Integrand size = 54, antiderivative size = 28 \begin {dmath*} \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {e^{2 e^{-x}}}{\log ^2\left (\frac {15}{\frac {x}{4}+e^2 x}\right )} \end {dmath*}
Time = 2.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \begin {dmath*} \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {e^{2 e^{-x}}}{\log ^2\left (\frac {60}{x+4 e^2 x}\right )} \end {dmath*}
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.
\(\displaystyle \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{4 e^2 x+x}\right )\right )}{x \log ^3\left (\frac {60}{4 e^2 x+x}\right )} \, dx\) |
\(\Big \downarrow \) 2894 |
\(\displaystyle \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{4 e^2 x+x}\right )\right )}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 e^{2 e^{-x}-x} \left (e^x-x \log \left (\frac {60}{4 e^2 x+x}\right )\right )}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {e^{2 e^{-x}-x} \left (e^x-x \log \left (\frac {60}{\left (1+4 e^2\right ) x}\right )\right )}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (\frac {e^{2 e^{-x}}}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}-\frac {e^{2 e^{-x}-x}}{\log ^2\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\int \frac {e^{2 e^{-x}}}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}dx-\int \frac {e^{2 e^{-x}-x}}{\log ^2\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}dx\right )\) |
3.3.56.3.1 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[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*( a + b*Log[c*ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, b, c, n, p}, x] && Lin earQ[v, x] && !LinearMatchQ[v, x] && !(EqQ[n, 1] && MatchQ[c*v, (e_.)*((f _) + (g_.)*x) /; FreeQ[{e, f, g}, x]])
Time = 2.61 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{2 \,{\mathrm e}^{-x}}}{{\ln \left (\frac {60}{x \left (4 \,{\mathrm e}^{2}+1\right )}\right )}^{2}}\) | \(27\) |
int((-2*x*ln(60/(4*x*exp(1)^2+x))+2*exp(x))*exp(1/exp(x))^2/x/exp(x)/ln(60 /(4*x*exp(1)^2+x))^3,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \begin {dmath*} \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {e^{\left (-{\left (x e^{x} - 2\right )} e^{\left (-x\right )} + x\right )}}{\log \left (\frac {60}{4 \, x e^{2} + x}\right )^{2}} \end {dmath*}
integrate((-2*x*log(60/(4*x*exp(1)^2+x))+2*exp(x))*exp(1/exp(x))^2/x/exp(x )/log(60/(4*x*exp(1)^2+x))^3,x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \begin {dmath*} \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {e^{2 e^{- x}}}{\log {\left (\frac {60}{x + 4 x e^{2}} \right )}^{2}} \end {dmath*}
integrate((-2*x*ln(60/(4*x*exp(1)**2+x))+2*exp(x))*exp(1/exp(x))**2/x/exp( x)/ln(60/(4*x*exp(1)**2+x))**3,x)
\begin {dmath*} \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\int { -\frac {2 \, {\left (x \log \left (\frac {60}{4 \, x e^{2} + x}\right ) - e^{x}\right )} e^{\left (-x + 2 \, e^{\left (-x\right )}\right )}}{x \log \left (\frac {60}{4 \, x e^{2} + x}\right )^{3}} \,d x } \end {dmath*}
integrate((-2*x*log(60/(4*x*exp(1)^2+x))+2*exp(x))*exp(1/exp(x))^2/x/exp(x )/log(60/(4*x*exp(1)^2+x))^3,x, algorithm=\
-2*integrate((x*log(60/(4*x*e^2 + x)) - e^x)*e^(-x + 2*e^(-x))/(x*log(60/( 4*x*e^2 + x))^3), x)
\begin {dmath*} \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\int { -\frac {2 \, {\left (x \log \left (\frac {60}{4 \, x e^{2} + x}\right ) - e^{x}\right )} e^{\left (-x + 2 \, e^{\left (-x\right )}\right )}}{x \log \left (\frac {60}{4 \, x e^{2} + x}\right )^{3}} \,d x } \end {dmath*}
integrate((-2*x*log(60/(4*x*exp(1)^2+x))+2*exp(x))*exp(1/exp(x))^2/x/exp(x )/log(60/(4*x*exp(1)^2+x))^3,x, algorithm=\
integrate(-2*(x*log(60/(4*x*e^2 + x)) - e^x)*e^(-x + 2*e^(-x))/(x*log(60/( 4*x*e^2 + x))^3), x)
Time = 14.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \begin {dmath*} \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}}}{{\ln \left (\frac {60}{x+4\,x\,{\mathrm {e}}^2}\right )}^2} \end {dmath*}