3.25.23 \(\int \frac {b^8+a^8 x^8}{\sqrt {b^4+a^4 x^4} (-b^8+a^8 x^8)} \, dx\)

Optimal. Leaf size=196 \[ -\frac {x}{2 \sqrt {a^4 x^4+b^4}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} a b x}{\sqrt {a^4 x^4+b^4}+a^2 x^2+b^2}\right )}{2 \sqrt {2} a b}+\frac {\tanh ^{-1}\left (\frac {\sqrt {6-4 \sqrt {2}} a b x}{\sqrt {a^4 x^4+b^4}+a^2 x^2+b^2}\right )}{4 \sqrt {2} a b}-\frac {\tanh ^{-1}\left (\frac {\sqrt {6+4 \sqrt {2}} a b x}{\sqrt {a^4 x^4+b^4}+a^2 x^2+b^2}\right )}{4 \sqrt {2} a b} \]

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Rubi [A]  time = 0.64, antiderivative size = 101, normalized size of antiderivative = 0.52, number of steps used = 16, number of rules used = 10, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {6725, 220, 1404, 414, 523, 409, 1211, 1699, 206, 203} \begin {gather*} -\frac {x}{2 \sqrt {a^4 x^4+b^4}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} a b x}{\sqrt {a^4 x^4+b^4}}\right )}{4 \sqrt {2} a b}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} a b x}{\sqrt {a^4 x^4+b^4}}\right )}{4 \sqrt {2} a b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b^8 + a^8*x^8)/(Sqrt[b^4 + a^4*x^4]*(-b^8 + a^8*x^8)),x]

[Out]

-1/2*x/Sqrt[b^4 + a^4*x^4] - ArcTan[(Sqrt[2]*a*b*x)/Sqrt[b^4 + a^4*x^4]]/(4*Sqrt[2]*a*b) - ArcTanh[(Sqrt[2]*a*
b*x)/Sqrt[b^4 + a^4*x^4]]/(4*Sqrt[2]*a*b)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 409

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1
- Rt[-(d/c), 2]*x^2)), x], x] + Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-(d/c), 2]*x^2)), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 523

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

Rule 1211

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[1/(2*d), Int[1/Sqrt[a + c*x^4], x],
x] + Dist[1/(2*d), Int[(d - e*x^2)/((d + e*x^2)*Sqrt[a + c*x^4]), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d
^2 + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0]

Rule 1404

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Int[(d + e*x^n)^(p + q)*(a/d
+ (c*x^n)/e)^p, x] /; FreeQ[{a, c, d, e, n, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && IntegerQ[p]

Rule 1699

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[A, Subst[Int[1/
(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ
[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {b^8+a^8 x^8}{\sqrt {b^4+a^4 x^4} \left (-b^8+a^8 x^8\right )} \, dx &=\int \left (\frac {1}{\sqrt {b^4+a^4 x^4}}+\frac {2 b^8}{\sqrt {b^4+a^4 x^4} \left (-b^8+a^8 x^8\right )}\right ) \, dx\\ &=\left (2 b^8\right ) \int \frac {1}{\sqrt {b^4+a^4 x^4} \left (-b^8+a^8 x^8\right )} \, dx+\int \frac {1}{\sqrt {b^4+a^4 x^4}} \, dx\\ &=\frac {\left (b^2+a^2 x^2\right ) \sqrt {\frac {b^4+a^4 x^4}{\left (b^2+a^2 x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {a x}{b}\right )|\frac {1}{2}\right )}{2 a b \sqrt {b^4+a^4 x^4}}+\left (2 b^8\right ) \int \frac {1}{\left (-b^4+a^4 x^4\right ) \left (b^4+a^4 x^4\right )^{3/2}} \, dx\\ &=-\frac {x}{2 \sqrt {b^4+a^4 x^4}}+\frac {\left (b^2+a^2 x^2\right ) \sqrt {\frac {b^4+a^4 x^4}{\left (b^2+a^2 x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {a x}{b}\right )|\frac {1}{2}\right )}{2 a b \sqrt {b^4+a^4 x^4}}+\frac {\int \frac {3 a^4 b^4-a^8 x^4}{\left (-b^4+a^4 x^4\right ) \sqrt {b^4+a^4 x^4}} \, dx}{2 a^4}\\ &=-\frac {x}{2 \sqrt {b^4+a^4 x^4}}+\frac {\left (b^2+a^2 x^2\right ) \sqrt {\frac {b^4+a^4 x^4}{\left (b^2+a^2 x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {a x}{b}\right )|\frac {1}{2}\right )}{2 a b \sqrt {b^4+a^4 x^4}}-\frac {1}{2} \int \frac {1}{\sqrt {b^4+a^4 x^4}} \, dx+b^4 \int \frac {1}{\left (-b^4+a^4 x^4\right ) \sqrt {b^4+a^4 x^4}} \, dx\\ &=-\frac {x}{2 \sqrt {b^4+a^4 x^4}}+\frac {\left (b^2+a^2 x^2\right ) \sqrt {\frac {b^4+a^4 x^4}{\left (b^2+a^2 x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {a x}{b}\right )|\frac {1}{2}\right )}{4 a b \sqrt {b^4+a^4 x^4}}-\frac {1}{2} \int \frac {1}{\left (1-\frac {a^2 x^2}{b^2}\right ) \sqrt {b^4+a^4 x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+\frac {a^2 x^2}{b^2}\right ) \sqrt {b^4+a^4 x^4}} \, dx\\ &=-\frac {x}{2 \sqrt {b^4+a^4 x^4}}+\frac {\left (b^2+a^2 x^2\right ) \sqrt {\frac {b^4+a^4 x^4}{\left (b^2+a^2 x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {a x}{b}\right )|\frac {1}{2}\right )}{4 a b \sqrt {b^4+a^4 x^4}}-2 \left (\frac {1}{4} \int \frac {1}{\sqrt {b^4+a^4 x^4}} \, dx\right )-\frac {1}{4} \int \frac {1-\frac {a^2 x^2}{b^2}}{\left (1+\frac {a^2 x^2}{b^2}\right ) \sqrt {b^4+a^4 x^4}} \, dx-\frac {1}{4} \int \frac {1+\frac {a^2 x^2}{b^2}}{\left (1-\frac {a^2 x^2}{b^2}\right ) \sqrt {b^4+a^4 x^4}} \, dx\\ &=-\frac {x}{2 \sqrt {b^4+a^4 x^4}}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1-2 a^2 b^2 x^2} \, dx,x,\frac {x}{\sqrt {b^4+a^4 x^4}}\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+2 a^2 b^2 x^2} \, dx,x,\frac {x}{\sqrt {b^4+a^4 x^4}}\right )\\ &=-\frac {x}{2 \sqrt {b^4+a^4 x^4}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} a b x}{\sqrt {b^4+a^4 x^4}}\right )}{4 \sqrt {2} a b}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} a b x}{\sqrt {b^4+a^4 x^4}}\right )}{4 \sqrt {2} a b}\\ \end {align*}

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Mathematica [C]  time = 0.58, size = 199, normalized size = 1.02 \begin {gather*} \frac {x \left (-\frac {5 \left (a^4 b^4 x^4+b^8\right ) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};-\frac {a^4 x^4}{b^4},\frac {a^4 x^4}{b^4}\right )}{\left (b^4-a^4 x^4\right ) \left (5 b^4 F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};-\frac {a^4 x^4}{b^4},\frac {a^4 x^4}{b^4}\right )+2 a^4 x^4 \left (2 F_1\left (\frac {5}{4};-\frac {1}{2},2;\frac {9}{4};-\frac {a^4 x^4}{b^4},\frac {a^4 x^4}{b^4}\right )+F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};-\frac {a^4 x^4}{b^4},\frac {a^4 x^4}{b^4}\right )\right )\right )}-1\right )}{2 \sqrt {a^4 x^4+b^4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(b^8 + a^8*x^8)/(Sqrt[b^4 + a^4*x^4]*(-b^8 + a^8*x^8)),x]

[Out]

(x*(-1 - (5*(b^8 + a^4*b^4*x^4)*AppellF1[1/4, -1/2, 1, 5/4, -((a^4*x^4)/b^4), (a^4*x^4)/b^4])/((b^4 - a^4*x^4)
*(5*b^4*AppellF1[1/4, -1/2, 1, 5/4, -((a^4*x^4)/b^4), (a^4*x^4)/b^4] + 2*a^4*x^4*(2*AppellF1[5/4, -1/2, 2, 9/4
, -((a^4*x^4)/b^4), (a^4*x^4)/b^4] + AppellF1[5/4, 1/2, 1, 9/4, -((a^4*x^4)/b^4), (a^4*x^4)/b^4])))))/(2*Sqrt[
b^4 + a^4*x^4])

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IntegrateAlgebraic [A]  time = 0.85, size = 114, normalized size = 0.58 \begin {gather*} -\frac {x}{2 \sqrt {b^4+a^4 x^4}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} a b x}{b^2+a^2 x^2+\sqrt {b^4+a^4 x^4}}\right )}{2 \sqrt {2} a b}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} a b x}{\sqrt {b^4+a^4 x^4}}\right )}{4 \sqrt {2} a b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b^8 + a^8*x^8)/(Sqrt[b^4 + a^4*x^4]*(-b^8 + a^8*x^8)),x]

[Out]

-1/2*x/Sqrt[b^4 + a^4*x^4] - ArcTan[(Sqrt[2]*a*b*x)/(b^2 + a^2*x^2 + Sqrt[b^4 + a^4*x^4])]/(2*Sqrt[2]*a*b) - A
rcTanh[(Sqrt[2]*a*b*x)/Sqrt[b^4 + a^4*x^4]]/(4*Sqrt[2]*a*b)

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fricas [A]  time = 0.78, size = 159, normalized size = 0.81 \begin {gather*} -\frac {8 \, \sqrt {a^{4} x^{4} + b^{4}} a b x + 2 \, \sqrt {2} {\left (a^{4} x^{4} + b^{4}\right )} \arctan \left (\frac {\sqrt {2} a b x}{\sqrt {a^{4} x^{4} + b^{4}}}\right ) - \sqrt {2} {\left (a^{4} x^{4} + b^{4}\right )} \log \left (\frac {a^{4} x^{4} + 2 \, a^{2} b^{2} x^{2} + b^{4} - 2 \, \sqrt {2} \sqrt {a^{4} x^{4} + b^{4}} a b x}{a^{4} x^{4} - 2 \, a^{2} b^{2} x^{2} + b^{4}}\right )}{16 \, {\left (a^{5} b x^{4} + a b^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^8*x^8+b^8)/(a^4*x^4+b^4)^(1/2)/(a^8*x^8-b^8),x, algorithm="fricas")

[Out]

-1/16*(8*sqrt(a^4*x^4 + b^4)*a*b*x + 2*sqrt(2)*(a^4*x^4 + b^4)*arctan(sqrt(2)*a*b*x/sqrt(a^4*x^4 + b^4)) - sqr
t(2)*(a^4*x^4 + b^4)*log((a^4*x^4 + 2*a^2*b^2*x^2 + b^4 - 2*sqrt(2)*sqrt(a^4*x^4 + b^4)*a*b*x)/(a^4*x^4 - 2*a^
2*b^2*x^2 + b^4)))/(a^5*b*x^4 + a*b^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a^{8} x^{8} + b^{8}}{{\left (a^{8} x^{8} - b^{8}\right )} \sqrt {a^{4} x^{4} + b^{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^8*x^8+b^8)/(a^4*x^4+b^4)^(1/2)/(a^8*x^8-b^8),x, algorithm="giac")

[Out]

integrate((a^8*x^8 + b^8)/((a^8*x^8 - b^8)*sqrt(a^4*x^4 + b^4)), x)

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maple [A]  time = 0.20, size = 131, normalized size = 0.67

method	result	size
elliptic	\(\frac {\left (-\frac {\sqrt {2}\, x}{2 \sqrt {a^{4} x^{4}+b^{4}}}+\frac {\ln \left (-a b +\frac {\sqrt {a^{4} x^{4}+b^{4}}\, \sqrt {2}}{2 x}\right )}{8 a b}-\frac {\ln \left (a b +\frac {\sqrt {a^{4} x^{4}+b^{4}}\, \sqrt {2}}{2 x}\right )}{8 a b}+\frac {\arctan \left (\frac {\sqrt {a^{4} x^{4}+b^{4}}\, \sqrt {2}}{2 x a b}\right )}{4 a b}\right ) \sqrt {2}}{2}\)	\(131\)
default	\(\frac {\sqrt {1-\frac {i a^{2} x^{2}}{b^{2}}}\, \sqrt {1+\frac {i a^{2} x^{2}}{b^{2}}}\, \EllipticF \left (x \sqrt {\frac {i a^{2}}{b^{2}}}, i\right )}{\sqrt {\frac {i a^{2}}{b^{2}}}\, \sqrt {a^{4} x^{4}+b^{4}}}-\frac {b \left (-\frac {\sqrt {2}\, \arctanh \left (\frac {\left (2 a^{2} b^{2} x^{2}+2 b^{4}\right ) \sqrt {2}}{4 \sqrt {b^{4}}\, \sqrt {a^{4} x^{4}+b^{4}}}\right )}{4 \sqrt {b^{4}}}+\frac {a \sqrt {1-\frac {i a^{2} x^{2}}{b^{2}}}\, \sqrt {1+\frac {i a^{2} x^{2}}{b^{2}}}\, \EllipticPi \left (x \sqrt {\frac {i a^{2}}{b^{2}}}, -i, \frac {\sqrt {-\frac {i a^{2}}{b^{2}}}}{\sqrt {\frac {i a^{2}}{b^{2}}}}\right )}{\sqrt {\frac {i a^{2}}{b^{2}}}\, b \sqrt {a^{4} x^{4}+b^{4}}}\right )}{4 a}+\frac {b \left (-\frac {\sqrt {2}\, \arctanh \left (\frac {\left (2 a^{2} b^{2} x^{2}+2 b^{4}\right ) \sqrt {2}}{4 \sqrt {b^{4}}\, \sqrt {a^{4} x^{4}+b^{4}}}\right )}{4 \sqrt {b^{4}}}-\frac {a \sqrt {1-\frac {i a^{2} x^{2}}{b^{2}}}\, \sqrt {1+\frac {i a^{2} x^{2}}{b^{2}}}\, \EllipticPi \left (x \sqrt {\frac {i a^{2}}{b^{2}}}, -i, \frac {\sqrt {-\frac {i a^{2}}{b^{2}}}}{\sqrt {\frac {i a^{2}}{b^{2}}}}\right )}{\sqrt {\frac {i a^{2}}{b^{2}}}\, b \sqrt {a^{4} x^{4}+b^{4}}}\right )}{4 a}-b^{4} \left (\frac {x}{2 b^{4} \sqrt {\left (x^{4}+\frac {b^{4}}{a^{4}}\right ) a^{4}}}+\frac {\sqrt {1-\frac {i a^{2} x^{2}}{b^{2}}}\, \sqrt {1+\frac {i a^{2} x^{2}}{b^{2}}}\, \EllipticF \left (x \sqrt {\frac {i a^{2}}{b^{2}}}, i\right )}{2 b^{4} \sqrt {\frac {i a^{2}}{b^{2}}}\, \sqrt {a^{4} x^{4}+b^{4}}}\right )-\frac {\sqrt {1-\frac {i a^{2} x^{2}}{b^{2}}}\, \sqrt {1+\frac {i a^{2} x^{2}}{b^{2}}}\, \EllipticPi \left (x \sqrt {\frac {i a^{2}}{b^{2}}}, i, \frac {\sqrt {-\frac {i a^{2}}{b^{2}}}}{\sqrt {\frac {i a^{2}}{b^{2}}}}\right )}{2 \sqrt {\frac {i a^{2}}{b^{2}}}\, \sqrt {a^{4} x^{4}+b^{4}}}\)	\(595\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^8*x^8+b^8)/(a^4*x^4+b^4)^(1/2)/(a^8*x^8-b^8),x,method=_RETURNVERBOSE)

[Out]

1/2*(-1/2/(a^4*x^4+b^4)^(1/2)*2^(1/2)*x+1/8/a/b*ln(-a*b+1/2*(a^4*x^4+b^4)^(1/2)*2^(1/2)/x)-1/8/a/b*ln(a*b+1/2*
(a^4*x^4+b^4)^(1/2)*2^(1/2)/x)+1/4/a/b*arctan(1/2*(a^4*x^4+b^4)^(1/2)*2^(1/2)/x/a/b))*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a^{8} x^{8} + b^{8}}{{\left (a^{8} x^{8} - b^{8}\right )} \sqrt {a^{4} x^{4} + b^{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^8*x^8+b^8)/(a^4*x^4+b^4)^(1/2)/(a^8*x^8-b^8),x, algorithm="maxima")

[Out]

integrate((a^8*x^8 + b^8)/((a^8*x^8 - b^8)*sqrt(a^4*x^4 + b^4)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {a^8\,x^8+b^8}{\sqrt {a^4\,x^4+b^4}\,\left (b^8-a^8\,x^8\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(b^8 + a^8*x^8)/((b^4 + a^4*x^4)^(1/2)*(b^8 - a^8*x^8)),x)

[Out]

int(-(b^8 + a^8*x^8)/((b^4 + a^4*x^4)^(1/2)*(b^8 - a^8*x^8)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a^{8} x^{8} + b^{8}}{\left (a x - b\right ) \left (a x + b\right ) \left (a^{2} x^{2} + b^{2}\right ) \left (a^{4} x^{4} + b^{4}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**8*x**8+b**8)/(a**4*x**4+b**4)**(1/2)/(a**8*x**8-b**8),x)

[Out]

Integral((a**8*x**8 + b**8)/((a*x - b)*(a*x + b)*(a**2*x**2 + b**2)*(a**4*x**4 + b**4)**(3/2)), x)

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