Optimal. Leaf size=23 \[ (-3+x) \log \left (\log \left (x-\left (e^x+e^x x^2\right ) \log (16)\right )\right ) \]
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Rubi [F] time = 5.44, 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 {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {(-3+x) \left (-1+e^x (1+x)^2 \log (16)\right )}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right )\right ) \, dx\\ &=\int \frac {(-3+x) \left (-1+e^x (1+x)^2 \log (16)\right )}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=\int \left (\frac {(-3+x) (1+x)^2}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {3-4 x-2 x^2-2 x^3+x^4}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=\int \frac {(-3+x) (1+x)^2}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {3-4 x-2 x^2-2 x^3+x^4}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=\int \left (-\frac {1}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {x}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}-\frac {2 (1+3 x)}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx+\int \left (-\frac {3}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}-\frac {2 x}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {x^2}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}-\frac {2 (-3+x)}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=-\left (2 \int \frac {1+3 x}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx\right )-2 \int \frac {x}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-2 \int \frac {-3+x}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-3 \int \frac {1}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-\int \frac {1}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x^2}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=-\left (2 \int \left (\frac {1}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {3 x}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx\right )-2 \int \left (-\frac {3}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {x}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx-2 \int \frac {x}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-3 \int \frac {1}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-\int \frac {1}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x^2}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=-\left (2 \int \frac {1}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx\right )-2 \int \frac {x}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-2 \int \frac {x}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-3 \int \frac {1}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-6 \int \frac {x}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+6 \int \frac {1}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-\int \frac {1}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x^2}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=-\left (2 \int \left (\frac {i}{2 (i-x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {i}{2 (i+x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx\right )-2 \int \left (-\frac {1}{2 (i-x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {1}{2 (i+x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx-2 \int \frac {x}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-3 \int \frac {1}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-6 \int \left (-\frac {1}{2 (i-x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {1}{2 (i+x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx+6 \int \left (\frac {i}{2 (i-x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {i}{2 (i+x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx-\int \frac {1}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x^2}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=-\left (i \int \frac {1}{(i-x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx\right )-i \int \frac {1}{(i+x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+3 i \int \frac {1}{(i-x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+3 i \int \frac {1}{(i+x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-2 \int \frac {x}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+3 \int \frac {1}{(i-x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-3 \int \frac {1}{(i+x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-3 \int \frac {1}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-\int \frac {1}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {1}{(i-x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x^2}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-\int \frac {1}{(i+x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.86, size = 37, normalized size = 1.61 \begin {gather*} -3 \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right )+x \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.79, size = 19, normalized size = 0.83 \begin {gather*} {\left (x - 3\right )} \log \left (\log \left (-4 \, {\left (x^{2} + 1\right )} e^{x} \log \relax (2) + x\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.53, size = 43, normalized size = 1.87 \begin {gather*} x \log \left (\log \left (-4 \, x^{2} e^{x} \log \relax (2) - 4 \, e^{x} \log \relax (2) + x\right )\right ) - 3 \, \log \left (\log \left (-4 \, x^{2} e^{x} \log \relax (2) - 4 \, e^{x} \log \relax (2) + x\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 40, normalized size = 1.74
method | result | size |
risch | \(\ln \left (\ln \left (4 \left (-x^{2}-1\right ) \ln \relax (2) {\mathrm e}^{x}+x \right )\right ) x -3 \ln \left (\ln \left (4 \left (-x^{2}-1\right ) \ln \relax (2) {\mathrm e}^{x}+x \right )\right )\) | \(40\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.55, size = 21, normalized size = 0.91 \begin {gather*} {\left (x - 3\right )} \log \left (\log \left (-4 \, {\left (x^{2} \log \relax (2) + \log \relax (2)\right )} e^{x} + x\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.87, size = 19, normalized size = 0.83 \begin {gather*} \ln \left (\ln \left (x-4\,{\mathrm {e}}^x\,\ln \relax (2)\,\left (x^2+1\right )\right )\right )\,\left (x-3\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.69, size = 44, normalized size = 1.91 \begin {gather*} \left (x - 1\right ) \log {\left (\log {\left (x + \left (- 4 x^{2} - 4\right ) e^{x} \log {\relax (2 )} \right )} \right )} - 2 \log {\left (\log {\left (x + \left (- 4 x^{2} - 4\right ) e^{x} \log {\relax (2 )} \right )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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