Optimal. Leaf size=28 \[ \frac {\log \left (x^2 \left (-4 x^2+\frac {1}{2} e^x (-3+\log (4))^2\right )\right )}{x} \]
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Rubi [A] time = 3.09, antiderivative size = 31, normalized size of antiderivative = 1.11, number of steps used = 18, number of rules used = 5, integrand size = 124, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {6688, 6742, 14, 43, 2551} \begin {gather*} \frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 43
Rule 2551
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32 x^2-e^x (2+x) (-3+\log (4))^2-\left (8 x^2-e^x (-3+\log (4))^2\right ) \log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )\right )}{x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )} \, dx\\ &=\int \left (\frac {8 (2-x)}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )}+\frac {2+x-\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )\right )}{x^2}\right ) \, dx\\ &=8 \int \frac {2-x}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx+\int \frac {2+x-\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )\right )}{x^2} \, dx\\ &=8 \int \left (\frac {2}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )}+\frac {x}{-8 x^2+9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )}\right ) \, dx+\int \left (\frac {2+x}{x^2}-\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )\right )}{x^2}\right ) \, dx\\ &=8 \int \frac {x}{-8 x^2+9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx+16 \int \frac {1}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx+\int \frac {2+x}{x^2} \, dx-\int \frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )\right )}{x^2} \, dx\\ &=\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x}+8 \int \frac {x}{-8 x^2+e^x (-3+\log (4))^2} \, dx+16 \int \frac {1}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx+\int \left (\frac {2}{x^2}+\frac {1}{x}\right ) \, dx-\int \frac {32 x^2-e^x (2+x) (-3+\log (4))^2}{x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )} \, dx\\ &=-\frac {2}{x}+\log (x)+\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x}+8 \int \frac {x}{-8 x^2+e^x (-3+\log (4))^2} \, dx+16 \int \frac {1}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx-\int \left (\frac {2+x}{x^2}+\frac {8 (2-x)}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )}\right ) \, dx\\ &=-\frac {2}{x}+\log (x)+\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x}+8 \int \frac {x}{-8 x^2+e^x (-3+\log (4))^2} \, dx-8 \int \frac {2-x}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx+16 \int \frac {1}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx-\int \frac {2+x}{x^2} \, dx\\ &=-\frac {2}{x}+\log (x)+\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x}+8 \int \frac {x}{-8 x^2+e^x (-3+\log (4))^2} \, dx-8 \int \left (\frac {2}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )}+\frac {x}{-8 x^2+9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )}\right ) \, dx+16 \int \frac {1}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx-\int \left (\frac {2}{x^2}+\frac {1}{x}\right ) \, dx\\ &=\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x}+8 \int \frac {x}{-8 x^2+e^x (-3+\log (4))^2} \, dx-8 \int \frac {x}{-8 x^2+9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx\\ &=\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 29, normalized size = 1.04 \begin {gather*} \frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.69, size = 37, normalized size = 1.32 \begin {gather*} \frac {\log \left (-4 \, x^{4} + \frac {1}{2} \, {\left (4 \, x^{2} \log \relax (2)^{2} - 12 \, x^{2} \log \relax (2) + 9 \, x^{2}\right )} e^{x}\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.65, size = 44, normalized size = 1.57 \begin {gather*} -\frac {\log \relax (2) - \log \left (4 \, x^{2} e^{x} \log \relax (2)^{2} - 8 \, x^{4} - 12 \, x^{2} e^{x} \log \relax (2) + 9 \, x^{2} e^{x}\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.27, size = 38, normalized size = 1.36
method | result | size |
norman | \(\frac {\ln \left (\frac {\left (4 x^{2} \ln \relax (2)^{2}-12 x^{2} \ln \relax (2)+9 x^{2}\right ) {\mathrm e}^{x}}{2}-4 x^{4}\right )}{x}\) | \(38\) |
risch | \(\frac {\ln \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )}{x}-\frac {i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right ) \mathrm {csgn}\left (i x^{2} \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right )-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right ) \mathrm {csgn}\left (i x^{2} \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right )^{2}+i \pi \mathrm {csgn}\left (i x^{2} \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right )^{3}+2 \ln \relax (2)-4 \ln \relax (x )}{2 x}\) | \(303\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.61, size = 36, normalized size = 1.29 \begin {gather*} -\frac {\log \relax (2) - \log \left (-8 \, x^{2} + {\left (4 \, \log \relax (2)^{2} - 12 \, \log \relax (2) + 9\right )} e^{x}\right ) - 2 \, \log \relax (x)}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.67, size = 37, normalized size = 1.32 \begin {gather*} \frac {\ln \left (\frac {{\mathrm {e}}^x\,\left (4\,x^2\,{\ln \relax (2)}^2-12\,x^2\,\ln \relax (2)+9\,x^2\right )}{2}-4\,x^4\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.66, size = 36, normalized size = 1.29 \begin {gather*} \frac {\log {\left (- 4 x^{4} + \left (- 6 x^{2} \log {\relax (2 )} + 2 x^{2} \log {\relax (2 )}^{2} + \frac {9 x^{2}}{2}\right ) e^{x} \right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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