3.44.8 \(\int \frac {e^{e^x} (-10+5 x-2 x^2-x^3+e^x (-10 x+9 x^2-4 x^3+x^4)+(-10+5 x+2 x^2-x^3+e^x (-10 x+9 x^2-4 x^3+x^4)) \log (4-4 x+x^2)+(-4 x+e^x (-4 x^2+2 x^3)+e^x (-4 x^2+2 x^3) \log (4-4 x+x^2)) \log (1+\log (4-4 x+x^2))+(-2+x+e^x (-2 x+x^2)+(-2+x+e^x (-2 x+x^2)) \log (4-4 x+x^2)) \log ^2(1+\log (4-4 x+x^2)))}{(5-2 x+x^2+2 x \log (1+\log (4-4 x+x^2))+\log ^2(1+\log (4-4 x+x^2))) (-10+9 x-4 x^2+x^3+(-10+9 x-4 x^2+x^3) \log (4-4 x+x^2)+(-4 x+2 x^2+(-4 x+2 x^2) \log (4-4 x+x^2)) \log (1+\log (4-4 x+x^2))+(-2+x+(-2+x) \log (4-4 x+x^2)) \log ^2(1+\log (4-4 x+x^2)))} \, dx\)
Optimal. Leaf size=29 \[ \frac {e^{e^x} x}{5-2 x+\left (x+\log \left (1+\log \left ((2-x)^2\right )\right )\right )^2} \]
________________________________________________________________________________________
Rubi [F] time = 180.00, antiderivative size = 0, normalized size of antiderivative = 0.00,
number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used =
{} \begin {gather*} \text {\$Aborted} \end {gather*}
Verification is not applicable to the result.
[In]
Int[(E^E^x*(-10 + 5*x - 2*x^2 - x^3 + E^x*(-10*x + 9*x^2 - 4*x^3 + x^4) + (-10 + 5*x + 2*x^2 - x^3 + E^x*(-10*
x + 9*x^2 - 4*x^3 + x^4))*Log[4 - 4*x + x^2] + (-4*x + E^x*(-4*x^2 + 2*x^3) + E^x*(-4*x^2 + 2*x^3)*Log[4 - 4*x
+ x^2])*Log[1 + Log[4 - 4*x + x^2]] + (-2 + x + E^x*(-2*x + x^2) + (-2 + x + E^x*(-2*x + x^2))*Log[4 - 4*x +
x^2])*Log[1 + Log[4 - 4*x + x^2]]^2))/((5 - 2*x + x^2 + 2*x*Log[1 + Log[4 - 4*x + x^2]] + Log[1 + Log[4 - 4*x
+ x^2]]^2)*(-10 + 9*x - 4*x^2 + x^3 + (-10 + 9*x - 4*x^2 + x^3)*Log[4 - 4*x + x^2] + (-4*x + 2*x^2 + (-4*x + 2
*x^2)*Log[4 - 4*x + x^2])*Log[1 + Log[4 - 4*x + x^2]] + (-2 + x + (-2 + x)*Log[4 - 4*x + x^2])*Log[1 + Log[4 -
4*x + x^2]]^2)),x]
[Out]
$Aborted
Rubi steps
Aborted
________________________________________________________________________________________
Mathematica [A] time = 0.31, size = 40, normalized size = 1.38 \begin {gather*} \frac {e^{e^x} x}{5-2 x+x^2+2 x \log \left (1+\log \left ((-2+x)^2\right )\right )+\log ^2\left (1+\log \left ((-2+x)^2\right )\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(E^E^x*(-10 + 5*x - 2*x^2 - x^3 + E^x*(-10*x + 9*x^2 - 4*x^3 + x^4) + (-10 + 5*x + 2*x^2 - x^3 + E^x
*(-10*x + 9*x^2 - 4*x^3 + x^4))*Log[4 - 4*x + x^2] + (-4*x + E^x*(-4*x^2 + 2*x^3) + E^x*(-4*x^2 + 2*x^3)*Log[4
- 4*x + x^2])*Log[1 + Log[4 - 4*x + x^2]] + (-2 + x + E^x*(-2*x + x^2) + (-2 + x + E^x*(-2*x + x^2))*Log[4 -
4*x + x^2])*Log[1 + Log[4 - 4*x + x^2]]^2))/((5 - 2*x + x^2 + 2*x*Log[1 + Log[4 - 4*x + x^2]] + Log[1 + Log[4
- 4*x + x^2]]^2)*(-10 + 9*x - 4*x^2 + x^3 + (-10 + 9*x - 4*x^2 + x^3)*Log[4 - 4*x + x^2] + (-4*x + 2*x^2 + (-4
*x + 2*x^2)*Log[4 - 4*x + x^2])*Log[1 + Log[4 - 4*x + x^2]] + (-2 + x + (-2 + x)*Log[4 - 4*x + x^2])*Log[1 + L
og[4 - 4*x + x^2]]^2)),x]
[Out]
(E^E^x*x)/(5 - 2*x + x^2 + 2*x*Log[1 + Log[(-2 + x)^2]] + Log[1 + Log[(-2 + x)^2]]^2)
________________________________________________________________________________________
fricas [A] time = 0.57, size = 46, normalized size = 1.59 \begin {gather*} x e^{\left (e^{x} - \log \left (x^{2} + 2 \, x \log \left (\log \left (x^{2} - 4 \, x + 4\right ) + 1\right ) + \log \left (\log \left (x^{2} - 4 \, x + 4\right ) + 1\right )^{2} - 2 \, x + 5\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((((x^2-2*x)*exp(x)+x-2)*log(x^2-4*x+4)+(x^2-2*x)*exp(x)+x-2)*log(log(x^2-4*x+4)+1)^2+((2*x^3-4*x^2)
*exp(x)*log(x^2-4*x+4)+(2*x^3-4*x^2)*exp(x)-4*x)*log(log(x^2-4*x+4)+1)+((x^4-4*x^3+9*x^2-10*x)*exp(x)-x^3+2*x^
2+5*x-10)*log(x^2-4*x+4)+(x^4-4*x^3+9*x^2-10*x)*exp(x)-x^3-2*x^2+5*x-10)*exp(-log(log(log(x^2-4*x+4)+1)^2+2*x*
log(log(x^2-4*x+4)+1)+x^2-2*x+5)+exp(x))/(((x-2)*log(x^2-4*x+4)+x-2)*log(log(x^2-4*x+4)+1)^2+((2*x^2-4*x)*log(
x^2-4*x+4)+2*x^2-4*x)*log(log(x^2-4*x+4)+1)+(x^3-4*x^2+9*x-10)*log(x^2-4*x+4)+x^3-4*x^2+9*x-10),x, algorithm="
fricas")
[Out]
x*e^(e^x - log(x^2 + 2*x*log(log(x^2 - 4*x + 4) + 1) + log(log(x^2 - 4*x + 4) + 1)^2 - 2*x + 5))
________________________________________________________________________________________
giac [B] time = 4.92, size = 106, normalized size = 3.66 \begin {gather*} \frac {2 \, x e^{\left (x + e^{x}\right )}}{x^{2} e^{x} + 2 \, x e^{x} \log \left (\log \left (x^{2} - 4 \, x + 4\right ) + 1\right ) + e^{x} \log \left (\log \left (x^{2} - 4 \, x + 4\right ) + 1\right )^{2} - 2 \, x e^{x} + 5 \, e^{x}} + \frac {2 \, x e^{\left (e^{x}\right )}}{x^{2} + 2 \, x \log \left (\log \left (x^{2} - 4 \, x + 4\right ) + 1\right ) + \log \left (\log \left (x^{2} - 4 \, x + 4\right ) + 1\right )^{2} - 2 \, x + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((((x^2-2*x)*exp(x)+x-2)*log(x^2-4*x+4)+(x^2-2*x)*exp(x)+x-2)*log(log(x^2-4*x+4)+1)^2+((2*x^3-4*x^2)
*exp(x)*log(x^2-4*x+4)+(2*x^3-4*x^2)*exp(x)-4*x)*log(log(x^2-4*x+4)+1)+((x^4-4*x^3+9*x^2-10*x)*exp(x)-x^3+2*x^
2+5*x-10)*log(x^2-4*x+4)+(x^4-4*x^3+9*x^2-10*x)*exp(x)-x^3-2*x^2+5*x-10)*exp(-log(log(log(x^2-4*x+4)+1)^2+2*x*
log(log(x^2-4*x+4)+1)+x^2-2*x+5)+exp(x))/(((x-2)*log(x^2-4*x+4)+x-2)*log(log(x^2-4*x+4)+1)^2+((2*x^2-4*x)*log(
x^2-4*x+4)+2*x^2-4*x)*log(log(x^2-4*x+4)+1)+(x^3-4*x^2+9*x-10)*log(x^2-4*x+4)+x^3-4*x^2+9*x-10),x, algorithm="
giac")
[Out]
2*x*e^(x + e^x)/(x^2*e^x + 2*x*e^x*log(log(x^2 - 4*x + 4) + 1) + e^x*log(log(x^2 - 4*x + 4) + 1)^2 - 2*x*e^x +
5*e^x) + 2*x*e^(e^x)/(x^2 + 2*x*log(log(x^2 - 4*x + 4) + 1) + log(log(x^2 - 4*x + 4) + 1)^2 - 2*x + 5)
________________________________________________________________________________________
maple [F] time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (\left (x^{2}-2 x \right ) {\mathrm e}^{x}+x -2\right ) \ln \left (x^{2}-4 x +4\right )+\left (x^{2}-2 x \right ) {\mathrm e}^{x}+x -2\right ) \ln \left (\ln \left (x^{2}-4 x +4\right )+1\right )^{2}+\left (\left (2 x^{3}-4 x^{2}\right ) {\mathrm e}^{x} \ln \left (x^{2}-4 x +4\right )+\left (2 x^{3}-4 x^{2}\right ) {\mathrm e}^{x}-4 x \right ) \ln \left (\ln \left (x^{2}-4 x +4\right )+1\right )+\left (\left (x^{4}-4 x^{3}+9 x^{2}-10 x \right ) {\mathrm e}^{x}-x^{3}+2 x^{2}+5 x -10\right ) \ln \left (x^{2}-4 x +4\right )+\left (x^{4}-4 x^{3}+9 x^{2}-10 x \right ) {\mathrm e}^{x}-x^{3}-2 x^{2}+5 x -10\right ) {\mathrm e}^{-\ln \left (\ln \left (\ln \left (x^{2}-4 x +4\right )+1\right )^{2}+2 x \ln \left (\ln \left (x^{2}-4 x +4\right )+1\right )+x^{2}-2 x +5\right )+{\mathrm e}^{x}}}{\left (\left (x -2\right ) \ln \left (x^{2}-4 x +4\right )+x -2\right ) \ln \left (\ln \left (x^{2}-4 x +4\right )+1\right )^{2}+\left (\left (2 x^{2}-4 x \right ) \ln \left (x^{2}-4 x +4\right )+2 x^{2}-4 x \right ) \ln \left (\ln \left (x^{2}-4 x +4\right )+1\right )+\left (x^{3}-4 x^{2}+9 x -10\right ) \ln \left (x^{2}-4 x +4\right )+x^{3}-4 x^{2}+9 x -10}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((((x^2-2*x)*exp(x)+x-2)*ln(x^2-4*x+4)+(x^2-2*x)*exp(x)+x-2)*ln(ln(x^2-4*x+4)+1)^2+((2*x^3-4*x^2)*exp(x)*l
n(x^2-4*x+4)+(2*x^3-4*x^2)*exp(x)-4*x)*ln(ln(x^2-4*x+4)+1)+((x^4-4*x^3+9*x^2-10*x)*exp(x)-x^3+2*x^2+5*x-10)*ln
(x^2-4*x+4)+(x^4-4*x^3+9*x^2-10*x)*exp(x)-x^3-2*x^2+5*x-10)*exp(-ln(ln(ln(x^2-4*x+4)+1)^2+2*x*ln(ln(x^2-4*x+4)
+1)+x^2-2*x+5)+exp(x))/(((x-2)*ln(x^2-4*x+4)+x-2)*ln(ln(x^2-4*x+4)+1)^2+((2*x^2-4*x)*ln(x^2-4*x+4)+2*x^2-4*x)*
ln(ln(x^2-4*x+4)+1)+(x^3-4*x^2+9*x-10)*ln(x^2-4*x+4)+x^3-4*x^2+9*x-10),x)
[Out]
int(((((x^2-2*x)*exp(x)+x-2)*ln(x^2-4*x+4)+(x^2-2*x)*exp(x)+x-2)*ln(ln(x^2-4*x+4)+1)^2+((2*x^3-4*x^2)*exp(x)*l
n(x^2-4*x+4)+(2*x^3-4*x^2)*exp(x)-4*x)*ln(ln(x^2-4*x+4)+1)+((x^4-4*x^3+9*x^2-10*x)*exp(x)-x^3+2*x^2+5*x-10)*ln
(x^2-4*x+4)+(x^4-4*x^3+9*x^2-10*x)*exp(x)-x^3-2*x^2+5*x-10)*exp(-ln(ln(ln(x^2-4*x+4)+1)^2+2*x*ln(ln(x^2-4*x+4)
+1)+x^2-2*x+5)+exp(x))/(((x-2)*ln(x^2-4*x+4)+x-2)*ln(ln(x^2-4*x+4)+1)^2+((2*x^2-4*x)*ln(x^2-4*x+4)+2*x^2-4*x)*
ln(ln(x^2-4*x+4)+1)+(x^3-4*x^2+9*x-10)*ln(x^2-4*x+4)+x^3-4*x^2+9*x-10),x)
________________________________________________________________________________________
maxima [A] time = 0.53, size = 38, normalized size = 1.31 \begin {gather*} \frac {x e^{\left (e^{x}\right )}}{x^{2} + 2 \, x \log \left (2 \, \log \left (x - 2\right ) + 1\right ) + \log \left (2 \, \log \left (x - 2\right ) + 1\right )^{2} - 2 \, x + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((((x^2-2*x)*exp(x)+x-2)*log(x^2-4*x+4)+(x^2-2*x)*exp(x)+x-2)*log(log(x^2-4*x+4)+1)^2+((2*x^3-4*x^2)
*exp(x)*log(x^2-4*x+4)+(2*x^3-4*x^2)*exp(x)-4*x)*log(log(x^2-4*x+4)+1)+((x^4-4*x^3+9*x^2-10*x)*exp(x)-x^3+2*x^
2+5*x-10)*log(x^2-4*x+4)+(x^4-4*x^3+9*x^2-10*x)*exp(x)-x^3-2*x^2+5*x-10)*exp(-log(log(log(x^2-4*x+4)+1)^2+2*x*
log(log(x^2-4*x+4)+1)+x^2-2*x+5)+exp(x))/(((x-2)*log(x^2-4*x+4)+x-2)*log(log(x^2-4*x+4)+1)^2+((2*x^2-4*x)*log(
x^2-4*x+4)+2*x^2-4*x)*log(log(x^2-4*x+4)+1)+(x^3-4*x^2+9*x-10)*log(x^2-4*x+4)+x^3-4*x^2+9*x-10),x, algorithm="
maxima")
[Out]
x*e^(e^x)/(x^2 + 2*x*log(2*log(x - 2) + 1) + log(2*log(x - 2) + 1)^2 - 2*x + 5)
________________________________________________________________________________________
mupad [B] time = 4.29, size = 44, normalized size = 1.52 \begin {gather*} \frac {x\,{\mathrm {e}}^{{\mathrm {e}}^x}}{{\ln \left (\ln \left (x^2-4\,x+4\right )+1\right )}^2+x^2+x\,\left (2\,\ln \left (\ln \left (x^2-4\,x+4\right )+1\right )-2\right )+5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(-(exp(exp(x) - log(2*x*log(log(x^2 - 4*x + 4) + 1) - 2*x + log(log(x^2 - 4*x + 4) + 1)^2 + x^2 + 5))*(exp(
x)*(10*x - 9*x^2 + 4*x^3 - x^4) - 5*x + log(x^2 - 4*x + 4)*(exp(x)*(10*x - 9*x^2 + 4*x^3 - x^4) - 5*x - 2*x^2
+ x^3 + 10) + log(log(x^2 - 4*x + 4) + 1)*(4*x + exp(x)*(4*x^2 - 2*x^3) + exp(x)*log(x^2 - 4*x + 4)*(4*x^2 - 2
*x^3)) + log(log(x^2 - 4*x + 4) + 1)^2*(exp(x)*(2*x - x^2) - x + log(x^2 - 4*x + 4)*(exp(x)*(2*x - x^2) - x +
2) + 2) + 2*x^2 + x^3 + 10))/(9*x + log(x^2 - 4*x + 4)*(9*x - 4*x^2 + x^3 - 10) - log(log(x^2 - 4*x + 4) + 1)*
(4*x + log(x^2 - 4*x + 4)*(4*x - 2*x^2) - 2*x^2) + log(log(x^2 - 4*x + 4) + 1)^2*(x + log(x^2 - 4*x + 4)*(x -
2) - 2) - 4*x^2 + x^3 - 10),x)
[Out]
(x*exp(exp(x)))/(log(log(x^2 - 4*x + 4) + 1)^2 + x^2 + x*(2*log(log(x^2 - 4*x + 4) + 1) - 2) + 5)
________________________________________________________________________________________
sympy [A] time = 1.21, size = 46, normalized size = 1.59 \begin {gather*} \frac {x e^{e^{x}}}{x^{2} + 2 x \log {\left (\log {\left (x^{2} - 4 x + 4 \right )} + 1 \right )} - 2 x + \log {\left (\log {\left (x^{2} - 4 x + 4 \right )} + 1 \right )}^{2} + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((((x**2-2*x)*exp(x)+x-2)*ln(x**2-4*x+4)+(x**2-2*x)*exp(x)+x-2)*ln(ln(x**2-4*x+4)+1)**2+((2*x**3-4*x
**2)*exp(x)*ln(x**2-4*x+4)+(2*x**3-4*x**2)*exp(x)-4*x)*ln(ln(x**2-4*x+4)+1)+((x**4-4*x**3+9*x**2-10*x)*exp(x)-
x**3+2*x**2+5*x-10)*ln(x**2-4*x+4)+(x**4-4*x**3+9*x**2-10*x)*exp(x)-x**3-2*x**2+5*x-10)*exp(-ln(ln(ln(x**2-4*x
+4)+1)**2+2*x*ln(ln(x**2-4*x+4)+1)+x**2-2*x+5)+exp(x))/(((x-2)*ln(x**2-4*x+4)+x-2)*ln(ln(x**2-4*x+4)+1)**2+((2
*x**2-4*x)*ln(x**2-4*x+4)+2*x**2-4*x)*ln(ln(x**2-4*x+4)+1)+(x**3-4*x**2+9*x-10)*ln(x**2-4*x+4)+x**3-4*x**2+9*x
-10),x)
[Out]
x*exp(exp(x))/(x**2 + 2*x*log(log(x**2 - 4*x + 4) + 1) - 2*x + log(log(x**2 - 4*x + 4) + 1)**2 + 5)
________________________________________________________________________________________