3.258 \(\int \frac{1}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=257 \[ \frac{193 \sin (c+d x)}{64 a^3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{109 \sin (c+d x)}{64 a^2 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{629 \sin (c+d x)}{64 a^3 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{1015 \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}-\frac{23 \sin (c+d x)}{48 a d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}-\frac{\sin (c+d x)}{6 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \]

[Out]

(1015*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(64*Sqrt[2]*a^(7/2
)*d) - Sin[c + d*x]/(6*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(7/2)) - (23*Sin[c + d*x])/(48*a*d*Cos[c + d*
x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)) - (109*Sin[c + d*x])/(64*a^2*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3
/2)) + (193*Sin[c + d*x])/(64*a^3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - (629*Sin[c + d*x])/(64*a^3*
d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.703447, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2766, 2978, 2984, 12, 2782, 205} \[ \frac{193 \sin (c+d x)}{64 a^3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{109 \sin (c+d x)}{64 a^2 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{629 \sin (c+d x)}{64 a^3 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{1015 \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}-\frac{23 \sin (c+d x)}{48 a d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}-\frac{\sin (c+d x)}{6 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(7/2)),x]

[Out]

(1015*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(64*Sqrt[2]*a^(7/2
)*d) - Sin[c + d*x]/(6*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(7/2)) - (23*Sin[c + d*x])/(48*a*d*Cos[c + d*
x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)) - (109*Sin[c + d*x])/(64*a^2*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3
/2)) + (193*Sin[c + d*x])/(64*a^3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - (629*Sin[c + d*x])/(64*a^3*
d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

Rule 2766

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

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{1}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx &=-\frac{\sin (c+d x)}{6 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}+\frac{\int \frac{\frac{15 a}{2}-4 a \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=-\frac{\sin (c+d x)}{6 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac{23 \sin (c+d x)}{48 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\frac{189 a^2}{4}-\frac{69}{2} a^2 \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=-\frac{\sin (c+d x)}{6 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac{23 \sin (c+d x)}{48 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{109 \sin (c+d x)}{64 a^2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\frac{1737 a^3}{8}-\frac{327}{2} a^3 \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=-\frac{\sin (c+d x)}{6 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac{23 \sin (c+d x)}{48 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{109 \sin (c+d x)}{64 a^2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{193 \sin (c+d x)}{64 a^3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{-\frac{5661 a^4}{16}+\frac{1737}{8} a^4 \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{72 a^7}\\ &=-\frac{\sin (c+d x)}{6 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac{23 \sin (c+d x)}{48 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{109 \sin (c+d x)}{64 a^2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{193 \sin (c+d x)}{64 a^3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{629 \sin (c+d x)}{64 a^3 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{9135 a^5}{32 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{36 a^8}\\ &=-\frac{\sin (c+d x)}{6 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac{23 \sin (c+d x)}{48 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{109 \sin (c+d x)}{64 a^2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{193 \sin (c+d x)}{64 a^3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{629 \sin (c+d x)}{64 a^3 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{1015 \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{128 a^3}\\ &=-\frac{\sin (c+d x)}{6 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac{23 \sin (c+d x)}{48 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{109 \sin (c+d x)}{64 a^2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{193 \sin (c+d x)}{64 a^3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{629 \sin (c+d x)}{64 a^3 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{1015 \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 )}{64 a^2 d}\\ &=\frac{1015 \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 )}{64 \sqrt{2} a^{7/2} d}-\frac{\sin (c+d x)}{6 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac{23 \sin (c+d x)}{48 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{109 \sin (c+d x)}{64 a^2 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{193 \sin (c+d x)}{64 a^3 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{629 \sin (c+d x)}{64 a^3 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 8.30114, size = 273, normalized size = 1.06 \[ \frac{i e^{-\frac{3}{2} i (c+d x)} \cos ^7\left (\frac{1}{2} (c+d x)\right ) \left (3045 \sqrt{2} \left (1+e^{i (c+d x)}\right )^6 \left (1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )-2 \left (8277 e^{i (c+d x)}+14388 e^{2 i (c+d x)}+13108 e^{3 i (c+d x)}+5622 e^{4 i (c+d x)}-5622 e^{5 i (c+d x)}-13108 e^{6 i (c+d x)}-14388 e^{7 i (c+d x)}-8277 e^{8 i (c+d x)}-1887 e^{9 i (c+d x)}+1887\right )\right )}{96 d \left (1+e^{i (c+d x)}\right )^6 \cos ^{\frac{3}{2}}(c+d x) (a (\cos (c+d x)+1))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(7/2)),x]

[Out]

((I/96)*(-2*(1887 + 8277*E^(I*(c + d*x)) + 14388*E^((2*I)*(c + d*x)) + 13108*E^((3*I)*(c + d*x)) + 5622*E^((4*
I)*(c + d*x)) - 5622*E^((5*I)*(c + d*x)) - 13108*E^((6*I)*(c + d*x)) - 14388*E^((7*I)*(c + d*x)) - 8277*E^((8*
I)*(c + d*x)) - 1887*E^((9*I)*(c + d*x))) + 3045*Sqrt[2]*(1 + E^(I*(c + d*x)))^6*(1 + E^((2*I)*(c + d*x)))^(3/
2)*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])*Cos[(c + d*x)/2]^7)/(d*E^(((3*I)/2)
*(c + d*x))*(1 + E^(I*(c + d*x)))^6*Cos[c + d*x]^(3/2)*(a*(1 + Cos[c + d*x]))^(7/2))

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Maple [B]  time = 0.408, size = 435, normalized size = 1.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(5/2)/(a+cos(d*x+c)*a)^(7/2),x)

[Out]

1/384/d*2^(1/2)/a^4*(-3045*cos(d*x+c)^5*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c
)))^(5/2)-15225*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)^4*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-3
0450*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-30450*arcsin
((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-15225*arcsin((-1+cos(d*
x+c))/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+1887*cos(d*x+c)^6*2^(1/2)-3045*arcsi
n((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+3195*cos(d*x+c)^5*2^(1/2)-831*2^(1/
2)*cos(d*x+c)^4-3355*cos(d*x+c)^3*2^(1/2)-1024*cos(d*x+c)^2*2^(1/2)+128*cos(d*x+c)*2^(1/2))*(a*(1+cos(d*x+c)))
^(1/2)/sin(d*x+c)/(1+cos(d*x+c))^3/cos(d*x+c)^(5/2)

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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(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.4316, size = 695, normalized size = 2.7 \begin{align*} \frac{3045 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{6} + 4 \, \cos \left (d x + c\right )^{5} + 6 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + \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 (1887 \, \cos \left (d x + c\right )^{4} + 5082 \, \cos \left (d x + c\right )^{3} + 4251 \, \cos \left (d x + c\right )^{2} + 896 \, \cos \left (d x + c\right ) - 128\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{384 \,{\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/384*(3045*sqrt(2)*(cos(d*x + c)^6 + 4*cos(d*x + c)^5 + 6*cos(d*x + c)^4 + 4*cos(d*x + c)^3 + cos(d*x + c)^2)
*sqrt(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*(1887*cos(d*x + c)^4 + 5082*cos(d*x + c)^3 + 4251*cos(d*x + c)^2 + 896*cos(d*x + c) -
128)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^4*d*cos(d*x + c)^6 + 4*a^4*d*cos(d*x + c)^5
+ 6*a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + a^4*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(1/cos(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{7}{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(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(1/((a*cos(d*x + c) + a)^(7/2)*cos(d*x + c)^(5/2)), x)