Optimal. Leaf size=28 \[ \frac {1}{4} \left (x-e^{-1+x} (-x+(5-x) x (x+\log (2)))\right ) \]
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Rubi [B] time = 0.17, antiderivative size = 66, normalized size of antiderivative = 2.36, number of steps used = 22, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 2196, 2194, 2176} \begin {gather*} \frac {1}{4} e^{x-1} x^3-\frac {5}{4} e^{x-1} x^2+\frac {1}{4} e^{x-1} x^2 \log (2)+\frac {1}{4} e^{x-1} x+\frac {x}{4}-\frac {5}{4} e^{x-1} x \log (2) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2176
Rule 2194
Rule 2196
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \left (1+e^{-1+x} \left (1-9 x-2 x^2+x^3+\left (-5-3 x+x^2\right ) \log (2)\right )\right ) \, dx\\ &=\frac {x}{4}+\frac {1}{4} \int e^{-1+x} \left (1-9 x-2 x^2+x^3+\left (-5-3 x+x^2\right ) \log (2)\right ) \, dx\\ &=\frac {x}{4}+\frac {1}{4} \int \left (e^{-1+x}-9 e^{-1+x} x-2 e^{-1+x} x^2+e^{-1+x} x^3+e^{-1+x} \left (-5-3 x+x^2\right ) \log (2)\right ) \, dx\\ &=\frac {x}{4}+\frac {1}{4} \int e^{-1+x} \, dx+\frac {1}{4} \int e^{-1+x} x^3 \, dx-\frac {1}{2} \int e^{-1+x} x^2 \, dx-\frac {9}{4} \int e^{-1+x} x \, dx+\frac {1}{4} \log (2) \int e^{-1+x} \left (-5-3 x+x^2\right ) \, dx\\ &=\frac {e^{-1+x}}{4}+\frac {x}{4}-\frac {9}{4} e^{-1+x} x-\frac {1}{2} e^{-1+x} x^2+\frac {1}{4} e^{-1+x} x^3-\frac {3}{4} \int e^{-1+x} x^2 \, dx+\frac {9}{4} \int e^{-1+x} \, dx+\frac {1}{4} \log (2) \int \left (-5 e^{-1+x}-3 e^{-1+x} x+e^{-1+x} x^2\right ) \, dx+\int e^{-1+x} x \, dx\\ &=\frac {5 e^{-1+x}}{2}+\frac {x}{4}-\frac {5}{4} e^{-1+x} x-\frac {5}{4} e^{-1+x} x^2+\frac {1}{4} e^{-1+x} x^3+\frac {3}{2} \int e^{-1+x} x \, dx+\frac {1}{4} \log (2) \int e^{-1+x} x^2 \, dx-\frac {1}{4} (3 \log (2)) \int e^{-1+x} x \, dx-\frac {1}{4} (5 \log (2)) \int e^{-1+x} \, dx-\int e^{-1+x} \, dx\\ &=\frac {3 e^{-1+x}}{2}+\frac {x}{4}+\frac {1}{4} e^{-1+x} x-\frac {5}{4} e^{-1+x} x^2+\frac {1}{4} e^{-1+x} x^3-\frac {5}{4} e^{-1+x} \log (2)-\frac {3}{4} e^{-1+x} x \log (2)+\frac {1}{4} e^{-1+x} x^2 \log (2)-\frac {3}{2} \int e^{-1+x} \, dx-\frac {1}{2} \log (2) \int e^{-1+x} x \, dx+\frac {1}{4} (3 \log (2)) \int e^{-1+x} \, dx\\ &=\frac {x}{4}+\frac {1}{4} e^{-1+x} x-\frac {5}{4} e^{-1+x} x^2+\frac {1}{4} e^{-1+x} x^3-\frac {1}{2} e^{-1+x} \log (2)-\frac {5}{4} e^{-1+x} x \log (2)+\frac {1}{4} e^{-1+x} x^2 \log (2)+\frac {1}{2} \log (2) \int e^{-1+x} \, dx\\ &=\frac {x}{4}+\frac {1}{4} e^{-1+x} x-\frac {5}{4} e^{-1+x} x^2+\frac {1}{4} e^{-1+x} x^3-\frac {5}{4} e^{-1+x} x \log (2)+\frac {1}{4} e^{-1+x} x^2 \log (2)\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 40, normalized size = 1.43 \begin {gather*} \frac {1}{4} \left (x+e^x \left (\frac {x^3}{e}+\frac {x (1-5 \log (2))}{e}+\frac {x^2 (-5+\log (2))}{e}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 30, normalized size = 1.07 \begin {gather*} \frac {1}{4} \, {\left (x^{3} - 5 \, x^{2} + {\left (x^{2} - 5 \, x\right )} \log \relax (2) + x\right )} e^{\left (x - 1\right )} + \frac {1}{4} \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 31, normalized size = 1.11 \begin {gather*} \frac {1}{4} \, {\left (x^{3} + x^{2} \log \relax (2) - 5 \, x^{2} - 5 \, x \log \relax (2) + x\right )} e^{\left (x - 1\right )} + \frac {1}{4} \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 32, normalized size = 1.14
method | result | size |
risch | \(\frac {\left (x^{2} \ln \relax (2)+x^{3}-5 x \ln \relax (2)-5 x^{2}+x \right ) {\mathrm e}^{x -1}}{4}+\frac {x}{4}\) | \(32\) |
norman | \(\left (-\frac {5}{4}+\frac {\ln \relax (2)}{4}\right ) x^{2} {\mathrm e}^{x -1}+\left (-\frac {5 \ln \relax (2)}{4}+\frac {1}{4}\right ) x \,{\mathrm e}^{x -1}+\frac {x}{4}+\frac {x^{3} {\mathrm e}^{x -1}}{4}\) | \(40\) |
default | \(\frac {x}{4}-\frac {{\mathrm e}^{x -1} \left (x -1\right )^{2}}{2}-\frac {3 \,{\mathrm e}^{x -1} \left (x -1\right )}{2}-\frac {3 \,{\mathrm e}^{x -1}}{4}+\frac {{\mathrm e}^{x -1} \left (x -1\right )^{3}}{4}+\frac {{\mathrm e}^{x -1} \ln \relax (2) \left (x -1\right )^{2}}{4}-\frac {3 \,{\mathrm e}^{x -1} \ln \relax (2) \left (x -1\right )}{4}-\ln \relax (2) {\mathrm e}^{x -1}\) | \(74\) |
derivativedivides | \(\frac {x}{4}-\frac {1}{4}-\frac {{\mathrm e}^{x -1} \left (x -1\right )^{2}}{2}-\frac {3 \,{\mathrm e}^{x -1} \left (x -1\right )}{2}-\frac {3 \,{\mathrm e}^{x -1}}{4}+\frac {{\mathrm e}^{x -1} \left (x -1\right )^{3}}{4}+\frac {{\mathrm e}^{x -1} \ln \relax (2) \left (x -1\right )^{2}}{4}-\frac {3 \,{\mathrm e}^{x -1} \ln \relax (2) \left (x -1\right )}{4}-\ln \relax (2) {\mathrm e}^{x -1}\) | \(75\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.49, size = 87, normalized size = 3.11 \begin {gather*} \frac {1}{4} \, {\left (x^{2} - 2 \, x + 2\right )} e^{\left (x - 1\right )} \log \relax (2) - \frac {3}{4} \, {\left (x - 1\right )} e^{\left (x - 1\right )} \log \relax (2) + \frac {1}{4} \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{\left (x - 1\right )} - \frac {1}{2} \, {\left (x^{2} - 2 \, x + 2\right )} e^{\left (x - 1\right )} - \frac {9}{4} \, {\left (x - 1\right )} e^{\left (x - 1\right )} - \frac {5}{4} \, e^{\left (x - 1\right )} \log \relax (2) + \frac {1}{4} \, x + \frac {1}{4} \, e^{\left (x - 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.08, size = 37, normalized size = 1.32 \begin {gather*} \frac {x}{4}+\frac {x^3\,{\mathrm {e}}^{x-1}}{4}+\frac {x^2\,{\mathrm {e}}^{x-1}\,\left (\ln \relax (2)-5\right )}{4}-\frac {x\,{\mathrm {e}}^{x-1}\,\left (\ln \left (32\right )-1\right )}{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.16, size = 32, normalized size = 1.14 \begin {gather*} \frac {x}{4} + \frac {\left (x^{3} - 5 x^{2} + x^{2} \log {\relax (2 )} - 5 x \log {\relax (2 )} + x\right ) e^{x - 1}}{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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