Integrand size = 109, antiderivative size = 26 \[ \int e^{x+e^{90+10 e^2} x+2 e^{45+5 e^2} x \log \left (e^{-x} x\right )+x \log ^2\left (e^{-x} x\right )} \left (1+e^{90+10 e^2}+e^{45+5 e^2} (2-2 x)+\left (2+2 e^{45+5 e^2}-2 x\right ) \log \left (e^{-x} x\right )+\log ^2\left (e^{-x} x\right )\right ) \, dx=e^{x+x \left (e^{5 \left (9+e^2\right )}+\log \left (e^{-x} x\right )\right )^2} \] Output:
exp(x*(ln(x/exp(x))+exp(5*exp(1)^2+45))^2+x)
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int e^{x+e^{90+10 e^2} x+2 e^{45+5 e^2} x \log \left (e^{-x} x\right )+x \log ^2\left (e^{-x} x\right )} \left (1+e^{90+10 e^2}+e^{45+5 e^2} (2-2 x)+\left (2+2 e^{45+5 e^2}-2 x\right ) \log \left (e^{-x} x\right )+\log ^2\left (e^{-x} x\right )\right ) \, dx=e^{x \left (1+e^{10 \left (9+e^2\right )}+2 e^{5 \left (9+e^2\right )} \log \left (e^{-x} x\right )+\log ^2\left (e^{-x} x\right )\right )} \] Input:
Integrate[E^(x + E^(90 + 10*E^2)*x + 2*E^(45 + 5*E^2)*x*Log[x/E^x] + x*Log [x/E^x]^2)*(1 + E^(90 + 10*E^2) + E^(45 + 5*E^2)*(2 - 2*x) + (2 + 2*E^(45 + 5*E^2) - 2*x)*Log[x/E^x] + Log[x/E^x]^2),x]
Output:
E^(x*(1 + E^(10*(9 + E^2)) + 2*E^(5*(9 + E^2))*Log[x/E^x] + Log[x/E^x]^2))
Time = 1.88 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {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.
\(\displaystyle \int \left (e^{45+5 e^2} (2-2 x)+\log ^2\left (e^{-x} x\right )+\left (-2 x+2 e^{45+5 e^2}+2\right ) \log \left (e^{-x} x\right )+e^{90+10 e^2}+1\right ) \exp \left (e^{90+10 e^2} x+x+x \log ^2\left (e^{-x} x\right )+2 e^{45+5 e^2} x \log \left (e^{-x} x\right )\right ) \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle \left (e^{-x} x\right )^{2 e^{45+5 e^2} x} e^{e^{10 \left (9+e^2\right )} x+x+x \log ^2\left (e^{-x} x\right )}\) |
Input:
Int[E^(x + E^(90 + 10*E^2)*x + 2*E^(45 + 5*E^2)*x*Log[x/E^x] + x*Log[x/E^x ]^2)*(1 + E^(90 + 10*E^2) + E^(45 + 5*E^2)*(2 - 2*x) + (2 + 2*E^(45 + 5*E^ 2) - 2*x)*Log[x/E^x] + Log[x/E^x]^2),x]
Output:
E^(x + E^(10*(9 + E^2))*x + x*Log[x/E^x]^2)*(x/E^x)^(2*E^(45 + 5*E^2)*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 = 0.63 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69
method | result | size |
parallelrisch | \({\mathrm e}^{x \left (1+\ln \left (x \,{\mathrm e}^{-x}\right )^{2}+2 \ln \left (x \,{\mathrm e}^{-x}\right ) {\mathrm e}^{5 \,{\mathrm e}^{2}+45}+{\mathrm e}^{10 \,{\mathrm e}^{2}+90}\right )}\) | \(44\) |
risch | \(\left ({\mathrm e}^{x}\right )^{i x \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \right )} x^{-i x \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \right )} x^{i \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \pi x} x^{i x \pi \,\operatorname {csgn}\left (i x \right )} \left ({\mathrm e}^{x}\right )^{-i \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \pi x} \left ({\mathrm e}^{x}\right )^{-i x \pi \,\operatorname {csgn}\left (i x \right )} \left ({\mathrm e}^{x}\right )^{-2 x \ln \left (x \right )} x^{2 \,{\mathrm e}^{5 \,{\mathrm e}^{2}+45} x} \left ({\mathrm e}^{x}\right )^{-2 \,{\mathrm e}^{5 \,{\mathrm e}^{2}+45} x} \left ({\mathrm e}^{x}\right )^{i x \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )} x^{-i x \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )} {\mathrm e}^{\frac {x \left (-\pi ^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{4}+2 \pi ^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3} \operatorname {csgn}\left (i x \right )-\pi ^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \right )^{2}+2 \pi ^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{5}-4 \pi ^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{4} \operatorname {csgn}\left (i x \right )+2 \pi ^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3} \operatorname {csgn}\left (i x \right )^{2}-\pi ^{2} \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{6}+2 \pi ^{2} \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{5} \operatorname {csgn}\left (i x \right )-\pi ^{2} \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{4} \operatorname {csgn}\left (i x \right )^{2}-4 i {\mathrm e}^{5 \,{\mathrm e}^{2}+45} \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}-4 i {\mathrm e}^{5 \,{\mathrm e}^{2}+45} \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \right )+4 i {\mathrm e}^{5 \,{\mathrm e}^{2}+45} \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}+4 i {\mathrm e}^{5 \,{\mathrm e}^{2}+45} \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \right )+4 \ln \left (x \right )^{2}+4 \ln \left ({\mathrm e}^{x}\right )^{2}+4 \,{\mathrm e}^{10 \,{\mathrm e}^{2}+90}+4\right )}{4}}\) | \(557\) |
Input:
int((ln(x/exp(x))^2+(2*exp(5*exp(1)^2+45)-2*x+2)*ln(x/exp(x))+exp(5*exp(1) ^2+45)^2+(2-2*x)*exp(5*exp(1)^2+45)+1)*exp(x*ln(x/exp(x))^2+2*x*exp(5*exp( 1)^2+45)*ln(x/exp(x))+x*exp(5*exp(1)^2+45)^2+x),x,method=_RETURNVERBOSE)
Output:
exp(x*(ln(x/exp(x))^2+2*exp(5*exp(1)^2+45)*ln(x/exp(x))+exp(5*exp(1)^2+45) ^2+1))
Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int e^{x+e^{90+10 e^2} x+2 e^{45+5 e^2} x \log \left (e^{-x} x\right )+x \log ^2\left (e^{-x} x\right )} \left (1+e^{90+10 e^2}+e^{45+5 e^2} (2-2 x)+\left (2+2 e^{45+5 e^2}-2 x\right ) \log \left (e^{-x} x\right )+\log ^2\left (e^{-x} x\right )\right ) \, dx=e^{\left (2 \, x e^{\left (5 \, e^{2} + 45\right )} \log \left (x e^{\left (-x\right )}\right ) + x \log \left (x e^{\left (-x\right )}\right )^{2} + x e^{\left (10 \, e^{2} + 90\right )} + x\right )} \] Input:
integrate((log(x/exp(x))^2+(2*exp(5*exp(1)^2+45)-2*x+2)*log(x/exp(x))+exp( 5*exp(1)^2+45)^2+(2-2*x)*exp(5*exp(1)^2+45)+1)*exp(x*log(x/exp(x))^2+2*x*e xp(5*exp(1)^2+45)*log(x/exp(x))+x*exp(5*exp(1)^2+45)^2+x),x, algorithm="fr icas")
Output:
e^(2*x*e^(5*e^2 + 45)*log(x*e^(-x)) + x*log(x*e^(-x))^2 + x*e^(10*e^2 + 90 ) + x)
Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 63.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int e^{x+e^{90+10 e^2} x+2 e^{45+5 e^2} x \log \left (e^{-x} x\right )+x \log ^2\left (e^{-x} x\right )} \left (1+e^{90+10 e^2}+e^{45+5 e^2} (2-2 x)+\left (2+2 e^{45+5 e^2}-2 x\right ) \log \left (e^{-x} x\right )+\log ^2\left (e^{-x} x\right )\right ) \, dx=e^{x \log {\left (x e^{- x} \right )}^{2} + 2 x e^{5 e^{2} + 45} \log {\left (x e^{- x} \right )} + x + x e^{10 e^{2} + 90}} \] Input:
integrate((ln(x/exp(x))**2+(2*exp(5*exp(1)**2+45)-2*x+2)*ln(x/exp(x))+exp( 5*exp(1)**2+45)**2+(2-2*x)*exp(5*exp(1)**2+45)+1)*exp(x*ln(x/exp(x))**2+2* x*exp(5*exp(1)**2+45)*ln(x/exp(x))+x*exp(5*exp(1)**2+45)**2+x),x)
Output:
exp(x*log(x*exp(-x))**2 + 2*x*exp(5*exp(2) + 45)*log(x*exp(-x)) + x + x*ex p(10*exp(2) + 90))
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (22) = 44\).
Time = 0.17 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int e^{x+e^{90+10 e^2} x+2 e^{45+5 e^2} x \log \left (e^{-x} x\right )+x \log ^2\left (e^{-x} x\right )} \left (1+e^{90+10 e^2}+e^{45+5 e^2} (2-2 x)+\left (2+2 e^{45+5 e^2}-2 x\right ) \log \left (e^{-x} x\right )+\log ^2\left (e^{-x} x\right )\right ) \, dx=e^{\left (x^{3} - 2 \, x^{2} e^{\left (5 \, e^{2} + 45\right )} - 2 \, x^{2} \log \left (x\right ) + 2 \, x e^{\left (5 \, e^{2} + 45\right )} \log \left (x\right ) + x \log \left (x\right )^{2} + x e^{\left (10 \, e^{2} + 90\right )} + x\right )} \] Input:
integrate((log(x/exp(x))^2+(2*exp(5*exp(1)^2+45)-2*x+2)*log(x/exp(x))+exp( 5*exp(1)^2+45)^2+(2-2*x)*exp(5*exp(1)^2+45)+1)*exp(x*log(x/exp(x))^2+2*x*e xp(5*exp(1)^2+45)*log(x/exp(x))+x*exp(5*exp(1)^2+45)^2+x),x, algorithm="ma xima")
Output:
e^(x^3 - 2*x^2*e^(5*e^2 + 45) - 2*x^2*log(x) + 2*x*e^(5*e^2 + 45)*log(x) + x*log(x)^2 + x*e^(10*e^2 + 90) + x)
Time = 0.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int e^{x+e^{90+10 e^2} x+2 e^{45+5 e^2} x \log \left (e^{-x} x\right )+x \log ^2\left (e^{-x} x\right )} \left (1+e^{90+10 e^2}+e^{45+5 e^2} (2-2 x)+\left (2+2 e^{45+5 e^2}-2 x\right ) \log \left (e^{-x} x\right )+\log ^2\left (e^{-x} x\right )\right ) \, dx=e^{\left (2 \, x e^{\left (5 \, e^{2} + 45\right )} \log \left (x e^{\left (-x\right )}\right ) + x \log \left (x e^{\left (-x\right )}\right )^{2} + x e^{\left (10 \, e^{2} + 90\right )} + x\right )} \] Input:
integrate((log(x/exp(x))^2+(2*exp(5*exp(1)^2+45)-2*x+2)*log(x/exp(x))+exp( 5*exp(1)^2+45)^2+(2-2*x)*exp(5*exp(1)^2+45)+1)*exp(x*log(x/exp(x))^2+2*x*e xp(5*exp(1)^2+45)*log(x/exp(x))+x*exp(5*exp(1)^2+45)^2+x),x, algorithm="gi ac")
Output:
e^(2*x*e^(5*e^2 + 45)*log(x*e^(-x)) + x*log(x*e^(-x))^2 + x*e^(10*e^2 + 90 ) + x)
Time = 1.75 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.12 \[ \int e^{x+e^{90+10 e^2} x+2 e^{45+5 e^2} x \log \left (e^{-x} x\right )+x \log ^2\left (e^{-x} x\right )} \left (1+e^{90+10 e^2}+e^{45+5 e^2} (2-2 x)+\left (2+2 e^{45+5 e^2}-2 x\right ) \log \left (e^{-x} x\right )+\log ^2\left (e^{-x} x\right )\right ) \, dx=x^{2\,x\,{\mathrm {e}}^{5\,{\mathrm {e}}^2}\,{\mathrm {e}}^{45}-2\,x^2}\,{\mathrm {e}}^{-2\,x^2\,{\mathrm {e}}^{5\,{\mathrm {e}}^2}\,{\mathrm {e}}^{45}}\,{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{x\,{\ln \left (x\right )}^2}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{10\,{\mathrm {e}}^2}\,{\mathrm {e}}^{90}}\,{\mathrm {e}}^x \] Input:
int(exp(x + x*exp(10*exp(2) + 90) + x*log(x*exp(-x))^2 + 2*x*exp(5*exp(2) + 45)*log(x*exp(-x)))*(exp(10*exp(2) + 90) + log(x*exp(-x))*(2*exp(5*exp(2 ) + 45) - 2*x + 2) - exp(5*exp(2) + 45)*(2*x - 2) + log(x*exp(-x))^2 + 1), x)
Output:
x^(2*x*exp(5*exp(2))*exp(45) - 2*x^2)*exp(-2*x^2*exp(5*exp(2))*exp(45))*ex p(x^3)*exp(x*log(x)^2)*exp(x*exp(10*exp(2))*exp(90))*exp(x)
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.88 \[ \int e^{x+e^{90+10 e^2} x+2 e^{45+5 e^2} x \log \left (e^{-x} x\right )+x \log ^2\left (e^{-x} x\right )} \left (1+e^{90+10 e^2}+e^{45+5 e^2} (2-2 x)+\left (2+2 e^{45+5 e^2}-2 x\right ) \log \left (e^{-x} x\right )+\log ^2\left (e^{-x} x\right )\right ) \, dx=e^{e^{10 e^{2}} e^{90} x +2 e^{5 e^{2}} \mathrm {log}\left (\frac {x}{e^{x}}\right ) e^{45} x +\mathrm {log}\left (\frac {x}{e^{x}}\right )^{2} x +x} \] Input:
int((log(x/exp(x))^2+(2*exp(5*exp(1)^2+45)-2*x+2)*log(x/exp(x))+exp(5*exp( 1)^2+45)^2+(2-2*x)*exp(5*exp(1)^2+45)+1)*exp(x*log(x/exp(x))^2+2*x*exp(5*e xp(1)^2+45)*log(x/exp(x))+x*exp(5*exp(1)^2+45)^2+x),x)
Output:
e**(e**(10*e**2)*e**90*x + 2*e**(5*e**2)*log(x/e**x)*e**45*x + log(x/e**x) **2*x + x)