Integrand size = 86, antiderivative size = 23 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{-5+\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x) \]
Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{-5+\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x) \]
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]
Leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(23)=46\).
Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.30, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {27, 25, 27, 2726}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (x \log \left (\frac {e}{4+\log ^2(5)+4 \log (5)}\right )-60 \log (x)\right ) \exp \left (\frac {60-5 x \log \left (\frac {e}{4+\log ^2(5)+4 \log (5)}\right )}{x \log \left (\frac {e}{4+\log ^2(5)+4 \log (5)}\right )}\right )}{x^2 \log \left (\frac {e}{4+\log ^2(5)+4 \log (5)}\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {\exp \left (\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}-\frac {5}{\log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}\right ) \left (4+4 \log (5)+\log ^2(5)\right )^{\frac {5}{\log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (60 \log (x)-x \log \left (\frac {e}{(2+\log (5))^2}\right )\right )}{x^2}dx}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\exp \left (\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}-\frac {5}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\right ) (2+\log (5))^{\frac {10}{\log \left (\frac {e}{(2+\log (5))^2}\right )}} \left (60 \log (x)-x \log \left (\frac {e}{(2+\log (5))^2}\right )\right )}{x^2}dx}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {(2+\log (5))^{\frac {10}{\log \left (\frac {e}{(2+\log (5))^2}\right )}} \int \frac {\exp \left (\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}-\frac {5}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\right ) \left (60 \log (x)-x \log \left (\frac {e}{(2+\log (5))^2}\right )\right )}{x^2}dx}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle (2+\log (5))^{\frac {10}{\log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x) \exp \left (\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}-\frac {5}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\right )\) |
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*Lo g[E/(4 + 4*Log[5] + Log[5]^2)]),x]
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]
3.2.69.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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 | \(-\frac {\left (2 \ln \left (2+\ln \left (5\right )\right )-1\right ) {\mathrm e}^{-\frac {5 \left (2 x \ln \left (2+\ln \left (5\right )\right )-x +12\right )}{x \left (2 \ln \left (2+\ln \left (5\right )\right )-1\right )}} \ln \left (x \right )}{-2 \ln \left (2+\ln \left (5\right )\right )+1}\) | \(55\) |
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)
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 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{\left (-\frac {5 \, {\left (x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right ) - 12\right )}}{x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right )}\right )} \log \left (x\right ) \]
integrate((-60*log(x)+x*log(exp(1)/(log(5)^2+4*log(5)+4)))*exp((-5*x*log(e xp(1)/(log(5)^2+4*log(5)+4))+60)/x/log(exp(1)/(log(5)^2+4*log(5)+4)))/x^2/ log(exp(1)/(log(5)^2+4*log(5)+4)),x, algorithm=\
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)
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 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{\frac {- 5 x \log {\left (\frac {e}{\log {\left (5 \right )}^{2} + 4 + 4 \log {\left (5 \right )}} \right )} + 60}{x \log {\left (\frac {e}{\log {\left (5 \right )}^{2} + 4 + 4 \log {\left (5 \right )}} \right )}}} \log {\left (x \right )} \]
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)
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)
Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{\left (\frac {60}{x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right )} - 5\right )} \log \left (x\right ) \]
integrate((-60*log(x)+x*log(exp(1)/(log(5)^2+4*log(5)+4)))*exp((-5*x*log(e xp(1)/(log(5)^2+4*log(5)+4))+60)/x/log(exp(1)/(log(5)^2+4*log(5)+4)))/x^2/ log(exp(1)/(log(5)^2+4*log(5)+4)),x, algorithm=\
\[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=\int { \frac {{\left (x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right ) - 60 \, \log \left (x\right )\right )} e^{\left (-\frac {5 \, {\left (x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right ) - 12\right )}}{x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right )}\right )}}{x^{2} \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right )} \,d x } \]
integrate((-60*log(x)+x*log(exp(1)/(log(5)^2+4*log(5)+4)))*exp((-5*x*log(e xp(1)/(log(5)^2+4*log(5)+4))+60)/x/log(exp(1)/(log(5)^2+4*log(5)+4)))/x^2/ log(exp(1)/(log(5)^2+4*log(5)+4)),x, algorithm=\
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)
Time = 10.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.13 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=\frac {{\mathrm {e}}^{\frac {60}{x-x\,\ln \left (4\,\ln \left (5\right )+{\ln \left (5\right )}^2+4\right )}}\,{\mathrm {e}}^{\frac {5}{\ln \left (4\,\ln \left (5\right )+{\ln \left (5\right )}^2+4\right )-1}}\,\ln \left (x\right )}{{\left (4\,\ln \left (5\right )+{\ln \left (5\right )}^2+4\right )}^{\frac {5}{\ln \left (4\,\ln \left (5\right )+{\ln \left (5\right )}^2+4\right )-1}}} \]
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)