Integrand size = 68, antiderivative size = 25 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3}{2} e^{\frac {5 (3+e) (2-x)}{\log (4)}} x \log \left (x^2\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(132\) vs. \(2(25)=50\).
Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 5.28, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 2259, 2225, 6, 2218, 2207, 2634} \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {e^{\frac {5 (3+e) (2-x)}{\log (4)}} (15 (3+e) x-\log (64)) \log \left (x^2\right )}{10 (3+e)}+\frac {3 \log (4) e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log \left (x^2\right )}{10 (3+e)}+\frac {3 \log (4) e^{\frac {10 (3+e)}{\log (4)}-\frac {5 (3+e) x}{\log (4)}}}{5 (3+e)}-\frac {3 \log (4) e^{\frac {5 (3+e) (2-x)}{\log (4)}}}{5 (3+e)} \]
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Rule 6
Rule 12
Rule 2207
Rule 2218
Rule 2225
Rule 2259
Rule 2634
Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )\right ) \, dx}{2 \log (4)} \\ & = 3 \int e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \, dx+\frac {\int e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right ) \, dx}{2 \log (4)} \\ & = 3 \int e^{\frac {10 (3+e)-5 (3+e) x}{\log (4)}} \, dx+\frac {\int e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} ((-45-15 e) x+3 \log (4)) \log \left (x^2\right ) \, dx}{2 \log (4)} \\ & = -\frac {3 e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log (4)}{5 (3+e)}+\frac {3 e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log (4) \log \left (x^2\right )}{10 (3+e)}+\frac {e^{\frac {5 (3+e) (2-x)}{\log (4)}} (15 (3+e) x-\log (64)) \log \left (x^2\right )}{10 (3+e)}-\frac {\int 6 e^{\frac {10 (3+e)}{\log (4)}-\frac {5 (3+e) x}{\log (4)}} \log (4) \, dx}{2 \log (4)} \\ & = -\frac {3 e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log (4)}{5 (3+e)}+\frac {3 e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log (4) \log \left (x^2\right )}{10 (3+e)}+\frac {e^{\frac {5 (3+e) (2-x)}{\log (4)}} (15 (3+e) x-\log (64)) \log \left (x^2\right )}{10 (3+e)}-3 \int e^{\frac {10 (3+e)}{\log (4)}-\frac {5 (3+e) x}{\log (4)}} \, dx \\ & = \frac {3 e^{\frac {10 (3+e)}{\log (4)}-\frac {5 (3+e) x}{\log (4)}} \log (4)}{5 (3+e)}-\frac {3 e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log (4)}{5 (3+e)}+\frac {3 e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log (4) \log \left (x^2\right )}{10 (3+e)}+\frac {e^{\frac {5 (3+e) (2-x)}{\log (4)}} (15 (3+e) x-\log (64)) \log \left (x^2\right )}{10 (3+e)} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3 e^{-\frac {5 (3+e) (-2+x)}{\log (4)}} x \log (4) \log \left (x^2\right )}{\log (16)} \]
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Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12
method | result | size |
norman | \(\frac {3 x \,{\mathrm e}^{\frac {\left (-5 x +10\right ) {\mathrm e}-15 x +30}{2 \ln \left (2\right )}} \ln \left (x^{2}\right )}{2}\) | \(28\) |
parallelrisch | \(\frac {3 \,{\mathrm e}^{-\frac {5 \left (x \,{\mathrm e}-2 \,{\mathrm e}+3 x -6\right )}{2 \ln \left (2\right )}} \ln \left (x^{2}\right ) x}{2}\) | \(28\) |
risch | \(3 x \,{\mathrm e}^{-\frac {5 \left (-2+x \right ) \left (3+{\mathrm e}\right )}{2 \ln \left (2\right )}} \ln \left (x \right )-\frac {3 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \left (\operatorname {csgn}\left (i x \right )^{2}-2 \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )+\operatorname {csgn}\left (i x^{2}\right )^{2}\right ) x \,{\mathrm e}^{-\frac {5 \left (-2+x \right ) \left (3+{\mathrm e}\right )}{2 \ln \left (2\right )}}}{4}\) | \(78\) |
default | \(\frac {\frac {24 \ln \left (2\right )^{2} {\mathrm e}^{\frac {\left (-5 x +10\right ) {\mathrm e}-15 x +30}{2 \ln \left (2\right )}}}{5 \left (3+{\mathrm e}\right )}+12 x \ln \left (2\right ) {\mathrm e}^{\frac {\left (-5 x +10\right ) {\mathrm e}-15 x +30}{2 \ln \left (2\right )}} \ln \left (x \right )+6 \ln \left (2\right ) \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) x \,{\mathrm e}^{\frac {\left (-5 x +10\right ) {\mathrm e}-15 x +30}{2 \ln \left (2\right )}}+\frac {24 \ln \left (2\right )^{2} {\mathrm e}^{\frac {\left (-5 \,{\mathrm e}-15\right ) x}{2 \ln \left (2\right )}+\frac {10 \,{\mathrm e}+30}{2 \ln \left (2\right )}}}{-5 \,{\mathrm e}-15}}{4 \ln \left (2\right )}\) | \(142\) |
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Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3}{2} \, x e^{\left (-\frac {5 \, {\left ({\left (x - 2\right )} e + 3 \, x - 6\right )}}{2 \, \log \left (2\right )}\right )} \log \left (x^{2}\right ) \]
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Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3 x e^{\frac {- \frac {15 x}{2} + \frac {e \left (10 - 5 x\right )}{2} + 15}{\log {\left (2 \right )}}} \log {\left (x^{2} \right )}}{2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (21) = 42\).
Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 5.20 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3 \, {\left (5 \, x e^{\left (-\frac {5 \, x e}{2 \, \log \left (2\right )} - \frac {15 \, x}{2 \, \log \left (2\right )} + \frac {5 \, e}{\log \left (2\right )} + \frac {15}{\log \left (2\right )}\right )} \log \left (2\right ) \log \left (x^{2}\right ) - \frac {4 \, e^{\left (-\frac {5 \, {\left ({\left (x - 2\right )} e + 3 \, x - 6\right )}}{2 \, \log \left (2\right )}\right )} \log \left (2\right )^{2}}{e + 3} + \frac {4 \, e^{\left (-\frac {5 \, x e}{2 \, \log \left (2\right )} - \frac {15 \, x}{2 \, \log \left (2\right )} + \frac {5 \, e}{\log \left (2\right )} + \frac {15}{\log \left (2\right )}\right )} \log \left (2\right )}{\frac {e}{\log \left (2\right )} + \frac {3}{\log \left (2\right )}}\right )}}{10 \, \log \left (2\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (21) = 42\).
Time = 0.34 (sec) , antiderivative size = 198, normalized size of antiderivative = 7.92 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3 \, {\left (\frac {4 \, e^{\left (-\frac {5 \, x e}{2 \, \log \left (2\right )} - \frac {15 \, x}{2 \, \log \left (2\right )} + \frac {5 \, e}{\log \left (2\right )} + \frac {15}{\log \left (2\right )} + 1\right )} \log \left (2\right )^{2}}{e^{2} + 3 \, e} - \frac {4 \, e^{\left (-\frac {5 \, {\left ({\left (x - 2\right )} e + 3 \, x - 6\right )}}{2 \, \log \left (2\right )}\right )} \log \left (2\right )^{2}}{e + 3} + {\left (\frac {{\left (5 \, x e \log \left (2\right ) + 15 \, x \log \left (2\right ) + 2 \, \log \left (2\right )^{2}\right )} e^{\left (-\frac {5 \, x e + 15 \, x - 10 \, e - 2 \, \log \left (2\right ) - 30}{2 \, \log \left (2\right )}\right )}}{e^{2} + 6 \, e + 9} + \frac {{\left (15 \, x e \log \left (2\right ) - 2 \, e \log \left (2\right )^{2} + 45 \, x \log \left (2\right )\right )} e^{\left (-\frac {5 \, {\left (x e + 3 \, x - 2 \, e - 6\right )}}{2 \, \log \left (2\right )}\right )}}{e^{2} + 6 \, e + 9}\right )} \log \left (x^{2}\right )\right )}}{10 \, \log \left (2\right )} \]
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Time = 8.45 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3\,x\,\ln \left (x^2\right )\,{\mathrm {e}}^{-\frac {15\,x}{2\,\ln \left (2\right )}}\,{\mathrm {e}}^{-\frac {5\,x\,\mathrm {e}}{2\,\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {15}{\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {5\,\mathrm {e}}{\ln \left (2\right )}}}{2} \]
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