Integrand size = 76, antiderivative size = 36 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=20+5 \left (x+\frac {2 (4-x)}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}\right ) \]
[Out]
Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.53, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6873, 12, 6874, 2395, 2334, 2336, 2209, 2339, 30} \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=5 x-\frac {10 x}{\log \left (\frac {\log \left (-2 \left (4-e^2\right )\right )+i \pi }{x}\right )}+\frac {40}{\log \left (\frac {\log \left (-2 \left (4-e^2\right )\right )+i \pi }{x}\right )} \]
[In]
[Out]
Rule 12
Rule 30
Rule 2209
Rule 2334
Rule 2336
Rule 2339
Rule 2395
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {5 \left (8-2 x-2 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx \\ & = 5 \int \frac {8-2 x-2 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx \\ & = 5 \int \left (1-\frac {2 (-4+x)}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}-\frac {2}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}\right ) \, dx \\ & = 5 x-10 \int \frac {-4+x}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx-10 \int \frac {1}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx \\ & = 5 x-10 \int \left (\frac {1}{\log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}-\frac {4}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}\right ) \, dx+\left (10 \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )\right ) \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )\right ) \\ & = 5 x+10 \operatorname {ExpIntegralEi}\left (-\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )\right ) \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )-10 \int \frac {1}{\log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx+40 \int \frac {1}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx \\ & = 5 x+10 \operatorname {ExpIntegralEi}\left (-\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )\right ) \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )-\frac {10 x}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}+10 \int \frac {1}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx-40 \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )\right ) \\ & = 5 x+10 \operatorname {ExpIntegralEi}\left (-\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )\right ) \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )+\frac {40}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}-\frac {10 x}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}-\left (10 \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )\right ) \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )\right ) \\ & = 5 x+\frac {40}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}-\frac {10 x}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.42 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=5 x+\frac {40}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}-\frac {10 x}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \]
[In]
[Out]
Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.75
method | result | size |
risch | \(5 x -\frac {10 \left (x -4\right )}{\ln \left (\frac {\ln \left (2\right )+\ln \left (-{\mathrm e}^{2}+4\right )}{x}\right )}\) | \(27\) |
norman | \(\frac {40-10 x +5 x \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}{\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}\) | \(36\) |
parallelrisch | \(\frac {40-10 x +5 x \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}{\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}\) | \(36\) |
parts | \(5 x -10 \ln \left (-2 \,{\mathrm e}^{2}+8\right ) \operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )+10 \ln \left (2\right ) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+10 \ln \left (-{\mathrm e}^{2}+4\right ) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+\frac {40}{\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}\) | \(139\) |
derivativedivides | \(\frac {5 \ln \left (2\right ) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}+\frac {5 \ln \left (-{\mathrm e}^{2}+4\right ) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}-10 \ln \left (2\right ) \operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )-10 \ln \left (-{\mathrm e}^{2}+4\right ) \operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )+10 \ln \left (2\right ) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+10 \ln \left (-{\mathrm e}^{2}+4\right ) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+\frac {40}{\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}\) | \(187\) |
default | \(\frac {5 \ln \left (2\right ) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}+\frac {5 \ln \left (-{\mathrm e}^{2}+4\right ) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}-10 \ln \left (2\right ) \operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )-10 \ln \left (-{\mathrm e}^{2}+4\right ) \operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )+10 \ln \left (2\right ) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+10 \ln \left (-{\mathrm e}^{2}+4\right ) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+\frac {40}{\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}\) | \(187\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=\frac {5 \, {\left (x \log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right ) - 2 \, x + 8\right )}}{\log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right )} \]
[In]
[Out]
Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.75 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=5 x + \frac {40 - 10 x}{\log {\left (\frac {\log {\left (2 \right )}}{x} + \frac {\log {\left (-4 + e^{2} \right )}}{x} + \frac {i \pi }{x} \right )}} \]
[In]
[Out]
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.92 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=5 \, x - \frac {10 \, {\left (x - 4\right )}}{\log \left (i \, \pi + \log \left (2\right ) + \log \left (e + 2\right ) + \log \left (e - 2\right )\right ) - \log \left (x\right )} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (24) = 48\).
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.06 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=\frac {5 \, {\left (\log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right ) \log \left (-2 \, e^{2} + 8\right )^{3} - 2 \, \log \left (-2 \, e^{2} + 8\right )^{3} + \frac {8 \, \log \left (-2 \, e^{2} + 8\right )^{3}}{x}\right )} x}{\log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right ) \log \left (-2 \, e^{2} + 8\right )^{3}} \]
[In]
[Out]
Time = 8.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=5\,x-\frac {10\,x-40}{\ln \left (\frac {\ln \left (8-2\,{\mathrm {e}}^2\right )}{x}\right )} \]
[In]
[Out]