3.203 \(\int \frac{(a+b \sin ^{-1}(c+d x))^3}{(c e+d e x)^2} \, dx\)

Optimal. Leaf size=190 \[ \frac{6 i b^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}-\frac{6 i b^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}-\frac{6 b^3 \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{6 b^3 \text{PolyLog}\left (3,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac{6 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2} \]

[Out]

-((a + b*ArcSin[c + d*x])^3/(d*e^2*(c + d*x))) - (6*b*(a + b*ArcSin[c + d*x])^2*ArcTanh[E^(I*ArcSin[c + d*x])]
)/(d*e^2) + ((6*I)*b^2*(a + b*ArcSin[c + d*x])*PolyLog[2, -E^(I*ArcSin[c + d*x])])/(d*e^2) - ((6*I)*b^2*(a + b
*ArcSin[c + d*x])*PolyLog[2, E^(I*ArcSin[c + d*x])])/(d*e^2) - (6*b^3*PolyLog[3, -E^(I*ArcSin[c + d*x])])/(d*e
^2) + (6*b^3*PolyLog[3, E^(I*ArcSin[c + d*x])])/(d*e^2)

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Rubi [A]  time = 0.247561, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {4805, 12, 4627, 4709, 4183, 2531, 2282, 6589} \[ \frac{6 i b^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}-\frac{6 i b^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}-\frac{6 b^3 \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{6 b^3 \text{PolyLog}\left (3,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac{6 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^3/(c*e + d*e*x)^2,x]

[Out]

-((a + b*ArcSin[c + d*x])^3/(d*e^2*(c + d*x))) - (6*b*(a + b*ArcSin[c + d*x])^2*ArcTanh[E^(I*ArcSin[c + d*x])]
)/(d*e^2) + ((6*I)*b^2*(a + b*ArcSin[c + d*x])*PolyLog[2, -E^(I*ArcSin[c + d*x])])/(d*e^2) - ((6*I)*b^2*(a + b
*ArcSin[c + d*x])*PolyLog[2, E^(I*ArcSin[c + d*x])])/(d*e^2) - (6*b^3*PolyLog[3, -E^(I*ArcSin[c + d*x])])/(d*e
^2) + (6*b^3*PolyLog[3, E^(I*ArcSin[c + d*x])])/(d*e^2)

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rule 12

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

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4709

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c^(m
+ 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*Sin[x]^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sin ^{-1}(c+d x)\right )^3}{(c e+d e x)^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^3}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^3}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^2}{x \sqrt{1-x^2}} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac{(3 b) \operatorname{Subst}\left (\int (a+b x)^2 \csc (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac{6 b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}+\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac{6 b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{6 i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{6 i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (6 i b^3\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}+\frac{\left (6 i b^3\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac{6 b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{6 i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{6 i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (6 b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{\left (6 b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac{6 b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{6 i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{6 i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{6 b^3 \text{Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{6 b^3 \text{Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}\\ \end{align*}

Mathematica [A]  time = 0.787842, size = 342, normalized size = 1.8 \[ -\frac{-6 i b^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )+6 i b^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )+6 b^3 \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c+d x)}\right )-6 b^3 \text{PolyLog}\left (3,e^{i \sin ^{-1}(c+d x)}\right )+3 a^2 b \log \left (\sqrt{-c^2-2 c d x-d^2 x^2+1}+1\right )-3 a^2 b \log (c+d x)+\frac{3 a^2 b \sin ^{-1}(c+d x)}{c+d x}+\frac{a^3}{c+d x}+\frac{3 a b^2 \sin ^{-1}(c+d x)^2}{c+d x}-6 a b^2 \sin ^{-1}(c+d x) \log \left (1-e^{i \sin ^{-1}(c+d x)}\right )+6 a b^2 \sin ^{-1}(c+d x) \log \left (1+e^{i \sin ^{-1}(c+d x)}\right )+\frac{b^3 \sin ^{-1}(c+d x)^3}{c+d x}-3 b^3 \sin ^{-1}(c+d x)^2 \log \left (1-e^{i \sin ^{-1}(c+d x)}\right )+3 b^3 \sin ^{-1}(c+d x)^2 \log \left (1+e^{i \sin ^{-1}(c+d x)}\right )}{d e^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSin[c + d*x])^3/(c*e + d*e*x)^2,x]

[Out]

-((a^3/(c + d*x) + (3*a^2*b*ArcSin[c + d*x])/(c + d*x) + (3*a*b^2*ArcSin[c + d*x]^2)/(c + d*x) + (b^3*ArcSin[c
+ d*x]^3)/(c + d*x) - 6*a*b^2*ArcSin[c + d*x]*Log[1 - E^(I*ArcSin[c + d*x])] - 3*b^3*ArcSin[c + d*x]^2*Log[1
- E^(I*ArcSin[c + d*x])] + 6*a*b^2*ArcSin[c + d*x]*Log[1 + E^(I*ArcSin[c + d*x])] + 3*b^3*ArcSin[c + d*x]^2*Lo
g[1 + E^(I*ArcSin[c + d*x])] - 3*a^2*b*Log[c + d*x] + 3*a^2*b*Log[1 + Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2]] - (6*
I)*b^2*(a + b*ArcSin[c + d*x])*PolyLog[2, -E^(I*ArcSin[c + d*x])] + (6*I)*b^2*(a + b*ArcSin[c + d*x])*PolyLog[
2, E^(I*ArcSin[c + d*x])] + 6*b^3*PolyLog[3, -E^(I*ArcSin[c + d*x])] - 6*b^3*PolyLog[3, E^(I*ArcSin[c + d*x])]
)/(d*e^2))

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Maple [B]  time = 0.067, size = 532, normalized size = 2.8 \begin{align*} -{\frac{{a}^{3}}{d{e}^{2} \left ( dx+c \right ) }}-{\frac{{b}^{3} \left ( \arcsin \left ( dx+c \right ) \right ) ^{3}}{d{e}^{2} \left ( dx+c \right ) }}-3\,{\frac{{b}^{3} \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}\ln \left ( 1+i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }{d{e}^{2}}}+{\frac{6\,i{b}^{3}\arcsin \left ( dx+c \right ) }{d{e}^{2}}{\it polylog} \left ( 2,-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }-6\,{\frac{{b}^{3}{\it polylog} \left ( 3,-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }{d{e}^{2}}}+3\,{\frac{{b}^{3} \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}\ln \left ( 1-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }{d{e}^{2}}}-{\frac{6\,i{b}^{3}\arcsin \left ( dx+c \right ) }{d{e}^{2}}{\it polylog} \left ( 2,i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }+6\,{\frac{{b}^{3}{\it polylog} \left ( 3,i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }{d{e}^{2}}}-3\,{\frac{a{b}^{2} \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}}{d{e}^{2} \left ( dx+c \right ) }}+6\,{\frac{a{b}^{2}\arcsin \left ( dx+c \right ) \ln \left ( 1-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }{d{e}^{2}}}-6\,{\frac{a{b}^{2}\arcsin \left ( dx+c \right ) \ln \left ( 1+i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }{d{e}^{2}}}+{\frac{6\,ia{b}^{2}}{d{e}^{2}}{\it polylog} \left ( 2,-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }-{\frac{6\,ia{b}^{2}}{d{e}^{2}}{\it polylog} \left ( 2,i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }-3\,{\frac{{a}^{2}b\arcsin \left ( dx+c \right ) }{d{e}^{2} \left ( dx+c \right ) }}-3\,{\frac{{a}^{2}b{\it Artanh} \left ({\frac{1}{\sqrt{1- \left ( dx+c \right ) ^{2}}}} \right ) }{d{e}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x+c))^3/(d*e*x+c*e)^2,x)

[Out]

-1/d*a^3/e^2/(d*x+c)-1/d*b^3/e^2/(d*x+c)*arcsin(d*x+c)^3-3/d*b^3/e^2*arcsin(d*x+c)^2*ln(1+I*(d*x+c)+(1-(d*x+c)
^2)^(1/2))+6*I/d*b^3/e^2*arcsin(d*x+c)*polylog(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-6*b^3*polylog(3,-I*(d*x+c)-(1
-(d*x+c)^2)^(1/2))/d/e^2+3/d*b^3/e^2*arcsin(d*x+c)^2*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-6*I/d*b^3/e^2*arcsin(
d*x+c)*polylog(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+6*b^3*polylog(3,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))/d/e^2-3/d*a*b^2
/e^2/(d*x+c)*arcsin(d*x+c)^2+6/d*a*b^2/e^2*arcsin(d*x+c)*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-6/d*a*b^2/e^2*arc
sin(d*x+c)*ln(1+I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+6*I/d*a*b^2/e^2*polylog(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-6*I/d
*a*b^2/e^2*polylog(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))-3/d*a^2*b/e^2/(d*x+c)*arcsin(d*x+c)-3/d*a^2*b/e^2*arctanh(
1/(1-(d*x+c)^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^3/(d*e*x+c*e)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^3/(d*e*x+c*e)^2,x, algorithm="fricas")

[Out]

integral((b^3*arcsin(d*x + c)^3 + 3*a*b^2*arcsin(d*x + c)^2 + 3*a^2*b*arcsin(d*x + c) + a^3)/(d^2*e^2*x^2 + 2*
c*d*e^2*x + c^2*e^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{3}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac{b^{3} \operatorname{asin}^{3}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac{3 a b^{2} \operatorname{asin}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac{3 a^{2} b \operatorname{asin}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x+c))**3/(d*e*x+c*e)**2,x)

[Out]

(Integral(a**3/(c**2 + 2*c*d*x + d**2*x**2), x) + Integral(b**3*asin(c + d*x)**3/(c**2 + 2*c*d*x + d**2*x**2),
x) + Integral(3*a*b**2*asin(c + d*x)**2/(c**2 + 2*c*d*x + d**2*x**2), x) + Integral(3*a**2*b*asin(c + d*x)/(c
**2 + 2*c*d*x + d**2*x**2), x))/e**2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^3/(d*e*x+c*e)^2,x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x + c) + a)^3/(d*e*x + c*e)^2, x)