Integrand size = 104, antiderivative size = 20 \[ \int \frac {e^{\log ^2\left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )} \left (8+8 e^x+8 e^{2 x}-8 x\right ) \log \left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )}{4 e^{2 x}+4 x+e^x (19+4 x)} \, dx=e^{\log ^2\left (\frac {19}{4}+e^x+x+e^{-x} x\right )} \] Output:
exp(ln(19/4+exp(x)+x+x/exp(x))^2)
Time = 0.96 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\log ^2\left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )} \left (8+8 e^x+8 e^{2 x}-8 x\right ) \log \left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )}{4 e^{2 x}+4 x+e^x (19+4 x)} \, dx=e^{\log ^2\left (\frac {19}{4}+e^x+x+e^{-x} x\right )} \] Input:
Integrate[(E^Log[(4*E^(2*x) + 4*x + E^x*(19 + 4*x))/(4*E^x)]^2*(8 + 8*E^x + 8*E^(2*x) - 8*x)*Log[(4*E^(2*x) + 4*x + E^x*(19 + 4*x))/(4*E^x)])/(4*E^( 2*x) + 4*x + E^x*(19 + 4*x)),x]
Output:
E^Log[19/4 + E^x + x + x/E^x]^2
Time = 12.62 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {7292, 27, 7239, 7257}
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.
\(\displaystyle \int \frac {\left (-8 x+8 e^x+8 e^{2 x}+8\right ) \log \left (\frac {1}{4} e^{-x} \left (4 x+4 e^{2 x}+e^x (4 x+19)\right )\right ) \exp \left (\log ^2\left (\frac {1}{4} e^{-x} \left (4 x+4 e^{2 x}+e^x (4 x+19)\right )\right )\right )}{4 x+4 e^{2 x}+e^x (4 x+19)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {8 \left (-x+e^x+e^{2 x}+1\right ) \log \left (\frac {1}{4} e^{-x} \left (4 x+4 e^{2 x}+e^x (4 x+19)\right )\right ) \exp \left (\log ^2\left (\frac {1}{4} e^{-x} \left (4 x+4 e^{2 x}+e^x (4 x+19)\right )\right )\right )}{4 x+4 e^{2 x}+e^x (4 x+19)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 8 \int \frac {\exp \left (\log ^2\left (\frac {1}{4} e^{-x} \left (4 x+4 e^{2 x}+e^x (4 x+19)\right )\right )\right ) \left (-x+e^x+e^{2 x}+1\right ) \log \left (\frac {1}{4} e^{-x} \left (4 x+4 e^{2 x}+e^x (4 x+19)\right )\right )}{4 x+4 e^{2 x}+e^x (4 x+19)}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle 8 \int \frac {e^{\log ^2\left (e^{-x} x+x+e^x+\frac {19}{4}\right )} \left (-x+e^x+e^{2 x}+1\right ) \log \left (e^{-x} x+x+e^x+\frac {19}{4}\right )}{4 x+4 e^{2 x}+e^x (4 x+19)}dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle e^{\log ^2\left (e^{-x} x+x+e^x+\frac {19}{4}\right )}\) |
Input:
Int[(E^Log[(4*E^(2*x) + 4*x + E^x*(19 + 4*x))/(4*E^x)]^2*(8 + 8*E^x + 8*E^ (2*x) - 8*x)*Log[(4*E^(2*x) + 4*x + E^x*(19 + 4*x))/(4*E^x)])/(4*E^(2*x) + 4*x + E^x*(19 + 4*x)),x]
Output:
E^Log[19/4 + E^x + x + x/E^x]^2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 1.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45
method | result | size |
parallelrisch | \({\mathrm e}^{{\ln \left (\frac {\left (4 \,{\mathrm e}^{2 x}+\left (4 x +19\right ) {\mathrm e}^{x}+4 x \right ) {\mathrm e}^{-x}}{4}\right )}^{2}}\) | \(29\) |
risch | \({\mathrm e}^{\frac {{\left (i \pi \,\operatorname {csgn}\left (i \left (\left ({\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left (\left ({\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right )^{2}-i \pi \,\operatorname {csgn}\left (i \left (\left ({\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left (\left ({\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )-i \pi \operatorname {csgn}\left (i {\mathrm e}^{-x} \left (\left ({\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right )^{3}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{-x} \left (\left ({\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )-2 \ln \left ({\mathrm e}^{x}\right )+2 \ln \left (\left ({\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}+\frac {19 \,{\mathrm e}^{x}}{4}\right )\right )}^{2}}{4}}\) | \(198\) |
Input:
int((8*exp(x)^2+8*exp(x)-8*x+8)*ln(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+4*x)/ex p(x))*exp(ln(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+4*x)/exp(x))^2)/(4*exp(x)^2+( 4*x+19)*exp(x)+4*x),x,method=_RETURNVERBOSE)
Output:
exp(ln(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+4*x)/exp(x))^2)
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\log ^2\left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )} \left (8+8 e^x+8 e^{2 x}-8 x\right ) \log \left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )}{4 e^{2 x}+4 x+e^x (19+4 x)} \, dx=e^{\left (\log \left (\frac {1}{4} \, {\left ({\left (4 \, x + 19\right )} e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )}\right )^{2}\right )} \] Input:
integrate((8*exp(x)^2+8*exp(x)-8*x+8)*log(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+ 4*x)/exp(x))*exp(log(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+4*x)/exp(x))^2)/(4*ex p(x)^2+(4*x+19)*exp(x)+4*x),x, algorithm="fricas")
Output:
e^(log(1/4*((4*x + 19)*e^x + 4*x + 4*e^(2*x))*e^(-x))^2)
Timed out. \[ \int \frac {e^{\log ^2\left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )} \left (8+8 e^x+8 e^{2 x}-8 x\right ) \log \left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )}{4 e^{2 x}+4 x+e^x (19+4 x)} \, dx=\text {Timed out} \] Input:
integrate((8*exp(x)**2+8*exp(x)-8*x+8)*ln(1/4*(4*exp(x)**2+(4*x+19)*exp(x) +4*x)/exp(x))*exp(ln(1/4*(4*exp(x)**2+(4*x+19)*exp(x)+4*x)/exp(x))**2)/(4* exp(x)**2+(4*x+19)*exp(x)+4*x),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (15) = 30\).
Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 4.10 \[ \int \frac {e^{\log ^2\left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )} \left (8+8 e^x+8 e^{2 x}-8 x\right ) \log \left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )}{4 e^{2 x}+4 x+e^x (19+4 x)} \, dx=e^{\left (x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2} - 2 \, x \log \left ({\left (4 \, x + 19\right )} e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )}\right ) - 4 \, \log \left (2\right ) \log \left ({\left (4 \, x + 19\right )} e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )}\right ) + \log \left ({\left (4 \, x + 19\right )} e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )}\right )^{2}\right )} \] Input:
integrate((8*exp(x)^2+8*exp(x)-8*x+8)*log(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+ 4*x)/exp(x))*exp(log(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+4*x)/exp(x))^2)/(4*ex p(x)^2+(4*x+19)*exp(x)+4*x),x, algorithm="maxima")
Output:
e^(x^2 + 4*x*log(2) + 4*log(2)^2 - 2*x*log((4*x + 19)*e^x + 4*x + 4*e^(2*x )) - 4*log(2)*log((4*x + 19)*e^x + 4*x + 4*e^(2*x)) + log((4*x + 19)*e^x + 4*x + 4*e^(2*x))^2)
Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (15) = 30\).
Time = 0.13 (sec) , antiderivative size = 85, normalized size of antiderivative = 4.25 \[ \int \frac {e^{\log ^2\left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )} \left (8+8 e^x+8 e^{2 x}-8 x\right ) \log \left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )}{4 e^{2 x}+4 x+e^x (19+4 x)} \, dx=e^{\left (x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2} - 2 \, x \log \left (4 \, x e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )} + 19 \, e^{x}\right ) - 4 \, \log \left (2\right ) \log \left (4 \, x e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )} + 19 \, e^{x}\right ) + \log \left (4 \, x e^{x} + 4 \, x + 4 \, e^{\left (2 \, x\right )} + 19 \, e^{x}\right )^{2}\right )} \] Input:
integrate((8*exp(x)^2+8*exp(x)-8*x+8)*log(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+ 4*x)/exp(x))*exp(log(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+4*x)/exp(x))^2)/(4*ex p(x)^2+(4*x+19)*exp(x)+4*x),x, algorithm="giac")
Output:
e^(x^2 + 4*x*log(2) + 4*log(2)^2 - 2*x*log(4*x*e^x + 4*x + 4*e^(2*x) + 19* e^x) - 4*log(2)*log(4*x*e^x + 4*x + 4*e^(2*x) + 19*e^x) + log(4*x*e^x + 4* x + 4*e^(2*x) + 19*e^x)^2)
Time = 2.35 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {e^{\log ^2\left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )} \left (8+8 e^x+8 e^{2 x}-8 x\right ) \log \left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )}{4 e^{2 x}+4 x+e^x (19+4 x)} \, dx={\mathrm {e}}^{{\ln \left (x+{\mathrm {e}}^x+x\,{\mathrm {e}}^{-x}+\frac {19}{4}\right )}^2} \] Input:
int((exp(log(exp(-x)*(x + exp(2*x) + (exp(x)*(4*x + 19))/4))^2)*log(exp(-x )*(x + exp(2*x) + (exp(x)*(4*x + 19))/4))*(8*exp(2*x) - 8*x + 8*exp(x) + 8 ))/(4*x + 4*exp(2*x) + exp(x)*(4*x + 19)),x)
Output:
exp(log(x + exp(x) + x*exp(-x) + 19/4)^2)
Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\log ^2\left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )} \left (8+8 e^x+8 e^{2 x}-8 x\right ) \log \left (\frac {1}{4} e^{-x} \left (4 e^{2 x}+4 x+e^x (19+4 x)\right )\right )}{4 e^{2 x}+4 x+e^x (19+4 x)} \, dx=e^{\mathrm {log}\left (\frac {4 e^{2 x}+4 e^{x} x +19 e^{x}+4 x}{4 e^{x}}\right )^{2}} \] Input:
int((8*exp(x)^2+8*exp(x)-8*x+8)*log(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+4*x)/e xp(x))*exp(log(1/4*(4*exp(x)^2+(4*x+19)*exp(x)+4*x)/exp(x))^2)/(4*exp(x)^2 +(4*x+19)*exp(x)+4*x),x)
Output:
e**(log((4*e**(2*x) + 4*e**x*x + 19*e**x + 4*x)/(4*e**x))**2)