3.471 \(\int \frac{A+B \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=368 \[ -\frac{2 \left (9 a^2 A b-3 a^3 B+2 a b^2 B-8 A b^3\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 b \left (8 a^2 A b-5 a^3 B+a b^2 B-4 A b^3\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}+\frac{2 b (A b-a B) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (-15 a^2 A b^2+3 a^4 A+6 a^3 b B-2 a b^3 B+8 A b^4\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

(-2*(9*a^2*A*b - 8*A*b^3 - 3*a^3*B + 2*a*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a
)/(a + b)]*Sqrt[Sec[c + d*x]])/(3*a^3*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(3*a^4*A - 15*a^2*A*b^2 + 8
*A*b^4 + 6*a^3*b*B - 2*a*b^3*B)*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a^3*(a^2 -
b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*b*(A*b - a*B)*Sqrt[Sec[c + d*x]]*Sin[c +
d*x])/(3*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (2*b*(8*a^2*A*b - 4*A*b^3 - 5*a^3*B + a*b^2*B)*Sqrt[Sec
[c + d*x]]*Sin[c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])

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Rubi [A]  time = 0.936826, antiderivative size = 368, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used = {4030, 4100, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 b \left (8 a^2 A b-5 a^3 B+a b^2 B-4 A b^3\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}+\frac{2 b (A b-a B) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (9 a^2 A b-3 a^3 B+2 a b^2 B-8 A b^3\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (-15 a^2 A b^2+3 a^4 A+6 a^3 b B-2 a b^3 B+8 A b^4\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x])/(Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)),x]

[Out]

(-2*(9*a^2*A*b - 8*A*b^3 - 3*a^3*B + 2*a*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a
)/(a + b)]*Sqrt[Sec[c + d*x]])/(3*a^3*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(3*a^4*A - 15*a^2*A*b^2 + 8
*A*b^4 + 6*a^3*b*B - 2*a*b^3*B)*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a^3*(a^2 -
b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*b*(A*b - a*B)*Sqrt[Sec[c + d*x]]*Sin[c +
d*x])/(3*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (2*b*(8*a^2*A*b - 4*A*b^3 - 5*a^3*B + a*b^2*B)*Sqrt[Sec
[c + d*x]]*Sin[c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])

Rule 4030

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(b*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/
(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*
x])^n*Simp[A*(a^2*(m + 1) - b^2*(m + n + 1)) + a*b*B*n - a*(A*b - a*B)*(m + 1)*Csc[e + f*x] + b*(A*b - a*B)*(m
+ n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b
^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] && ILtQ[n, 0])

Rule 4100

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a +
b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), I
nt[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*
(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] &
& ILtQ[n, 0])

Rule 4035

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]

Rule 3856

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{A+B \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{5/2}} \, dx &=\frac{2 b (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 \int \frac{\frac{1}{2} \left (-3 a^2 A+4 A b^2-a b B\right )+\frac{3}{2} a (A b-a B) \sec (c+d x)-b (A b-a B) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )}\\ &=\frac{2 b (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{4 \int \frac{\frac{1}{4} \left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right )-\frac{1}{4} a \left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 b (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}-\frac{\left (9 a^2 A b-8 A b^3-3 a^3 B+2 a b^2 B\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right )}+\frac{\left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{2 b (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (9 a^2 A b-8 A b^3-3 a^3 B+2 a b^2 B\right ) \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{3 a^3 \left (a^2-b^2\right )^2 \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=\frac{2 b (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (9 a^2 A b-8 A b^3-3 a^3 B+2 a b^2 B\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{3 a^3 \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{3 a^3 \left (a^2-b^2\right )^2 \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=-\frac{2 \left (9 a^2 A b-8 A b^3-3 a^3 B+2 a b^2 B\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{\sec (c+d x)}}{3 a^3 \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}+\frac{2 b (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.35227, size = 297, normalized size = 0.81 \[ \frac{2 \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b) \left (-\frac{\left (\frac{a \cos (c+d x)+b}{a+b}\right )^{3/2} \left (a^2 \left (-\left (-6 a^2 A b+3 a^3 B+a b^2 B+2 A b^3\right )\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )-\left (-15 a^2 A b^2+3 a^4 A+6 a^3 b B-2 a b^3 B+8 A b^4\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )-b \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )\right )\right )}{(a-b)^2 (a+b)}-\frac{a b \sin (c+d x) \left (a \left (-9 a^2 A b+6 a^3 B-2 a b^2 B+5 A b^3\right ) \cos (c+d x)+b \left (-8 a^2 A b+5 a^3 B-a b^2 B+4 A b^3\right )\right )}{\left (a^2-b^2\right )^2}\right )}{3 a^3 d (a+b \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sec[c + d*x])/(Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)),x]

[Out]

(2*(b + a*Cos[c + d*x])*Sec[c + d*x]^(5/2)*(-((((b + a*Cos[c + d*x])/(a + b))^(3/2)*(-(a^2*(-6*a^2*A*b + 2*A*b
^3 + 3*a^3*B + a*b^2*B)*EllipticF[(c + d*x)/2, (2*a)/(a + b)]) - (3*a^4*A - 15*a^2*A*b^2 + 8*A*b^4 + 6*a^3*b*B
- 2*a*b^3*B)*((a + b)*EllipticE[(c + d*x)/2, (2*a)/(a + b)] - b*EllipticF[(c + d*x)/2, (2*a)/(a + b)])))/((a
- b)^2*(a + b))) - (a*b*(b*(-8*a^2*A*b + 4*A*b^3 + 5*a^3*B - a*b^2*B) + a*(-9*a^2*A*b + 5*A*b^3 + 6*a^3*B - 2*
a*b^2*B)*Cos[c + d*x])*Sin[c + d*x])/(a^2 - b^2)^2))/(3*a^3*d*(a + b*Sec[c + d*x])^(5/2))

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Maple [B]  time = 0.464, size = 5169, normalized size = 14.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x)

[Out]

result too large to display

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{b \sec \left (d x + c\right ) + a} \sqrt{\sec \left (d x + c\right )}}{b^{3} \sec \left (d x + c\right )^{4} + 3 \, a b^{2} \sec \left (d x + c\right )^{3} + 3 \, a^{2} b \sec \left (d x + c\right )^{2} + a^{3} \sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)*sqrt(sec(d*x + c))/(b^3*sec(d*x + c)^4 + 3*a*b^2*sec(d*
x + c)^3 + 3*a^2*b*sec(d*x + c)^2 + a^3*sec(d*x + c)), x)

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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*sec(d*x+c))/(a+b*sec(d*x+c))**(5/2)/sec(d*x+c)**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^(5/2)*sqrt(sec(d*x + c))), x)