Integrand size = 63, antiderivative size = 28 \[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {1}{4} \log ^2\left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right ) \]
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Time = 0.85 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6873, 12, 6874, 2339, 29, 6820, 6818} \[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {1}{4} \log ^2\left (\frac {3}{4 x \log \left (e^5-2\right ) \log (x)}+4\right ) \]
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Rule 12
Rule 29
Rule 2339
Rule 6818
Rule 6820
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {3 (-1-\log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{2 x \log (x) \left (3+16 x \log \left (-2+e^5\right ) \log (x)\right )} \, dx \\ & = \frac {3}{2} \int \frac {(-1-\log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \log (x) \left (3+16 x \log \left (-2+e^5\right ) \log (x)\right )} \, dx \\ & = \frac {3}{2} \int \frac {(-1-\log (x)) \log \left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \log (x) \left (3+16 x \log \left (-2+e^5\right ) \log (x)\right )} \, dx \\ & = \frac {1}{4} \log ^2\left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right ) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {1}{4} \log ^2\left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(88\) vs. \(2(23)=46\).
Time = 4.44 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.18
method | result | size |
default | \(\frac {3 \ln \left (\ln \left ({\mathrm e}^{5}-2\right )\right ) \left (\frac {\ln \left (x \right )}{3}+\frac {\ln \left (\ln \left (x \right )\right )}{3}-\frac {\ln \left (16 x \ln \left (x \right ) \ln \left ({\mathrm e}^{5}-2\right )+3\right )}{3}\right )}{2}+3 \ln \left (2\right ) \left (\frac {\ln \left (x \right )}{3}+\frac {\ln \left (\ln \left (x \right )\right )}{3}-\frac {\ln \left (16 x \ln \left (x \right ) \ln \left ({\mathrm e}^{5}-2\right )+3\right )}{3}\right )+\frac {{\ln \left (\frac {16 x \ln \left (x \right ) \ln \left ({\mathrm e}^{5}-2\right )+3}{x \ln \left (x \right )}\right )}^{2}}{4}\) | \(89\) |
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none
Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {1}{4} \, \log \left (\frac {16 \, x \log \left (x\right ) \log \left (e^{5} - 2\right ) + 3}{4 \, x \log \left (x\right ) \log \left (e^{5} - 2\right )}\right )^{2} \]
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Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {\log {\left (\frac {4 x \log {\left (x \right )} \log {\left (-2 + e^{5} \right )} + \frac {3}{4}}{x \log {\left (x \right )} \log {\left (-2 + e^{5} \right )}} \right )}^{2}}{4} \]
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\[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\int { -\frac {3 \, {\left (\log \left (x\right ) + 1\right )} \log \left (\frac {16 \, x \log \left (x\right ) \log \left (e^{5} - 2\right ) + 3}{4 \, x \log \left (x\right ) \log \left (e^{5} - 2\right )}\right )}{2 \, {\left (16 \, x^{2} \log \left (x\right )^{2} \log \left (e^{5} - 2\right ) + 3 \, x \log \left (x\right )\right )}} \,d x } \]
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\[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\int { -\frac {3 \, {\left (\log \left (x\right ) + 1\right )} \log \left (\frac {16 \, x \log \left (x\right ) \log \left (e^{5} - 2\right ) + 3}{4 \, x \log \left (x\right ) \log \left (e^{5} - 2\right )}\right )}{2 \, {\left (16 \, x^{2} \log \left (x\right )^{2} \log \left (e^{5} - 2\right ) + 3 \, x \log \left (x\right )\right )}} \,d x } \]
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Time = 14.53 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {{\ln \left (\frac {4\,x\,\ln \left ({\mathrm {e}}^5-2\right )\,\ln \left (x\right )+\frac {3}{4}}{x\,\ln \left ({\mathrm {e}}^5-2\right )\,\ln \left (x\right )}\right )}^2}{4} \]
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