Optimal. Leaf size=24 \[ \frac {e^x \left (-\frac {11}{2}+4 x\right ) (1+x-\log (4))}{5 x} \]
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Rubi [A]
time = 0.10, antiderivative size = 45, normalized size of antiderivative = 1.88, number of steps
used = 9, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 2230, 2207,
2225, 2208, 2209} \begin {gather*} \frac {4 e^x x}{5}-\frac {4 e^x}{5}+\frac {1}{10} e^x (5-8 \log (4))-\frac {11 e^x (1-\log (4))}{10 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2207
Rule 2208
Rule 2209
Rule 2225
Rule 2230
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{10} \int \frac {e^x \left (11-11 x+5 x^2+8 x^3+\left (-11+11 x-8 x^2\right ) \log (4)\right )}{x^2} \, dx\\ &=\frac {1}{10} \int \left (8 e^x x+5 e^x \left (1-\frac {8 \log (4)}{5}\right )-\frac {11 e^x (-1+\log (4))}{x^2}+\frac {11 e^x (-1+\log (4))}{x}\right ) \, dx\\ &=\frac {4}{5} \int e^x x \, dx+\frac {1}{10} (5-8 \log (4)) \int e^x \, dx+\frac {1}{10} (11 (1-\log (4))) \int \frac {e^x}{x^2} \, dx-\frac {1}{10} (11 (1-\log (4))) \int \frac {e^x}{x} \, dx\\ &=\frac {4 e^x x}{5}+\frac {1}{10} e^x (5-8 \log (4))-\frac {11 e^x (1-\log (4))}{10 x}-\frac {11}{10} \text {Ei}(x) (1-\log (4))-\frac {4 \int e^x \, dx}{5}+\frac {1}{10} (11 (1-\log (4))) \int \frac {e^x}{x} \, dx\\ &=-\frac {4 e^x}{5}+\frac {4 e^x x}{5}+\frac {1}{10} e^x (5-8 \log (4))-\frac {11 e^x (1-\log (4))}{10 x}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.06, size = 22, normalized size = 0.92 \begin {gather*} \frac {e^x (-11+8 x) (1+x-\log (4))}{10 x} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.13, size = 33, normalized size = 1.38
method | result | size |
gosper | \(-\frac {{\mathrm e}^{x} \left (16 x \ln \left (2\right )-8 x^{2}-22 \ln \left (2\right )+3 x +11\right )}{10 x}\) | \(27\) |
risch | \(-\frac {{\mathrm e}^{x} \left (16 x \ln \left (2\right )-8 x^{2}-22 \ln \left (2\right )+3 x +11\right )}{10 x}\) | \(27\) |
norman | \(\frac {\left (\frac {11 \ln \left (2\right )}{5}-\frac {11}{10}\right ) {\mathrm e}^{x}+\left (-\frac {8 \ln \left (2\right )}{5}-\frac {3}{10}\right ) x \,{\mathrm e}^{x}+\frac {4 \,{\mathrm e}^{x} x^{2}}{5}}{x}\) | \(32\) |
default | \(-\frac {11 \,{\mathrm e}^{x}}{10 x}+\frac {4 \,{\mathrm e}^{x} x}{5}-\frac {3 \,{\mathrm e}^{x}}{10}-\frac {8 \,{\mathrm e}^{x} \ln \left (2\right )}{5}+\frac {11 \ln \left (2\right ) {\mathrm e}^{x}}{5 x}\) | \(33\) |
meijerg | \(\frac {\frac {11 x}{10}+\frac {11}{10}}{x}-\frac {11 \,{\mathrm e}^{x}}{10 x}-\frac {11 \ln \left (-x \right )}{10}-\frac {11 \expIntegral \left (1, -x \right )}{10}-\frac {3}{10}+\frac {11 \ln \left (x \right )}{10}+\frac {11 i \pi }{10}-\frac {11}{10 x}-\left (-\frac {11 \ln \left (2\right )}{5}+\frac {11}{10}\right ) \left (-\ln \left (-x \right )-\expIntegral \left (1, -x \right )+\ln \left (x \right )+i \pi \right )-\left (-\frac {8 \ln \left (2\right )}{5}+\frac {1}{2}\right ) \left (1-{\mathrm e}^{x}\right )-\frac {2 \left (-2 x +2\right ) {\mathrm e}^{x}}{5}+\frac {11 \ln \left (2\right ) \left (-\frac {2 x +2}{2 x}+\frac {{\mathrm e}^{x}}{x}+\ln \left (-x \right )+\expIntegral \left (1, -x \right )+1-\ln \left (x \right )-i \pi +\frac {1}{x}\right )}{5}\) | \(139\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 0.30, size = 44, normalized size = 1.83 \begin {gather*} \frac {4}{5} \, {\left (x - 1\right )} e^{x} + \frac {11}{5} \, {\rm Ei}\left (x\right ) \log \left (2\right ) - \frac {8}{5} \, e^{x} \log \left (2\right ) - \frac {11}{5} \, \Gamma \left (-1, -x\right ) \log \left (2\right ) - \frac {11}{10} \, {\rm Ei}\left (x\right ) + \frac {1}{2} \, e^{x} + \frac {11}{10} \, \Gamma \left (-1, -x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 26, normalized size = 1.08 \begin {gather*} \frac {{\left (8 \, x^{2} - 2 \, {\left (8 \, x - 11\right )} \log \left (2\right ) - 3 \, x - 11\right )} e^{x}}{10 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.06, size = 27, normalized size = 1.12 \begin {gather*} \frac {\left (8 x^{2} - 16 x \log {\left (2 \right )} - 3 x - 11 + 22 \log {\left (2 \right )}\right ) e^{x}}{10 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 35, normalized size = 1.46 \begin {gather*} \frac {8 \, x^{2} e^{x} - 16 \, x e^{x} \log \left (2\right ) - 3 \, x e^{x} + 22 \, e^{x} \log \left (2\right ) - 11 \, e^{x}}{10 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.07, size = 19, normalized size = 0.79 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (8\,x-11\right )\,\left (x-\ln \left (4\right )+1\right )}{10\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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