Optimal. Leaf size=22 \[ -4+x+\frac {1}{16} \left (e^x-x^2\right )^2 (x+\log (x)) \]
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Rubi [B] time = 0.23, antiderivative size = 54, normalized size of antiderivative = 2.45, number of steps used = 6, number of rules used = 4, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {12, 14, 2304, 2288} \begin {gather*} \frac {x^5}{16}+\frac {1}{16} x^4 \log (x)-\frac {1}{8} e^x x \left (x^2+x \log (x)\right )+\frac {e^{2 x} \left (x^2+x \log (x)\right )}{16 x}+x \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 2288
Rule 2304
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{16} \int \frac {16 x+x^4+5 x^5+e^{2 x} \left (1+x+2 x^2\right )+e^x \left (-2 x^2-6 x^3-2 x^4\right )+\left (2 e^{2 x} x+4 x^4+e^x \left (-4 x^2-2 x^3\right )\right ) \log (x)}{x} \, dx\\ &=\frac {1}{16} \int \left (16+x^3+5 x^4+4 x^3 \log (x)-2 e^x x \left (1+3 x+x^2+2 \log (x)+x \log (x)\right )+\frac {e^{2 x} \left (1+x+2 x^2+2 x \log (x)\right )}{x}\right ) \, dx\\ &=x+\frac {x^4}{64}+\frac {x^5}{16}+\frac {1}{16} \int \frac {e^{2 x} \left (1+x+2 x^2+2 x \log (x)\right )}{x} \, dx-\frac {1}{8} \int e^x x \left (1+3 x+x^2+2 \log (x)+x \log (x)\right ) \, dx+\frac {1}{4} \int x^3 \log (x) \, dx\\ &=x+\frac {x^5}{16}+\frac {1}{16} x^4 \log (x)+\frac {e^{2 x} \left (x^2+x \log (x)\right )}{16 x}-\frac {1}{8} e^x x \left (x^2+x \log (x)\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.09, size = 39, normalized size = 1.77 \begin {gather*} \frac {1}{16} \left (x \left (16+e^{2 x}-2 e^x x^2+x^4\right )+\left (e^x-x^2\right )^2 \log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.66, size = 40, normalized size = 1.82 \begin {gather*} \frac {1}{16} \, x^{5} - \frac {1}{8} \, x^{3} e^{x} + \frac {1}{16} \, x e^{\left (2 \, x\right )} + \frac {1}{16} \, {\left (x^{4} - 2 \, x^{2} e^{x} + e^{\left (2 \, x\right )}\right )} \log \relax (x) + x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.42, size = 45, normalized size = 2.05 \begin {gather*} \frac {1}{16} \, x^{5} + \frac {1}{16} \, x^{4} \log \relax (x) - \frac {1}{8} \, x^{3} e^{x} - \frac {1}{8} \, x^{2} e^{x} \log \relax (x) + \frac {1}{16} \, x e^{\left (2 \, x\right )} + \frac {1}{16} \, e^{\left (2 \, x\right )} \log \relax (x) + x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 41, normalized size = 1.86
method | result | size |
risch | \(\frac {\left ({\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x^{2}+x^{4}\right ) \ln \relax (x )}{16}+\frac {x^{5}}{16}-\frac {{\mathrm e}^{x} x^{3}}{8}+\frac {x \,{\mathrm e}^{2 x}}{16}+x\) | \(41\) |
default | \(x -\frac {{\mathrm e}^{x} x^{3}}{8}-\frac {x^{2} {\mathrm e}^{x} \ln \relax (x )}{8}+\frac {x \,{\mathrm e}^{2 x}}{16}+\frac {{\mathrm e}^{2 x} \ln \relax (x )}{16}+\frac {x^{5}}{16}+\frac {x^{4} \ln \relax (x )}{16}\) | \(46\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{16} \, x^{5} + \frac {1}{16} \, x^{4} \log \relax (x) + \frac {1}{32} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - \frac {1}{8} \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{x} - \frac {1}{8} \, {\left (x^{2} \log \relax (x) - x + 1\right )} e^{x} - \frac {3}{8} \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} - \frac {1}{8} \, {\left (x - 1\right )} e^{x} + \frac {1}{16} \, e^{\left (2 \, x\right )} \log \relax (x) + x + \frac {1}{16} \, {\rm Ei}\left (2 \, x\right ) + \frac {1}{32} \, e^{\left (2 \, x\right )} - \frac {1}{16} \, \int \frac {e^{\left (2 \, x\right )}}{x}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.46, size = 45, normalized size = 2.05 \begin {gather*} x+\frac {x\,{\mathrm {e}}^{2\,x}}{16}-\frac {x^3\,{\mathrm {e}}^x}{8}+\frac {x^4\,\ln \relax (x)}{16}+\frac {{\mathrm {e}}^{2\,x}\,\ln \relax (x)}{16}+\frac {x^5}{16}-\frac {x^2\,{\mathrm {e}}^x\,\ln \relax (x)}{8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.47, size = 49, normalized size = 2.23 \begin {gather*} \frac {x^{5}}{16} + \frac {x^{4} \log {\relax (x )}}{16} + x + \frac {\left (8 x + 8 \log {\relax (x )}\right ) e^{2 x}}{128} + \frac {\left (- 16 x^{3} - 16 x^{2} \log {\relax (x )}\right ) e^{x}}{128} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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