Optimal. Leaf size=28 \[ x \left (-x+\log ^2\left (4 \left (\left (4+e^{x/16}\right ) (-3+x)-x\right )\right )\right ) \]
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Rubi [F] time = 6.88, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {192 x-48 x^2+e^{x/16} \left (48 x-16 x^2\right )+\left (48 x+e^{x/16} \left (13 x+x^2\right )\right ) \log \left (-48+12 x+e^{x/16} (-12+4 x)\right )+\left (-96+24 x+e^{x/16} (-24+8 x)\right ) \log ^2\left (-48+12 x+e^{x/16} (-12+4 x)\right )}{-96+24 x+e^{x/16} (-24+8 x)} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-2 x+\frac {x \left (48+e^{x/16} (13+x)\right ) \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )}{8 \left (3 (-4+x)+e^{x/16} (-3+x)\right )}+\log ^2\left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \, dx\\ &=-x^2+\frac {1}{8} \int \frac {x \left (48+e^{x/16} (13+x)\right ) \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )}{3 (-4+x)+e^{x/16} (-3+x)} \, dx+\int \log ^2\left (12 (-4+x)+4 e^{x/16} (-3+x)\right ) \, dx\\ &=-x^2+\frac {1}{8} \int \left (\frac {x (13+x) \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )}{-3+x}-\frac {3 x \left (-4-7 x+x^2\right ) \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )}{(-3+x) \left (-12-3 e^{x/16}+3 x+e^{x/16} x\right )}\right ) \, dx+\int \log ^2\left (12 (-4+x)+4 e^{x/16} (-3+x)\right ) \, dx\\ &=-x^2+\frac {1}{8} \int \frac {x (13+x) \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )}{-3+x} \, dx-\frac {3}{8} \int \frac {x \left (-4-7 x+x^2\right ) \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )}{(-3+x) \left (-12-3 e^{x/16}+3 x+e^{x/16} x\right )} \, dx+\int \log ^2\left (12 (-4+x)+4 e^{x/16} (-3+x)\right ) \, dx\\ &=-x^2+\frac {1}{8} \int \left (16 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )+\frac {48 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )}{-3+x}+x \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \, dx+\frac {3}{8} \int \frac {\left (48+e^{x/16} (13+x)\right ) \left (-16 \int \frac {1}{3 (-4+x)+e^{x/16} (-3+x)} \, dx-48 \int \frac {1}{\left (3 (-4+x)+e^{x/16} (-3+x)\right ) (-3+x)} \, dx-4 \int \frac {x}{3 (-4+x)+e^{x/16} (-3+x)} \, dx+\int \frac {x^2}{3 (-4+x)+e^{x/16} (-3+x)} \, dx\right )}{16 \left (3 (-4+x)+e^{x/16} (-3+x)\right )} \, dx-\frac {1}{8} \left (3 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {x^2}{-12-3 e^{x/16}+3 x+e^{x/16} x} \, dx+\frac {1}{2} \left (3 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {x}{-12-3 e^{x/16}+3 x+e^{x/16} x} \, dx+\left (6 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {1}{-12-3 e^{x/16}+3 x+e^{x/16} x} \, dx+\left (18 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {1}{(-3+x) \left (-12-3 e^{x/16}+3 x+e^{x/16} x\right )} \, dx+\int \log ^2\left (12 (-4+x)+4 e^{x/16} (-3+x)\right ) \, dx\\ &=-x^2+\frac {3}{128} \int \frac {\left (48+e^{x/16} (13+x)\right ) \left (-16 \int \frac {1}{3 (-4+x)+e^{x/16} (-3+x)} \, dx-48 \int \frac {1}{\left (3 (-4+x)+e^{x/16} (-3+x)\right ) (-3+x)} \, dx-4 \int \frac {x}{3 (-4+x)+e^{x/16} (-3+x)} \, dx+\int \frac {x^2}{3 (-4+x)+e^{x/16} (-3+x)} \, dx\right )}{3 (-4+x)+e^{x/16} (-3+x)} \, dx+\frac {1}{8} \int x \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right ) \, dx+2 \int \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right ) \, dx+6 \int \frac {\log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )}{-3+x} \, dx-\frac {1}{8} \left (3 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {x^2}{-12-3 e^{x/16}+3 x+e^{x/16} x} \, dx+\frac {1}{2} \left (3 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {x}{-12-3 e^{x/16}+3 x+e^{x/16} x} \, dx+\left (6 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {1}{-12-3 e^{x/16}+3 x+e^{x/16} x} \, dx+\left (18 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {1}{(-3+x) \left (-12-3 e^{x/16}+3 x+e^{x/16} x\right )} \, dx+\int \log ^2\left (12 (-4+x)+4 e^{x/16} (-3+x)\right ) \, dx\\ &=-x^2+2 x \log \left (-4 e^{x/16} (3-x)-12 (4-x)\right )+\frac {1}{16} x^2 \log \left (-4 e^{x/16} (3-x)-12 (4-x)\right )+\frac {3}{128} \int \left (-\frac {(13+x) \left (16 \int \frac {1}{3 (-4+x)+e^{x/16} (-3+x)} \, dx+48 \int \frac {1}{\left (3 (-4+x)+e^{x/16} (-3+x)\right ) (-3+x)} \, dx+4 \int \frac {x}{3 (-4+x)+e^{x/16} (-3+x)} \, dx-\int \frac {x^2}{3 (-4+x)+e^{x/16} (-3+x)} \, dx\right )}{-3+x}+\frac {3 \left (-4-7 x+x^2\right ) \left (16 \int \frac {1}{3 (-4+x)+e^{x/16} (-3+x)} \, dx+48 \int \frac {1}{\left (3 (-4+x)+e^{x/16} (-3+x)\right ) (-3+x)} \, dx+4 \int \frac {x}{3 (-4+x)+e^{x/16} (-3+x)} \, dx-\int \frac {x^2}{3 (-4+x)+e^{x/16} (-3+x)} \, dx\right )}{(-3+x) \left (-12-3 e^{x/16}+3 x+e^{x/16} x\right )}\right ) \, dx-\frac {1}{16} \int \frac {x^2 \left (48+e^{x/16} (13+x)\right )}{16 \left (3 (-4+x)+e^{x/16} (-3+x)\right )} \, dx-2 \int \frac {x \left (48+e^{x/16} (13+x)\right )}{16 \left (3 (-4+x)+e^{x/16} (-3+x)\right )} \, dx+6 \int \frac {\log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )}{-3+x} \, dx-\frac {1}{8} \left (3 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {x^2}{-12-3 e^{x/16}+3 x+e^{x/16} x} \, dx+\frac {1}{2} \left (3 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {x}{-12-3 e^{x/16}+3 x+e^{x/16} x} \, dx+\left (6 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {1}{-12-3 e^{x/16}+3 x+e^{x/16} x} \, dx+\left (18 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {1}{(-3+x) \left (-12-3 e^{x/16}+3 x+e^{x/16} x\right )} \, dx+\int \log ^2\left (12 (-4+x)+4 e^{x/16} (-3+x)\right ) \, dx\\ &=-x^2+2 x \log \left (-4 e^{x/16} (3-x)-12 (4-x)\right )+\frac {1}{16} x^2 \log \left (-4 e^{x/16} (3-x)-12 (4-x)\right )-\frac {1}{256} \int \frac {x^2 \left (48+e^{x/16} (13+x)\right )}{3 (-4+x)+e^{x/16} (-3+x)} \, dx-\frac {3}{128} \int \frac {(13+x) \left (16 \int \frac {1}{3 (-4+x)+e^{x/16} (-3+x)} \, dx+48 \int \frac {1}{\left (3 (-4+x)+e^{x/16} (-3+x)\right ) (-3+x)} \, dx+4 \int \frac {x}{3 (-4+x)+e^{x/16} (-3+x)} \, dx-\int \frac {x^2}{3 (-4+x)+e^{x/16} (-3+x)} \, dx\right )}{-3+x} \, dx+\frac {9}{128} \int \frac {\left (-4-7 x+x^2\right ) \left (16 \int \frac {1}{3 (-4+x)+e^{x/16} (-3+x)} \, dx+48 \int \frac {1}{\left (3 (-4+x)+e^{x/16} (-3+x)\right ) (-3+x)} \, dx+4 \int \frac {x}{3 (-4+x)+e^{x/16} (-3+x)} \, dx-\int \frac {x^2}{3 (-4+x)+e^{x/16} (-3+x)} \, dx\right )}{(-3+x) \left (-12-3 e^{x/16}+3 x+e^{x/16} x\right )} \, dx-\frac {1}{8} \int \frac {x \left (48+e^{x/16} (13+x)\right )}{3 (-4+x)+e^{x/16} (-3+x)} \, dx+6 \int \frac {\log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )}{-3+x} \, dx-\frac {1}{8} \left (3 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {x^2}{-12-3 e^{x/16}+3 x+e^{x/16} x} \, dx+\frac {1}{2} \left (3 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {x}{-12-3 e^{x/16}+3 x+e^{x/16} x} \, dx+\left (6 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {1}{-12-3 e^{x/16}+3 x+e^{x/16} x} \, dx+\left (18 \log \left (12 (-4+x)+4 e^{x/16} (-3+x)\right )\right ) \int \frac {1}{(-3+x) \left (-12-3 e^{x/16}+3 x+e^{x/16} x\right )} \, dx+\int \log ^2\left (12 (-4+x)+4 e^{x/16} (-3+x)\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [F] time = 1.17, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {192 x-48 x^2+e^{x/16} \left (48 x-16 x^2\right )+\left (48 x+e^{x/16} \left (13 x+x^2\right )\right ) \log \left (-48+12 x+e^{x/16} (-12+4 x)\right )+\left (-96+24 x+e^{x/16} (-24+8 x)\right ) \log ^2\left (-48+12 x+e^{x/16} (-12+4 x)\right )}{-96+24 x+e^{x/16} (-24+8 x)} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.62, size = 25, normalized size = 0.89 \begin {gather*} x \log \left (4 \, {\left (x - 3\right )} e^{\left (\frac {1}{16} \, x\right )} + 12 \, x - 48\right )^{2} - x^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.26, size = 29, normalized size = 1.04 \begin {gather*} x \log \left (4 \, x e^{\left (\frac {1}{16} \, x\right )} + 12 \, x - 12 \, e^{\left (\frac {1}{16} \, x\right )} - 48\right )^{2} - x^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 27, normalized size = 0.96
method | result | size |
norman | \(x \ln \left (\left (4 x -12\right ) {\mathrm e}^{\frac {x}{16}}+12 x -48\right )^{2}-x^{2}\) | \(27\) |
risch | \(x \ln \left (\left (4 x -12\right ) {\mathrm e}^{\frac {x}{16}}+12 x -48\right )^{2}-x^{2}\) | \(27\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.91, size = 50, normalized size = 1.79 \begin {gather*} 4 \, x \log \relax (2)^{2} + 4 \, x \log \relax (2) \log \left ({\left (x - 3\right )} e^{\left (\frac {1}{16} \, x\right )} + 3 \, x - 12\right ) + x \log \left ({\left (x - 3\right )} e^{\left (\frac {1}{16} \, x\right )} + 3 \, x - 12\right )^{2} - x^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.67, size = 25, normalized size = 0.89 \begin {gather*} -x\,\left (x-{\ln \left (12\,x+{\mathrm {e}}^{x/16}\,\left (4\,x-12\right )-48\right )}^2\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.45, size = 22, normalized size = 0.79 \begin {gather*} - x^{2} + x \log {\left (12 x + \left (4 x - 12\right ) e^{\frac {x}{16}} - 48 \right )}^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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