3.12.67 \(\int \frac {-5+\log (3)+e^{\frac {e^{e^x}}{-5+\log (3)}} (-5+\log (3)+e^{e^x+x} x \log (x)+(-5+\log (3)) \log (x))}{-5+\log (3)} \, dx\)

Optimal. Leaf size=23 \[ x+e^{-\frac {e^{e^x}}{5-\log (3)}} x \log (x) \]

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {12, 2288} \begin {gather*} x+x e^{-\frac {e^{e^x}}{5-\log (3)}} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + Log[3] + E^(E^E^x/(-5 + Log[3]))*(-5 + Log[3] + E^(E^x + x)*x*Log[x] + (-5 + Log[3])*Log[x]))/(-5 +
Log[3]),x]

[Out]

x + (x*Log[x])/E^(E^E^x/(5 - Log[3]))

Rule 12

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (-5+\log (3)+e^{\frac {e^{e^x}}{-5+\log (3)}} \left (-5+\log (3)+e^{e^x+x} x \log (x)+(-5+\log (3)) \log (x)\right )\right ) \, dx}{-5+\log (3)}\\ &=x+\frac {\int e^{\frac {e^{e^x}}{-5+\log (3)}} \left (-5+\log (3)+e^{e^x+x} x \log (x)+(-5+\log (3)) \log (x)\right ) \, dx}{-5+\log (3)}\\ &=x+e^{-\frac {e^{e^x}}{5-\log (3)}} x \log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.52, size = 37, normalized size = 1.61 \begin {gather*} \frac {-5 x+x \log (3)+e^{\frac {e^{e^x}}{-5+\log (3)}} x (-5+\log (3)) \log (x)}{-5+\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + Log[3] + E^(E^E^x/(-5 + Log[3]))*(-5 + Log[3] + E^(E^x + x)*x*Log[x] + (-5 + Log[3])*Log[x]))/
(-5 + Log[3]),x]

[Out]

(-5*x + x*Log[3] + E^(E^E^x/(-5 + Log[3]))*x*(-5 + Log[3])*Log[x])/(-5 + Log[3])

________________________________________________________________________________________

fricas [A]  time = 0.90, size = 17, normalized size = 0.74 \begin {gather*} x e^{\left (\frac {e^{\left (e^{x}\right )}}{\log \relax (3) - 5}\right )} \log \relax (x) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)*log(x)*exp(exp(x))+(log(3)-5)*log(x)+log(3)-5)*exp(exp(exp(x))/(log(3)-5))+log(3)-5)/(log
(3)-5),x, algorithm="fricas")

[Out]

x*e^(e^(e^x)/(log(3) - 5))*log(x) + x

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x e^{\left (x + e^{x}\right )} \log \relax (x) + {\left (\log \relax (3) - 5\right )} \log \relax (x) + \log \relax (3) - 5\right )} e^{\left (\frac {e^{\left (e^{x}\right )}}{\log \relax (3) - 5}\right )} + \log \relax (3) - 5}{\log \relax (3) - 5}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)*log(x)*exp(exp(x))+(log(3)-5)*log(x)+log(3)-5)*exp(exp(exp(x))/(log(3)-5))+log(3)-5)/(log
(3)-5),x, algorithm="giac")

[Out]

integrate(((x*e^(x + e^x)*log(x) + (log(3) - 5)*log(x) + log(3) - 5)*e^(e^(e^x)/(log(3) - 5)) + log(3) - 5)/(l
og(3) - 5), x)

________________________________________________________________________________________

maple [A]  time = 0.08, size = 18, normalized size = 0.78

method	result	size
risch	\(x +x \ln \relax (x ) {\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{x}}}{\ln \relax (3)-5}}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(x)*ln(x)*exp(exp(x))+(ln(3)-5)*ln(x)+ln(3)-5)*exp(exp(exp(x))/(ln(3)-5))+ln(3)-5)/(ln(3)-5),x,meth
od=_RETURNVERBOSE)

[Out]

x+x*ln(x)*exp(exp(exp(x))/(ln(3)-5))

________________________________________________________________________________________

maxima [A]  time = 0.50, size = 34, normalized size = 1.48 \begin {gather*} \frac {x {\left (\log \relax (3) - 5\right )} e^{\left (\frac {e^{\left (e^{x}\right )}}{\log \relax (3) - 5}\right )} \log \relax (x) + x \log \relax (3) - 5 \, x}{\log \relax (3) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)*log(x)*exp(exp(x))+(log(3)-5)*log(x)+log(3)-5)*exp(exp(exp(x))/(log(3)-5))+log(3)-5)/(log
(3)-5),x, algorithm="maxima")

[Out]

(x*(log(3) - 5)*e^(e^(e^x)/(log(3) - 5))*log(x) + x*log(3) - 5*x)/(log(3) - 5)

________________________________________________________________________________________

mupad [B]  time = 0.88, size = 17, normalized size = 0.74 \begin {gather*} x+x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^x}}{\ln \relax (3)-5}}\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3) + exp(exp(exp(x))/(log(3) - 5))*(log(3) + log(x)*(log(3) - 5) + x*exp(exp(x))*exp(x)*log(x) - 5) -
5)/(log(3) - 5),x)

[Out]

x + x*exp(exp(exp(x))/(log(3) - 5))*log(x)

________________________________________________________________________________________

sympy [A]  time = 9.85, size = 17, normalized size = 0.74 \begin {gather*} x e^{\frac {e^{e^{x}}}{-5 + \log {\relax (3 )}}} \log {\relax (x )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)*ln(x)*exp(exp(x))+(ln(3)-5)*ln(x)+ln(3)-5)*exp(exp(exp(x))/(ln(3)-5))+ln(3)-5)/(ln(3)-5),
x)

[Out]

x*exp(exp(exp(x))/(-5 + log(3)))*log(x) + x

________________________________________________________________________________________