Integrand size = 118, antiderivative size = 25 \begin {dmath*} \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx=e^{\left (-\frac {1}{4}-\log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )^2} \end {dmath*}
Time = 0.10 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \begin {dmath*} \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx=\sqrt {2} e^{\frac {1}{16}+\log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )} \sqrt {\log \left (\frac {e^5+e^x}{x}\right )} \end {dmath*}
Integrate[(E^((1 + 8*Log[2*Log[(E^5 + E^x)/x]] + 16*Log[2*Log[(E^5 + E^x)/ x]]^2)/16)*(-E^5 + E^x*(-1 + x) + (-4*E^5 + E^x*(-4 + 4*x))*Log[2*Log[(E^5 + E^x)/x]]))/((2*E^5*x + 2*E^x*x)*Log[(E^5 + E^x)/x]),x]
Leaf count is larger than twice the leaf count of optimal. \(107\) vs. \(2(25)=50\).
Time = 0.90 (sec) , antiderivative size = 107, normalized size of antiderivative = 4.28, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {2704, 27, 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.
\(\displaystyle \int \frac {\left (e^x (x-1)+\left (e^x (4 x-4)-4 e^5\right ) \log \left (2 \log \left (\frac {e^x+e^5}{x}\right )\right )-e^5\right ) \exp \left (\frac {1}{16} \left (16 \log ^2\left (2 \log \left (\frac {e^x+e^5}{x}\right )\right )+8 \log \left (2 \log \left (\frac {e^x+e^5}{x}\right )\right )+1\right )\right )}{\left (2 e^x x+2 e^5 x\right ) \log \left (\frac {e^x+e^5}{x}\right )} \, dx\) |
\(\Big \downarrow \) 2704 |
\(\displaystyle \int \frac {\sqrt {2} e^{\frac {1}{16} \left (16 \log ^2\left (2 \log \left (\frac {e^x+e^5}{x}\right )\right )+1\right )} \left (e^x (x-1)+\left (e^x (4 x-4)-4 e^5\right ) \log \left (2 \log \left (\frac {e^x+e^5}{x}\right )\right )-e^5\right )}{\left (2 e^x x+2 e^5 x\right ) \sqrt {\log \left (\frac {e^x+e^5}{x}\right )}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {2} \int -\frac {e^{\frac {1}{16} \left (16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+1\right )} \left (e^x (1-x)+4 \left (e^x (1-x)+e^5\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+e^5\right )}{2 \left (e^x x+e^5 x\right ) \sqrt {\log \left (\frac {e^5+e^x}{x}\right )}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {e^{\frac {1}{16} \left (16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+1\right )} \left (e^x (1-x)+4 \left (e^x (1-x)+e^5\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+e^5\right )}{\left (e^x x+e^5 x\right ) \sqrt {\log \left (\frac {e^5+e^x}{x}\right )}}dx}{\sqrt {2}}\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {\sqrt {2} \left (e^x+e^5\right ) \left (e^x (1-x)+e^5\right ) e^{\frac {1}{16} \left (16 \log ^2\left (2 \log \left (\frac {e^x+e^5}{x}\right )\right )+1\right )} \sqrt {\log \left (\frac {e^x+e^5}{x}\right )}}{\left (\frac {e^x+e^5}{x^2}-\frac {e^x}{x}\right ) x \left (e^x x+e^5 x\right )}\) |
Int[(E^((1 + 8*Log[2*Log[(E^5 + E^x)/x]] + 16*Log[2*Log[(E^5 + E^x)/x]]^2) /16)*(-E^5 + E^x*(-1 + x) + (-4*E^5 + E^x*(-4 + 4*x))*Log[2*Log[(E^5 + E^x )/x]]))/((2*E^5*x + 2*E^x*x)*Log[(E^5 + E^x)/x]),x]
(Sqrt[2]*E^((1 + 16*Log[2*Log[(E^5 + E^x)/x]]^2)/16)*(E^5 + E^x)*(E^5 + E^ x*(1 - x))*Sqrt[Log[(E^5 + E^x)/x]])/(((E^5 + E^x)/x^2 - E^x/x)*x*(E^5*x + E^x*x))
3.4.30.3.1 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[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)* z^(a*b*Log[F]), x] /; FreeQ[{F, a, 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]
Time = 44.64 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36
method | result | size |
parallelrisch | \({\mathrm e}^{{\ln \left (2 \ln \left (\frac {{\mathrm e}^{5}+{\mathrm e}^{x}}{x}\right )\right )}^{2}+\frac {\ln \left (2 \ln \left (\frac {{\mathrm e}^{5}+{\mathrm e}^{x}}{x}\right )\right )}{2}+\frac {1}{16}}\) | \(34\) |
risch | \(\sqrt {-2 \ln \left (x \right )+2 \ln \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right ) \left (-\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right )+\operatorname {csgn}\left (i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )\right )\right )}\, {\mathrm e}^{{\ln \left (-2 \ln \left (x \right )+2 \ln \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right ) \left (-\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right )+\operatorname {csgn}\left (i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )\right )\right )\right )}^{2}+\frac {1}{16}}\) | \(160\) |
int((((-4+4*x)*exp(x)-4*exp(5))*ln(2*ln((exp(5)+exp(x))/x))+(-1+x)*exp(x)- exp(5))*exp(ln(2*ln((exp(5)+exp(x))/x))^2+1/2*ln(2*ln((exp(5)+exp(x))/x))+ 1/16)/(2*exp(x)*x+2*x*exp(5))/ln((exp(5)+exp(x))/x),x,method=_RETURNVERBOS E)
Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \begin {dmath*} \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx=e^{\left (\log \left (2 \, \log \left (\frac {e^{5} + e^{x}}{x}\right )\right )^{2} + \frac {1}{2} \, \log \left (2 \, \log \left (\frac {e^{5} + e^{x}}{x}\right )\right ) + \frac {1}{16}\right )} \end {dmath*}
integrate((((-4+4*x)*exp(x)-4*exp(5))*log(2*log((exp(5)+exp(x))/x))+(-1+x) *exp(x)-exp(5))*exp(log(2*log((exp(5)+exp(x))/x))^2+1/2*log(2*log((exp(5)+ exp(x))/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/log((exp(5)+exp(x))/x),x, algori thm=\
Timed out. \begin {dmath*} \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx=\text {Timed out} \end {dmath*}
integrate((((-4+4*x)*exp(x)-4*exp(5))*ln(2*ln((exp(5)+exp(x))/x))+(-1+x)*e xp(x)-exp(5))*exp(ln(2*ln((exp(5)+exp(x))/x))**2+1/2*ln(2*ln((exp(5)+exp(x ))/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/ln((exp(5)+exp(x))/x),x)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).
Time = 0.57 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \begin {dmath*} \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx=\sqrt {2} \sqrt {-\log \left (x\right ) + \log \left (e^{5} + e^{x}\right )} e^{\left (\log \left (2\right )^{2} + 2 \, \log \left (2\right ) \log \left (-\log \left (x\right ) + \log \left (e^{5} + e^{x}\right )\right ) + \log \left (-\log \left (x\right ) + \log \left (e^{5} + e^{x}\right )\right )^{2} + \frac {1}{16}\right )} \end {dmath*}
integrate((((-4+4*x)*exp(x)-4*exp(5))*log(2*log((exp(5)+exp(x))/x))+(-1+x) *exp(x)-exp(5))*exp(log(2*log((exp(5)+exp(x))/x))^2+1/2*log(2*log((exp(5)+ exp(x))/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/log((exp(5)+exp(x))/x),x, algori thm=\
sqrt(2)*sqrt(-log(x) + log(e^5 + e^x))*e^(log(2)^2 + 2*log(2)*log(-log(x) + log(e^5 + e^x)) + log(-log(x) + log(e^5 + e^x))^2 + 1/16)
\begin {dmath*} \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx=\int { \frac {{\left ({\left (x - 1\right )} e^{x} + 4 \, {\left ({\left (x - 1\right )} e^{x} - e^{5}\right )} \log \left (2 \, \log \left (\frac {e^{5} + e^{x}}{x}\right )\right ) - e^{5}\right )} e^{\left (\log \left (2 \, \log \left (\frac {e^{5} + e^{x}}{x}\right )\right )^{2} + \frac {1}{2} \, \log \left (2 \, \log \left (\frac {e^{5} + e^{x}}{x}\right )\right ) + \frac {1}{16}\right )}}{2 \, {\left (x e^{5} + x e^{x}\right )} \log \left (\frac {e^{5} + e^{x}}{x}\right )} \,d x } \end {dmath*}
integrate((((-4+4*x)*exp(x)-4*exp(5))*log(2*log((exp(5)+exp(x))/x))+(-1+x) *exp(x)-exp(5))*exp(log(2*log((exp(5)+exp(x))/x))^2+1/2*log(2*log((exp(5)+ exp(x))/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/log((exp(5)+exp(x))/x),x, algori thm=\
Time = 13.48 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \begin {dmath*} \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx={\mathrm {e}}^{1/16}\,{\mathrm {e}}^{{\ln \left (2\,\ln \left (\frac {1}{x}\right )+\ln \left ({\left ({\mathrm {e}}^5+{\mathrm {e}}^x\right )}^2\right )\right )}^2}\,\sqrt {2\,\ln \left (\frac {1}{x}\right )+\ln \left ({\left ({\mathrm {e}}^5+{\mathrm {e}}^x\right )}^2\right )} \end {dmath*}
int(-(exp(log(2*log((exp(5) + exp(x))/x))/2 + log(2*log((exp(5) + exp(x))/ x))^2 + 1/16)*(exp(5) + log(2*log((exp(5) + exp(x))/x))*(4*exp(5) - exp(x) *(4*x - 4)) - exp(x)*(x - 1)))/(log((exp(5) + exp(x))/x)*(2*x*exp(5) + 2*x *exp(x))),x)