Integrand size = 76, antiderivative size = 36 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=20+5 \left (x+\frac {2 (4-x)}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}\right ) \]
Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.42 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=5 x+\frac {40}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}-\frac {10 x}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \]
Integrate[(40 - 10*x - 10*x*Log[(I*Pi + Log[-8 + 2*E^2])/x] + 5*x*Log[(I*P i + Log[-8 + 2*E^2])/x]^2)/(x*Log[(I*Pi + Log[-8 + 2*E^2])/x]^2),x]
Time = 0.64 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.53, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {7292, 27, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-10 x+5 x \log ^2\left (\frac {\log \left (2 e^2-8\right )+i \pi }{x}\right )-10 x \log \left (\frac {\log \left (2 e^2-8\right )+i \pi }{x}\right )+40}{x \log ^2\left (\frac {\log \left (2 e^2-8\right )+i \pi }{x}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 \left (-2 x+x \log ^2\left (\frac {\log \left (2 e^2-8\right )+i \pi }{x}\right )-2 x \log \left (\frac {\log \left (2 e^2-8\right )+i \pi }{x}\right )+8\right )}{x \log ^2\left (\frac {\log \left (2 e^2-8\right )+i \pi }{x}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int \frac {x \log ^2\left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )-2 x \log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )-2 x+8}{x \log ^2\left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 5 \int \left (-\frac {2 (x-4)}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}-\frac {2}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \left (x-\frac {2 x}{\log \left (\frac {\log \left (-2 \left (4-e^2\right )\right )+i \pi }{x}\right )}+\frac {8}{\log \left (\frac {\log \left (-2 \left (4-e^2\right )\right )+i \pi }{x}\right )}\right )\) |
Int[(40 - 10*x - 10*x*Log[(I*Pi + Log[-8 + 2*E^2])/x] + 5*x*Log[(I*Pi + Lo g[-8 + 2*E^2])/x]^2)/(x*Log[(I*Pi + Log[-8 + 2*E^2])/x]^2),x]
3.8.98.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 | \(\frac {5 \ln \left (2\right ) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}+\frac {5 \ln \left (-{\mathrm e}^{2}+4\right ) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}-10 \ln \left (2\right ) \operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )-10 \ln \left (-{\mathrm e}^{2}+4\right ) \operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )+10 \ln \left (2\right ) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+10 \ln \left (-{\mathrm e}^{2}+4\right ) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\operatorname {Ei}_{1}\left (\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+\frac {40}{\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}\) | \(187\) |
int((5*x*ln(ln(-2*exp(2)+8)/x)^2-10*x*ln(ln(-2*exp(2)+8)/x)-10*x+40)/x/ln( ln(-2*exp(2)+8)/x)^2,x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=\frac {5 \, {\left (x \log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right ) - 2 \, x + 8\right )}}{\log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right )} \]
integrate((5*x*log(log(-2*exp(2)+8)/x)^2-10*x*log(log(-2*exp(2)+8)/x)-10*x +40)/x/log(log(-2*exp(2)+8)/x)^2,x, algorithm=\
Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.75 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=5 x + \frac {40 - 10 x}{\log {\left (\frac {\log {\left (2 \right )}}{x} + \frac {\log {\left (-4 + e^{2} \right )}}{x} + \frac {i \pi }{x} \right )}} \]
integrate((5*x*ln(ln(-2*exp(2)+8)/x)**2-10*x*ln(ln(-2*exp(2)+8)/x)-10*x+40 )/x/ln(ln(-2*exp(2)+8)/x)**2,x)
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.92 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=5 \, x - \frac {10 \, {\left (x - 4\right )}}{\log \left (i \, \pi + \log \left (2\right ) + \log \left (e + 2\right ) + \log \left (e - 2\right )\right ) - \log \left (x\right )} \]
integrate((5*x*log(log(-2*exp(2)+8)/x)^2-10*x*log(log(-2*exp(2)+8)/x)-10*x +40)/x/log(log(-2*exp(2)+8)/x)^2,x, algorithm=\
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 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=\frac {5 \, {\left (\log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right ) \log \left (-2 \, e^{2} + 8\right )^{3} - 2 \, \log \left (-2 \, e^{2} + 8\right )^{3} + \frac {8 \, \log \left (-2 \, e^{2} + 8\right )^{3}}{x}\right )} x}{\log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right ) \log \left (-2 \, e^{2} + 8\right )^{3}} \]
integrate((5*x*log(log(-2*exp(2)+8)/x)^2-10*x*log(log(-2*exp(2)+8)/x)-10*x +40)/x/log(log(-2*exp(2)+8)/x)^2,x, algorithm=\
5*(log(log(-2*e^2 + 8)/x)*log(-2*e^2 + 8)^3 - 2*log(-2*e^2 + 8)^3 + 8*log( -2*e^2 + 8)^3/x)*x/(log(log(-2*e^2 + 8)/x)*log(-2*e^2 + 8)^3)
Time = 8.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69 \[ \int \frac {40-10 x-10 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+5 x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx=5\,x-\frac {10\,x-40}{\ln \left (\frac {\ln \left (8-2\,{\mathrm {e}}^2\right )}{x}\right )} \]