Optimal. Leaf size=28 \[ \frac {1}{2} x \left (e^{2 x} x^2+\log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\right ) \]
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Rubi [B] time = 0.95, antiderivative size = 61, normalized size of antiderivative = 2.18, number of steps used = 21, number of rules used = 11, integrand size = 131, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.084, Rules used = {6688, 12, 2196, 2176, 2194, 2411, 2353, 2298, 2302, 29, 2520} \begin {gather*} \frac {1}{2} e^{2 x} x^3-\frac {1}{2} \left (e^{e^9}-x\right ) \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )+\frac {1}{2} e^{e^9} \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 29
Rule 2176
Rule 2194
Rule 2196
Rule 2298
Rule 2302
Rule 2353
Rule 2411
Rule 2520
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{2} \left (e^{2 x} x^2 (3+2 x)-\frac {x}{\left (e^{e^9}-x\right ) \log \left (4 \left (e^{e^9}-x\right )\right )}+\log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\right ) \, dx\\ &=\frac {1}{2} \int \left (e^{2 x} x^2 (3+2 x)-\frac {x}{\left (e^{e^9}-x\right ) \log \left (4 \left (e^{e^9}-x\right )\right )}+\log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\right ) \, dx\\ &=\frac {1}{2} \int e^{2 x} x^2 (3+2 x) \, dx-\frac {1}{2} \int \frac {x}{\left (e^{e^9}-x\right ) \log \left (4 \left (e^{e^9}-x\right )\right )} \, dx+\frac {1}{2} \int \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right ) \, dx\\ &=-\left (\frac {1}{8} \operatorname {Subst}\left (\int \log (\log (x)) \, dx,x,4 e^{e^9}-4 x\right )\right )+\frac {1}{2} \int \left (3 e^{2 x} x^2+2 e^{2 x} x^3\right ) \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \frac {e^{e^9}-x}{x \log (4 x)} \, dx,x,e^{e^9}-x\right )\\ &=-\frac {1}{2} \left (e^{e^9}-x\right ) \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,4 e^{e^9}-4 x\right )+\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {1}{\log (4 x)}+\frac {e^{e^9}}{x \log (4 x)}\right ) \, dx,x,e^{e^9}-x\right )+\frac {3}{2} \int e^{2 x} x^2 \, dx+\int e^{2 x} x^3 \, dx\\ &=\frac {3}{4} e^{2 x} x^2+\frac {1}{2} e^{2 x} x^3-\frac {1}{2} \left (e^{e^9}-x\right ) \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )+\frac {1}{8} \text {li}\left (4 \left (e^{e^9}-x\right )\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\log (4 x)} \, dx,x,e^{e^9}-x\right )-\frac {3}{2} \int e^{2 x} x \, dx-\frac {3}{2} \int e^{2 x} x^2 \, dx+\frac {1}{2} e^{e^9} \operatorname {Subst}\left (\int \frac {1}{x \log (4 x)} \, dx,x,e^{e^9}-x\right )\\ &=-\frac {3}{4} e^{2 x} x+\frac {1}{2} e^{2 x} x^3-\frac {1}{2} \left (e^{e^9}-x\right ) \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )+\frac {3}{4} \int e^{2 x} \, dx+\frac {3}{2} \int e^{2 x} x \, dx+\frac {1}{2} e^{e^9} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (4 \left (e^{e^9}-x\right )\right )\right )\\ &=\frac {3 e^{2 x}}{8}+\frac {1}{2} e^{2 x} x^3+\frac {1}{2} e^{e^9} \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )-\frac {1}{2} \left (e^{e^9}-x\right ) \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )-\frac {3}{4} \int e^{2 x} \, dx\\ &=\frac {1}{2} e^{2 x} x^3+\frac {1}{2} e^{e^9} \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )-\frac {1}{2} \left (e^{e^9}-x\right ) \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.15, size = 28, normalized size = 1.00 \begin {gather*} \frac {1}{2} x \left (e^{2 x} x^2+\log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 39, normalized size = 1.39 \begin {gather*} \frac {1}{2} \, {\left (x^{3} e^{\left (e^{9}\right )} + x e^{\left (-2 \, x + e^{9}\right )} \log \left (\log \left (-4 \, x + 4 \, e^{\left (e^{9}\right )}\right )\right )\right )} e^{\left (2 \, x - e^{9}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.54, size = 25, normalized size = 0.89 \begin {gather*} \frac {1}{2} \, x^{3} e^{\left (2 \, x\right )} + \frac {1}{2} \, x \log \left ({\left | \log \left (-4 \, x + 4 \, e^{\left (e^{9}\right )}\right ) \right |}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 25, normalized size = 0.89
method | result | size |
risch | \(\frac {x \ln \left (\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{9}}-4 x \right )\right )}{2}+\frac {{\mathrm e}^{2 x} x^{3}}{2}\) | \(25\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.68, size = 30, normalized size = 1.07 \begin {gather*} \frac {1}{2} \, x^{3} e^{\left (2 \, x\right )} + \frac {1}{2} \, x \log \left (i \, \pi + 2 \, \log \relax (2) + \log \left (x - e^{\left (e^{9}\right )}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.48, size = 24, normalized size = 0.86 \begin {gather*} \frac {x\,\ln \left (\ln \left (4\,{\mathrm {e}}^{{\mathrm {e}}^9}-4\,x\right )\right )}{2}+\frac {x^3\,{\mathrm {e}}^{2\,x}}{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.89, size = 53, normalized size = 1.89 \begin {gather*} \frac {x^{3} e^{2 x}}{2} + \left (\frac {x}{2} - \frac {e^{e^{9}}}{4}\right ) \log {\left (\log {\left (- 4 x + 4 e^{e^{9}} \right )} \right )} + \frac {e^{e^{9}} \log {\left (\log {\left (- 4 x + 4 e^{e^{9}} \right )} \right )}}{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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