Optimal. Leaf size=28 \[ x-16 (-1+x) \left (-4-e^x+\frac {1}{x}+\frac {2 (4+x)}{-2+\log (4)}\right ) \]
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Rubi [A] time = 0.07, antiderivative size = 47, normalized size of antiderivative = 1.68, number of steps used = 8, number of rules used = 5, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {6, 12, 14, 2176, 2194} \begin {gather*} \frac {32 x^2}{2-\log (4)}+16 e^x x-16 e^x+\frac {16}{x}+\frac {2 x (113-65 \log (2))}{2-\log (4)} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 14
Rule 2176
Rule 2194
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32-226 x^2-64 x^3+\left (-16+65 x^2\right ) \log (4)+e^x \left (-32 x^3+16 x^3 \log (4)\right )}{x^2 (-2+\log (4))} \, dx\\ &=\frac {\int \frac {32-226 x^2-64 x^3+\left (-16+65 x^2\right ) \log (4)+e^x \left (-32 x^3+16 x^3 \log (4)\right )}{x^2} \, dx}{-2+\log (4)}\\ &=\frac {\int \left (\frac {-64 x^3-x^2 (226-65 \log (4))+16 (2-\log (4))}{x^2}+16 e^x x (-2+\log (4))\right ) \, dx}{-2+\log (4)}\\ &=16 \int e^x x \, dx+\frac {\int \frac {-64 x^3-x^2 (226-65 \log (4))+16 (2-\log (4))}{x^2} \, dx}{-2+\log (4)}\\ &=16 e^x x-16 \int e^x \, dx+\frac {\int \left (-64 x-226 \left (1-\frac {65 \log (2)}{113}\right )-\frac {16 (-2+\log (4))}{x^2}\right ) \, dx}{-2+\log (4)}\\ &=-16 e^x+\frac {16}{x}+16 e^x x+\frac {32 x^2}{2-\log (4)}+\frac {2 x (113-65 \log (2))}{2-\log (4)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.06, size = 39, normalized size = 1.39 \begin {gather*} 16 e^x (-1+x)+\frac {16}{x}-\frac {32 x^2}{-2+\log (4)}+\frac {x (-226+65 \log (4))}{-2+\log (4)} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 3.08, size = 57, normalized size = 2.04 \begin {gather*} -\frac {16 \, x^{3} + 113 \, x^{2} + 16 \, {\left (x^{2} - {\left (x^{2} - x\right )} \log \relax (2) - x\right )} e^{x} - {\left (65 \, x^{2} + 16\right )} \log \relax (2) + 16}{x \log \relax (2) - x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.18, size = 62, normalized size = 2.21 \begin {gather*} \frac {16 \, x^{2} e^{x} \log \relax (2) - 16 \, x^{3} - 16 \, x^{2} e^{x} + 65 \, x^{2} \log \relax (2) - 16 \, x e^{x} \log \relax (2) - 113 \, x^{2} + 16 \, x e^{x} + 16 \, \log \relax (2) - 16}{x \log \relax (2) - x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 46, normalized size = 1.64
method | result | size |
norman | \(\frac {16+\frac {\left (65 \ln \relax (2)-113\right ) x^{2}}{\ln \relax (2)-1}-\frac {16 x^{3}}{\ln \relax (2)-1}-16 \,{\mathrm e}^{x} x +16 \,{\mathrm e}^{x} x^{2}}{x}\) | \(46\) |
risch | \(\frac {65 \ln \relax (2) x}{\ln \relax (2)-1}-\frac {16 x^{2}}{\ln \relax (2)-1}-\frac {113 x}{\ln \relax (2)-1}+\frac {16 \ln \relax (2)^{2}}{\left (\ln \relax (2)-1\right )^{2} x}-\frac {32 \ln \relax (2)}{\left (\ln \relax (2)-1\right )^{2} x}+\frac {16}{\left (\ln \relax (2)-1\right )^{2} x}+\left (16 x -16\right ) {\mathrm e}^{x}\) | \(80\) |
default | \(-\frac {16}{\left (\ln \relax (2)-1\right ) x}-\frac {113 x}{\ln \relax (2)-1}-\frac {16 x^{2}}{\ln \relax (2)-1}+\frac {16 \ln \relax (2)}{\left (\ln \relax (2)-1\right ) x}+\frac {65 \ln \relax (2) x}{\ln \relax (2)-1}-\frac {16 \,{\mathrm e}^{x} x}{\ln \relax (2)-1}+\frac {16 \,{\mathrm e}^{x}}{\ln \relax (2)-1}+\frac {16 \ln \relax (2) {\mathrm e}^{x} x}{\ln \relax (2)-1}-\frac {16 \ln \relax (2) {\mathrm e}^{x}}{\ln \relax (2)-1}\) | \(103\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.47, size = 84, normalized size = 3.00 \begin {gather*} \frac {16 \, {\left (x - 1\right )} e^{x} \log \relax (2)}{\log \relax (2) - 1} - \frac {16 \, x^{2}}{\log \relax (2) - 1} - \frac {16 \, {\left (x - 1\right )} e^{x}}{\log \relax (2) - 1} + \frac {65 \, x \log \relax (2)}{\log \relax (2) - 1} - \frac {113 \, x}{\log \relax (2) - 1} + \frac {16 \, \log \relax (2)}{x {\left (\log \relax (2) - 1\right )}} - \frac {16}{x {\left (\log \relax (2) - 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.11, size = 54, normalized size = 1.93 \begin {gather*} \frac {16\,\ln \relax (2)+x^2\,\left (65\,\ln \relax (2)-113\right )-16\,x^3+x^2\,{\mathrm {e}}^x\,\left (16\,\ln \relax (2)-16\right )-x\,{\mathrm {e}}^x\,\left (16\,\ln \relax (2)-16\right )-16}{x\,\left (\ln \relax (2)-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.17, size = 36, normalized size = 1.29 \begin {gather*} \left (16 x - 16\right ) e^{x} + \frac {- 16 x^{2} - x \left (113 - 65 \log {\relax (2 )}\right ) - \frac {16 - 16 \log {\relax (2 )}}{x}}{-1 + \log {\relax (2 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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