3.201 \(\int \frac{A+B \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=203 \[ \frac{(11 A-7 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(7 A-3 B) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{(19 A-15 B) \sin (c+d x)}{6 a d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}} \]

[Out]

((11*A - 7*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]
*a^(3/2)*d) - ((A - B)*Sin[c + d*x])/(2*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)) + ((7*A - 3*B)*Sin[c
+ d*x])/(6*a*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((19*A - 15*B)*Sin[c + d*x])/(6*a*d*Sqrt[Cos[c +
d*x]]*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.555461, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2978, 2984, 12, 2782, 205} \[ \frac{(11 A-7 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(7 A-3 B) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{(19 A-15 B) \sin (c+d x)}{6 a d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(3/2)),x]

[Out]

((11*A - 7*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]
*a^(3/2)*d) - ((A - B)*Sin[c + d*x])/(2*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)) + ((7*A - 3*B)*Sin[c
+ d*x])/(6*a*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((19*A - 15*B)*Sin[c + d*x])/(6*a*d*Sqrt[Cos[c +
d*x]]*Sqrt[a + a*Cos[c + d*x]])

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 2984

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 12

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

Rule 2782

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

Rule 205

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

Rubi steps

\begin{align*} \int \frac{A+B \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx &=-\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\frac{1}{2} a (7 A-3 B)-2 a (A-B) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{-\frac{1}{4} a^2 (19 A-15 B)+\frac{1}{2} a^2 (7 A-3 B) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{3 a^3}\\ &=-\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(19 A-15 B) \sin (c+d x)}{6 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{2 \int \frac{3 a^3 (11 A-7 B)}{8 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{3 a^4}\\ &=-\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(19 A-15 B) \sin (c+d x)}{6 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{(11 A-7 B) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(19 A-15 B) \sin (c+d x)}{6 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{(11 A-7 B) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{2 d}\\ &=\frac{(11 A-7 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B) \sin (c+d x)}{6 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(19 A-15 B) \sin (c+d x)}{6 a d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.80029, size = 1054, normalized size = 5.19 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(3/2)),x]

[Out]

-((A - B)*Cos[c/2 + (d*x)/2]^3*(1 - 2*Sin[c/2 + (d*x)/2]))/(6*d*(a*(1 + Cos[c + d*x]))^(3/2)*(1 + Sin[c/2 + (d
*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + ((A - B)*Cos[c/2 + (d*x)/2]^3*(1 + 2*Sin[c/2 + (d*x)/2]))/(6*d*(
a*(1 + Cos[c + d*x]))^(3/2)*(1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) - ((A - B)*Cos[c/2 +
(d*x)/2]^3*(5*ArcTan[(1 - 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (1 + Sin[c/2 + (d*x)/2])/(
(1 - Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (3*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 - Sin[c/2
+ (d*x)/2])))/(d*(a*(1 + Cos[c + d*x]))^(3/2)) + ((A - B)*Cos[c/2 + (d*x)/2]^3*(5*ArcTan[(1 + 2*Sin[c/2 + (d*
x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (1 - Sin[c/2 + (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c
/2 + (d*x)/2]^2]) + (3*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 + Sin[c/2 + (d*x)/2])))/(d*(a*(1 + Cos[c + d*x]))^
(3/2)) + ((A + 3*B)*Cot[c/2 + (d*x)/2]^3*Csc[c/2 + (d*x)/2]^2*(-12*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2,
7/2}, {1, 9/2}, -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^8 - 12*Hypergeometri
c2F1[2, 7/2, 9/2, -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^8*(4 - 7*Sin[c/2 +
(d*x)/2]^2 + 3*Sin[c/2 + (d*x)/2]^4) + 7*Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*(1 - 2*Sin
[c/2 + (d*x)/2]^2)^3*(15 - 20*Sin[c/2 + (d*x)/2]^2 + 8*Sin[c/2 + (d*x)/2]^4)*((3 - 7*Sin[c/2 + (d*x)/2]^2)*Sqr
t[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))] - 3*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2
+ (d*x)/2]^2))]]*(1 - 2*Sin[c/2 + (d*x)/2]^2))))/(63*d*(a*(1 + Cos[c + d*x]))^(3/2)*(1 - 2*Sin[c/2 + (d*x)/2]
^2)^(7/2))

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Maple [B]  time = 0.623, size = 443, normalized size = 2.2 \begin{align*}{\frac{\sin \left ( dx+c \right ) }{12\,{a}^{2}d \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 1+\cos \left ( dx+c \right ) \right ) ^{2}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( 33\,A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}-21\,B\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}+66\,A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \cos \left ( dx+c \right ) \sqrt{2}\sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}-42\,B\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \cos \left ( dx+c \right ) \sqrt{2}\sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}+33\,A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sqrt{2}\sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}-21\,B\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sqrt{2}\sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}-38\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+30\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+14\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}-6\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+32\,A\cos \left ( dx+c \right ) -24\,B\cos \left ( dx+c \right ) -8\,A \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+cos(d*x+c)*a)^(3/2),x)

[Out]

1/12/d*sin(d*x+c)*(a*(1+cos(d*x+c)))^(1/2)*(33*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*2^(1/2)*sin(d
*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-21*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*2^(1/2)*sin(d*x+c
)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+66*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*2^(1/2)*sin(d*x+c)*(cos
(d*x+c)/(1+cos(d*x+c)))^(3/2)-42*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*2^(1/2)*sin(d*x+c)*(cos(d*x+c
)/(1+cos(d*x+c)))^(3/2)+33*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))
^(3/2)-21*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-38*A*cos(d
*x+c)^3+30*B*cos(d*x+c)^3+14*A*cos(d*x+c)^2-6*B*cos(d*x+c)^2+32*A*cos(d*x+c)-24*B*cos(d*x+c)-8*A)/a^2/(-1+cos(
d*x+c))/(1+cos(d*x+c))^2/cos(d*x+c)^(3/2)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.72612, size = 586, normalized size = 2.89 \begin{align*} \frac{3 \, \sqrt{2}{\left ({\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \,{\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \,{\left ({\left (19 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{2} + 12 \,{\left (A - B\right )} \cos \left (d x + c\right ) - 4 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/12*(3*sqrt(2)*((11*A - 7*B)*cos(d*x + c)^4 + 2*(11*A - 7*B)*cos(d*x + c)^3 + (11*A - 7*B)*cos(d*x + c)^2)*sq
rt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 +
a*cos(d*x + c))) - 2*((19*A - 15*B)*cos(d*x + c)^2 + 12*(A - B)*cos(d*x + c) - 4*A)*sqrt(a*cos(d*x + c) + a)*s
qrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^(3/2)*cos(d*x + c)^(5/2)), x)