3.127 \(\int \frac{x^3}{\sqrt{-3-4 x-x^2} (3+4 x+2 x^2)} \, dx\)

Optimal. Leaf size=115 \[ -\frac{1}{2} \sqrt{-x^2-4 x-3}+\frac{\tan ^{-1}\left (\frac{1-\frac{x+3}{\sqrt{-x^2-4 x-3}}}{\sqrt{2}}\right )}{2 \sqrt{2}}-\frac{\tan ^{-1}\left (\frac{\frac{x+3}{\sqrt{-x^2-4 x-3}}+1}{\sqrt{2}}\right )}{2 \sqrt{2}}+\tanh ^{-1}\left (\frac{x}{\sqrt{-x^2-4 x-3}}\right )-2 \sin ^{-1}(x+2) \]

[Out]

-Sqrt[-3 - 4*x - x^2]/2 - 2*ArcSin[2 + x] + ArcTan[(1 - (3 + x)/Sqrt[-3 - 4*x - x^2])/Sqrt[2]]/(2*Sqrt[2]) - A
rcTan[(1 + (3 + x)/Sqrt[-3 - 4*x - x^2])/Sqrt[2]]/(2*Sqrt[2]) + ArcTanh[x/Sqrt[-3 - 4*x - x^2]]

________________________________________________________________________________________

Rubi [A]  time = 0.420103, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 13, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.433, Rules used = {6728, 619, 216, 640, 1028, 986, 12, 1026, 1161, 618, 204, 1027, 206} \[ -\frac{1}{2} \sqrt{-x^2-4 x-3}+\frac{\tan ^{-1}\left (\frac{1-\frac{x+3}{\sqrt{-x^2-4 x-3}}}{\sqrt{2}}\right )}{2 \sqrt{2}}-\frac{\tan ^{-1}\left (\frac{\frac{x+3}{\sqrt{-x^2-4 x-3}}+1}{\sqrt{2}}\right )}{2 \sqrt{2}}+\tanh ^{-1}\left (\frac{x}{\sqrt{-x^2-4 x-3}}\right )-2 \sin ^{-1}(x+2) \]

Antiderivative was successfully verified.

[In]

Int[x^3/(Sqrt[-3 - 4*x - x^2]*(3 + 4*x + 2*x^2)),x]

[Out]

-Sqrt[-3 - 4*x - x^2]/2 - 2*ArcSin[2 + x] + ArcTan[(1 - (3 + x)/Sqrt[-3 - 4*x - x^2])/Sqrt[2]]/(2*Sqrt[2]) - A
rcTan[(1 + (3 + x)/Sqrt[-3 - 4*x - x^2])/Sqrt[2]]/(2*Sqrt[2]) + ArcTanh[x/Sqrt[-3 - 4*x - x^2]]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 1028

Int[((g_) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol]
:> -Dist[(2*h*d - g*e)/e, Int[1/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[h/e, Int[(2*d + e*x)
/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[e^2 - 4*d*f, 0] && EqQ[b*d - a*e, 0] && NeQ[2*h*d - g*e, 0]

Rule 986

Int[1/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q = Rt
[(c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f), 2]}, Dist[1/(2*q), Int[(c*d - a*f + q + (c*e - b*f)*x)/((a + b*x + c
*x^2)*Sqrt[d + e*x + f*x^2]), x], x] - Dist[1/(2*q), Int[(c*d - a*f - q + (c*e - b*f)*x)/((a + b*x + c*x^2)*Sq
rt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] &&
NeQ[c*e - b*f, 0] && NegQ[b^2 - 4*a*c]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1026

Int[(x_)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[-2*e,
Subst[Int[(1 - d*x^2)/(c*e - b*f - e*(2*c*d - b*e + 2*a*f)*x^2 + d^2*(c*e - b*f)*x^4), x], x, (1 + ((e + Sqrt[
e^2 - 4*d*f])*x)/(2*d))/Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[e^2 - 4*d*f, 0] && EqQ[b*d - a*e, 0]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1027

Int[((g_) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol]
:> Dist[g, Subst[Int[1/(a + (c*d - a*f)*x^2), x], x, x/Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c, d, e, f,
g, h}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && EqQ[b*d - a*e, 0] && EqQ[2*h*d - g*e, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^3}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx &=\int \left (-\frac{1}{\sqrt{-3-4 x-x^2}}+\frac{x}{2 \sqrt{-3-4 x-x^2}}+\frac{6+5 x}{2 \sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )}\right ) \, dx\\ &=\frac{1}{2} \int \frac{x}{\sqrt{-3-4 x-x^2}} \, dx+\frac{1}{2} \int \frac{6+5 x}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\int \frac{1}{\sqrt{-3-4 x-x^2}} \, dx\\ &=-\frac{1}{2} \sqrt{-3-4 x-x^2}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{4}}} \, dx,x,-4-2 x\right )-\frac{5}{8} \int \frac{-6-4 x}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac{3}{4} \int \frac{1}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\int \frac{1}{\sqrt{-3-4 x-x^2}} \, dx\\ &=-\frac{1}{2} \sqrt{-3-4 x-x^2}-\sin ^{-1}(2+x)+\frac{1}{8} \int \frac{-6-4 x}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac{1}{8} \int -\frac{4 x}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{4}}} \, dx,x,-4-2 x\right )+\frac{15}{4} \operatorname{Subst}\left (\int \frac{1}{3-3 x^2} \, dx,x,\frac{x}{\sqrt{-3-4 x-x^2}}\right )\\ &=-\frac{1}{2} \sqrt{-3-4 x-x^2}-2 \sin ^{-1}(2+x)+\frac{5}{4} \tanh ^{-1}\left (\frac{x}{\sqrt{-3-4 x-x^2}}\right )+\frac{1}{2} \int \frac{x}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{3-3 x^2} \, dx,x,\frac{x}{\sqrt{-3-4 x-x^2}}\right )\\ &=-\frac{1}{2} \sqrt{-3-4 x-x^2}-2 \sin ^{-1}(2+x)+\tanh ^{-1}\left (\frac{x}{\sqrt{-3-4 x-x^2}}\right )+4 \operatorname{Subst}\left (\int \frac{1+3 x^2}{-4-8 x^2-36 x^4} \, dx,x,\frac{1+\frac{x}{3}}{\sqrt{-3-4 x-x^2}}\right )\\ &=-\frac{1}{2} \sqrt{-3-4 x-x^2}-2 \sin ^{-1}(2+x)+\tanh ^{-1}\left (\frac{x}{\sqrt{-3-4 x-x^2}}\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{\frac{1}{3}-\frac{2 x}{3}+x^2} \, dx,x,\frac{1+\frac{x}{3}}{\sqrt{-3-4 x-x^2}}\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{\frac{1}{3}+\frac{2 x}{3}+x^2} \, dx,x,\frac{1+\frac{x}{3}}{\sqrt{-3-4 x-x^2}}\right )\\ &=-\frac{1}{2} \sqrt{-3-4 x-x^2}-2 \sin ^{-1}(2+x)+\tanh ^{-1}\left (\frac{x}{\sqrt{-3-4 x-x^2}}\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-\frac{8}{9}-x^2} \, dx,x,\frac{2}{3} \left (-1+\frac{3+x}{\sqrt{-3-4 x-x^2}}\right )\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-\frac{8}{9}-x^2} \, dx,x,\frac{2}{3} \left (1+\frac{3+x}{\sqrt{-3-4 x-x^2}}\right )\right )\\ &=-\frac{1}{2} \sqrt{-3-4 x-x^2}-2 \sin ^{-1}(2+x)+\frac{\tan ^{-1}\left (\frac{1-\frac{3+x}{\sqrt{-3-4 x-x^2}}}{\sqrt{2}}\right )}{2 \sqrt{2}}-\frac{\tan ^{-1}\left (\frac{1+\frac{3+x}{\sqrt{-3-4 x-x^2}}}{\sqrt{2}}\right )}{2 \sqrt{2}}+\tanh ^{-1}\left (\frac{x}{\sqrt{-3-4 x-x^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.436546, size = 192, normalized size = 1.67 \[ \frac{1}{8} \left (-4 \left (\sqrt{-x^2-4 x-3}+4 \sin ^{-1}(x+2)\right )+\frac{\left (5 \sqrt{2}-2 i\right ) \tanh ^{-1}\left (\frac{i \sqrt{2} x+2 x+2 i \sqrt{2}+2}{\sqrt{2-4 i \sqrt{2}} \sqrt{-x^2-4 x-3}}\right )}{\sqrt{1-2 i \sqrt{2}}}+\frac{\left (5 \sqrt{2}+2 i\right ) \tanh ^{-1}\left (\frac{\left (2-i \sqrt{2}\right ) x-2 i \sqrt{2}+2}{\sqrt{2+4 i \sqrt{2}} \sqrt{-x^2-4 x-3}}\right )}{\sqrt{1+2 i \sqrt{2}}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(Sqrt[-3 - 4*x - x^2]*(3 + 4*x + 2*x^2)),x]

[Out]

(-4*(Sqrt[-3 - 4*x - x^2] + 4*ArcSin[2 + x]) + ((-2*I + 5*Sqrt[2])*ArcTanh[(2 + (2*I)*Sqrt[2] + 2*x + I*Sqrt[2
]*x)/(Sqrt[2 - (4*I)*Sqrt[2]]*Sqrt[-3 - 4*x - x^2])])/Sqrt[1 - (2*I)*Sqrt[2]] + ((2*I + 5*Sqrt[2])*ArcTanh[(2
- (2*I)*Sqrt[2] + (2 - I*Sqrt[2])*x)/(Sqrt[2 + (4*I)*Sqrt[2]]*Sqrt[-3 - 4*x - x^2])])/Sqrt[1 + (2*I)*Sqrt[2]])
/8

________________________________________________________________________________________

Maple [A]  time = 0.135, size = 144, normalized size = 1.3 \begin{align*} -{\frac{1}{2}\sqrt{-{x}^{2}-4\,x-3}}-2\,\arcsin \left ( 2+x \right ) +{\frac{\sqrt{4}\sqrt{3}}{24}\sqrt{3\,{\frac{{x}^{2}}{ \left ( -3/2-x \right ) ^{2}}}-12} \left ( \sqrt{2}\arctan \left ({\frac{\sqrt{2}}{6}\sqrt{3\,{\frac{{x}^{2}}{ \left ( -3/2-x \right ) ^{2}}}-12}} \right ) -4\,{\it Artanh} \left ( 3\,{\frac{x}{-3/2-x}{\frac{1}{\sqrt{3\,{\frac{{x}^{2}}{ \left ( -3/2-x \right ) ^{2}}}-12}}}} \right ) \right ){\frac{1}{\sqrt{{ \left ({{x}^{2} \left ( -{\frac{3}{2}}-x \right ) ^{-2}}-4 \right ) \left ( 1+{x \left ( -{\frac{3}{2}}-x \right ) ^{-1}} \right ) ^{-2}}}}} \left ( 1+{x \left ( -{\frac{3}{2}}-x \right ) ^{-1}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(2*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x)

[Out]

-1/2*(-x^2-4*x-3)^(1/2)-2*arcsin(2+x)+1/24*3^(1/2)*4^(1/2)*(3*x^2/(-3/2-x)^2-12)^(1/2)*(2^(1/2)*arctan(1/6*(3*
x^2/(-3/2-x)^2-12)^(1/2)*2^(1/2))-4*arctanh(3*x/(-3/2-x)/(3*x^2/(-3/2-x)^2-12)^(1/2)))/((x^2/(-3/2-x)^2-4)/(1+
x/(-3/2-x))^2)^(1/2)/(1+x/(-3/2-x))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\left (2 \, x^{2} + 4 \, x + 3\right )} \sqrt{-x^{2} - 4 \, x - 3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(2*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^3/((2*x^2 + 4*x + 3)*sqrt(-x^2 - 4*x - 3)), x)

________________________________________________________________________________________

Fricas [A]  time = 2.20015, size = 474, normalized size = 4.12 \begin{align*} \frac{1}{8} \, \sqrt{2} \arctan \left (\frac{\sqrt{2} x + 3 \, \sqrt{2} \sqrt{-x^{2} - 4 \, x - 3}}{2 \,{\left (2 \, x + 3\right )}}\right ) + \frac{1}{8} \, \sqrt{2} \arctan \left (-\frac{\sqrt{2} x - 3 \, \sqrt{2} \sqrt{-x^{2} - 4 \, x - 3}}{2 \,{\left (2 \, x + 3\right )}}\right ) - \frac{1}{2} \, \sqrt{-x^{2} - 4 \, x - 3} + 2 \, \arctan \left (\frac{\sqrt{-x^{2} - 4 \, x - 3}{\left (x + 2\right )}}{x^{2} + 4 \, x + 3}\right ) - \frac{1}{4} \, \log \left (-\frac{2 \, \sqrt{-x^{2} - 4 \, x - 3} x + 4 \, x + 3}{x^{2}}\right ) + \frac{1}{4} \, \log \left (\frac{2 \, \sqrt{-x^{2} - 4 \, x - 3} x - 4 \, x - 3}{x^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(2*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x, algorithm="fricas")

[Out]

1/8*sqrt(2)*arctan(1/2*(sqrt(2)*x + 3*sqrt(2)*sqrt(-x^2 - 4*x - 3))/(2*x + 3)) + 1/8*sqrt(2)*arctan(-1/2*(sqrt
(2)*x - 3*sqrt(2)*sqrt(-x^2 - 4*x - 3))/(2*x + 3)) - 1/2*sqrt(-x^2 - 4*x - 3) + 2*arctan(sqrt(-x^2 - 4*x - 3)*
(x + 2)/(x^2 + 4*x + 3)) - 1/4*log(-(2*sqrt(-x^2 - 4*x - 3)*x + 4*x + 3)/x^2) + 1/4*log((2*sqrt(-x^2 - 4*x - 3
)*x - 4*x - 3)/x^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\sqrt{- \left (x + 1\right ) \left (x + 3\right )} \left (2 x^{2} + 4 x + 3\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(2*x**2+4*x+3)/(-x**2-4*x-3)**(1/2),x)

[Out]

Integral(x**3/(sqrt(-(x + 1)*(x + 3))*(2*x**2 + 4*x + 3)), x)

________________________________________________________________________________________

Giac [A]  time = 1.29332, size = 250, normalized size = 2.17 \begin{align*} \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\frac{3 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + 1\right )}\right ) + \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\frac{\sqrt{-x^{2} - 4 \, x - 3} - 1}{x + 2} + 1\right )}\right ) - \frac{1}{2} \, \sqrt{-x^{2} - 4 \, x - 3} - 2 \, \arcsin \left (x + 2\right ) + \frac{1}{2} \, \log \left (\frac{2 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + \frac{3 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}^{2}}{{\left (x + 2\right )}^{2}} + 1\right ) - \frac{1}{2} \, \log \left (\frac{2 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + \frac{{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}^{2}}{{\left (x + 2\right )}^{2}} + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(2*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x, algorithm="giac")

[Out]

1/4*sqrt(2)*arctan(1/2*sqrt(2)*(3*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) + 1/4*sqrt(2)*arctan(1/2*sqrt(2)*((
sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) - 1/2*sqrt(-x^2 - 4*x - 3) - 2*arcsin(x + 2) + 1/2*log(2*(sqrt(-x^2 -
4*x - 3) - 1)/(x + 2) + 3*(sqrt(-x^2 - 4*x - 3) - 1)^2/(x + 2)^2 + 1) - 1/2*log(2*(sqrt(-x^2 - 4*x - 3) - 1)/(
x + 2) + (sqrt(-x^2 - 4*x - 3) - 1)^2/(x + 2)^2 + 3)