3.34.86 \(\int \frac {e^{e^{\frac {1}{2 x \log (x)}}+\frac {1}{2 x \log (x)}} (-3-3 \log (x))-6 x^2 \log ^2(x)}{8 e^{4+e^{\frac {1}{2 x \log (x)}}} x^2 \log ^2(x)-8 e^4 x^3 \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {3 \log \left (e^{e^{\frac {1}{2 x \log (x)}}}-x\right )}{4 e^4} \]

________________________________________________________________________________________

Rubi [A]  time = 0.65, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6741, 12, 6684} \begin {gather*} \frac {3 \log \left (e^{e^{\frac {1}{2 x \log (x)}}}-x\right )}{4 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E^(1/(2*x*Log[x])) + 1/(2*x*Log[x]))*(-3 - 3*Log[x]) - 6*x^2*Log[x]^2)/(8*E^(4 + E^(1/(2*x*Log[x])))*x
^2*Log[x]^2 - 8*E^4*x^3*Log[x]^2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^{\frac {1}{2 x \log (x)}}+\frac {1}{2 x \log (x)}} (-3-3 \log (x))-6 x^2 \log ^2(x)}{8 e^4 \left (e^{e^{\frac {1}{2 x \log (x)}}}-x\right ) x^2 \log ^2(x)} \, dx\\ &=\frac {\int \frac {e^{e^{\frac {1}{2 x \log (x)}}+\frac {1}{2 x \log (x)}} (-3-3 \log (x))-6 x^2 \log ^2(x)}{\left (e^{e^{\frac {1}{2 x \log (x)}}}-x\right ) x^2 \log ^2(x)} \, dx}{8 e^4}\\ &=\frac {3 \log \left (e^{e^{\frac {1}{2 x \log (x)}}}-x\right )}{4 e^4}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.53, size = 27, normalized size = 1.00 \begin {gather*} \frac {3 \log \left (e^{e^{\frac {1}{2 x \log (x)}}}-x\right )}{4 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^(1/(2*x*Log[x])) + 1/(2*x*Log[x]))*(-3 - 3*Log[x]) - 6*x^2*Log[x]^2)/(8*E^(4 + E^(1/(2*x*Log[x
])))*x^2*Log[x]^2 - 8*E^4*x^3*Log[x]^2),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.77, size = 24, normalized size = 0.89 \begin {gather*} \frac {3}{4} \, e^{\left (-4\right )} \log \left (-x e^{4} + e^{\left (e^{\left (\frac {1}{2 \, x \log \relax (x)}\right )} + 4\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*log(x)-3)*exp(1/2/x/log(x))*exp(exp(1/2/x/log(x)))-6*x^2*log(x)^2)/(8*x^2*exp(4)*log(x)^2*exp(e
xp(1/2/x/log(x)))-8*x^3*exp(4)*log(x)^2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*log(x)-3)*exp(1/2/x/log(x))*exp(exp(1/2/x/log(x)))-6*x^2*log(x)^2)/(8*x^2*exp(4)*log(x)^2*exp(e
xp(1/2/x/log(x)))-8*x^3*exp(4)*log(x)^2),x, algorithm="giac")

[Out]

undef

________________________________________________________________________________________

maple [A]  time = 0.08, size = 21, normalized size = 0.78

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{-4} \ln \left ({\mathrm e}^{{\mathrm e}^{\frac {1}{2 x \ln \relax (x )}}}-x \right )}{4}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*ln(x)-3)*exp(1/2/x/ln(x))*exp(exp(1/2/x/ln(x)))-6*x^2*ln(x)^2)/(8*x^2*exp(4)*ln(x)^2*exp(exp(1/2/x/ln
(x)))-8*x^3*exp(4)*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*log(x)-3)*exp(1/2/x/log(x))*exp(exp(1/2/x/log(x)))-6*x^2*log(x)^2)/(8*x^2*exp(4)*log(x)^2*exp(e
xp(1/2/x/log(x)))-8*x^3*exp(4)*log(x)^2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

________________________________________________________________________________________

mupad [B]  time = 2.53, size = 20, normalized size = 0.74 \begin {gather*} \frac {3\,{\mathrm {e}}^{-4}\,\ln \left ({\mathrm {e}}^{{\mathrm {e}}^{\frac {1}{2\,x\,\ln \relax (x)}}}-x\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x^2*log(x)^2 + exp(exp(1/(2*x*log(x))))*exp(1/(2*x*log(x)))*(3*log(x) + 3))/(8*x^3*exp(4)*log(x)^2 - 8*
x^2*exp(exp(1/(2*x*log(x))))*exp(4)*log(x)^2),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.73, size = 20, normalized size = 0.74 \begin {gather*} \frac {3 \log {\left (- x + e^{e^{\frac {1}{2 x \log {\relax (x )}}}} \right )}}{4 e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*ln(x)-3)*exp(1/2/x/ln(x))*exp(exp(1/2/x/ln(x)))-6*x**2*ln(x)**2)/(8*x**2*exp(4)*ln(x)**2*exp(ex
p(1/2/x/ln(x)))-8*x**3*exp(4)*ln(x)**2),x)

[Out]

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

________________________________________________________________________________________