Integrand size = 124, antiderivative size = 28 \[ \int \frac {-32 x^2+e^x \left (18+9 x+(-12-6 x) \log (4)+(2+x) \log ^2(4)\right )+\left (8 x^2+e^x \left (-9+6 \log (4)-\log ^2(4)\right )\right ) \log \left (\frac {1}{2} \left (-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )\right )\right )}{-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )} \, dx=\frac {\log \left (x^2 \left (-4 x^2+\frac {1}{2} e^x (-3+\log (4))^2\right )\right )}{x} \]
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Time = 1.54 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11, number of steps used = 18, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {6820, 6874, 14, 45, 2631} \[ \int \frac {-32 x^2+e^x \left (18+9 x+(-12-6 x) \log (4)+(2+x) \log ^2(4)\right )+\left (8 x^2+e^x \left (-9+6 \log (4)-\log ^2(4)\right )\right ) \log \left (\frac {1}{2} \left (-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )\right )\right )}{-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )} \, dx=\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x} \]
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Rule 14
Rule 45
Rule 2631
Rule 6820
Rule 6874
Rubi steps \begin{align*} \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{align*}
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-32 x^2+e^x \left (18+9 x+(-12-6 x) \log (4)+(2+x) \log ^2(4)\right )+\left (8 x^2+e^x \left (-9+6 \log (4)-\log ^2(4)\right )\right ) \log \left (\frac {1}{2} \left (-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )\right )\right )}{-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )} \, dx=\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )\right )}{x} \]
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Time = 0.85 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36
method | result | size |
norman | \(\frac {\ln \left (\frac {\left (4 x^{2} \ln \left (2\right )^{2}-12 x^{2} \ln \left (2\right )+9 x^{2}\right ) {\mathrm e}^{x}}{2}-4 x^{4}\right )}{x}\) | \(38\) |
parallelrisch | \(\frac {\ln \left (\frac {\left (4 x^{2} \ln \left (2\right )^{2}-12 x^{2} \ln \left (2\right )+9 x^{2}\right ) {\mathrm e}^{x}}{2}-4 x^{4}\right )}{x}\) | \(38\) |
risch | \(\frac {\ln \left (\ln \left (2\right )^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \left (2\right )-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )}{x}-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i \left (\ln \left (2\right )^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \left (2\right )-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right ) \operatorname {csgn}\left (i x^{2} \left (\ln \left (2\right )^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \left (2\right )-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right )-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{2} \left (\ln \left (2\right )^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \left (2\right )-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right )^{2}-i \pi \,\operatorname {csgn}\left (i \left (\ln \left (2\right )^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \left (2\right )-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right ) \operatorname {csgn}\left (i x^{2} \left (\ln \left (2\right )^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \left (2\right )-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right )^{2}+i \pi \operatorname {csgn}\left (i x^{2} \left (\ln \left (2\right )^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \left (2\right )-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right )^{3}+2 \ln \left (2\right )-4 \ln \left (x \right )}{2 x}\) | \(303\) |
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Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-32 x^2+e^x \left (18+9 x+(-12-6 x) \log (4)+(2+x) \log ^2(4)\right )+\left (8 x^2+e^x \left (-9+6 \log (4)-\log ^2(4)\right )\right ) \log \left (\frac {1}{2} \left (-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )\right )\right )}{-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )} \, dx=\frac {\log \left (-4 \, x^{4} + \frac {1}{2} \, {\left (4 \, x^{2} \log \left (2\right )^{2} - 12 \, x^{2} \log \left (2\right ) + 9 \, x^{2}\right )} e^{x}\right )}{x} \]
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Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-32 x^2+e^x \left (18+9 x+(-12-6 x) \log (4)+(2+x) \log ^2(4)\right )+\left (8 x^2+e^x \left (-9+6 \log (4)-\log ^2(4)\right )\right ) \log \left (\frac {1}{2} \left (-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )\right )\right )}{-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )} \, dx=\frac {\log {\left (- 4 x^{4} + \left (- 6 x^{2} \log {\left (2 \right )} + 2 x^{2} \log {\left (2 \right )}^{2} + \frac {9 x^{2}}{2}\right ) e^{x} \right )}}{x} \]
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Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-32 x^2+e^x \left (18+9 x+(-12-6 x) \log (4)+(2+x) \log ^2(4)\right )+\left (8 x^2+e^x \left (-9+6 \log (4)-\log ^2(4)\right )\right ) \log \left (\frac {1}{2} \left (-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )\right )\right )}{-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )} \, dx=-\frac {\log \left (2\right ) - \log \left (-8 \, x^{2} + {\left (4 \, \log \left (2\right )^{2} - 12 \, \log \left (2\right ) + 9\right )} e^{x}\right ) - 2 \, \log \left (x\right )}{x} \]
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Time = 0.33 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {-32 x^2+e^x \left (18+9 x+(-12-6 x) \log (4)+(2+x) \log ^2(4)\right )+\left (8 x^2+e^x \left (-9+6 \log (4)-\log ^2(4)\right )\right ) \log \left (\frac {1}{2} \left (-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )\right )\right )}{-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )} \, dx=-\frac {\log \left (2\right ) - \log \left (4 \, x^{2} e^{x} \log \left (2\right )^{2} - 8 \, x^{4} - 12 \, x^{2} e^{x} \log \left (2\right ) + 9 \, x^{2} e^{x}\right )}{x} \]
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Time = 10.79 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-32 x^2+e^x \left (18+9 x+(-12-6 x) \log (4)+(2+x) \log ^2(4)\right )+\left (8 x^2+e^x \left (-9+6 \log (4)-\log ^2(4)\right )\right ) \log \left (\frac {1}{2} \left (-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )\right )\right )}{-8 x^4+e^x \left (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4)\right )} \, dx=\frac {\ln \left (\frac {{\mathrm {e}}^x\,\left (4\,x^2\,{\ln \left (2\right )}^2-12\,x^2\,\ln \left (2\right )+9\,x^2\right )}{2}-4\,x^4\right )}{x} \]
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