Integrand size = 75, antiderivative size = 33 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=\left (5+e^{2 x}\right ) \left (-x+e^4 x-\log (2+x)\right ) \log (\log (i \pi +\log (2))) \]
[Out]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.54 (sec) , antiderivative size = 124, normalized size of antiderivative = 3.76, number of steps used = 16, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {12, 6874, 45, 6820, 2230, 2225, 2209, 2207, 2634} \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=-\frac {\log (\log (\log (2)+i \pi )) \operatorname {ExpIntegralEi}(2 (x+2))}{e^4}+\frac {\log (\log (\log (2)+i \pi )) \operatorname {ExpIntegralEi}(2 x+4)}{e^4}+\left (1-e^4\right ) \left (-e^{2 x}\right ) x \log (\log (\log (2)+i \pi ))-5 \left (1-e^4\right ) x \log (\log (\log (2)+i \pi ))-e^{2 x} \log (\log (\log (2)+i \pi )) \log (x+2)-5 \log (\log (\log (2)+i \pi )) \log (x+2) \]
[In]
[Out]
Rule 12
Rule 45
Rule 2207
Rule 2209
Rule 2225
Rule 2230
Rule 2634
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \log (\log (i \pi +\log (2))) \int \frac {-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)}{2+x} \, dx \\ & = \log (\log (i \pi +\log (2))) \int \left (\frac {5 \left (-3+2 e^4-\left (1-e^4\right ) x\right )}{2+x}+\frac {e^{2 x} \left (-3 \left (1-\frac {2 e^4}{3}\right )-5 \left (1-e^4\right ) x-2 \left (1-e^4\right ) x^2-4 \log (2+x)-2 x \log (2+x)\right )}{2+x}\right ) \, dx \\ & = \log (\log (i \pi +\log (2))) \int \frac {e^{2 x} \left (-3 \left (1-\frac {2 e^4}{3}\right )-5 \left (1-e^4\right ) x-2 \left (1-e^4\right ) x^2-4 \log (2+x)-2 x \log (2+x)\right )}{2+x} \, dx+(5 \log (\log (i \pi +\log (2)))) \int \frac {-3+2 e^4-\left (1-e^4\right ) x}{2+x} \, dx \\ & = \log (\log (i \pi +\log (2))) \int \frac {e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )-2 (2+x) \log (2+x)\right )}{2+x} \, dx+(5 \log (\log (i \pi +\log (2)))) \int \left (-1+e^4+\frac {1}{-2-x}\right ) \, dx \\ & = -5 \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-5 \log (2+x) \log (\log (i \pi +\log (2)))+\log (\log (i \pi +\log (2))) \int \left (\frac {e^{2 x} \left (-3+2 e^4-5 \left (1-e^4\right ) x-2 \left (1-e^4\right ) x^2\right )}{2+x}-2 e^{2 x} \log (2+x)\right ) \, dx \\ & = -5 \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-5 \log (2+x) \log (\log (i \pi +\log (2)))+\log (\log (i \pi +\log (2))) \int \frac {e^{2 x} \left (-3+2 e^4-5 \left (1-e^4\right ) x-2 \left (1-e^4\right ) x^2\right )}{2+x} \, dx-(2 \log (\log (i \pi +\log (2)))) \int e^{2 x} \log (2+x) \, dx \\ & = -5 \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-5 \log (2+x) \log (\log (i \pi +\log (2)))-e^{2 x} \log (2+x) \log (\log (i \pi +\log (2)))+\log (\log (i \pi +\log (2))) \int \left (e^{2 x} \left (-1+e^4\right )+\frac {e^{2 x}}{-2-x}+2 e^{2 x} \left (-1+e^4\right ) x\right ) \, dx+(2 \log (\log (i \pi +\log (2)))) \int \frac {e^{2 x}}{4+2 x} \, dx \\ & = -5 \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))+\frac {\text {Ei}(4+2 x) \log (\log (i \pi +\log (2)))}{e^4}-5 \log (2+x) \log (\log (i \pi +\log (2)))-e^{2 x} \log (2+x) \log (\log (i \pi +\log (2)))+\log (\log (i \pi +\log (2))) \int \frac {e^{2 x}}{-2-x} \, dx-\left (\left (1-e^4\right ) \log (\log (i \pi +\log (2)))\right ) \int e^{2 x} \, dx-\left (2 \left (1-e^4\right ) \log (\log (i \pi +\log (2)))\right ) \int e^{2 x} x \, dx \\ & = -\frac {1}{2} e^{2 x} \left (1-e^4\right ) \log (\log (i \pi +\log (2)))-5 \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-e^{2 x} \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-\frac {\text {Ei}(2 (2+x)) \log (\log (i \pi +\log (2)))}{e^4}+\frac {\text {Ei}(4+2 x) \log (\log (i \pi +\log (2)))}{e^4}-5 \log (2+x) \log (\log (i \pi +\log (2)))-e^{2 x} \log (2+x) \log (\log (i \pi +\log (2)))+\left (\left (1-e^4\right ) \log (\log (i \pi +\log (2)))\right ) \int e^{2 x} \, dx \\ & = -5 \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-e^{2 x} \left (1-e^4\right ) x \log (\log (i \pi +\log (2)))-\frac {\text {Ei}(2 (2+x)) \log (\log (i \pi +\log (2)))}{e^4}+\frac {\text {Ei}(4+2 x) \log (\log (i \pi +\log (2)))}{e^4}-5 \log (2+x) \log (\log (i \pi +\log (2)))-e^{2 x} \log (2+x) \log (\log (i \pi +\log (2))) \\ \end{align*}
Time = 2.57 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=\left (\left (-1+e^4\right ) \left (10+\left (5+e^{2 x}\right ) x\right )-\left (5+e^{2 x}\right ) \log (2+x)\right ) \log (\log (i \pi +\log (2))) \]
[In]
[Out]
Time = 2.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39
method | result | size |
default | \(\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \left (\left ({\mathrm e}^{4}-1\right ) x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x} \ln \left (2+x \right )+5 x \,{\mathrm e}^{4}-5 x -5 \ln \left (2+x \right )\right )\) | \(46\) |
parallelrisch | \(\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \left ({\mathrm e}^{4} {\mathrm e}^{2 x} x +5 x \,{\mathrm e}^{4}-x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x} \ln \left (2+x \right )-20 \,{\mathrm e}^{4}-5 x -5 \ln \left (2+x \right )+20\right )\) | \(56\) |
parts | \(\left (\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{4}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right )\right ) x \,{\mathrm e}^{2 x}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{2 x} \ln \left (2+x \right )+5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \left (x \,{\mathrm e}^{4}-x -\ln \left (2+x \right )\right )\) | \(76\) |
norman | \(\left (5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{4}-5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right )\right ) x -5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \ln \left (2+x \right )+\left (\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{4}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right )\right ) x \,{\mathrm e}^{2 x}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{2 x} \ln \left (2+x \right )\) | \(93\) |
risch | \(-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{2 x} \ln \left (2+x \right )+\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) x \,{\mathrm e}^{4+2 x}+5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) x \,{\mathrm e}^{4}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) x \,{\mathrm e}^{2 x}-5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \ln \left (2+x \right )-5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) x\) | \(95\) |
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx={\left (5 \, x e^{4} + {\left (x e^{4} - x\right )} e^{\left (2 \, x\right )} - {\left (e^{\left (2 \, x\right )} + 5\right )} \log \left (x + 2\right ) - 5 \, x\right )} \log \left (\log \left (i \, \pi + \log \left (2\right )\right )\right ) \]
[In]
[Out]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (27) = 54\).
Time = 0.51 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.79 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=x \left (- 5 \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )} + 5 e^{4} \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )}\right ) + \left (- x \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )} + x e^{4} \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )} - \log {\left (x + 2 \right )} \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )}\right ) e^{2 x} - 5 \log {\left (x + 2 \right )} \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )} \]
[In]
[Out]
\[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=\int { -\frac {{\left (2 \, {\left (x + 2\right )} e^{\left (2 \, x\right )} \log \left (x + 2\right ) - 5 \, {\left (x + 2\right )} e^{4} + {\left (2 \, x^{2} - {\left (2 \, x^{2} + 5 \, x + 2\right )} e^{4} + 5 \, x + 3\right )} e^{\left (2 \, x\right )} + 5 \, x + 15\right )} \log \left (\log \left (i \, \pi + \log \left (2\right )\right )\right )}{x + 2} \,d x } \]
[In]
[Out]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (29) = 58\).
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.52 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx={\left (5 \, {\left (x + 2\right )} e^{8} - 5 \, {\left (x + 2\right )} e^{4} + {\left (x + 2\right )} e^{\left (2 \, x + 8\right )} - {\left (x + 2\right )} e^{\left (2 \, x + 4\right )} - 5 \, e^{4} \log \left (x + 2\right ) - e^{\left (2 \, x + 4\right )} \log \left (x + 2\right ) - 2 \, e^{\left (2 \, x + 8\right )} + 2 \, e^{\left (2 \, x + 4\right )}\right )} e^{\left (-4\right )} \log \left (\log \left (i \, \pi + \log \left (2\right )\right )\right ) \]
[In]
[Out]
Time = 9.78 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=-\ln \left (\ln \left (\ln \left (2\right )+\Pi \,1{}\mathrm {i}\right )\right )\,\left (5\,x+5\,\ln \left (x+2\right )+x\,{\mathrm {e}}^{2\,x}-5\,x\,{\mathrm {e}}^4-x\,{\mathrm {e}}^{2\,x+4}+\ln \left (x+2\right )\,{\mathrm {e}}^{2\,x}\right ) \]
[In]
[Out]