3.21.4 \(\int \frac {e^{\frac {x}{\log (x)}} \log (\frac {1}{2} (4-e^5+2 e^{\frac {x}{\log (x)}})) (-4+4 \log (x))}{2 e^{\frac {x}{\log (x)}} \log ^2(x)+(4-e^5) \log ^2(x)} \, dx\)

Optimal. Leaf size=23 \[ \log ^2\left (e^{\frac {x}{\log (x)}}+\frac {1}{2} \left (4-e^5\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.82, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 1, number of rules used = 4, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6741, 12, 6684, 6686} \begin {gather*} \log ^2\left (\frac {1}{2} \left (2 e^{\frac {x}{\log (x)}}+4-e^5\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(x/Log[x])*Log[(4 - E^5 + 2*E^(x/Log[x]))/2]*(-4 + 4*Log[x]))/(2*E^(x/Log[x])*Log[x]^2 + (4 - E^5)*Log[
x]^2),x]

[Out]

Log[(4 - E^5 + 2*E^(x/Log[x]))/2]^2

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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log ^2\left (\frac {1}{2} \left (4-e^5+2 e^{\frac {x}{\log (x)}}\right )\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 20, normalized size = 0.87 \begin {gather*} \log ^2\left (2-\frac {e^5}{2}+e^{\frac {x}{\log (x)}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x/Log[x])*Log[(4 - E^5 + 2*E^(x/Log[x]))/2]*(-4 + 4*Log[x]))/(2*E^(x/Log[x])*Log[x]^2 + (4 - E^5
)*Log[x]^2),x]

[Out]

Log[2 - E^5/2 + E^(x/Log[x])]^2

________________________________________________________________________________________

fricas [A]  time = 1.17, size = 16, normalized size = 0.70 \begin {gather*} \log \left (-\frac {1}{2} \, e^{5} + e^{\frac {x}{\log \relax (x)}} + 2\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(x)-4)*exp(x/log(x))*log(exp(x/log(x))+2-1/2*exp(5))/(2*log(x)^2*exp(x/log(x))+(4-exp(5))*log(
x)^2),x, algorithm="fricas")

[Out]

log(-1/2*e^5 + e^(x/log(x)) + 2)^2

________________________________________________________________________________________

giac [B]  time = 0.30, size = 39, normalized size = 1.70 \begin {gather*} -2 \, \log \relax (2) \log \left (-e^{5} + 2 \, e^{\frac {x}{\log \relax (x)}} + 4\right ) + \log \left (-e^{5} + 2 \, e^{\frac {x}{\log \relax (x)}} + 4\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(x)-4)*exp(x/log(x))*log(exp(x/log(x))+2-1/2*exp(5))/(2*log(x)^2*exp(x/log(x))+(4-exp(5))*log(
x)^2),x, algorithm="giac")

[Out]

-2*log(2)*log(-e^5 + 2*e^(x/log(x)) + 4) + log(-e^5 + 2*e^(x/log(x)) + 4)^2

________________________________________________________________________________________

maple [A]  time = 0.05, size = 17, normalized size = 0.74

method	result	size
risch	\(\ln \left ({\mathrm e}^{\frac {x}{\ln \relax (x )}}+2-\frac {{\mathrm e}^{5}}{2}\right )^{2}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*ln(x)-4)*exp(x/ln(x))*ln(exp(x/ln(x))+2-1/2*exp(5))/(2*ln(x)^2*exp(x/ln(x))+(4-exp(5))*ln(x)^2),x,metho
d=_RETURNVERBOSE)

[Out]

ln(exp(x/ln(x))+2-1/2*exp(5))^2

________________________________________________________________________________________

maxima [B]  time = 0.73, size = 53, normalized size = 2.30 \begin {gather*} 2 \, \log \left (-\frac {1}{2} \, e^{5} + e^{\frac {x}{\log \relax (x)}} + 2\right ) \log \left (-e^{5} + 2 \, e^{\frac {x}{\log \relax (x)}} + 4\right ) - \log \left (-e^{5} + 2 \, e^{\frac {x}{\log \relax (x)}} + 4\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(x)-4)*exp(x/log(x))*log(exp(x/log(x))+2-1/2*exp(5))/(2*log(x)^2*exp(x/log(x))+(4-exp(5))*log(
x)^2),x, algorithm="maxima")

[Out]

2*log(-1/2*e^5 + e^(x/log(x)) + 2)*log(-e^5 + 2*e^(x/log(x)) + 4) - log(-e^5 + 2*e^(x/log(x)) + 4)^2

________________________________________________________________________________________

mupad [B]  time = 1.88, size = 16, normalized size = 0.70 \begin {gather*} {\ln \left ({\mathrm {e}}^{\frac {x}{\ln \relax (x)}}-\frac {{\mathrm {e}}^5}{2}+2\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(x/log(x)) - exp(5)/2 + 2)*exp(x/log(x))*(4*log(x) - 4))/(2*exp(x/log(x))*log(x)^2 - log(x)^2*(exp
(5) - 4)),x)

[Out]

log(exp(x/log(x)) - exp(5)/2 + 2)^2

________________________________________________________________________________________

sympy [A]  time = 1.63, size = 15, normalized size = 0.65 \begin {gather*} \log {\left (e^{\frac {x}{\log {\relax (x )}}} - \frac {e^{5}}{2} + 2 \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*ln(x)-4)*exp(x/ln(x))*ln(exp(x/ln(x))+2-1/2*exp(5))/(2*ln(x)**2*exp(x/ln(x))+(4-exp(5))*ln(x)**2)
,x)

[Out]

log(exp(x/log(x)) - exp(5)/2 + 2)**2

________________________________________________________________________________________