\(\int \frac {e^{\frac {60-5 x \log (\frac {e}{4+4 \log (5)+\log ^2(5)})}{x \log (\frac {e}{4+4 \log (5)+\log ^2(5)})}} (-60 \log (x)+x \log (\frac {e}{4+4 \log (5)+\log ^2(5)}))}{x^2 \log (\frac {e}{4+4 \log (5)+\log ^2(5)})} \, dx\) [3169]

Optimal result

Integrand size = 86, antiderivative size = 23 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{-5+\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x) \]

[Out]

ln(x)*exp(60/x/ln(exp(1)/(2+ln(5))^2)-5)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(103\) vs. \(2(23)=46\).

Time = 0.06 (sec) , antiderivative size = 103, normalized size of antiderivative = 4.48, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {12, 2326} \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=\frac {12 (2+\log (5))^{\frac {10}{\log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x) \exp \left (\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}-\frac {5}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\right )}{x^2 \log \left (\frac {e}{(2+\log (5))^2}\right ) \left (\frac {12-x \log \left (\frac {e}{(2+\log (5))^2}\right )}{x^2 \log \left (\frac {e}{(2+\log (5))^2}\right )}+\frac {1}{x}\right )} \]

[In]

Int[(E^((60 - 5*x*Log[E/(4 + 4*Log[5] + Log[5]^2)])/(x*Log[E/(4 + 4*Log[5] + Log[5]^2)]))*(-60*Log[x] + x*Log[
E/(4 + 4*Log[5] + Log[5]^2)]))/(x^2*Log[E/(4 + 4*Log[5] + Log[5]^2)]),x]

[Out]

(12*E^(-5/Log[E/(2 + Log[5])^2] + 60/(x*Log[E/(2 + Log[5])^2]))*(2 + Log[5])^(10/Log[E/(2 + Log[5])^2])*Log[x]
)/(x^2*Log[E/(2 + Log[5])^2]*(x^(-1) + (12 - x*Log[E/(2 + Log[5])^2])/(x^2*Log[E/(2 + Log[5])^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 2326

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{align*} \text {integral}& = \frac {\int \frac {\exp \left (\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}\right ) \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2} \, dx}{\log \left (\frac {e}{(2+\log (5))^2}\right )} \\ & = \frac {12 \exp \left (-\frac {5}{\log \left (\frac {e}{(2+\log (5))^2}\right )}+\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}\right ) (2+\log (5))^{\frac {10}{\log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x)}{x^2 \log \left (\frac {e}{(2+\log (5))^2}\right ) \left (\frac {1}{x}+\frac {12-x \log \left (\frac {e}{(2+\log (5))^2}\right )}{x^2 \log \left (\frac {e}{(2+\log (5))^2}\right )}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{-5+\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x) \]

[In]

Integrate[(E^((60 - 5*x*Log[E/(4 + 4*Log[5] + Log[5]^2)])/(x*Log[E/(4 + 4*Log[5] + Log[5]^2)]))*(-60*Log[x] +
x*Log[E/(4 + 4*Log[5] + Log[5]^2)]))/(x^2*Log[E/(4 + 4*Log[5] + Log[5]^2)]),x]

[Out]

E^(-5 + 60/(x*Log[E/(2 + Log[5])^2]))*Log[x]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(47\) vs. \(2(23)=46\).

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.09

method	result	size
default	\(\ln \left (x \right ) {\mathrm e}^{\frac {-5 x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )+60}{x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )}}\)	\(48\)
norman	\(\ln \left (x \right ) {\mathrm e}^{\frac {-5 x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )+60}{x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )}}\)	\(48\)
risch	\(-\frac {\left (2 \ln \left (2+\ln \left (5\right )\right )-1\right ) {\mathrm e}^{-\frac {5 \left (2 x \ln \left (2+\ln \left (5\right )\right )-x +12\right )}{x \left (2 \ln \left (2+\ln \left (5\right )\right )-1\right )}} \ln \left (x \right )}{-2 \ln \left (2+\ln \left (5\right )\right )+1}\)	\(55\)

[In]

int((-60*ln(x)+x*ln(exp(1)/(ln(5)^2+4*ln(5)+4)))*exp((-5*x*ln(exp(1)/(ln(5)^2+4*ln(5)+4))+60)/x/ln(exp(1)/(ln(
5)^2+4*ln(5)+4)))/x^2/ln(exp(1)/(ln(5)^2+4*ln(5)+4)),x,method=_RETURNVERBOSE)

[Out]

ln(x)*exp((-5*x*ln(exp(1)/(ln(5)^2+4*ln(5)+4))+60)/x/ln(exp(1)/(ln(5)^2+4*ln(5)+4)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{\left (-\frac {5 \, {\left (x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right ) - 12\right )}}{x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right )}\right )} \log \left (x\right ) \]

[In]

integrate((-60*log(x)+x*log(exp(1)/(log(5)^2+4*log(5)+4)))*exp((-5*x*log(exp(1)/(log(5)^2+4*log(5)+4))+60)/x/l
og(exp(1)/(log(5)^2+4*log(5)+4)))/x^2/log(exp(1)/(log(5)^2+4*log(5)+4)),x, algorithm="fricas")

[Out]

e^(-5*(x*log(e/(log(5)^2 + 4*log(5) + 4)) - 12)/(x*log(e/(log(5)^2 + 4*log(5) + 4))))*log(x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (20) = 40\).

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{\frac {- 5 x \log {\left (\frac {e}{\log {\left (5 \right )}^{2} + 4 + 4 \log {\left (5 \right )}} \right )} + 60}{x \log {\left (\frac {e}{\log {\left (5 \right )}^{2} + 4 + 4 \log {\left (5 \right )}} \right )}}} \log {\left (x \right )} \]

[In]

integrate((-60*ln(x)+x*ln(exp(1)/(ln(5)**2+4*ln(5)+4)))*exp((-5*x*ln(exp(1)/(ln(5)**2+4*ln(5)+4))+60)/x/ln(exp
(1)/(ln(5)**2+4*ln(5)+4)))/x**2/ln(exp(1)/(ln(5)**2+4*ln(5)+4)),x)

[Out]

exp((-5*x*log(E/(log(5)**2 + 4 + 4*log(5))) + 60)/(x*log(E/(log(5)**2 + 4 + 4*log(5)))))*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{\left (\frac {60}{x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right )} - 5\right )} \log \left (x\right ) \]

[In]

integrate((-60*log(x)+x*log(exp(1)/(log(5)^2+4*log(5)+4)))*exp((-5*x*log(exp(1)/(log(5)^2+4*log(5)+4))+60)/x/l
og(exp(1)/(log(5)^2+4*log(5)+4)))/x^2/log(exp(1)/(log(5)^2+4*log(5)+4)),x, algorithm="maxima")

[Out]

e^(60/(x*log(e/(log(5)^2 + 4*log(5) + 4))) - 5)*log(x)

Giac [F]

\[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=\int { \frac {{\left (x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right ) - 60 \, \log \left (x\right )\right )} e^{\left (-\frac {5 \, {\left (x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right ) - 12\right )}}{x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right )}\right )}}{x^{2} \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right )} \,d x } \]

[In]

integrate((-60*log(x)+x*log(exp(1)/(log(5)^2+4*log(5)+4)))*exp((-5*x*log(exp(1)/(log(5)^2+4*log(5)+4))+60)/x/l
og(exp(1)/(log(5)^2+4*log(5)+4)))/x^2/log(exp(1)/(log(5)^2+4*log(5)+4)),x, algorithm="giac")

[Out]

integrate((x*log(e/(log(5)^2 + 4*log(5) + 4)) - 60*log(x))*e^(-5*(x*log(e/(log(5)^2 + 4*log(5) + 4)) - 12)/(x*
log(e/(log(5)^2 + 4*log(5) + 4))))/(x^2*log(e/(log(5)^2 + 4*log(5) + 4))), x)

Mupad [B] (verification not implemented)

Time = 10.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.13 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=\frac {{\mathrm {e}}^{\frac {60}{x-x\,\ln \left (4\,\ln \left (5\right )+{\ln \left (5\right )}^2+4\right )}}\,{\mathrm {e}}^{\frac {5}{\ln \left (4\,\ln \left (5\right )+{\ln \left (5\right )}^2+4\right )-1}}\,\ln \left (x\right )}{{\left (4\,\ln \left (5\right )+{\ln \left (5\right )}^2+4\right )}^{\frac {5}{\ln \left (4\,\ln \left (5\right )+{\ln \left (5\right )}^2+4\right )-1}}} \]

[In]

int(-(exp(-(5*x*log(exp(1)/(4*log(5) + log(5)^2 + 4)) - 60)/(x*log(exp(1)/(4*log(5) + log(5)^2 + 4))))*(60*log
(x) - x*log(exp(1)/(4*log(5) + log(5)^2 + 4))))/(x^2*log(exp(1)/(4*log(5) + log(5)^2 + 4))),x)

[Out]

(exp(60/(x - x*log(4*log(5) + log(5)^2 + 4)))*exp(5/(log(4*log(5) + log(5)^2 + 4) - 1))*log(x))/(4*log(5) + lo
g(5)^2 + 4)^(5/(log(4*log(5) + log(5)^2 + 4) - 1))