Integrand size = 49, antiderivative size = 24 \[ \int e^{16-4 e^3-4 x} \left ((2-2 x-2 x \log (x)) \log \left (x^{1-x}\right )+(1-4 x) \log ^2\left (x^{1-x}\right )\right ) \, dx=e^{16-4 e^3-4 x} x \log ^2\left (x^{1-x}\right ) \] Output:
x*ln(x/exp(x*ln(x)))^2/exp(exp(3)+x-4)^4
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int e^{16-4 e^3-4 x} \left ((2-2 x-2 x \log (x)) \log \left (x^{1-x}\right )+(1-4 x) \log ^2\left (x^{1-x}\right )\right ) \, dx=e^{16-4 e^3-4 x} x \log ^2\left (x^{1-x}\right ) \] Input:
Integrate[E^(16 - 4*E^3 - 4*x)*((2 - 2*x - 2*x*Log[x])*Log[x^(1 - x)] + (1 - 4*x)*Log[x^(1 - x)]^2),x]
Output:
E^(16 - 4*E^3 - 4*x)*x*Log[x^(1 - x)]^2
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.
\(\displaystyle \int e^{-4 x-4 e^3+16} \left ((1-4 x) \log ^2\left (x^{1-x}\right )+(-2 x-2 x \log (x)+2) \log \left (x^{1-x}\right )\right ) \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-e^{-4 x-4 e^3+16} (4 x-1) \log ^2\left (x^{1-x}\right )-2 e^{-4 x-4 e^3+16} (x+x \log (x)-1) \log \left (x^{1-x}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int e^{-4 x-4 e^3+16} \log ^2\left (x^{1-x}\right )dx-4 \int e^{-4 x-4 e^3+16} x \log ^2\left (x^{1-x}\right )dx+\frac {1}{8} \int e^{4 \left (4-e^3\right )-4 x} \log ^2(x)dx+\frac {1}{2} \int e^{4 \left (4-e^3\right )-4 x} x \log ^2(x)dx-e^{16-4 e^3} x \, _3F_3(1,1,1;2,2,2;-4 x)+\frac {1}{4} e^{16-4 e^3} \log (x) (\operatorname {ExpIntegralE}(1,4 x)+\operatorname {ExpIntegralEi}(-4 x))-\frac {1}{8} e^{16-4 e^3} \operatorname {ExpIntegralEi}(-4 x) \log \left (x^{1-x}\right )+\frac {3}{16} e^{16-4 e^3} \operatorname {ExpIntegralEi}(-4 x)-\frac {1}{8} e^{16-4 e^3} x \operatorname {ExpIntegralEi}(-4 x) \log (x)-\frac {1}{8} e^{16-4 e^3} \operatorname {ExpIntegralEi}(-4 x) \log (x)-\frac {1}{4} e^{4 \left (4-e^3\right )-4 x} \log \left (x^{1-x}\right )+\frac {1}{2} e^{4 \left (4-e^3\right )-4 x} x \log \left (x^{1-x}\right )+\frac {1}{8} e^{4 \left (4-e^3\right )-4 x} \log (x) \log \left (x^{1-x}\right )+\frac {1}{2} e^{4 \left (4-e^3\right )-4 x} x \log (x) \log \left (x^{1-x}\right )+\frac {3}{32} e^{4 \left (4-e^3\right )-4 x}-\frac {1}{8} e^{4 \left (4-e^3\right )-4 x} x+\frac {1}{8} e^{16-4 e^3} \log ^2(4 x)+\frac {3}{32} e^{4 \left (4-e^3\right )-4 x} \log (x)-\frac {1}{8} e^{4 \left (4-e^3\right )-4 x} x \log (x)-\frac {1}{32} e^{4 \left (4-e^3\right )-4 x} (4 x+1) \log (x)+\frac {1}{4} e^{16-4 e^3} \gamma \log (x)\) |
Input:
Int[E^(16 - 4*E^3 - 4*x)*((2 - 2*x - 2*x*Log[x])*Log[x^(1 - x)] + (1 - 4*x )*Log[x^(1 - x)]^2),x]
Output:
$Aborted
Time = 0.45 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(x \ln \left (x \,{\mathrm e}^{-x \ln \left (x \right )}\right )^{2} {\mathrm e}^{-4 \,{\mathrm e}^{3}-4 x +16}\) | \(23\) |
risch | \(x \,{\mathrm e}^{-4 \,{\mathrm e}^{3}-4 x +16} \ln \left (x^{x}\right )^{2}-x \left (i \pi \,\operatorname {csgn}\left (i x^{-x}\right ) \operatorname {csgn}\left (i x \,x^{-x}\right )^{2}-i \pi \,\operatorname {csgn}\left (i x^{-x}\right ) \operatorname {csgn}\left (i x \,x^{-x}\right ) \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x \,x^{-x}\right )^{3}+i \pi \operatorname {csgn}\left (i x \,x^{-x}\right )^{2} \operatorname {csgn}\left (i x \right )+2 \ln \left (x \right )\right ) {\mathrm e}^{-4 \,{\mathrm e}^{3}-4 x +16} \ln \left (x^{x}\right )+\frac {x \left (4 i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i x^{-x}\right ) \operatorname {csgn}\left (i x \,x^{-x}\right )^{2}+4 i \ln \left (x \right ) \pi \operatorname {csgn}\left (i x \,x^{-x}\right )^{2} \operatorname {csgn}\left (i x \right )-\pi ^{2} \operatorname {csgn}\left (i x \,x^{-x}\right )^{4} \operatorname {csgn}\left (i x \right )^{2}-\pi ^{2} \operatorname {csgn}\left (i x^{-x}\right )^{2} \operatorname {csgn}\left (i x \,x^{-x}\right )^{4}+2 \pi ^{2} \operatorname {csgn}\left (i x^{-x}\right ) \operatorname {csgn}\left (i x \,x^{-x}\right )^{5}+2 \pi ^{2} \operatorname {csgn}\left (i x \,x^{-x}\right )^{5} \operatorname {csgn}\left (i x \right )+4 \ln \left (x \right )^{2}-4 i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i x^{-x}\right ) \operatorname {csgn}\left (i x \,x^{-x}\right ) \operatorname {csgn}\left (i x \right )-\pi ^{2} \operatorname {csgn}\left (i x \,x^{-x}\right )^{6}-4 i \ln \left (x \right ) \pi \operatorname {csgn}\left (i x \,x^{-x}\right )^{3}+2 \pi ^{2} \operatorname {csgn}\left (i x^{-x}\right )^{2} \operatorname {csgn}\left (i x \,x^{-x}\right )^{3} \operatorname {csgn}\left (i x \right )-4 \pi ^{2} \operatorname {csgn}\left (i x^{-x}\right ) \operatorname {csgn}\left (i x \,x^{-x}\right )^{4} \operatorname {csgn}\left (i x \right )-\pi ^{2} \operatorname {csgn}\left (i x^{-x}\right )^{2} \operatorname {csgn}\left (i x \,x^{-x}\right )^{2} \operatorname {csgn}\left (i x \right )^{2}+2 \pi ^{2} \operatorname {csgn}\left (i x^{-x}\right ) \operatorname {csgn}\left (i x \,x^{-x}\right )^{3} \operatorname {csgn}\left (i x \right )^{2}\right ) {\mathrm e}^{-4 \,{\mathrm e}^{3}-4 x +16}}{4}\) | \(499\) |
Input:
int(((-4*x+1)*ln(x/exp(x*ln(x)))^2+(-2*x*ln(x)-2*x+2)*ln(x/exp(x*ln(x))))/ exp(exp(3)+x-4)^4,x,method=_RETURNVERBOSE)
Output:
x*ln(x/exp(x*ln(x)))^2/exp(exp(3)+x-4)^4
Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int e^{16-4 e^3-4 x} \left ((2-2 x-2 x \log (x)) \log \left (x^{1-x}\right )+(1-4 x) \log ^2\left (x^{1-x}\right )\right ) \, dx={\left (x^{3} - 2 \, x^{2} + x\right )} e^{\left (-4 \, x - 4 \, e^{3} + 16\right )} \log \left (x\right )^{2} \] Input:
integrate(((-4*x+1)*log(x/exp(x*log(x)))^2+(-2*x*log(x)-2*x+2)*log(x/exp(x *log(x))))/exp(exp(3)+x-4)^4,x, algorithm="fricas")
Output:
(x^3 - 2*x^2 + x)*e^(-4*x - 4*e^3 + 16)*log(x)^2
Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int e^{16-4 e^3-4 x} \left ((2-2 x-2 x \log (x)) \log \left (x^{1-x}\right )+(1-4 x) \log ^2\left (x^{1-x}\right )\right ) \, dx=\left (x^{3} \log {\left (x \right )}^{2} - 2 x^{2} \log {\left (x \right )}^{2} + x \log {\left (x \right )}^{2}\right ) e^{- 4 x - 4 e^{3} + 16} \] Input:
integrate(((-4*x+1)*ln(x/exp(x*ln(x)))**2+(-2*x*ln(x)-2*x+2)*ln(x/exp(x*ln (x))))/exp(exp(3)+x-4)**4,x)
Output:
(x**3*log(x)**2 - 2*x**2*log(x)**2 + x*log(x)**2)*exp(-4*x - 4*exp(3) + 16 )
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).
Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.12 \[ \int e^{16-4 e^3-4 x} \left ((2-2 x-2 x \log (x)) \log \left (x^{1-x}\right )+(1-4 x) \log ^2\left (x^{1-x}\right )\right ) \, dx=-{\left (2 \, x e^{\left (-4 \, x + 16\right )} \log \left (x\right ) \log \left (x^{x}\right ) - x e^{\left (-4 \, x + 16\right )} \log \left (x^{x}\right )^{2} - x e^{\left (-4 \, x + 16\right )} \log \left (x\right )^{2}\right )} e^{\left (-4 \, e^{3}\right )} \] Input:
integrate(((-4*x+1)*log(x/exp(x*log(x)))^2+(-2*x*log(x)-2*x+2)*log(x/exp(x *log(x))))/exp(exp(3)+x-4)^4,x, algorithm="maxima")
Output:
-(2*x*e^(-4*x + 16)*log(x)*log(x^x) - x*e^(-4*x + 16)*log(x^x)^2 - x*e^(-4 *x + 16)*log(x)^2)*e^(-4*e^3)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.25 \[ \int e^{16-4 e^3-4 x} \left ((2-2 x-2 x \log (x)) \log \left (x^{1-x}\right )+(1-4 x) \log ^2\left (x^{1-x}\right )\right ) \, dx=x^{3} e^{\left (-4 \, x - 4 \, e^{3} + 16\right )} \log \left (x\right )^{2} - 2 \, x^{2} e^{\left (-4 \, x - 4 \, e^{3} + 16\right )} \log \left (x\right )^{2} + x e^{\left (-4 \, x - 4 \, e^{3} + 16\right )} \log \left (x\right )^{2} \] Input:
integrate(((-4*x+1)*log(x/exp(x*log(x)))^2+(-2*x*log(x)-2*x+2)*log(x/exp(x *log(x))))/exp(exp(3)+x-4)^4,x, algorithm="giac")
Output:
x^3*e^(-4*x - 4*e^3 + 16)*log(x)^2 - 2*x^2*e^(-4*x - 4*e^3 + 16)*log(x)^2 + x*e^(-4*x - 4*e^3 + 16)*log(x)^2
Timed out. \[ \int e^{16-4 e^3-4 x} \left ((2-2 x-2 x \log (x)) \log \left (x^{1-x}\right )+(1-4 x) \log ^2\left (x^{1-x}\right )\right ) \, dx=\int -{\mathrm {e}}^{16-4\,{\mathrm {e}}^3-4\,x}\,\left (\left (4\,x-1\right )\,{\ln \left (x\,{\mathrm {e}}^{-x\,\ln \left (x\right )}\right )}^2+\left (2\,x+2\,x\,\ln \left (x\right )-2\right )\,\ln \left (x\,{\mathrm {e}}^{-x\,\ln \left (x\right )}\right )\right ) \,d x \] Input:
int(-exp(16 - 4*exp(3) - 4*x)*(log(x*exp(-x*log(x)))^2*(4*x - 1) + log(x*e xp(-x*log(x)))*(2*x + 2*x*log(x) - 2)),x)
Output:
int(-exp(16 - 4*exp(3) - 4*x)*(log(x*exp(-x*log(x)))^2*(4*x - 1) + log(x*e xp(-x*log(x)))*(2*x + 2*x*log(x) - 2)), x)
Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int e^{16-4 e^3-4 x} \left ((2-2 x-2 x \log (x)) \log \left (x^{1-x}\right )+(1-4 x) \log ^2\left (x^{1-x}\right )\right ) \, dx=\frac {\mathrm {log}\left (\frac {x}{x^{x}}\right )^{2} e^{16} x}{e^{4 e^{3}+4 x}} \] Input:
int(((-4*x+1)*log(x/exp(x*log(x)))^2+(-2*x*log(x)-2*x+2)*log(x/exp(x*log(x ))))/exp(exp(3)+x-4)^4,x)
Output:
(log(x/x**x)**2*e**16*x)/e**(4*e**3 + 4*x)