3.594 \(\int \frac {\cos ^2(c+d x) (1-\cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=150 \[ -\frac {2 a^2 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d}+\frac {a x \left (2 a^2-b^2\right )}{2 b^4}-\frac {\left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^3 d}+\frac {a \sin (c+d x) \cos (c+d x)}{2 b^2 d}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d} \]

[Out]

1/2*a*(2*a^2-b^2)*x/b^4-1/3*(3*a^2-b^2)*sin(d*x+c)/b^3/d+1/2*a*cos(d*x+c)*sin(d*x+c)/b^2/d-1/3*cos(d*x+c)^2*si
n(d*x+c)/b/d-2*a^2*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))*(a-b)^(1/2)*(a+b)^(1/2)/b^4/d

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Rubi [A]  time = 0.39, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3050, 3049, 3023, 2735, 2659, 205} \[ -\frac {\left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^3 d}-\frac {2 a^2 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d}+\frac {a x \left (2 a^2-b^2\right )}{2 b^4}+\frac {a \sin (c+d x) \cos (c+d x)}{2 b^2 d}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*(1 - Cos[c + d*x]^2))/(a + b*Cos[c + d*x]),x]

[Out]

(a*(2*a^2 - b^2)*x)/(2*b^4) - (2*a^2*Sqrt[a - b]*Sqrt[a + b]*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]
])/(b^4*d) - ((3*a^2 - b^2)*Sin[c + d*x])/(3*b^3*d) + (a*Cos[c + d*x]*Sin[c + d*x])/(2*b^2*d) - (Cos[c + d*x]^
2*Sin[c + d*x])/(3*b*d)

Rule 205

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 2735

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)
*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n
+ 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n
*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0
] && NeQ[c, 0])))

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac {\int \frac {\cos (c+d x) \left (-2 a+b \cos (c+d x)+3 a \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{3 b}\\ &=\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac {\int \frac {3 a^2-a b \cos (c+d x)-2 \left (3 a^2-b^2\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^2}\\ &=-\frac {\left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^3 d}+\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac {\int \frac {3 a^2 b+3 a \left (2 a^2-b^2\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^3}\\ &=\frac {a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac {\left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^3 d}+\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d}-\frac {\left (a^2 \left (a^2-b^2\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{b^4}\\ &=\frac {a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac {\left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^3 d}+\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d}-\frac {\left (2 a^2 \left (a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^4 d}\\ &=\frac {a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac {2 a^2 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d}-\frac {\left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^3 d}+\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 b d}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 125, normalized size = 0.83 \[ -\frac {-6 a \left (2 a^2-b^2\right ) (c+d x)+24 a^2 \sqrt {b^2-a^2} \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )-3 a b^2 \sin (2 (c+d x))+3 b (2 a-b) (2 a+b) \sin (c+d x)+b^3 \sin (3 (c+d x))}{12 b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*(1 - Cos[c + d*x]^2))/(a + b*Cos[c + d*x]),x]

[Out]

-1/12*(-6*a*(2*a^2 - b^2)*(c + d*x) + 24*a^2*Sqrt[-a^2 + b^2]*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b
^2]] + 3*(2*a - b)*b*(2*a + b)*Sin[c + d*x] - 3*a*b^2*Sin[2*(c + d*x)] + b^3*Sin[3*(c + d*x)])/(b^4*d)

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fricas [A]  time = 0.50, size = 304, normalized size = 2.03 \[ \left [\frac {3 \, \sqrt {-a^{2} + b^{2}} a^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 3 \, {\left (2 \, a^{3} - a b^{2}\right )} d x - {\left (2 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, a b^{2} \cos \left (d x + c\right ) + 6 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} d}, -\frac {6 \, \sqrt {a^{2} - b^{2}} a^{2} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - 3 \, {\left (2 \, a^{3} - a b^{2}\right )} d x + {\left (2 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, a b^{2} \cos \left (d x + c\right ) + 6 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(1-cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

[1/6*(3*sqrt(-a^2 + b^2)*a^2*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*co
s(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) + 3*(2*a^3 - a*b^
2)*d*x - (2*b^3*cos(d*x + c)^2 - 3*a*b^2*cos(d*x + c) + 6*a^2*b - 2*b^3)*sin(d*x + c))/(b^4*d), -1/6*(6*sqrt(a
^2 - b^2)*a^2*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - 3*(2*a^3 - a*b^2)*d*x + (2*b^3*co
s(d*x + c)^2 - 3*a*b^2*cos(d*x + c) + 6*a^2*b - 2*b^3)*sin(d*x + c))/(b^4*d)]

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giac [A]  time = 1.38, size = 229, normalized size = 1.53 \[ \frac {\frac {3 \, {\left (2 \, a^{3} - a b^{2}\right )} {\left (d x + c\right )}}{b^{4}} + \frac {12 \, {\left (a^{4} - a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} - \frac {2 \, {\left (6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(1-cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

1/6*(3*(2*a^3 - a*b^2)*(d*x + c)/b^4 + 12*(a^4 - a^2*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) +
arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*b^4) - 2*(6*a^2*t
an(1/2*d*x + 1/2*c)^5 + 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 12*a^2*tan(1/2*d*x + 1/2*c)^3 - 8*b^2*tan(1/2*d*x + 1/2
*c)^3 + 6*a^2*tan(1/2*d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^3))/d

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maple [B]  time = 0.09, size = 350, normalized size = 2.33 \[ -\frac {2 a^{4} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 a^{2} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(1-cos(d*x+c)^2)/(a+b*cos(d*x+c)),x)

[Out]

-2/d*a^4/b^4/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))+2/d*a^2/b^2/((a-b)*(a+b)
)^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/
2*c)^5*a^2-1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*a-4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/
2*d*x+1/2*c)^3*a^2+8/3/d/b/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3-2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*
tan(1/2*d*x+1/2*c)*a^2+1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*a+2/d/b^4*arctan(tan(1/2*d*x+1/2*
c))*a^3-1/d/b^2*arctan(tan(1/2*d*x+1/2*c))*a

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(1-cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 1.95, size = 183, normalized size = 1.22 \[ \frac {\frac {\sin \left (c+d\,x\right )}{4}-\frac {\sin \left (3\,c+3\,d\,x\right )}{12}}{b\,d}-\frac {a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4}}{b^2\,d}-\frac {a^2\,\sin \left (c+d\,x\right )}{b^3\,d}+\frac {2\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^4\,d}+\frac {2\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a+b\right )}\right )\,\sqrt {b^2-a^2}}{b^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(cos(c + d*x)^2*(cos(c + d*x)^2 - 1))/(a + b*cos(c + d*x)),x)

[Out]

(sin(c + d*x)/4 - sin(3*c + 3*d*x)/12)/(b*d) - (a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - (a*sin(2*c + 2
*d*x))/4)/(b^2*d) - (a^2*sin(c + d*x))/(b^3*d) + (2*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^4*d) +
(2*a^2*atanh((sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(cos(c/2 + (d*x)/2)*(a + b)))*(b^2 - a^2)^(1/2))/(b^4*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(1-cos(d*x+c)**2)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

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