Integrand size = 137, antiderivative size = 28 \[ \int \frac {e^{\frac {1}{4} \left (4 x+\left (e^{4+4 x} x-x^2\right ) \log (\log (x+\log (x)))\right )} \left (-x-x^2+e^{4+4 x} (1+x)+(4 x+4 \log (x)) \log (x+\log (x))+\left (-2 x^2+e^{4+4 x} \left (x+4 x^2\right )+\left (-2 x+e^{4+4 x} (1+4 x)\right ) \log (x)\right ) \log (x+\log (x)) \log (\log (x+\log (x)))\right )}{(4 x+4 \log (x)) \log (x+\log (x))} \, dx=e^{\frac {1}{4} x \left (4-\left (-e^{4+4 x}+x\right ) \log (\log (x+\log (x)))\right )} \]
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {1}{4} \left (4 x+\left (e^{4+4 x} x-x^2\right ) \log (\log (x+\log (x)))\right )} \left (-x-x^2+e^{4+4 x} (1+x)+(4 x+4 \log (x)) \log (x+\log (x))+\left (-2 x^2+e^{4+4 x} \left (x+4 x^2\right )+\left (-2 x+e^{4+4 x} (1+4 x)\right ) \log (x)\right ) \log (x+\log (x)) \log (\log (x+\log (x)))\right )}{(4 x+4 \log (x)) \log (x+\log (x))} \, dx=e^x \log ^{\frac {1}{4} \left (e^{4+4 x}-x\right ) x}(x+\log (x)) \]
Integrate[(E^((4*x + (E^(4 + 4*x)*x - x^2)*Log[Log[x + Log[x]]])/4)*(-x - x^2 + E^(4 + 4*x)*(1 + x) + (4*x + 4*Log[x])*Log[x + Log[x]] + (-2*x^2 + E ^(4 + 4*x)*(x + 4*x^2) + (-2*x + E^(4 + 4*x)*(1 + 4*x))*Log[x])*Log[x + Lo g[x]]*Log[Log[x + Log[x]]]))/((4*x + 4*Log[x])*Log[x + Log[x]]),x]
Time = 4.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {7292, 27, 25, 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.
\(\displaystyle \int \frac {\left (-x^2+\left (-2 x^2+e^{4 x+4} \left (4 x^2+x\right )+\left (e^{4 x+4} (4 x+1)-2 x\right ) \log (x)\right ) \log (x+\log (x)) \log (\log (x+\log (x)))-x+e^{4 x+4} (x+1)+(4 x+4 \log (x)) \log (x+\log (x))\right ) \exp \left (\frac {1}{4} \left (\left (e^{4 x+4} x-x^2\right ) \log (\log (x+\log (x)))+4 x\right )\right )}{(4 x+4 \log (x)) \log (x+\log (x))} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (-x^2+\left (-2 x^2+e^{4 x+4} \left (4 x^2+x\right )+\left (e^{4 x+4} (4 x+1)-2 x\right ) \log (x)\right ) \log (x+\log (x)) \log (\log (x+\log (x)))-x+e^{4 x+4} (x+1)+(4 x+4 \log (x)) \log (x+\log (x))\right ) \exp \left (\frac {1}{4} \left (\left (e^{4 x+4} x-x^2\right ) \log (\log (x+\log (x)))+4 x\right )\right )}{4 (x+\log (x)) \log (x+\log (x))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {e^x \log ^{\frac {1}{4} \left (e^{4 x+4} x-x^2\right )-1}(x+\log (x)) \left (x^2+x-e^{4 x+4} (x+1)-4 (x+\log (x)) \log (x+\log (x))+\left (2 x^2-e^{4 x+4} \left (4 x^2+x\right )+\left (2 x-e^{4 x+4} (4 x+1)\right ) \log (x)\right ) \log (x+\log (x)) \log (\log (x+\log (x)))\right )}{x+\log (x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {e^x \log ^{\frac {1}{4} \left (e^{4 x+4} x-x^2\right )-1}(x+\log (x)) \left (x^2+x-e^{4 x+4} (x+1)-4 (x+\log (x)) \log (x+\log (x))+\left (2 x^2-e^{4 x+4} \left (4 x^2+x\right )+\left (2 x-e^{4 x+4} (4 x+1)\right ) \log (x)\right ) \log (x+\log (x)) \log (\log (x+\log (x)))\right )}{x+\log (x)}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle e^x \log ^{\frac {1}{4} \left (e^{4 x+4} x-x^2\right )}(x+\log (x))\) |
Int[(E^((4*x + (E^(4 + 4*x)*x - x^2)*Log[Log[x + Log[x]]])/4)*(-x - x^2 + E^(4 + 4*x)*(1 + x) + (4*x + 4*Log[x])*Log[x + Log[x]] + (-2*x^2 + E^(4 + 4*x)*(x + 4*x^2) + (-2*x + E^(4 + 4*x)*(1 + 4*x))*Log[x])*Log[x + Log[x]]* Log[Log[x + Log[x]]]))/((4*x + 4*Log[x])*Log[x + Log[x]]),x]
3.21.100.3.1 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]
Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
\[\ln \left (x +\ln \left (x \right )\right )^{\frac {x \left ({\mathrm e}^{4+4 x}-x \right )}{4}} {\mathrm e}^{x}\]
int(((((1+4*x)*exp(1+x)^4-2*x)*ln(x)+(4*x^2+x)*exp(1+x)^4-2*x^2)*ln(x+ln(x ))*ln(ln(x+ln(x)))+(4*x+4*ln(x))*ln(x+ln(x))+(1+x)*exp(1+x)^4-x^2-x)*exp(1 /4*(x*exp(1+x)^4-x^2)*ln(ln(x+ln(x)))+x)/(4*x+4*ln(x))/ln(x+ln(x)),x)
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\frac {1}{4} \left (4 x+\left (e^{4+4 x} x-x^2\right ) \log (\log (x+\log (x)))\right )} \left (-x-x^2+e^{4+4 x} (1+x)+(4 x+4 \log (x)) \log (x+\log (x))+\left (-2 x^2+e^{4+4 x} \left (x+4 x^2\right )+\left (-2 x+e^{4+4 x} (1+4 x)\right ) \log (x)\right ) \log (x+\log (x)) \log (\log (x+\log (x)))\right )}{(4 x+4 \log (x)) \log (x+\log (x))} \, dx=e^{\left (-\frac {1}{4} \, {\left (x^{2} - x e^{\left (4 \, x + 4\right )}\right )} \log \left (\log \left (x + \log \left (x\right )\right )\right ) + x\right )} \]
integrate(((((1+4*x)*exp(1+x)^4-2*x)*log(x)+(4*x^2+x)*exp(1+x)^4-2*x^2)*lo g(x+log(x))*log(log(x+log(x)))+(4*x+4*log(x))*log(x+log(x))+(1+x)*exp(1+x) ^4-x^2-x)*exp(1/4*(x*exp(1+x)^4-x^2)*log(log(x+log(x)))+x)/(4*x+4*log(x))/ log(x+log(x)),x, algorithm=\
Timed out. \[ \int \frac {e^{\frac {1}{4} \left (4 x+\left (e^{4+4 x} x-x^2\right ) \log (\log (x+\log (x)))\right )} \left (-x-x^2+e^{4+4 x} (1+x)+(4 x+4 \log (x)) \log (x+\log (x))+\left (-2 x^2+e^{4+4 x} \left (x+4 x^2\right )+\left (-2 x+e^{4+4 x} (1+4 x)\right ) \log (x)\right ) \log (x+\log (x)) \log (\log (x+\log (x)))\right )}{(4 x+4 \log (x)) \log (x+\log (x))} \, dx=\text {Timed out} \]
integrate(((((1+4*x)*exp(1+x)**4-2*x)*ln(x)+(4*x**2+x)*exp(1+x)**4-2*x**2) *ln(x+ln(x))*ln(ln(x+ln(x)))+(4*x+4*ln(x))*ln(x+ln(x))+(1+x)*exp(1+x)**4-x **2-x)*exp(1/4*(x*exp(1+x)**4-x**2)*ln(ln(x+ln(x)))+x)/(4*x+4*ln(x))/ln(x+ ln(x)),x)
Time = 0.34 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {1}{4} \left (4 x+\left (e^{4+4 x} x-x^2\right ) \log (\log (x+\log (x)))\right )} \left (-x-x^2+e^{4+4 x} (1+x)+(4 x+4 \log (x)) \log (x+\log (x))+\left (-2 x^2+e^{4+4 x} \left (x+4 x^2\right )+\left (-2 x+e^{4+4 x} (1+4 x)\right ) \log (x)\right ) \log (x+\log (x)) \log (\log (x+\log (x)))\right )}{(4 x+4 \log (x)) \log (x+\log (x))} \, dx=e^{\left (-\frac {1}{4} \, x^{2} \log \left (\log \left (x + \log \left (x\right )\right )\right ) + \frac {1}{4} \, x e^{\left (4 \, x + 4\right )} \log \left (\log \left (x + \log \left (x\right )\right )\right ) + x\right )} \]
integrate(((((1+4*x)*exp(1+x)^4-2*x)*log(x)+(4*x^2+x)*exp(1+x)^4-2*x^2)*lo g(x+log(x))*log(log(x+log(x)))+(4*x+4*log(x))*log(x+log(x))+(1+x)*exp(1+x) ^4-x^2-x)*exp(1/4*(x*exp(1+x)^4-x^2)*log(log(x+log(x)))+x)/(4*x+4*log(x))/ log(x+log(x)),x, algorithm=\
\[ \int \frac {e^{\frac {1}{4} \left (4 x+\left (e^{4+4 x} x-x^2\right ) \log (\log (x+\log (x)))\right )} \left (-x-x^2+e^{4+4 x} (1+x)+(4 x+4 \log (x)) \log (x+\log (x))+\left (-2 x^2+e^{4+4 x} \left (x+4 x^2\right )+\left (-2 x+e^{4+4 x} (1+4 x)\right ) \log (x)\right ) \log (x+\log (x)) \log (\log (x+\log (x)))\right )}{(4 x+4 \log (x)) \log (x+\log (x))} \, dx=\int { -\frac {{\left ({\left (2 \, x^{2} - {\left (4 \, x^{2} + x\right )} e^{\left (4 \, x + 4\right )} - {\left ({\left (4 \, x + 1\right )} e^{\left (4 \, x + 4\right )} - 2 \, x\right )} \log \left (x\right )\right )} \log \left (x + \log \left (x\right )\right ) \log \left (\log \left (x + \log \left (x\right )\right )\right ) + x^{2} - {\left (x + 1\right )} e^{\left (4 \, x + 4\right )} - 4 \, {\left (x + \log \left (x\right )\right )} \log \left (x + \log \left (x\right )\right ) + x\right )} e^{\left (-\frac {1}{4} \, {\left (x^{2} - x e^{\left (4 \, x + 4\right )}\right )} \log \left (\log \left (x + \log \left (x\right )\right )\right ) + x\right )}}{4 \, {\left (x + \log \left (x\right )\right )} \log \left (x + \log \left (x\right )\right )} \,d x } \]
integrate(((((1+4*x)*exp(1+x)^4-2*x)*log(x)+(4*x^2+x)*exp(1+x)^4-2*x^2)*lo g(x+log(x))*log(log(x+log(x)))+(4*x+4*log(x))*log(x+log(x))+(1+x)*exp(1+x) ^4-x^2-x)*exp(1/4*(x*exp(1+x)^4-x^2)*log(log(x+log(x)))+x)/(4*x+4*log(x))/ log(x+log(x)),x, algorithm=\
Time = 13.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\frac {1}{4} \left (4 x+\left (e^{4+4 x} x-x^2\right ) \log (\log (x+\log (x)))\right )} \left (-x-x^2+e^{4+4 x} (1+x)+(4 x+4 \log (x)) \log (x+\log (x))+\left (-2 x^2+e^{4+4 x} \left (x+4 x^2\right )+\left (-2 x+e^{4+4 x} (1+4 x)\right ) \log (x)\right ) \log (x+\log (x)) \log (\log (x+\log (x)))\right )}{(4 x+4 \log (x)) \log (x+\log (x))} \, dx={\ln \left (x+\ln \left (x\right )\right )}^{\frac {x\,{\mathrm {e}}^{4\,x+4}}{4}-\frac {x^2}{4}}\,{\mathrm {e}}^x \]