Integrand size = 114, antiderivative size = 24 \begin {dmath*} \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 ) \end {dmath*}
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \begin {dmath*} \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 ) \end {dmath*}
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]
Time = 1.09 (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 |
\(\displaystyle -\frac {1}{2} e^x \left (\log \left (\log \left (e^{-x} x-4\right )\right )-2 \log (\log (\log (16)))\right )\) |
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]
3.12.82.3.1 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]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 3.92 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29
method | result | size |
parallelrisch | \(\ln \left (\ln \left (4 \ln \left (2\right )\right )\right ) {\mathrm e}^{x}-\frac {{\mathrm e}^{x} \ln \left (\ln \left (-\left (4 \,{\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}\right )\right )}{2}\) | \(31\) |
risch | \(-\frac {{\mathrm e}^{x} \ln \left (-\ln \left ({\mathrm e}^{x}\right )+\ln \left (-4 \,{\mathrm e}^{x}+x \right )+\frac {i \pi \,\operatorname {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}\right ) \left (\operatorname {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{-x}\right )\right ) \left (\operatorname {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}\right )-\operatorname {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right )\right )\right )}{2}\right )}{2}+{\mathrm e}^{x} \ln \left (2 \ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right )\) | \(108\) |
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)
Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \begin {dmath*} \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 ) \end {dmath*}
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=\
Timed out. \begin {dmath*} \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=\text {Timed out} \end {dmath*}
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)
Time = 0.35 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \begin {dmath*} \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 ) \end {dmath*}
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=\
Time = 0.35 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \begin {dmath*} \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 ) \end {dmath*}
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=\
Time = 15.77 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \begin {dmath*} \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} \end {dmath*}