\(\int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} (2-2 \log (x)+3 \log ^2(x))}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx\) [10088]

Optimal result

Integrand size = 100, antiderivative size = 33 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=\log \left (3-\frac {e^{4-i \pi +\frac {3 x}{2}-\frac {x}{\log (x)}}}{-1+2 e}\right ) \]

[Out]

ln(3-exp(4+3/2*x-x/ln(x)-ln(-2*exp(1)+1)))

Rubi [F]

\[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=\int \frac {\exp \left (\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}\right ) \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 \exp \left (\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}\right ) \log ^2(x)} \, dx \]

[In]

Int[(E^((-2*x + (8 + 3*x - 2*(I*Pi + Log[-1 + 2*E]))*Log[x])/(2*Log[x]))*(2 - 2*Log[x] + 3*Log[x]^2))/(-6*Log[
x]^2 + 2*E^((-2*x + (8 + 3*x - 2*(I*Pi + Log[-1 + 2*E]))*Log[x])/(2*Log[x]))*Log[x]^2),x]

[Out]

(3*Defer[Int][E^(4 + (3*x)/2)/(E^(4 + (3*x)/2) + 6*(1 - 1/(2*E))*E^(1 + x/Log[x])), x])/2 + Defer[Int][E^(4 +
(3*x)/2)/((E^(4 + (3*x)/2) + 6*(1 - 1/(2*E))*E^(1 + x/Log[x]))*Log[x]^2), x] + Defer[Int][E^(4 + (3*x)/2)/((-E
^(4 + (3*x)/2) - 6*(1 - 1/(2*E))*E^(1 + x/Log[x]))*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{4+\frac {3 x}{2}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{2 \left (e^{4+\frac {3 x}{2}}+3 e^{\frac {x}{\log (x)}} (-1+2 e)\right ) \log ^2(x)} \, dx \\ & = \frac {1}{2} \int \frac {e^{4+\frac {3 x}{2}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{\left (e^{4+\frac {3 x}{2}}+3 e^{\frac {x}{\log (x)}} (-1+2 e)\right ) \log ^2(x)} \, dx \\ & = \frac {1}{2} \int \left (\frac {3 e^{4+\frac {3 x}{2}}}{e^{4+\frac {3 x}{2}}+6 \left (1-\frac {1}{2 e}\right ) e^{1+\frac {x}{\log (x)}}}+\frac {2 e^{4+\frac {3 x}{2}}}{\left (e^{4+\frac {3 x}{2}}+6 \left (1-\frac {1}{2 e}\right ) e^{1+\frac {x}{\log (x)}}\right ) \log ^2(x)}+\frac {2 e^{4+\frac {3 x}{2}}}{\left (-e^{4+\frac {3 x}{2}}-6 \left (1-\frac {1}{2 e}\right ) e^{1+\frac {x}{\log (x)}}\right ) \log (x)}\right ) \, dx \\ & = \frac {3}{2} \int \frac {e^{4+\frac {3 x}{2}}}{e^{4+\frac {3 x}{2}}+6 \left (1-\frac {1}{2 e}\right ) e^{1+\frac {x}{\log (x)}}} \, dx+\int \frac {e^{4+\frac {3 x}{2}}}{\left (e^{4+\frac {3 x}{2}}+6 \left (1-\frac {1}{2 e}\right ) e^{1+\frac {x}{\log (x)}}\right ) \log ^2(x)} \, dx+\int \frac {e^{4+\frac {3 x}{2}}}{\left (-e^{4+\frac {3 x}{2}}-6 \left (1-\frac {1}{2 e}\right ) e^{1+\frac {x}{\log (x)}}\right ) \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=\frac {1}{2} \left (2 \log \left (e^{4+\frac {3 x}{2}}+6 e^{1+\frac {x}{\log (x)}}-3 e^{\frac {x}{\log (x)}}\right )-\frac {2 x}{\log (x)}\right ) \]

[In]

Integrate[(E^((-2*x + (8 + 3*x - 2*(I*Pi + Log[-1 + 2*E]))*Log[x])/(2*Log[x]))*(2 - 2*Log[x] + 3*Log[x]^2))/(-
6*Log[x]^2 + 2*E^((-2*x + (8 + 3*x - 2*(I*Pi + Log[-1 + 2*E]))*Log[x])/(2*Log[x]))*Log[x]^2),x]

[Out]

(2*Log[E^(4 + (3*x)/2) + 6*E^(1 + x/Log[x]) - 3*E^(x/Log[x])] - (2*x)/Log[x])/2

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97

method	result	size
norman	\(\ln \left ({\mathrm e}^{\frac {\left (-2 \ln \left (-2 \,{\mathrm e}+1\right )+3 x +8\right ) \ln \left (x \right )-2 x}{2 \ln \left (x \right )}}-3\right )\)	\(32\)
parallelrisch	\(\ln \left ({\mathrm e}^{\frac {\left (-2 \ln \left (-2 \,{\mathrm e}+1\right )+3 x +8\right ) \ln \left (x \right )-2 x}{2 \ln \left (x \right )}}-3\right )\)	\(32\)
risch	\(\frac {3 x}{2}-\frac {x}{\ln \left (x \right )}-\frac {\left (-2 \ln \left (-2 \,{\mathrm e}+1\right )+3 x +8\right ) \ln \left (x \right )-2 x}{2 \ln \left (x \right )}+\ln \left (\frac {{\mathrm e}^{\frac {3 x \ln \left (x \right )+8 \ln \left (x \right )-2 x}{2 \ln \left (x \right )}}}{-2 \,{\mathrm e}+1}-3\right )\)	\(71\)

[In]

int((3*ln(x)^2-2*ln(x)+2)*exp(1/2*((-2*ln(-2*exp(1)+1)+3*x+8)*ln(x)-2*x)/ln(x))/(2*ln(x)^2*exp(1/2*((-2*ln(-2*
exp(1)+1)+3*x+8)*ln(x)-2*x)/ln(x))-6*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

ln(exp(1/2*((-2*ln(-2*exp(1)+1)+3*x+8)*ln(x)-2*x)/ln(x))-3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=\log \left (e^{\left (\frac {{\left (3 \, x - 2 \, \log \left (-2 \, e + 1\right ) + 8\right )} \log \left (x\right ) - 2 \, x}{2 \, \log \left (x\right )}\right )} - 3\right ) \]

[In]

integrate((3*log(x)^2-2*log(x)+2)*exp(1/2*((-2*log(-2*exp(1)+1)+3*x+8)*log(x)-2*x)/log(x))/(2*log(x)^2*exp(1/2
*((-2*log(-2*exp(1)+1)+3*x+8)*log(x)-2*x)/log(x))-6*log(x)^2),x, algorithm="fricas")

[Out]

log(e^(1/2*((3*x - 2*log(-2*e + 1) + 8)*log(x) - 2*x)/log(x)) - 3)

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=- \frac {x}{\log {\left (x \right )}} + \log {\left (e^{\frac {x}{\log {\left (x \right )}}} + \frac {e^{4}}{- 3 e^{- \frac {3 x}{2}} + 6 e e^{- \frac {3 x}{2}}} \right )} \]

[In]

integrate((3*ln(x)**2-2*ln(x)+2)*exp(1/2*((-2*ln(-2*exp(1)+1)+3*x+8)*ln(x)-2*x)/ln(x))/(2*ln(x)**2*exp(1/2*((-
2*ln(-2*exp(1)+1)+3*x+8)*ln(x)-2*x)/ln(x))-6*ln(x)**2),x)

[Out]

-x/log(x) + log(exp(x/log(x)) + exp(4)/(-3*exp(-3*x/2) + 6*E*exp(-3*x/2)))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=-\frac {x}{\log \left (x\right )} + \log \left (\frac {3 \, {\left (2 \, e - 1\right )} e^{\frac {x}{\log \left (x\right )}} + e^{\left (\frac {3}{2} \, x + 4\right )}}{3 \, {\left (2 \, e - 1\right )}}\right ) \]

[In]

integrate((3*log(x)^2-2*log(x)+2)*exp(1/2*((-2*log(-2*exp(1)+1)+3*x+8)*log(x)-2*x)/log(x))/(2*log(x)^2*exp(1/2
*((-2*log(-2*exp(1)+1)+3*x+8)*log(x)-2*x)/log(x))-6*log(x)^2),x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((3*log(x)^2-2*log(x)+2)*exp(1/2*((-2*log(-2*exp(1)+1)+3*x+8)*log(x)-2*x)/log(x))/(2*log(x)^2*exp(1/2
*((-2*log(-2*exp(1)+1)+3*x+8)*log(x)-2*x)/log(x))-6*log(x)^2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{-128,[0,11,0]%%%} / %%%{256,[0,11,0]%%%} Error: Bad Argu
ment Value

Mupad [B] (verification not implemented)

Time = 17.45 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=\ln \left (-\frac {{\mathrm {e}}^4\,{\mathrm {e}}^{-\frac {x}{\ln \left (x\right )}}\,{\left ({\mathrm {e}}^x\right )}^{3/2}}{2\,\mathrm {e}-1}-3\right ) \]

[In]

int(-(exp(-(x - (log(x)*(3*x - 2*log(1 - 2*exp(1)) + 8))/2)/log(x))*(3*log(x)^2 - 2*log(x) + 2))/(6*log(x)^2 -
2*exp(-(x - (log(x)*(3*x - 2*log(1 - 2*exp(1)) + 8))/2)/log(x))*log(x)^2),x)

[Out]

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