3.9.66 \(\int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^{5/2}} \, dx\) [866]

Optimal. Leaf size=325 \[ -\frac {4 \sqrt {2} b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))^{3/2}}-\frac {32 \sqrt {2} a b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right )^2 d \sqrt {2 a+b \sinh (2 c+2 d x)}}-\frac {32 i \sqrt {2} a E\left (\frac {1}{2} \left (2 i c-\frac {\pi }{2}+2 i d x\right )|\frac {2 b}{2 i a+b}\right ) \sqrt {2 a+b \sinh (2 c+2 d x)}}{3 \left (4 a^2+b^2\right )^2 d \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}+\frac {4 i \sqrt {2} F\left (\frac {1}{2} \left (2 i c-\frac {\pi }{2}+2 i d x\right )|\frac {2 b}{2 i a+b}\right ) \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}{3 \left (4 a^2+b^2\right ) d \sqrt {2 a+b \sinh (2 c+2 d x)}} \]

[Out]

-4/3*b*cosh(2*d*x+2*c)*2^(1/2)/(4*a^2+b^2)/d/(2*a+b*sinh(2*d*x+2*c))^(3/2)-32/3*a*b*cosh(2*d*x+2*c)*2^(1/2)/(4
*a^2+b^2)^2/d/(2*a+b*sinh(2*d*x+2*c))^(1/2)+32/3*I*a*(sin(I*c+1/4*Pi+I*d*x)^2)^(1/2)/sin(I*c+1/4*Pi+I*d*x)*Ell
ipticE(cos(I*c+1/4*Pi+I*d*x),2^(1/2)*(b/(2*I*a+b))^(1/2))*2^(1/2)*(2*a+b*sinh(2*d*x+2*c))^(1/2)/(4*a^2+b^2)^2/
d/((2*a+b*sinh(2*d*x+2*c))/(2*a-I*b))^(1/2)-4/3*I*(sin(I*c+1/4*Pi+I*d*x)^2)^(1/2)/sin(I*c+1/4*Pi+I*d*x)*Ellipt
icF(cos(I*c+1/4*Pi+I*d*x),2^(1/2)*(b/(2*I*a+b))^(1/2))*2^(1/2)*((2*a+b*sinh(2*d*x+2*c))/(2*a-I*b))^(1/2)/(4*a^
2+b^2)/d/(2*a+b*sinh(2*d*x+2*c))^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 325, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2745, 2743, 2833, 2831, 2742, 2740, 2734, 2732} \begin {gather*} -\frac {32 \sqrt {2} a b \cosh (2 c+2 d x)}{3 d \left (4 a^2+b^2\right )^2 \sqrt {2 a+b \sinh (2 c+2 d x)}}-\frac {4 \sqrt {2} b \cosh (2 c+2 d x)}{3 d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))^{3/2}}+\frac {4 i \sqrt {2} \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}} F\left (\frac {1}{2} \left (2 i c+2 i d x-\frac {\pi }{2}\right )|\frac {2 b}{2 i a+b}\right )}{3 d \left (4 a^2+b^2\right ) \sqrt {2 a+b \sinh (2 c+2 d x)}}-\frac {32 i \sqrt {2} a \sqrt {2 a+b \sinh (2 c+2 d x)} E\left (\frac {1}{2} \left (2 i c+2 i d x-\frac {\pi }{2}\right )|\frac {2 b}{2 i a+b}\right )}{3 d \left (4 a^2+b^2\right )^2 \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^(-5/2),x]

[Out]

(-4*Sqrt[2]*b*Cosh[2*c + 2*d*x])/(3*(4*a^2 + b^2)*d*(2*a + b*Sinh[2*c + 2*d*x])^(3/2)) - (32*Sqrt[2]*a*b*Cosh[
2*c + 2*d*x])/(3*(4*a^2 + b^2)^2*d*Sqrt[2*a + b*Sinh[2*c + 2*d*x]]) - (((32*I)/3)*Sqrt[2]*a*EllipticE[((2*I)*c
- Pi/2 + (2*I)*d*x)/2, (2*b)/((2*I)*a + b)]*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])/((4*a^2 + b^2)^2*d*Sqrt[(2*a + b
*Sinh[2*c + 2*d*x])/(2*a - I*b)]) + (((4*I)/3)*Sqrt[2]*EllipticF[((2*I)*c - Pi/2 + (2*I)*d*x)/2, (2*b)/((2*I)*
a + b)]*Sqrt[(2*a + b*Sinh[2*c + 2*d*x])/(2*a - I*b)])/((4*a^2 + b^2)*d*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])

Rule 2732

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

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

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

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2743

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
+ 1)/(d*(n + 1)*(a^2 - b^2))), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n +
1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ
erQ[2*n]

Rule 2745

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + b*(Sin[2*c + 2*
d*x]/2))^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rule 2831

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

Rule 2833

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

Rubi steps

\begin {align*} \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^{5/2}} \, dx &=\int \frac {1}{\left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^{5/2}} \, dx\\ &=-\frac {4 \sqrt {2} b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))^{3/2}}-\frac {8 \int \frac {-\frac {3 a}{2}+\frac {1}{4} b \sinh (2 c+2 d x)}{\left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^{3/2}} \, dx}{3 \left (4 a^2+b^2\right )}\\ &=-\frac {4 \sqrt {2} b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))^{3/2}}-\frac {32 \sqrt {2} a b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right )^2 d \sqrt {2 a+b \sinh (2 c+2 d x)}}+\frac {64 \int \frac {\frac {1}{16} \left (12 a^2-b^2\right )+\frac {1}{2} a b \sinh (2 c+2 d x)}{\sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)}} \, dx}{3 \left (4 a^2+b^2\right )^2}\\ &=-\frac {4 \sqrt {2} b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))^{3/2}}-\frac {32 \sqrt {2} a b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right )^2 d \sqrt {2 a+b \sinh (2 c+2 d x)}}+\frac {(64 a) \int \sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)} \, dx}{3 \left (4 a^2+b^2\right )^2}-\frac {4 \int \frac {1}{\sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)}} \, dx}{3 \left (4 a^2+b^2\right )}\\ &=-\frac {4 \sqrt {2} b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))^{3/2}}-\frac {32 \sqrt {2} a b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right )^2 d \sqrt {2 a+b \sinh (2 c+2 d x)}}+\frac {\left (64 a \sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)}\right ) \int \sqrt {\frac {a}{a-\frac {i b}{2}}+\frac {b \sinh (2 c+2 d x)}{2 \left (a-\frac {i b}{2}\right )}} \, dx}{3 \left (4 a^2+b^2\right )^2 \sqrt {\frac {a+\frac {1}{2} b \sinh (2 c+2 d x)}{a-\frac {i b}{2}}}}-\frac {\left (4 \sqrt {\frac {a+\frac {1}{2} b \sinh (2 c+2 d x)}{a-\frac {i b}{2}}}\right ) \int \frac {1}{\sqrt {\frac {a}{a-\frac {i b}{2}}+\frac {b \sinh (2 c+2 d x)}{2 \left (a-\frac {i b}{2}\right )}}} \, dx}{3 \left (4 a^2+b^2\right ) \sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)}}\\ &=-\frac {4 \sqrt {2} b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))^{3/2}}-\frac {32 \sqrt {2} a b \cosh (2 c+2 d x)}{3 \left (4 a^2+b^2\right )^2 d \sqrt {2 a+b \sinh (2 c+2 d x)}}-\frac {32 i \sqrt {2} a E\left (\frac {1}{2} \left (2 i c-\frac {\pi }{2}+2 i d x\right )|\frac {2 b}{2 i a+b}\right ) \sqrt {2 a+b \sinh (2 c+2 d x)}}{3 \left (4 a^2+b^2\right )^2 d \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}+\frac {4 i \sqrt {2} F\left (\frac {1}{2} \left (2 i c-\frac {\pi }{2}+2 i d x\right )|\frac {2 b}{2 i a+b}\right ) \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}{3 \left (4 a^2+b^2\right ) d \sqrt {2 a+b \sinh (2 c+2 d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.14, size = 237, normalized size = 0.73 \begin {gather*} \frac {4 \sqrt {2} \left (8 i a (2 a-i b)^2 E\left (\frac {1}{4} (-4 i c+\pi -4 i d x)|-\frac {2 i b}{2 a-i b}\right ) \left (\frac {2 a+b \sinh (2 (c+d x))}{2 a-i b}\right )^{3/2}+(2 a-i b)^2 (-2 i a+b) F\left (\frac {1}{4} (-4 i c+\pi -4 i d x)|-\frac {2 i b}{2 a-i b}\right ) \left (\frac {2 a+b \sinh (2 (c+d x))}{2 a-i b}\right )^{3/2}-b \cosh (2 (c+d x)) \left (20 a^2+b^2+8 a b \sinh (2 (c+d x))\right )\right )}{3 \left (4 a^2+b^2\right )^2 d (2 a+b \sinh (2 (c+d x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^(-5/2),x]

[Out]

(4*Sqrt[2]*((8*I)*a*(2*a - I*b)^2*EllipticE[((-4*I)*c + Pi - (4*I)*d*x)/4, ((-2*I)*b)/(2*a - I*b)]*((2*a + b*S
inh[2*(c + d*x)])/(2*a - I*b))^(3/2) + (2*a - I*b)^2*((-2*I)*a + b)*EllipticF[((-4*I)*c + Pi - (4*I)*d*x)/4, (
(-2*I)*b)/(2*a - I*b)]*((2*a + b*Sinh[2*(c + d*x)])/(2*a - I*b))^(3/2) - b*Cosh[2*(c + d*x)]*(20*a^2 + b^2 + 8
*a*b*Sinh[2*(c + d*x)])))/(3*(4*a^2 + b^2)^2*d*(2*a + b*Sinh[2*(c + d*x)])^(3/2))

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Maple [A]
time = 5.72, size = 641, normalized size = 1.97

method	result	size
default	\(\frac {4 \sqrt {\left (\cosh ^{2}\left (2 d x +2 c \right )\right ) \left (2 a +b \sinh \left (2 d x +2 c \right )\right )}\, \left (-\frac {2 \sqrt {\left (\cosh ^{2}\left (2 d x +2 c \right )\right ) \left (2 a +b \sinh \left (2 d x +2 c \right )\right )}}{3 b \left (4 a^{2}+b^{2}\right ) \left (\sinh \left (2 d x +2 c \right )+\frac {2 a}{b}\right )^{2}}-\frac {16 b \left (\cosh ^{2}\left (2 d x +2 c \right )\right ) a}{3 \left (4 a^{2}+b^{2}\right )^{2} \sqrt {\left (\cosh ^{2}\left (2 d x +2 c \right )\right ) \left (2 a +b \sinh \left (2 d x +2 c \right )\right )}}+\frac {2 \left (12 a^{2}-b^{2}\right ) \left (\frac {2 a}{b}-i\right ) \sqrt {\frac {-b \sinh \left (2 d x +2 c \right )-2 a}{i b -2 a}}\, \sqrt {\frac {\left (-\sinh \left (2 d x +2 c \right )+i\right ) b}{i b +2 a}}\, \sqrt {\frac {\left (\sinh \left (2 d x +2 c \right )+i\right ) b}{i b -2 a}}\, \EllipticF \left (\sqrt {\frac {-b \sinh \left (2 d x +2 c \right )-2 a}{i b -2 a}}, \sqrt {\frac {-i b +2 a}{i b +2 a}}\right )}{\left (48 a^{4}+24 a^{2} b^{2}+3 b^{4}\right ) \sqrt {\left (\cosh ^{2}\left (2 d x +2 c \right )\right ) \left (2 a +b \sinh \left (2 d x +2 c \right )\right )}}+\frac {16 a b \left (\frac {2 a}{b}-i\right ) \sqrt {\frac {-b \sinh \left (2 d x +2 c \right )-2 a}{i b -2 a}}\, \sqrt {\frac {\left (-\sinh \left (2 d x +2 c \right )+i\right ) b}{i b +2 a}}\, \sqrt {\frac {\left (\sinh \left (2 d x +2 c \right )+i\right ) b}{i b -2 a}}\, \left (\left (-\frac {2 a}{b}-i\right ) \EllipticE \left (\sqrt {\frac {-b \sinh \left (2 d x +2 c \right )-2 a}{i b -2 a}}, \sqrt {\frac {-i b +2 a}{i b +2 a}}\right )+i \EllipticF \left (\sqrt {\frac {-b \sinh \left (2 d x +2 c \right )-2 a}{i b -2 a}}, \sqrt {\frac {-i b +2 a}{i b +2 a}}\right )\right )}{3 \left (4 a^{2}+b^{2}\right )^{2} \sqrt {\left (\cosh ^{2}\left (2 d x +2 c \right )\right ) \left (2 a +b \sinh \left (2 d x +2 c \right )\right )}}\right )}{\cosh \left (2 d x +2 c \right ) \sqrt {4 a +2 b \sinh \left (2 d x +2 c \right )}\, d}\)	\(641\)
risch	\(\text {Expression too large to display}\)	\(6261\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cosh(d*x+c)*sinh(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

4*(cosh(2*d*x+2*c)^2*(2*a+b*sinh(2*d*x+2*c)))^(1/2)*(-2/3/b/(4*a^2+b^2)*(cosh(2*d*x+2*c)^2*(2*a+b*sinh(2*d*x+2
*c)))^(1/2)/(sinh(2*d*x+2*c)+2*a/b)^2-16/3*b*cosh(2*d*x+2*c)^2/(4*a^2+b^2)^2*a/(cosh(2*d*x+2*c)^2*(2*a+b*sinh(
2*d*x+2*c)))^(1/2)+2*(12*a^2-b^2)/(48*a^4+24*a^2*b^2+3*b^4)*(2*a/b-I)*((-b*sinh(2*d*x+2*c)-2*a)/(I*b-2*a))^(1/
2)*((-sinh(2*d*x+2*c)+I)*b/(I*b+2*a))^(1/2)*((sinh(2*d*x+2*c)+I)*b/(I*b-2*a))^(1/2)/(cosh(2*d*x+2*c)^2*(2*a+b*
sinh(2*d*x+2*c)))^(1/2)*EllipticF(((-b*sinh(2*d*x+2*c)-2*a)/(I*b-2*a))^(1/2),((2*a-I*b)/(I*b+2*a))^(1/2))+16/3
*a*b/(4*a^2+b^2)^2*(2*a/b-I)*((-b*sinh(2*d*x+2*c)-2*a)/(I*b-2*a))^(1/2)*((-sinh(2*d*x+2*c)+I)*b/(I*b+2*a))^(1/
2)*((sinh(2*d*x+2*c)+I)*b/(I*b-2*a))^(1/2)/(cosh(2*d*x+2*c)^2*(2*a+b*sinh(2*d*x+2*c)))^(1/2)*((-2*a/b-I)*Ellip
ticE(((-b*sinh(2*d*x+2*c)-2*a)/(I*b-2*a))^(1/2),((2*a-I*b)/(I*b+2*a))^(1/2))+I*EllipticF(((-b*sinh(2*d*x+2*c)-
2*a)/(I*b-2*a))^(1/2),((2*a-I*b)/(I*b+2*a))^(1/2))))/cosh(2*d*x+2*c)/(4*a+2*b*sinh(2*d*x+2*c))^(1/2)/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*cosh(d*x + c)*sinh(d*x + c) + a)^(-5/2), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(d*x+c)*sinh(d*x+c))**(5/2),x)

[Out]

Integral((a + b*sinh(c + d*x)*cosh(c + d*x))**(-5/2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Error: Bad Argume
nt Type

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (a+b\,\mathrm {cosh}\left (c+d\,x\right )\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cosh(c + d*x)*sinh(c + d*x))^(5/2),x)

[Out]

int(1/(a + b*cosh(c + d*x)*sinh(c + d*x))^(5/2), x)

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