x^6\right )+12 x^2 \log (2)+2 x \log ^2(2)+e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right )\right ) \, dx=\frac {1}{75} x^2 \left (12 x^2+3 e^{4 x} x^4-6 e^{2 x} x^2 (2 x+\log (2))+3 \left (5+\log ^2(2)\right )+2 x \log (64)\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(55\) vs. \(2(25)=50\).
Time = 0.32 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07
method | result | size |
risch | \(\frac {{\mathrm e}^{4 x} x^{6}}{25}+\frac {\left (-2 x^{4} \ln \left (2\right )-4 x^{5}\right ) {\mathrm e}^{2 x}}{25}+\frac {x^{2} \ln \left (2\right )^{2}}{25}+\frac {4 x^{3} \ln \left (2\right )}{25}+\frac {4 x^{4}}{25}+\frac {x^{2}}{5}\) | \(56\) |
default | \(\frac {{\mathrm e}^{4 x} x^{6}}{25}-\frac {2 \ln \left (2\right ) {\mathrm e}^{2 x} x^{4}}{25}-\frac {4 x^{5} {\mathrm e}^{2 x}}{25}+\frac {x^{2} \ln \left (2\right )^{2}}{25}+\frac {4 x^{3} \ln \left (2\right )}{25}+\frac {4 x^{4}}{25}+\frac {x^{2}}{5}\) | \(57\) |
parallelrisch | \(\frac {{\mathrm e}^{4 x} x^{6}}{25}-\frac {2 \ln \left (2\right ) {\mathrm e}^{2 x} x^{4}}{25}-\frac {4 x^{5} {\mathrm e}^{2 x}}{25}+\frac {x^{2} \ln \left (2\right )^{2}}{25}+\frac {4 x^{3} \ln \left (2\right )}{25}+\frac {4 x^{4}}{25}+\frac {x^{2}}{5}\) | \(57\) |
parts | \(\frac {{\mathrm e}^{4 x} x^{6}}{25}-\frac {2 \ln \left (2\right ) {\mathrm e}^{2 x} x^{4}}{25}-\frac {4 x^{5} {\mathrm e}^{2 x}}{25}+\frac {x^{2} \ln \left (2\right )^{2}}{25}+\frac {4 x^{3} \ln \left (2\right )}{25}+\frac {4 x^{4}}{25}+\frac {x^{2}}{5}\) | \(57\) |
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Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (25) = 50\).
Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {1}{25} \left (10 x+16 x^3+e^{4 x} \left (6 x^5+4 x^6\right )+12 x^2 \log (2)+2 x \log ^2(2)+e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right )\right ) \, dx=\frac {1}{25} \, x^{6} e^{\left (4 \, x\right )} + \frac {4}{25} \, x^{4} + \frac {4}{25} \, x^{3} \log \left (2\right ) + \frac {1}{25} \, x^{2} \log \left (2\right )^{2} + \frac {1}{5} \, x^{2} - \frac {2}{25} \, {\left (2 \, x^{5} + x^{4} \log \left (2\right )\right )} e^{\left (2 \, x\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (24) = 48\).
Time = 0.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.26 \[ \int \frac {1}{25} \left (10 x+16 x^3+e^{4 x} \left (6 x^5+4 x^6\right )+12 x^2 \log (2)+2 x \log ^2(2)+e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right )\right ) \, dx=\frac {x^{6} e^{4 x}}{25} + \frac {4 x^{4}}{25} + \frac {4 x^{3} \log {\left (2 \right )}}{25} + x^{2} \left (\frac {\log {\left (2 \right )}^{2}}{25} + \frac {1}{5}\right ) + \frac {\left (- 100 x^{5} - 50 x^{4} \log {\left (2 \right )}\right ) e^{2 x}}{625} \]
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Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (25) = 50\).
Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {1}{25} \left (10 x+16 x^3+e^{4 x} \left (6 x^5+4 x^6\right )+12 x^2 \log (2)+2 x \log ^2(2)+e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right )\right ) \, dx=\frac {1}{25} \, x^{6} e^{\left (4 \, x\right )} + \frac {4}{25} \, x^{4} + \frac {4}{25} \, x^{3} \log \left (2\right ) + \frac {1}{25} \, x^{2} \log \left (2\right )^{2} + \frac {1}{5} \, x^{2} - \frac {2}{25} \, {\left (2 \, x^{5} + x^{4} \log \left (2\right )\right )} e^{\left (2 \, x\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (25) = 50\).
Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {1}{25} \left (10 x+16 x^3+e^{4 x} \left (6 x^5+4 x^6\right )+12 x^2 \log (2)+2 x \log ^2(2)+e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right )\right ) \, dx=\frac {1}{25} \, x^{6} e^{\left (4 \, x\right )} + \frac {4}{25} \, x^{4} + \frac {4}{25} \, x^{3} \log \left (2\right ) + \frac {1}{25} \, x^{2} \log \left (2\right )^{2} + \frac {1}{5} \, x^{2} - \frac {2}{25} \, {\left (2 \, x^{5} + x^{4} \log \left (2\right )\right )} e^{\left (2 \, x\right )} \]
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Time = 15.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {1}{25} \left (10 x+16 x^3+e^{4 x} \left (6 x^5+4 x^6\right )+12 x^2 \log (2)+2 x \log ^2(2)+e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right )\right ) \, dx=\frac {x^6\,{\mathrm {e}}^{4\,x}}{25}-\frac {4\,x^5\,{\mathrm {e}}^{2\,x}}{25}+\frac {x^3\,\ln \left (16\right )}{25}+x^2\,\left (\frac {{\ln \left (2\right )}^2}{25}+\frac {1}{5}\right )+\frac {4\,x^4}{25}-\frac {x^4\,{\mathrm {e}}^{2\,x}\,\ln \left (4\right )}{25} \]
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