3.5.70 \(\int \frac {A+B x}{\sqrt {e x} (a+c x^2)^{3/2}} \, dx\) [470]

3.5.70.1 Optimal result
3.5.70.2 Mathematica [C] (verified)
3.5.70.3 Rubi [A] (verified)
3.5.70.4 Maple [A] (verified)
3.5.70.5 Fricas [C] (verification not implemented)
3.5.70.6 Sympy [C] (verification not implemented)
3.5.70.7 Maxima [F]
3.5.70.8 Giac [F]
3.5.70.9 Mupad [F(-1)]

3.5.70.1 Optimal result

Integrand size = 24, antiderivative size = 290 \[ \int \frac {A+B x}{\sqrt {e x} \left (a+c x^2\right )^{3/2}} \, dx=\frac {\sqrt {e x} (A+B x)}{a e \sqrt {a+c x^2}}-\frac {B x \sqrt {a+c x^2}}{a \sqrt {c} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {B \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} c^{3/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 a^{5/4} c^{3/4} \sqrt {e x} \sqrt {a+c x^2}} \]

output
(B*x+A)*(e*x)^(1/2)/a/e/(c*x^2+a)^(1/2)-B*x*(c*x^2+a)^(1/2)/a/c^(1/2)/(a^( 
1/2)+x*c^(1/2))/(e*x)^(1/2)+B*(cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))^2)^( 
1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))*EllipticE(sin(2*arctan(c^(1/4) 
*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x*c^(1/2))*x^(1/2)*((c*x^2+a)/(a^ 
(1/2)+x*c^(1/2))^2)^(1/2)/a^(3/4)/c^(3/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)-1/2* 
(cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1 
/2)/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2) 
)*(B*a^(1/2)-A*c^(1/2))*(a^(1/2)+x*c^(1/2))*x^(1/2)*((c*x^2+a)/(a^(1/2)+x* 
c^(1/2))^2)^(1/2)/a^(5/4)/c^(3/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)
 
3.5.70.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.05 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.37 \[ \int \frac {A+B x}{\sqrt {e x} \left (a+c x^2\right )^{3/2}} \, dx=\frac {x \left (3 (A+B x)+3 A \sqrt {1+\frac {c x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{a}\right )-B x \sqrt {1+\frac {c x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^2}{a}\right )\right )}{3 a \sqrt {e x} \sqrt {a+c x^2}} \]

input
Integrate[(A + B*x)/(Sqrt[e*x]*(a + c*x^2)^(3/2)),x]
 
output
(x*(3*(A + B*x) + 3*A*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, 
 -((c*x^2)/a)] - B*x*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, 
-((c*x^2)/a)]))/(3*a*Sqrt[e*x]*Sqrt[a + c*x^2])
 
3.5.70.3 Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {551, 27, 556, 555, 1512, 27, 761, 1510}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+B x}{\sqrt {e x} \left (a+c x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 551

\(\displaystyle \frac {\sqrt {e x} (A+B x)}{a e \sqrt {a+c x^2}}-\frac {\int -\frac {A-B x}{2 \sqrt {e x} \sqrt {c x^2+a}}dx}{a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {A-B x}{\sqrt {e x} \sqrt {c x^2+a}}dx}{2 a}+\frac {\sqrt {e x} (A+B x)}{a e \sqrt {a+c x^2}}\)

\(\Big \downarrow \) 556

\(\displaystyle \frac {\sqrt {x} \int \frac {A-B x}{\sqrt {x} \sqrt {c x^2+a}}dx}{2 a \sqrt {e x}}+\frac {\sqrt {e x} (A+B x)}{a e \sqrt {a+c x^2}}\)

\(\Big \downarrow \) 555

\(\displaystyle \frac {\sqrt {x} \int \frac {A-B x}{\sqrt {c x^2+a}}d\sqrt {x}}{a \sqrt {e x}}+\frac {\sqrt {e x} (A+B x)}{a e \sqrt {a+c x^2}}\)

\(\Big \downarrow \) 1512

\(\displaystyle \frac {\sqrt {x} \left (\left (A-\frac {\sqrt {a} B}{\sqrt {c}}\right ) \int \frac {1}{\sqrt {c x^2+a}}d\sqrt {x}+\frac {\sqrt {a} B \int \frac {\sqrt {a}-\sqrt {c} x}{\sqrt {a} \sqrt {c x^2+a}}d\sqrt {x}}{\sqrt {c}}\right )}{a \sqrt {e x}}+\frac {\sqrt {e x} (A+B x)}{a e \sqrt {a+c x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {x} \left (\left (A-\frac {\sqrt {a} B}{\sqrt {c}}\right ) \int \frac {1}{\sqrt {c x^2+a}}d\sqrt {x}+\frac {B \int \frac {\sqrt {a}-\sqrt {c} x}{\sqrt {c x^2+a}}d\sqrt {x}}{\sqrt {c}}\right )}{a \sqrt {e x}}+\frac {\sqrt {e x} (A+B x)}{a e \sqrt {a+c x^2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {\sqrt {x} \left (\frac {B \int \frac {\sqrt {a}-\sqrt {c} x}{\sqrt {c x^2+a}}d\sqrt {x}}{\sqrt {c}}+\frac {\left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (A-\frac {\sqrt {a} B}{\sqrt {c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^2}}\right )}{a \sqrt {e x}}+\frac {\sqrt {e x} (A+B x)}{a e \sqrt {a+c x^2}}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {\sqrt {x} \left (\frac {\left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (A-\frac {\sqrt {a} B}{\sqrt {c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^2}}+\frac {B \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {a+c x^2}}-\frac {\sqrt {x} \sqrt {a+c x^2}}{\sqrt {a}+\sqrt {c} x}\right )}{\sqrt {c}}\right )}{a \sqrt {e x}}+\frac {\sqrt {e x} (A+B x)}{a e \sqrt {a+c x^2}}\)

input
Int[(A + B*x)/(Sqrt[e*x]*(a + c*x^2)^(3/2)),x]
 
output
(Sqrt[e*x]*(A + B*x))/(a*e*Sqrt[a + c*x^2]) + (Sqrt[x]*((B*(-((Sqrt[x]*Sqr 
t[a + c*x^2])/(Sqrt[a] + Sqrt[c]*x)) + (a^(1/4)*(Sqrt[a] + Sqrt[c]*x)*Sqrt 
[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x]) 
/a^(1/4)], 1/2])/(c^(1/4)*Sqrt[a + c*x^2])))/Sqrt[c] + ((A - (Sqrt[a]*B)/S 
qrt[c])*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*El 
lipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(2*a^(1/4)*c^(1/4)*Sqrt 
[a + c*x^2])))/(a*Sqrt[e*x])
 

3.5.70.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 551
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym 
bol] :> Simp[(-(e*x)^(m + 1))*(c + d*x)*((a + b*x^2)^(p + 1)/(2*a*e*(p + 1) 
)), x] + Simp[1/(2*a*(p + 1))   Int[(e*x)^m*(c*(m + 2*p + 3) + d*(m + 2*p + 
 4)*x)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p 
, -1] && LtQ[m, 0]
 

rule 555
Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> 
Simp[2   Subst[Int[(f + g*x^2)/Sqrt[a + c*x^4], x], x, Sqrt[x]], x] /; Free 
Q[{a, c, f, g}, x]
 

rule 556
Int[((c_) + (d_.)*(x_))/(Sqrt[(e_)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symb 
ol] :> Simp[Sqrt[x]/Sqrt[e*x]   Int[(c + d*x)/(Sqrt[x]*Sqrt[a + b*x^2]), x] 
, x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 761
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1510
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

rule 1512
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + c*x^4], x], x] - Simp[e/q 
 Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, c 
, d, e}, x] && PosQ[c/a]
 
3.5.70.4 Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 288, normalized size of antiderivative = 0.99

method result size
default \(\frac {A \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {x c}{\sqrt {-a c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a c}+B \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {x c}{\sqrt {-a c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right ) a -2 B \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {x c}{\sqrt {-a c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right ) a +2 B c \,x^{2}+2 A c x}{2 \sqrt {c \,x^{2}+a}\, c a \sqrt {e x}}\) \(288\)
elliptic \(\frac {\sqrt {\left (c \,x^{2}+a \right ) e x}\, \left (-\frac {2 x c e \left (-\frac {B x}{2 a e c}-\frac {A}{2 a e c}\right )}{\sqrt {\left (x^{2}+\frac {a}{c}\right ) x e c}}+\frac {A \sqrt {-a c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}\, \sqrt {-\frac {x c}{\sqrt {-a c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{2 a c \sqrt {c e \,x^{3}+a e x}}-\frac {B \sqrt {-a c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}\, \sqrt {-\frac {x c}{\sqrt {-a c}}}\, \left (-\frac {2 \sqrt {-a c}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-a c}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a c}}{c}\right ) c}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right )}{2 a c \sqrt {c e \,x^{3}+a e x}}\right )}{\sqrt {e x}\, \sqrt {c \,x^{2}+a}}\) \(353\)

input
int((B*x+A)/(e*x)^(1/2)/(c*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
 
output
1/2*(A*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2) 
)/(-a*c)^(1/2))^(1/2)*(-x/(-a*c)^(1/2)*c)^(1/2)*EllipticF(((c*x+(-a*c)^(1/ 
2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*(-a*c)^(1/2)+B*((c*x+(-a*c)^(1/2))/(- 
a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x/(-a 
*c)^(1/2)*c)^(1/2)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2 
^(1/2))*a-2*B*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c 
)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x/(-a*c)^(1/2)*c)^(1/2)*EllipticE(((c*x+(-a 
*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a+2*B*c*x^2+2*A*c*x)/(c*x^2+a) 
^(1/2)/c/a/(e*x)^(1/2)
 
3.5.70.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.35 \[ \int \frac {A+B x}{\sqrt {e x} \left (a+c x^2\right )^{3/2}} \, dx=\frac {{\left (A c x^{2} + A a\right )} \sqrt {c e} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{c}, 0, x\right ) + {\left (B c x^{2} + B a\right )} \sqrt {c e} {\rm weierstrassZeta}\left (-\frac {4 \, a}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{c}, 0, x\right )\right ) + {\left (B c x + A c\right )} \sqrt {c x^{2} + a} \sqrt {e x}}{a c^{2} e x^{2} + a^{2} c e} \]

input
integrate((B*x+A)/(e*x)^(1/2)/(c*x^2+a)^(3/2),x, algorithm="fricas")
 
output
((A*c*x^2 + A*a)*sqrt(c*e)*weierstrassPInverse(-4*a/c, 0, x) + (B*c*x^2 + 
B*a)*sqrt(c*e)*weierstrassZeta(-4*a/c, 0, weierstrassPInverse(-4*a/c, 0, x 
)) + (B*c*x + A*c)*sqrt(c*x^2 + a)*sqrt(e*x))/(a*c^2*e*x^2 + a^2*c*e)
 
3.5.70.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 12.81 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.32 \[ \int \frac {A+B x}{\sqrt {e x} \left (a+c x^2\right )^{3/2}} \, dx=\frac {A \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \sqrt {e} \Gamma \left (\frac {5}{4}\right )} + \frac {B x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \sqrt {e} \Gamma \left (\frac {7}{4}\right )} \]

input
integrate((B*x+A)/(e*x)**(1/2)/(c*x**2+a)**(3/2),x)
 
output
A*sqrt(x)*gamma(1/4)*hyper((1/4, 3/2), (5/4,), c*x**2*exp_polar(I*pi)/a)/( 
2*a**(3/2)*sqrt(e)*gamma(5/4)) + B*x**(3/2)*gamma(3/4)*hyper((3/4, 3/2), ( 
7/4,), c*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*sqrt(e)*gamma(7/4))
 
3.5.70.7 Maxima [F]

\[ \int \frac {A+B x}{\sqrt {e x} \left (a+c x^2\right )^{3/2}} \, dx=\int { \frac {B x + A}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} \sqrt {e x}} \,d x } \]

input
integrate((B*x+A)/(e*x)^(1/2)/(c*x^2+a)^(3/2),x, algorithm="maxima")
 
output
integrate((B*x + A)/((c*x^2 + a)^(3/2)*sqrt(e*x)), x)
 
3.5.70.8 Giac [F]

\[ \int \frac {A+B x}{\sqrt {e x} \left (a+c x^2\right )^{3/2}} \, dx=\int { \frac {B x + A}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} \sqrt {e x}} \,d x } \]

input
integrate((B*x+A)/(e*x)^(1/2)/(c*x^2+a)^(3/2),x, algorithm="giac")
 
output
integrate((B*x + A)/((c*x^2 + a)^(3/2)*sqrt(e*x)), x)
 
3.5.70.9 Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{\sqrt {e x} \left (a+c x^2\right )^{3/2}} \, dx=\int \frac {A+B\,x}{\sqrt {e\,x}\,{\left (c\,x^2+a\right )}^{3/2}} \,d x \]

input
int((A + B*x)/((e*x)^(1/2)*(a + c*x^2)^(3/2)),x)
 
output
int((A + B*x)/((e*x)^(1/2)*(a + c*x^2)^(3/2)), x)