3.11.17 \(\int \frac {-4 x+e^x x+(-12 x-4 x^2+e^x (3 x+3 x^2)) \log (x)+(-8 x+2 e^x x) \log (x) \log (\frac {3}{(16 x^3-8 e^x x^3+e^{2 x} x^3) \log (x)})}{(-12 x^2+3 e^x x^2) \log (x)+(-24 x+6 e^x x) \log (x) \log (\frac {3}{(16 x^3-8 e^x x^3+e^{2 x} x^3) \log (x)})+(-12+3 e^x) \log (x) \log ^2(\frac {3}{(16 x^3-8 e^x x^3+e^{2 x} x^3) \log (x)})} \, dx\) [1017]

Optimal. Leaf size=30 \[ \frac {x^2}{3 \left (x+\log \left (\frac {3}{\left (4-e^x\right )^2 x^3 \log (x)}\right )\right )} \]

[Out]

1/3*x^2/(x+ln(3/ln(x)/(-exp(x)+4)^2/x^3))

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Rubi [F]
time = 1.68, 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*} \int \frac {-4 x+e^x x+\left (-12 x-4 x^2+e^x \left (3 x+3 x^2\right )\right ) \log (x)+\left (-8 x+2 e^x x\right ) \log (x) \log \left (\frac {3}{\left (16 x^3-8 e^x x^3+e^{2 x} x^3\right ) \log (x)}\right )}{\left (-12 x^2+3 e^x x^2\right ) \log (x)+\left (-24 x+6 e^x x\right ) \log (x) \log \left (\frac {3}{\left (16 x^3-8 e^x x^3+e^{2 x} x^3\right ) \log (x)}\right )+\left (-12+3 e^x\right ) \log (x) \log ^2\left (\frac {3}{\left (16 x^3-8 e^x x^3+e^{2 x} x^3\right ) \log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*x + E^x*x + (-12*x - 4*x^2 + E^x*(3*x + 3*x^2))*Log[x] + (-8*x + 2*E^x*x)*Log[x]*Log[3/((16*x^3 - 8*E^
x*x^3 + E^(2*x)*x^3)*Log[x])])/((-12*x^2 + 3*E^x*x^2)*Log[x] + (-24*x + 6*E^x*x)*Log[x]*Log[3/((16*x^3 - 8*E^x
*x^3 + E^(2*x)*x^3)*Log[x])] + (-12 + 3*E^x)*Log[x]*Log[3/((16*x^3 - 8*E^x*x^3 + E^(2*x)*x^3)*Log[x])]^2),x]

[Out]

Defer[Int][x/(x + Log[3/((-4 + E^x)^2*x^3*Log[x])])^2, x] + Defer[Int][x^2/(x + Log[3/((-4 + E^x)^2*x^3*Log[x]
)])^2, x]/3 + (8*Defer[Int][x^2/((-4 + E^x)*(x + Log[3/((-4 + E^x)^2*x^3*Log[x])])^2), x])/3 + Defer[Int][x/(L
og[x]*(x + Log[3/((-4 + E^x)^2*x^3*Log[x])])^2), x]/3 + (2*Defer[Int][x/(x + Log[3/((-4 + E^x)^2*x^3*Log[x])])
, x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x \left (4-e^x-\log (x) \left (3 e^x (1+x)-4 (3+x)+2 \left (-4+e^x\right ) \log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )\right )}{3 \left (4-e^x\right ) \log (x) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {x \left (4-e^x-\log (x) \left (3 e^x (1+x)-4 (3+x)+2 \left (-4+e^x\right ) \log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )\right )}{\left (4-e^x\right ) \log (x) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx\\ &=\frac {1}{3} \int \left (\frac {8 x^2}{\left (-4+e^x\right ) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2}+\frac {x+3 x \log (x)+3 x^2 \log (x)+2 x \log (x) \log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )}{\log (x) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2}\right ) \, dx\\ &=\frac {1}{3} \int \frac {x+3 x \log (x)+3 x^2 \log (x)+2 x \log (x) \log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )}{\log (x) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx+\frac {8}{3} \int \frac {x^2}{\left (-4+e^x\right ) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx\\ &=\frac {1}{3} \int \left (\frac {x+3 x \log (x)+x^2 \log (x)}{\log (x) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2}+\frac {2 x}{x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )}\right ) \, dx+\frac {8}{3} \int \frac {x^2}{\left (-4+e^x\right ) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {x+3 x \log (x)+x^2 \log (x)}{\log (x) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx+\frac {2}{3} \int \frac {x}{x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )} \, dx+\frac {8}{3} \int \frac {x^2}{\left (-4+e^x\right ) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {x+x (3+x) \log (x)}{\log (x) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx+\frac {2}{3} \int \frac {x}{x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )} \, dx+\frac {8}{3} \int \frac {x^2}{\left (-4+e^x\right ) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx\\ &=\frac {1}{3} \int \left (\frac {3 x}{\left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2}+\frac {x^2}{\left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2}+\frac {x}{\log (x) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2}\right ) \, dx+\frac {2}{3} \int \frac {x}{x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )} \, dx+\frac {8}{3} \int \frac {x^2}{\left (-4+e^x\right ) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {x^2}{\left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx+\frac {1}{3} \int \frac {x}{\log (x) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx+\frac {2}{3} \int \frac {x}{x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )} \, dx+\frac {8}{3} \int \frac {x^2}{\left (-4+e^x\right ) \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx+\int \frac {x}{\left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.11, size = 28, normalized size = 0.93 \begin {gather*} \frac {x^2}{3 \left (x+\log \left (\frac {3}{\left (-4+e^x\right )^2 x^3 \log (x)}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x + E^x*x + (-12*x - 4*x^2 + E^x*(3*x + 3*x^2))*Log[x] + (-8*x + 2*E^x*x)*Log[x]*Log[3/((16*x^3
- 8*E^x*x^3 + E^(2*x)*x^3)*Log[x])])/((-12*x^2 + 3*E^x*x^2)*Log[x] + (-24*x + 6*E^x*x)*Log[x]*Log[3/((16*x^3 -
8*E^x*x^3 + E^(2*x)*x^3)*Log[x])] + (-12 + 3*E^x)*Log[x]*Log[3/((16*x^3 - 8*E^x*x^3 + E^(2*x)*x^3)*Log[x])]^2
),x]

[Out]

x^2/(3*(x + Log[3/((-4 + E^x)^2*x^3*Log[x])]))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.67, size = 465, normalized size = 15.50

method	result	size
risch	\(\frac {2 x^{2}}{3 \left (2 x +i \pi \,\mathrm {csgn}\left (\frac {i}{x^{3}}\right ) \mathrm {csgn}\left (\frac {i}{x^{3} \ln \left (x \right ) \left ({\mathrm e}^{x}-4\right )^{2}}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{3}\right )^{3}-2 \ln \left (\ln \left (x \right )\right )+2 \ln \left (3\right )-6 \ln \left (x \right )+i \pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}-4\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-4\right )^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-i \pi \mathrm {csgn}\left (\frac {i}{\ln \left (x \right ) \left ({\mathrm e}^{x}-4\right )^{2}}\right )^{3}-i \pi \mathrm {csgn}\left (i x^{3}\right )^{2} \mathrm {csgn}\left (i x \right )+i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+i \pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-4\right )^{2}\right )^{3}-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-4 \ln \left ({\mathrm e}^{x}-4\right )+i \pi \,\mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{x}-4\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (x \right ) \left ({\mathrm e}^{x}-4\right )^{2}}\right )^{2}-i \pi \mathrm {csgn}\left (\frac {i}{x^{3} \ln \left (x \right ) \left ({\mathrm e}^{x}-4\right )^{2}}\right )^{3}-i \pi \mathrm {csgn}\left (i x^{3}\right )^{2} \mathrm {csgn}\left (i x^{2}\right )+i \pi \,\mathrm {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (x \right ) \left ({\mathrm e}^{x}-4\right )^{2}}\right )^{2}+i \pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-4\right )\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-4\right )^{2}\right )-i \pi \,\mathrm {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{x}-4\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (x \right ) \left ({\mathrm e}^{x}-4\right )^{2}}\right )-i \pi \,\mathrm {csgn}\left (\frac {i}{x^{3}}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (x \right ) \left ({\mathrm e}^{x}-4\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i}{x^{3} \ln \left (x \right ) \left ({\mathrm e}^{x}-4\right )^{2}}\right )+i \pi \,\mathrm {csgn}\left (\frac {i}{\ln \left (x \right ) \left ({\mathrm e}^{x}-4\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i}{x^{3} \ln \left (x \right ) \left ({\mathrm e}^{x}-4\right )^{2}}\right )^{2}\right )}\)	\(465\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)*x-8*x)*ln(x)*ln(3/(exp(x)^2*x^3-8*exp(x)*x^3+16*x^3)/ln(x))+((3*x^2+3*x)*exp(x)-4*x^2-12*x)*ln(
x)+exp(x)*x-4*x)/((3*exp(x)-12)*ln(x)*ln(3/(exp(x)^2*x^3-8*exp(x)*x^3+16*x^3)/ln(x))^2+(6*exp(x)*x-24*x)*ln(x)
*ln(3/(exp(x)^2*x^3-8*exp(x)*x^3+16*x^3)/ln(x))+(3*exp(x)*x^2-12*x^2)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

2/3*x^2/(2*x+I*Pi*csgn(I/x^3)*csgn(I/x^3/ln(x)/(exp(x)-4)^2)^2+I*Pi*csgn(I*x^3)^3-2*ln(ln(x))+2*ln(3)-6*ln(x)+
I*Pi*csgn(I*x^3)*csgn(I*x)*csgn(I*x^2)-2*I*Pi*csgn(I*(exp(x)-4))*csgn(I*(exp(x)-4)^2)^2+I*Pi*csgn(I*x^2)^3-I*P
i*csgn(I/ln(x)/(exp(x)-4)^2)^3-I*Pi*csgn(I*x^3)^2*csgn(I*x)+I*Pi*csgn(I*x^2)*csgn(I*x)^2+I*Pi*csgn(I*(exp(x)-4
)^2)^3-2*I*Pi*csgn(I*x^2)^2*csgn(I*x)-4*ln(exp(x)-4)+I*Pi*csgn(I/(exp(x)-4)^2)*csgn(I/ln(x)/(exp(x)-4)^2)^2-I*
Pi*csgn(I/x^3/ln(x)/(exp(x)-4)^2)^3-I*Pi*csgn(I*x^3)^2*csgn(I*x^2)+I*Pi*csgn(I/ln(x))*csgn(I/ln(x)/(exp(x)-4)^
2)^2+I*Pi*csgn(I*(exp(x)-4))^2*csgn(I*(exp(x)-4)^2)-I*Pi*csgn(I/ln(x))*csgn(I/(exp(x)-4)^2)*csgn(I/ln(x)/(exp(
x)-4)^2)-I*Pi*csgn(I/x^3)*csgn(I/ln(x)/(exp(x)-4)^2)*csgn(I/x^3/ln(x)/(exp(x)-4)^2)+I*Pi*csgn(I/ln(x)/(exp(x)-
4)^2)*csgn(I/x^3/ln(x)/(exp(x)-4)^2)^2)

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Maxima [A]
time = 0.57, size = 27, normalized size = 0.90 \begin {gather*} \frac {x^{2}}{3 \, {\left (x + \log \left (3\right ) - 3 \, \log \left (x\right ) - 2 \, \log \left (e^{x} - 4\right ) - \log \left (\log \left (x\right )\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x-8*x)*log(x)*log(3/(exp(x)^2*x^3-8*exp(x)*x^3+16*x^3)/log(x))+((3*x^2+3*x)*exp(x)-4*x^2-
12*x)*log(x)+exp(x)*x-4*x)/((3*exp(x)-12)*log(x)*log(3/(exp(x)^2*x^3-8*exp(x)*x^3+16*x^3)/log(x))^2+(6*exp(x)*
x-24*x)*log(x)*log(3/(exp(x)^2*x^3-8*exp(x)*x^3+16*x^3)/log(x))+(3*exp(x)*x^2-12*x^2)*log(x)),x, algorithm="ma
xima")

[Out]

1/3*x^2/(x + log(3) - 3*log(x) - 2*log(e^x - 4) - log(log(x)))

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Fricas [A]
time = 0.39, size = 39, normalized size = 1.30 \begin {gather*} \frac {x^{2}}{3 \, {\left (x + \log \left (\frac {3}{{\left (x^{3} e^{\left (2 \, x\right )} - 8 \, x^{3} e^{x} + 16 \, x^{3}\right )} \log \left (x\right )}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x-8*x)*log(x)*log(3/(exp(x)^2*x^3-8*exp(x)*x^3+16*x^3)/log(x))+((3*x^2+3*x)*exp(x)-4*x^2-
12*x)*log(x)+exp(x)*x-4*x)/((3*exp(x)-12)*log(x)*log(3/(exp(x)^2*x^3-8*exp(x)*x^3+16*x^3)/log(x))^2+(6*exp(x)*
x-24*x)*log(x)*log(3/(exp(x)^2*x^3-8*exp(x)*x^3+16*x^3)/log(x))+(3*exp(x)*x^2-12*x^2)*log(x)),x, algorithm="fr
icas")

[Out]

1/3*x^2/(x + log(3/((x^3*e^(2*x) - 8*x^3*e^x + 16*x^3)*log(x))))

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Sympy [A]
time = 0.33, size = 36, normalized size = 1.20 \begin {gather*} \frac {x^{2}}{3 x + 3 \log {\left (\frac {3}{\left (x^{3} e^{2 x} - 8 x^{3} e^{x} + 16 x^{3}\right ) \log {\left (x \right )}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x-8*x)*ln(x)*ln(3/(exp(x)**2*x**3-8*exp(x)*x**3+16*x**3)/ln(x))+((3*x**2+3*x)*exp(x)-4*x*
*2-12*x)*ln(x)+exp(x)*x-4*x)/((3*exp(x)-12)*ln(x)*ln(3/(exp(x)**2*x**3-8*exp(x)*x**3+16*x**3)/ln(x))**2+(6*exp
(x)*x-24*x)*ln(x)*ln(3/(exp(x)**2*x**3-8*exp(x)*x**3+16*x**3)/ln(x))+(3*exp(x)*x**2-12*x**2)*ln(x)),x)

[Out]

x**2/(3*x + 3*log(3/((x**3*exp(2*x) - 8*x**3*exp(x) + 16*x**3)*log(x))))

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Giac [A]
time = 1.52, size = 36, normalized size = 1.20 \begin {gather*} \frac {x^{2}}{3 \, {\left (x + \log \left (3\right ) - \log \left (e^{\left (2 \, x\right )} \log \left (x\right ) - 8 \, e^{x} \log \left (x\right ) + 16 \, \log \left (x\right )\right ) - 3 \, \log \left (x\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x-8*x)*log(x)*log(3/(exp(x)^2*x^3-8*exp(x)*x^3+16*x^3)/log(x))+((3*x^2+3*x)*exp(x)-4*x^2-
12*x)*log(x)+exp(x)*x-4*x)/((3*exp(x)-12)*log(x)*log(3/(exp(x)^2*x^3-8*exp(x)*x^3+16*x^3)/log(x))^2+(6*exp(x)*
x-24*x)*log(x)*log(3/(exp(x)^2*x^3-8*exp(x)*x^3+16*x^3)/log(x))+(3*exp(x)*x^2-12*x^2)*log(x)),x, algorithm="gi
ac")

[Out]

1/3*x^2/(x + log(3) - log(e^(2*x)*log(x) - 8*e^x*log(x) + 16*log(x)) - 3*log(x))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {4\,x+\ln \left (x\right )\,\left (12\,x-{\mathrm {e}}^x\,\left (3\,x^2+3\,x\right )+4\,x^2\right )-x\,{\mathrm {e}}^x+\ln \left (\frac {3}{\ln \left (x\right )\,\left (x^3\,{\mathrm {e}}^{2\,x}-8\,x^3\,{\mathrm {e}}^x+16\,x^3\right )}\right )\,\ln \left (x\right )\,\left (8\,x-2\,x\,{\mathrm {e}}^x\right )}{\ln \left (x\right )\,\left (3\,{\mathrm {e}}^x-12\right )\,{\ln \left (\frac {3}{\ln \left (x\right )\,\left (x^3\,{\mathrm {e}}^{2\,x}-8\,x^3\,{\mathrm {e}}^x+16\,x^3\right )}\right )}^2-\ln \left (x\right )\,\left (24\,x-6\,x\,{\mathrm {e}}^x\right )\,\ln \left (\frac {3}{\ln \left (x\right )\,\left (x^3\,{\mathrm {e}}^{2\,x}-8\,x^3\,{\mathrm {e}}^x+16\,x^3\right )}\right )+\ln \left (x\right )\,\left (3\,x^2\,{\mathrm {e}}^x-12\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x + log(x)*(12*x - exp(x)*(3*x + 3*x^2) + 4*x^2) - x*exp(x) + log(3/(log(x)*(x^3*exp(2*x) - 8*x^3*exp(
x) + 16*x^3)))*log(x)*(8*x - 2*x*exp(x)))/(log(x)*(3*x^2*exp(x) - 12*x^2) - log(3/(log(x)*(x^3*exp(2*x) - 8*x^
3*exp(x) + 16*x^3)))*log(x)*(24*x - 6*x*exp(x)) + log(3/(log(x)*(x^3*exp(2*x) - 8*x^3*exp(x) + 16*x^3)))^2*log
(x)*(3*exp(x) - 12)),x)

[Out]

int(-(4*x + log(x)*(12*x - exp(x)*(3*x + 3*x^2) + 4*x^2) - x*exp(x) + log(3/(log(x)*(x^3*exp(2*x) - 8*x^3*exp(
x) + 16*x^3)))*log(x)*(8*x - 2*x*exp(x)))/(log(x)*(3*x^2*exp(x) - 12*x^2) - log(3/(log(x)*(x^3*exp(2*x) - 8*x^
3*exp(x) + 16*x^3)))*log(x)*(24*x - 6*x*exp(x)) + log(3/(log(x)*(x^3*exp(2*x) - 8*x^3*exp(x) + 16*x^3)))^2*log
(x)*(3*exp(x) - 12)), x)

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