Integrand size = 114, antiderivative size = 24 \[ \int \frac {e^x (1-x)+\left (-4 e^{2 x}+e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log \left (\log \left (e^{-x} \left (-4 e^x+x\right )\right )\right )+\left (8 e^{2 x}-2 e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log (\log (\log (16)))}{\left (8 e^x-2 x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right )} \, dx=e^x \left (-\frac {1}{2} \log \left (\log \left (-4+e^{-x} x\right )\right )+\log (\log (\log (16)))\right ) \] Output:
(ln(ln(4*ln(2)))-1/2*ln(ln(x/exp(x)-4)))*exp(x)
Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^x (1-x)+\left (-4 e^{2 x}+e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log \left (\log \left (e^{-x} \left (-4 e^x+x\right )\right )\right )+\left (8 e^{2 x}-2 e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log (\log (\log (16)))}{\left (8 e^x-2 x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right )} \, dx=-\frac {1}{2} e^x \left (\log \left (\log \left (-4+e^{-x} x\right )\right )-2 \log (\log (\log (16)))\right ) \] Input:
Integrate[(E^x*(1 - x) + (-4*E^(2*x) + E^x*x)*Log[(-4*E^x + x)/E^x]*Log[Lo g[(-4*E^x + x)/E^x]] + (8*E^(2*x) - 2*E^x*x)*Log[(-4*E^x + x)/E^x]*Log[Log [Log[16]]])/((8*E^x - 2*x)*Log[(-4*E^x + x)/E^x]),x]
Output:
-1/2*(E^x*(Log[Log[-4 + x/E^x]] - 2*Log[Log[Log[16]]]))
Time = 1.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {7292, 27, 7239, 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.
\(\displaystyle \int \frac {e^x (1-x)+\left (e^x x-4 e^{2 x}\right ) \log \left (e^{-x} \left (x-4 e^x\right )\right ) \log \left (\log \left (e^{-x} \left (x-4 e^x\right )\right )\right )+\left (8 e^{2 x}-2 e^x x\right ) \log (\log (\log (16))) \log \left (e^{-x} \left (x-4 e^x\right )\right )}{\left (8 e^x-2 x\right ) \log \left (e^{-x} \left (x-4 e^x\right )\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^x (1-x)+\left (e^x x-4 e^{2 x}\right ) \log \left (e^{-x} \left (x-4 e^x\right )\right ) \log \left (\log \left (e^{-x} \left (x-4 e^x\right )\right )\right )+\left (8 e^{2 x}-2 e^x x\right ) \log (\log (\log (16))) \log \left (e^{-x} \left (x-4 e^x\right )\right )}{2 \left (4 e^x-x\right ) \log \left (e^{-x} \left (x-4 e^x\right )\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {e^x (1-x)-\left (4 e^{2 x}-e^x x\right ) \log \left (-e^{-x} \left (4 e^x-x\right )\right ) \log \left (\log \left (-e^{-x} \left (4 e^x-x\right )\right )\right )+2 \left (4 e^{2 x}-e^x x\right ) \log \left (-e^{-x} \left (4 e^x-x\right )\right ) \log (\log (\log (16)))}{\left (4 e^x-x\right ) \log \left (-e^{-x} \left (4 e^x-x\right )\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \frac {1}{2} \int \frac {e^x \left (-x-\left (4 e^x-x\right ) \log \left (e^{-x} x-4\right ) \left (\log \left (\log \left (e^{-x} x-4\right )\right )-2 \log (\log (\log (16)))\right )+1\right )}{\left (4 e^x-x\right ) \log \left (e^{-x} x-4\right )}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle -\frac {1}{2} e^x \left (\log \left (\log \left (e^{-x} x-4\right )\right )-2 \log (\log (\log (16)))\right )\) |
Input:
Int[(E^x*(1 - x) + (-4*E^(2*x) + E^x*x)*Log[(-4*E^x + x)/E^x]*Log[Log[(-4* E^x + x)/E^x]] + (8*E^(2*x) - 2*E^x*x)*Log[(-4*E^x + x)/E^x]*Log[Log[Log[1 6]]])/((8*E^x - 2*x)*Log[(-4*E^x + x)/E^x]),x]
Output:
-1/2*(E^x*(Log[Log[-4 + x/E^x]] - 2*Log[Log[Log[16]]]))
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]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.97 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29
method | result | size |
parallelrisch | \(\ln \left (\ln \left (4 \ln \left (2\right )\right )\right ) {\mathrm e}^{x}-\frac {\ln \left (\ln \left (-\left (4 \,{\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}\right )\right ) {\mathrm e}^{x}}{2}\) | \(31\) |
risch | \(-\frac {{\mathrm e}^{x} \ln \left (-\ln \left ({\mathrm e}^{x}\right )+\ln \left (-4 \,{\mathrm e}^{x}+x \right )+\frac {i \pi \,\operatorname {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}\right ) \left (\operatorname {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{-x}\right )\right ) \left (\operatorname {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}\right )-\operatorname {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right )\right )\right )}{2}\right )}{2}+{\mathrm e}^{x} \ln \left (2 \ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right )\) | \(108\) |
Input:
int(((-4*exp(x)^2+exp(x)*x)*ln((-4*exp(x)+x)/exp(x))*ln(ln((-4*exp(x)+x)/e xp(x)))+(8*exp(x)^2-2*exp(x)*x)*ln((-4*exp(x)+x)/exp(x))*ln(ln(4*ln(2)))+( 1-x)*exp(x))/(8*exp(x)-2*x)/ln((-4*exp(x)+x)/exp(x)),x,method=_RETURNVERBO SE)
Output:
ln(ln(4*ln(2)))*exp(x)-1/2*ln(ln(-(4*exp(x)-x)/exp(x)))*exp(x)
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^x (1-x)+\left (-4 e^{2 x}+e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log \left (\log \left (e^{-x} \left (-4 e^x+x\right )\right )\right )+\left (8 e^{2 x}-2 e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log (\log (\log (16)))}{\left (8 e^x-2 x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right )} \, dx=-\frac {1}{2} \, e^{x} \log \left (\log \left ({\left (x - 4 \, e^{x}\right )} e^{\left (-x\right )}\right )\right ) + e^{x} \log \left (\log \left (4 \, \log \left (2\right )\right )\right ) \] Input:
integrate(((-4*exp(x)^2+exp(x)*x)*log((-4*exp(x)+x)/exp(x))*log(log((-4*ex p(x)+x)/exp(x)))+(8*exp(x)^2-2*exp(x)*x)*log((-4*exp(x)+x)/exp(x))*log(log (4*log(2)))+(1-x)*exp(x))/(8*exp(x)-2*x)/log((-4*exp(x)+x)/exp(x)),x, algo rithm="fricas")
Output:
-1/2*e^x*log(log((x - 4*e^x)*e^(-x))) + e^x*log(log(4*log(2)))
Time = 0.39 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^x (1-x)+\left (-4 e^{2 x}+e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log \left (\log \left (e^{-x} \left (-4 e^x+x\right )\right )\right )+\left (8 e^{2 x}-2 e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log (\log (\log (16)))}{\left (8 e^x-2 x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right )} \, dx=- \frac {e^{x} \log {\left (\log {\left (\left (x - 4 e^{x}\right ) e^{- x} \right )} \right )}}{2} + e^{x} \log {\left (\log {\left (\log {\left (2 \right )} \right )} + 2 \log {\left (2 \right )} \right )} \] Input:
integrate(((-4*exp(x)**2+exp(x)*x)*ln((-4*exp(x)+x)/exp(x))*ln(ln((-4*exp( x)+x)/exp(x)))+(8*exp(x)**2-2*exp(x)*x)*ln((-4*exp(x)+x)/exp(x))*ln(ln(4*l n(2)))+(1-x)*exp(x))/(8*exp(x)-2*x)/ln((-4*exp(x)+x)/exp(x)),x)
Output:
-exp(x)*log(log((x - 4*exp(x))*exp(-x)))/2 + exp(x)*log(log(log(2)) + 2*lo g(2))
Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^x (1-x)+\left (-4 e^{2 x}+e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log \left (\log \left (e^{-x} \left (-4 e^x+x\right )\right )\right )+\left (8 e^{2 x}-2 e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log (\log (\log (16)))}{\left (8 e^x-2 x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right )} \, dx=-\frac {1}{2} \, e^{x} \log \left (-x + \log \left (x - 4 \, e^{x}\right )\right ) + e^{x} \log \left (2 \, \log \left (2\right ) + \log \left (\log \left (2\right )\right )\right ) \] Input:
integrate(((-4*exp(x)^2+exp(x)*x)*log((-4*exp(x)+x)/exp(x))*log(log((-4*ex p(x)+x)/exp(x)))+(8*exp(x)^2-2*exp(x)*x)*log((-4*exp(x)+x)/exp(x))*log(log (4*log(2)))+(1-x)*exp(x))/(8*exp(x)-2*x)/log((-4*exp(x)+x)/exp(x)),x, algo rithm="maxima")
Output:
-1/2*e^x*log(-x + log(x - 4*e^x)) + e^x*log(2*log(2) + log(log(2)))
Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^x (1-x)+\left (-4 e^{2 x}+e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log \left (\log \left (e^{-x} \left (-4 e^x+x\right )\right )\right )+\left (8 e^{2 x}-2 e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log (\log (\log (16)))}{\left (8 e^x-2 x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right )} \, dx=e^{x} \log \left (2 \, \log \left (2\right ) + \log \left (\log \left (2\right )\right )\right ) - \frac {1}{2} \, e^{x} \log \left (\log \left ({\left (x - 4 \, e^{x}\right )} e^{\left (-x\right )}\right )\right ) \] Input:
integrate(((-4*exp(x)^2+exp(x)*x)*log((-4*exp(x)+x)/exp(x))*log(log((-4*ex p(x)+x)/exp(x)))+(8*exp(x)^2-2*exp(x)*x)*log((-4*exp(x)+x)/exp(x))*log(log (4*log(2)))+(1-x)*exp(x))/(8*exp(x)-2*x)/log((-4*exp(x)+x)/exp(x)),x, algo rithm="giac")
Output:
e^x*log(2*log(2) + log(log(2))) - 1/2*e^x*log(log((x - 4*e^x)*e^(-x)))
Time = 7.63 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (1-x)+\left (-4 e^{2 x}+e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log \left (\log \left (e^{-x} \left (-4 e^x+x\right )\right )\right )+\left (8 e^{2 x}-2 e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log (\log (\log (16)))}{\left (8 e^x-2 x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right )} \, dx=-\frac {{\mathrm {e}}^x\,\left (\ln \left (\ln \left (x\,{\mathrm {e}}^{-x}-4\right )\right )-2\,\ln \left (2\,\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right )\right )}{2} \] Input:
int((exp(x)*(x - 1) + log(exp(-x)*(x - 4*exp(x)))*log(log(exp(-x)*(x - 4*e xp(x))))*(4*exp(2*x) - x*exp(x)) - log(exp(-x)*(x - 4*exp(x)))*log(log(4*l og(2)))*(8*exp(2*x) - 2*x*exp(x)))/(log(exp(-x)*(x - 4*exp(x)))*(2*x - 8*e xp(x))),x)
Output:
-(exp(x)*(log(log(x*exp(-x) - 4)) - 2*log(2*log(2) + log(log(2)))))/2
Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^x (1-x)+\left (-4 e^{2 x}+e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log \left (\log \left (e^{-x} \left (-4 e^x+x\right )\right )\right )+\left (8 e^{2 x}-2 e^x x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right ) \log (\log (\log (16)))}{\left (8 e^x-2 x\right ) \log \left (e^{-x} \left (-4 e^x+x\right )\right )} \, dx=\frac {e^{x} \left (-\mathrm {log}\left (\mathrm {log}\left (\frac {-4 e^{x}+x}{e^{x}}\right )\right )+2 \,\mathrm {log}\left (\mathrm {log}\left (4 \,\mathrm {log}\left (2\right )\right )\right )\right )}{2} \] Input:
int(((-4*exp(x)^2+exp(x)*x)*log((-4*exp(x)+x)/exp(x))*log(log((-4*exp(x)+x )/exp(x)))+(8*exp(x)^2-2*exp(x)*x)*log((-4*exp(x)+x)/exp(x))*log(log(4*log (2)))+(1-x)*exp(x))/(8*exp(x)-2*x)/log((-4*exp(x)+x)/exp(x)),x)
Output:
(e**x*( - log(log(( - 4*e**x + x)/e**x)) + 2*log(log(4*log(2)))))/2