Integrand size = 110, antiderivative size = 33 \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=\frac {(i \pi +\log (-\log (\log (2))))^2}{-3+2^{\frac {1}{2} x^3 \left (e^x+x\right )}} \]
Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=-\frac {(\pi -i \log (-\log (\log (2))))^2}{-3+2^{\frac {1}{2} x^3 \left (e^x+x\right )}} \]
Integrate[(E^((E^x*x^3*Log[2] + x^4*Log[2])/2)*(-4*x^3*Log[2] + E^x*(-3*x^ 2 - x^3)*Log[2])*(I*Pi + Log[-Log[Log[2]]])^2)/(18 - 12*E^((E^x*x^3*Log[2] + x^4*Log[2])/2) + 2*E^(E^x*x^3*Log[2] + x^4*Log[2])),x]
Time = 1.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {27, 25, 7292, 27, 7237}
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.
\(\displaystyle \int \frac {(\log (-\log (\log (2)))+i \pi )^2 e^{\frac {1}{2} \left (x^4 \log (2)+e^x x^3 \log (2)\right )} \left (e^x \left (-x^3-3 x^2\right ) \log (2)-4 x^3 \log (2)\right )}{-12 e^{\frac {1}{2} \left (x^4 \log (2)+e^x x^3 \log (2)\right )}+2 e^{x^4 \log (2)+e^x x^3 \log (2)}+18} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle (\log (-\log (\log (2)))+i \pi )^2 \int -\frac {2^{\frac {x^4}{2}+\frac {e^x x^3}{2}} \left (4 \log (2) x^3+e^x \left (x^3+3 x^2\right ) \log (2)\right )}{18-3\ 2^{\frac {x^4}{2}+\frac {e^x x^3}{2}+2}+2^{x^4+e^x x^3+1}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -(\log (-\log (\log (2)))+i \pi )^2 \int \frac {2^{\frac {x^4}{2}+\frac {e^x x^3}{2}} \left (4 \log (2) x^3+e^x \left (x^3+3 x^2\right ) \log (2)\right )}{18-3\ 2^{\frac {x^4}{2}+\frac {e^x x^3}{2}+2}+2^{x^4+e^x x^3+1}}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -(\log (-\log (\log (2)))+i \pi )^2 \int \frac {2^{\frac {1}{2} x^3 \left (x+e^x\right )-1} x^2 \left (e^x x+4 x+3 e^x\right ) \log (2)}{\left (3-2^{\frac {1}{2} x^3 \left (x+e^x\right )}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\log (2) (\log (-\log (\log (2)))+i \pi )^2 \int \frac {2^{\frac {1}{2} x^3 \left (x+e^x\right )-1} x^2 \left (e^x x+4 x+3 e^x\right )}{\left (3-2^{\frac {1}{2} x^3 \left (x+e^x\right )}\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle -\frac {(\log (-\log (\log (2)))+i \pi )^2}{3-2^{\frac {1}{2} x^3 \left (x+e^x\right )}}\) |
Int[(E^((E^x*x^3*Log[2] + x^4*Log[2])/2)*(-4*x^3*Log[2] + E^x*(-3*x^2 - x^ 3)*Log[2])*(I*Pi + Log[-Log[Log[2]]])^2)/(18 - 12*E^((E^x*x^3*Log[2] + x^4 *Log[2])/2) + 2*E^(E^x*x^3*Log[2] + x^4*Log[2])),x]
3.2.7.3.1 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]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Time = 1.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73
method | result | size |
parallelrisch | \(\frac {\ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}}{{\mathrm e}^{\frac {\ln \left (2\right ) \left ({\mathrm e}^{x}+x \right ) x^{3}}{2}}-3}\) | \(24\) |
risch | \(\frac {\ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}}{2^{\frac {{\mathrm e}^{x} x^{3}}{2}} 2^{\frac {x^{4}}{2}}-3}\) | \(29\) |
norman | \(\frac {\ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}}{{\mathrm e}^{\frac {x^{3} \ln \left (2\right ) {\mathrm e}^{x}}{2}+\frac {x^{4} \ln \left (2\right )}{2}}-3}\) | \(30\) |
int(((-x^3-3*x^2)*ln(2)*exp(x)-4*x^3*ln(2))*exp(1/2*x^3*ln(2)*exp(x)+1/2*x ^4*ln(2))*ln(ln(ln(2)))^2/(2*exp(1/2*x^3*ln(2)*exp(x)+1/2*x^4*ln(2))^2-12* exp(1/2*x^3*ln(2)*exp(x)+1/2*x^4*ln(2))+18),x,method=_RETURNVERBOSE)
Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=\frac {\log \left (\log \left (\log \left (2\right )\right )\right )^{2}}{e^{\left (\frac {1}{2} \, x^{4} \log \left (2\right ) + \frac {1}{2} \, x^{3} e^{x} \log \left (2\right )\right )} - 3} \]
integrate(((-x^3-3*x^2)*log(2)*exp(x)-4*x^3*log(2))*exp(1/2*x^3*log(2)*exp (x)+1/2*x^4*log(2))*log(log(log(2)))^2/(2*exp(1/2*x^3*log(2)*exp(x)+1/2*x^ 4*log(2))^2-12*exp(1/2*x^3*log(2)*exp(x)+1/2*x^4*log(2))+18),x, algorithm= \
Time = 0.33 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.55 \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=\frac {- \pi ^{2} + \log {\left (- \log {\left (\log {\left (2 \right )} \right )} \right )}^{2} + 2 i \pi \log {\left (- \log {\left (\log {\left (2 \right )} \right )} \right )}}{e^{\frac {x^{4} \log {\left (2 \right )}}{2}} e^{\frac {x^{3} e^{x} \log {\left (2 \right )}}{2}} - 3} \]
integrate(((-x**3-3*x**2)*ln(2)*exp(x)-4*x**3*ln(2))*exp(1/2*x**3*ln(2)*ex p(x)+1/2*x**4*ln(2))*ln(ln(ln(2)))**2/(2*exp(1/2*x**3*ln(2)*exp(x)+1/2*x** 4*ln(2))**2-12*exp(1/2*x**3*ln(2)*exp(x)+1/2*x**4*ln(2))+18),x)
(-pi**2 + log(-log(log(2)))**2 + 2*I*pi*log(-log(log(2))))/(exp(x**4*log(2 )/2)*exp(x**3*exp(x)*log(2)/2) - 3)
Time = 0.38 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=\frac {\log \left (\log \left (\log \left (2\right )\right )\right )^{2}}{e^{\left (\frac {1}{2} \, x^{4} \log \left (2\right ) + \frac {1}{2} \, x^{3} e^{x} \log \left (2\right )\right )} - 3} \]
integrate(((-x^3-3*x^2)*log(2)*exp(x)-4*x^3*log(2))*exp(1/2*x^3*log(2)*exp (x)+1/2*x^4*log(2))*log(log(log(2)))^2/(2*exp(1/2*x^3*log(2)*exp(x)+1/2*x^ 4*log(2))^2-12*exp(1/2*x^3*log(2)*exp(x)+1/2*x^4*log(2))+18),x, algorithm= \
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=\frac {\log \left (\log \left (\log \left (2\right )\right )\right )^{2}}{e^{\left (\frac {1}{2} \, x^{4} \log \left (2\right ) + \frac {1}{2} \, x^{3} e^{x} \log \left (2\right )\right )} - 3} \]
integrate(((-x^3-3*x^2)*log(2)*exp(x)-4*x^3*log(2))*exp(1/2*x^3*log(2)*exp (x)+1/2*x^4*log(2))*log(log(log(2)))^2/(2*exp(1/2*x^3*log(2)*exp(x)+1/2*x^ 4*log(2))^2-12*exp(1/2*x^3*log(2)*exp(x)+1/2*x^4*log(2))+18),x, algorithm= \
Timed out. \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=\int -\frac {{\ln \left (\ln \left (\ln \left (2\right )\right )\right )}^2\,{\mathrm {e}}^{\frac {x^4\,\ln \left (2\right )}{2}+\frac {x^3\,{\mathrm {e}}^x\,\ln \left (2\right )}{2}}\,\left (4\,x^3\,\ln \left (2\right )+{\mathrm {e}}^x\,\ln \left (2\right )\,\left (x^3+3\,x^2\right )\right )}{2\,{\mathrm {e}}^{x^4\,\ln \left (2\right )+x^3\,{\mathrm {e}}^x\,\ln \left (2\right )}-12\,{\mathrm {e}}^{\frac {x^4\,\ln \left (2\right )}{2}+\frac {x^3\,{\mathrm {e}}^x\,\ln \left (2\right )}{2}}+18} \,d x \]
int(-(log(log(log(2)))^2*exp((x^4*log(2))/2 + (x^3*exp(x)*log(2))/2)*(4*x^ 3*log(2) + exp(x)*log(2)*(3*x^2 + x^3)))/(2*exp(x^4*log(2) + x^3*exp(x)*lo g(2)) - 12*exp((x^4*log(2))/2 + (x^3*exp(x)*log(2))/2) + 18),x)