Integrand size = 42, antiderivative size = 21 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {1}{2} \left (7-x-\frac {8}{5} e^{-8 x} x\right )^2 \]
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Leaf count is larger than twice the leaf count of optimal. \(45\) vs. \(2(21)=42\).
Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.14, number of steps used = 18, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {12, 2227, 2225, 2207, 1607} \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {32}{25} e^{-16 x} x^2+\frac {8}{5} e^{-8 x} x^2+\frac {x^2}{2}-\frac {56}{5} e^{-8 x} x-7 x \]
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Rule 12
Rule 1607
Rule 2207
Rule 2225
Rule 2227
Rubi steps \begin{align*} \text {integral}& = \frac {1}{25} \int \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx \\ & = -7 x+\frac {x^2}{2}+\frac {4}{25} \int e^{-8 x} \left (-70+580 x-80 x^2\right ) \, dx+\frac {16}{25} \int e^{-16 x} \left (4 x-32 x^2\right ) \, dx \\ & = -7 x+\frac {x^2}{2}+\frac {4}{25} \int \left (-70 e^{-8 x}+580 e^{-8 x} x-80 e^{-8 x} x^2\right ) \, dx+\frac {16}{25} \int e^{-16 x} (4-32 x) x \, dx \\ & = -7 x+\frac {x^2}{2}+\frac {16}{25} \int \left (4 e^{-16 x} x-32 e^{-16 x} x^2\right ) \, dx-\frac {56}{5} \int e^{-8 x} \, dx-\frac {64}{5} \int e^{-8 x} x^2 \, dx+\frac {464}{5} \int e^{-8 x} x \, dx \\ & = \frac {7 e^{-8 x}}{5}-7 x-\frac {58}{5} e^{-8 x} x+\frac {x^2}{2}+\frac {8}{5} e^{-8 x} x^2+\frac {64}{25} \int e^{-16 x} x \, dx-\frac {16}{5} \int e^{-8 x} x \, dx+\frac {58}{5} \int e^{-8 x} \, dx-\frac {512}{25} \int e^{-16 x} x^2 \, dx \\ & = -\frac {1}{20} e^{-8 x}-7 x-\frac {4}{25} e^{-16 x} x-\frac {56}{5} e^{-8 x} x+\frac {x^2}{2}+\frac {32}{25} e^{-16 x} x^2+\frac {8}{5} e^{-8 x} x^2+\frac {4}{25} \int e^{-16 x} \, dx-\frac {2}{5} \int e^{-8 x} \, dx-\frac {64}{25} \int e^{-16 x} x \, dx \\ & = -\frac {1}{100} e^{-16 x}-7 x-\frac {56}{5} e^{-8 x} x+\frac {x^2}{2}+\frac {32}{25} e^{-16 x} x^2+\frac {8}{5} e^{-8 x} x^2-\frac {4}{25} \int e^{-16 x} \, dx \\ & = -7 x-\frac {56}{5} e^{-8 x} x+\frac {x^2}{2}+\frac {32}{25} e^{-16 x} x^2+\frac {8}{5} e^{-8 x} x^2 \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=-7 x+\frac {x^2}{2}+\frac {32}{25} e^{-16 x} x^2-\frac {8}{5} e^{-8 x} \left (7 x-x^2\right ) \]
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Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62
method | result | size |
risch | \(\frac {32 \,{\mathrm e}^{-16 x} x^{2}}{25}+\frac {4 \left (10 x^{2}-70 x \right ) {\mathrm e}^{-8 x}}{25}+\frac {x^{2}}{2}-7 x\) | \(34\) |
norman | \(\frac {32 \,{\mathrm e}^{-16 x} x^{2}}{25}+\frac {8 \,{\mathrm e}^{-8 x} x^{2}}{5}-\frac {56 \,{\mathrm e}^{-8 x} x}{5}+\frac {x^{2}}{2}-7 x\) | \(50\) |
parallelrisch | \(\frac {32 \,{\mathrm e}^{-16 x} x^{2}}{25}+\frac {8 \,{\mathrm e}^{-8 x} x^{2}}{5}-\frac {56 \,{\mathrm e}^{-8 x} x}{5}+\frac {x^{2}}{2}-7 x\) | \(50\) |
default | \(-7 x +\frac {2 \,{\mathrm e}^{-16 x} \left (\ln \left (2\right )-4 x \right )^{2}}{25}+\frac {2 \ln \left (2\right )^{2} {\mathrm e}^{-16 x}}{25}-\frac {4 \ln \left (2\right ) {\mathrm e}^{-16 x} \left (\ln \left (2\right )-4 x \right )}{25}+\frac {14 \,{\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )}{5}+\frac {{\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )^{2}}{10}-\frac {14 \ln \left (2\right ) {\mathrm e}^{-8 x}}{5}+\frac {\ln \left (2\right )^{2} {\mathrm e}^{-8 x}}{10}-\frac {\ln \left (2\right ) {\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )}{5}+\frac {x^{2}}{2}\) | \(146\) |
parts | \(-7 x +\frac {2 \,{\mathrm e}^{-16 x} \left (\ln \left (2\right )-4 x \right )^{2}}{25}+\frac {2 \ln \left (2\right )^{2} {\mathrm e}^{-16 x}}{25}-\frac {4 \ln \left (2\right ) {\mathrm e}^{-16 x} \left (\ln \left (2\right )-4 x \right )}{25}+\frac {14 \,{\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )}{5}+\frac {{\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )^{2}}{10}-\frac {14 \ln \left (2\right ) {\mathrm e}^{-8 x}}{5}+\frac {\ln \left (2\right )^{2} {\mathrm e}^{-8 x}}{10}-\frac {\ln \left (2\right ) {\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )}{5}+\frac {x^{2}}{2}\) | \(146\) |
derivativedivides | \(\frac {7 \ln \left (2\right )}{4}-7 x +\frac {\left (\ln \left (2\right )-4 x \right )^{2}}{32}+\frac {2 \,{\mathrm e}^{-16 x} \left (\ln \left (2\right )-4 x \right )^{2}}{25}+\frac {2 \ln \left (2\right )^{2} {\mathrm e}^{-16 x}}{25}-\frac {4 \ln \left (2\right ) {\mathrm e}^{-16 x} \left (\ln \left (2\right )-4 x \right )}{25}+\frac {14 \,{\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )}{5}+\frac {{\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )^{2}}{10}-\frac {14 \ln \left (2\right ) {\mathrm e}^{-8 x}}{5}+\frac {\ln \left (2\right )^{2} {\mathrm e}^{-8 x}}{10}-\frac {\ln \left (2\right ) {\mathrm e}^{-8 x} \left (\ln \left (2\right )-4 x \right )}{5}-\frac {\left (\ln \left (2\right )-4 x \right ) \ln \left (2\right )}{16}\) | \(165\) |
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Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {2}{25} \, x^{2} e^{\left (-16 \, x + 4 \, \log \left (2\right )\right )} + \frac {1}{2} \, x^{2} + \frac {2}{5} \, {\left (x^{2} - 7 \, x\right )} e^{\left (-8 \, x + 2 \, \log \left (2\right )\right )} - 7 \, x \]
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Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (15) = 30\).
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {x^{2}}{2} + \frac {32 x^{2} e^{- 16 x}}{25} - 7 x + \frac {\left (200 x^{2} - 1400 x\right ) e^{- 8 x}}{125} \]
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Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {32}{25} \, x^{2} e^{\left (-16 \, x\right )} + \frac {1}{2} \, x^{2} + \frac {8}{5} \, {\left (x^{2} - 7 \, x\right )} e^{\left (-8 \, x\right )} - 7 \, x \]
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Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {2}{25} \, x^{2} e^{\left (-16 \, x + 4 \, \log \left (2\right )\right )} + \frac {1}{2} \, x^{2} + \frac {2}{5} \, {\left (x^{2} - 7 \, x\right )} e^{\left (-8 \, x + 2 \, \log \left (2\right )\right )} - 7 \, x \]
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Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {1}{25} \left (-175+25 x+4 e^{-8 x} \left (-70+580 x-80 x^2\right )+16 e^{-16 x} \left (4 x-32 x^2\right )\right ) \, dx=\frac {x\,{\mathrm {e}}^{-16\,x}\,\left (5\,{\mathrm {e}}^{8\,x}+8\right )\,\left (8\,x-70\,{\mathrm {e}}^{8\,x}+5\,x\,{\mathrm {e}}^{8\,x}\right )}{50} \]
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