\(\int \frac {e^{\frac {1}{2} (e^x x^3 \log (2)+x^4 \log (2))} (-4 x^3 \log (2)+e^x (-3 x^2-x^3) \log (2)) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} (e^x x^3 \log (2)+x^4 \log (2))}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx\) [107]

Optimal result

Integrand size = 110, antiderivative size = 33 \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=\frac {(i \pi +\log (-\log (\log (2))))^2}{-3+2^{\frac {1}{2} x^3 \left (e^x+x\right )}} \] Output:

ln(ln(ln(2)))^2/(exp(1/2*ln(2)*(exp(x)+x)*x^3)-3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=-\frac {(\pi -i \log (-\log (\log (2))))^2}{-3+2^{\frac {1}{2} x^3 \left (e^x+x\right )}} \] Input:

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]
 

Output:

-((Pi - I*Log[-Log[Log[2]]])^2/(-3 + 2^((x^3*(E^x + x))/2)))
 

Rubi [A] (verified)

Time = 1.98 (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

\(\displaystyle -\frac {(\log (-\log (\log (2)))+i \pi )^2}{3-2^{\frac {1}{2} x^3 \left (x+e^x\right )}}\)

Input:

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]
 

Output:

-((I*Pi + Log[-Log[Log[2]]])^2/(3 - 2^((x^3*(E^x + x))/2)))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 7237

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]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

Maple [A] (verified)

Time = 0.51 (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	\(\frac {\ln \left (\ln \left (\ln \left (2\right )\right )\right )^{2}}{{\mathrm e}^{\frac {x^{3} \ln \left (2\right ) {\mathrm e}^{x}}{2}+\frac {x^{4} \ln \left (2\right )}{2}}-3}\)	\(30\)

Input:

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)
 

Output:

ln(ln(ln(2)))^2/(exp(1/2*ln(2)*(exp(x)+x)*x^3)-3)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=\frac {\log \left (\log \left (\log \left (2\right )\right )\right )^{2}}{e^{\left (\frac {1}{2} \, x^{4} \log \left (2\right ) + \frac {1}{2} \, x^{3} e^{x} \log \left (2\right )\right )} - 3} \] Input:

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= 
"fricas")
 

Output:

log(log(log(2)))^2/(e^(1/2*x^4*log(2) + 1/2*x^3*e^x*log(2)) - 3)
 

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.55 \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=\frac {- \pi ^{2} + \log {\left (- \log {\left (\log {\left (2 \right )} \right )} \right )}^{2} + 2 i \pi \log {\left (- \log {\left (\log {\left (2 \right )} \right )} \right )}}{e^{\frac {x^{4} \log {\left (2 \right )}}{2}} e^{\frac {x^{3} e^{x} \log {\left (2 \right )}}{2}} - 3} \] Input:

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)
 

Output:

(-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)
 

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=\frac {\log \left (\log \left (\log \left (2\right )\right )\right )^{2}}{e^{\left (\frac {1}{2} \, x^{4} \log \left (2\right ) + \frac {1}{2} \, x^{3} e^{x} \log \left (2\right )\right )} - 3} \] Input:

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= 
"maxima")
 

Output:

log(log(log(2)))^2/(e^(1/2*x^4*log(2) + 1/2*x^3*e^x*log(2)) - 3)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=\frac {\log \left (\log \left (\log \left (2\right )\right )\right )^{2}}{e^{\left (\frac {1}{2} \, x^{4} \log \left (2\right ) + \frac {1}{2} \, x^{3} e^{x} \log \left (2\right )\right )} - 3} \] Input:

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= 
"giac")
 

Output:

log(log(log(2)))^2/(e^(1/2*x^4*log(2) + 1/2*x^3*e^x*log(2)) - 3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=\int -\frac {{\ln \left (\ln \left (\ln \left (2\right )\right )\right )}^2\,{\mathrm {e}}^{\frac {x^4\,\ln \left (2\right )}{2}+\frac {x^3\,{\mathrm {e}}^x\,\ln \left (2\right )}{2}}\,\left (4\,x^3\,\ln \left (2\right )+{\mathrm {e}}^x\,\ln \left (2\right )\,\left (x^3+3\,x^2\right )\right )}{2\,{\mathrm {e}}^{x^4\,\ln \left (2\right )+x^3\,{\mathrm {e}}^x\,\ln \left (2\right )}-12\,{\mathrm {e}}^{\frac {x^4\,\ln \left (2\right )}{2}+\frac {x^3\,{\mathrm {e}}^x\,\ln \left (2\right )}{2}}+18} \,d x \] Input:

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)
 

Output:

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)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.55 \[ \int \frac {e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )} \left (-4 x^3 \log (2)+e^x \left (-3 x^2-x^3\right ) \log (2)\right ) (i \pi +\log (-\log (\log (2))))^2}{18-12 e^{\frac {1}{2} \left (e^x x^3 \log (2)+x^4 \log (2)\right )}+2 e^{e^x x^3 \log (2)+x^4 \log (2)}} \, dx=\frac {e^{\frac {e^{x} \mathrm {log}\left (2\right ) x^{3}}{2}} 2^{\frac {x^{4}}{2}} \mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (2\right )\right )\right )^{2}}{3 e^{\frac {e^{x} \mathrm {log}\left (2\right ) x^{3}}{2}} 2^{\frac {x^{4}}{2}}-9} \] Input:

int(((-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)
 

Output:

(e**((e**x*log(2)*x**3)/2)*2**(x**4/2)*log(log(log(2)))**2)/(3*(e**((e**x* 
log(2)*x**3)/2)*2**(x**4/2) - 3))