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

Optimal. Leaf size=33 \[ \log \left (3-\frac {e^{4-i \pi +\frac {3 x}{2}-\frac {x}{\log (x)}}}{-1+2 e}\right ) \]

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Rubi [F]  time = 2.37, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.69, size = 47, normalized size = 1.42 \begin {gather*} \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 ) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.68, size = 31, normalized size = 0.94 \begin {gather*} \log \left (e^{\left (\frac {{\left (3 \, x - 2 \, \log \left (-2 \, e + 1\right ) + 8\right )} \log \relax (x) - 2 \, x}{2 \, \log \relax (x)}\right )} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.04, size = 32, normalized size = 0.97

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.42, size = 41, normalized size = 1.24 \begin {gather*} -\frac {x}{\log \relax (x)} + \log \left (\frac {3 \, {\left (2 \, e - 1\right )} e^{\frac {x}{\log \relax (x)}} + e^{\left (\frac {3}{2} \, x + 4\right )}}{3 \, {\left (2 \, e - 1\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 7.09, size = 27, normalized size = 0.82 \begin {gather*} \ln \left (-\frac {{\mathrm {e}}^4\,{\mathrm {e}}^{-\frac {x}{\ln \relax (x)}}\,{\left ({\mathrm {e}}^x\right )}^{3/2}}{2\,\mathrm {e}-1}-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.68, size = 29, normalized size = 0.88 \begin {gather*} - \frac {x}{\log {\relax (x )}} + \log {\left (\frac {e^{4} e^{\frac {3 x}{2}}}{-3 + 6 e} + e^{\frac {x}{\log {\relax (x )}}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

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