3.241 \(\int \frac {(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac {11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=241 \[ \frac {284 a^3 \sin (c+d x)}{231 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {710 a^3 \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {46 a^3 \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {2272 a^3 \sin (c+d x) \sqrt {\sec (c+d x)}}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {1136 a^3 \sin (c+d x)}{693 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)} \]

[Out]

46/99*a^3*sin(d*x+c)/d/sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(1/2)+710/693*a^3*sin(d*x+c)/d/sec(d*x+c)^(5/2)/(a+a*
sec(d*x+c))^(1/2)+284/231*a^3*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+1136/693*a^3*sin(d*x+c)/d/s
ec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+2272/693*a^3*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/11*
a^2*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(9/2)

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Rubi [A]  time = 0.40, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3813, 4015, 3805, 3804} \[ \frac {284 a^3 \sin (c+d x)}{231 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {710 a^3 \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {46 a^3 \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2272 a^3 \sin (c+d x) \sqrt {\sec (c+d x)}}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {1136 a^3 \sin (c+d x)}{693 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(5/2)/Sec[c + d*x]^(11/2),x]

[Out]

(46*a^3*Sin[c + d*x])/(99*d*Sec[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]) + (710*a^3*Sin[c + d*x])/(693*d*Sec[c
+ d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (284*a^3*Sin[c + d*x])/(231*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c +
d*x]]) + (1136*a^3*Sin[c + d*x])/(693*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2272*a^3*Sqrt[Sec[c +
d*x]]*Sin[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(11*d*Sec
[c + d*x]^(9/2))

Rule 3804

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[(-2*a*Co
t[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^
2, 0]

Rule 3805

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(a*Cot[
e + f*x]*(d*Csc[e + f*x])^n)/(f*n*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[(a*(2*n + 1))/(2*b*d*n), Int[Sqrt[a + b
*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2
^(-1)] && IntegerQ[2*n]

Rule 3813

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[a/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 2)*(d*Csc[e + f*x])^(n + 1)*(b*(m - 2*n - 2) - a*(m + 2*n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d,
e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && (LtQ[n, -1] || (EqQ[m, 3/2] && EqQ[n, -2^(-1)])) && IntegerQ[2
*m]

Rule 4015

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(A*b^2*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(a*f*n*Sqrt[a + b*Csc[e + f*x]]), x] +
Dist[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; Fr
eeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] &&
LtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac {11}{2}}(c+d x)} \, dx &=\frac {2 a^2 \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {1}{11} (2 a) \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {23 a}{2}+\frac {19}{2} a \sec (c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {46 a^3 \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {1}{99} \left (355 a^2\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {46 a^3 \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {710 a^3 \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {1}{231} \left (710 a^2\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {46 a^3 \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {710 a^3 \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {284 a^3 \sin (c+d x)}{231 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {1}{231} \left (568 a^2\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {46 a^3 \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {710 a^3 \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {284 a^3 \sin (c+d x)}{231 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {1136 a^3 \sin (c+d x)}{693 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {1}{693} \left (1136 a^2\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {46 a^3 \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {710 a^3 \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {284 a^3 \sin (c+d x)}{231 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {1136 a^3 \sin (c+d x)}{693 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2272 a^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 90, normalized size = 0.37 \[ \frac {2 a^3 \sin (c+d x) \left (1136 \sec ^5(c+d x)+568 \sec ^4(c+d x)+426 \sec ^3(c+d x)+355 \sec ^2(c+d x)+224 \sec (c+d x)+63\right )}{693 d \sec ^{\frac {9}{2}}(c+d x) \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^(5/2)/Sec[c + d*x]^(11/2),x]

[Out]

(2*a^3*(63 + 224*Sec[c + d*x] + 355*Sec[c + d*x]^2 + 426*Sec[c + d*x]^3 + 568*Sec[c + d*x]^4 + 1136*Sec[c + d*
x]^5)*Sin[c + d*x])/(693*d*Sec[c + d*x]^(9/2)*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 1.51, size = 126, normalized size = 0.52 \[ \frac {2 \, {\left (63 \, a^{2} \cos \left (d x + c\right )^{6} + 224 \, a^{2} \cos \left (d x + c\right )^{5} + 355 \, a^{2} \cos \left (d x + c\right )^{4} + 426 \, a^{2} \cos \left (d x + c\right )^{3} + 568 \, a^{2} \cos \left (d x + c\right )^{2} + 1136 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{693 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

2/693*(63*a^2*cos(d*x + c)^6 + 224*a^2*cos(d*x + c)^5 + 355*a^2*cos(d*x + c)^4 + 426*a^2*cos(d*x + c)^3 + 568*
a^2*cos(d*x + c)^2 + 1136*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/((d*cos(d*x +
c) + d)*sqrt(cos(d*x + c)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sec \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(11/2), x)

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maple [A]  time = 1.74, size = 115, normalized size = 0.48 \[ -\frac {2 \left (63 \left (\cos ^{6}\left (d x +c \right )\right )+161 \left (\cos ^{5}\left (d x +c \right )\right )+131 \left (\cos ^{4}\left (d x +c \right )\right )+71 \left (\cos ^{3}\left (d x +c \right )\right )+142 \left (\cos ^{2}\left (d x +c \right )\right )+568 \cos \left (d x +c \right )-1136\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {11}{2}} \left (\cos ^{6}\left (d x +c \right )\right ) a^{2}}{693 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(11/2),x)

[Out]

-2/693/d*(63*cos(d*x+c)^6+161*cos(d*x+c)^5+131*cos(d*x+c)^4+71*cos(d*x+c)^3+142*cos(d*x+c)^2+568*cos(d*x+c)-11
36)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(1/cos(d*x+c))^(11/2)*cos(d*x+c)^6/sin(d*x+c)*a^2

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maxima [B]  time = 0.76, size = 521, normalized size = 2.16 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

1/22176*sqrt(2)*(31878*a^2*cos(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 1
1/2*c) + 8778*a^2*cos(8/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 3
465*a^2*cos(6/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 1287*a^2*co
s(4/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 385*a^2*cos(2/11*arct
an2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) - 31878*a^2*cos(11/2*d*x + 11/2*c)
*sin(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 8778*a^2*cos(11/2*d*x + 11/2*c)*sin(8/11
*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 3465*a^2*cos(11/2*d*x + 11/2*c)*sin(6/11*arctan2(s
in(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 1287*a^2*cos(11/2*d*x + 11/2*c)*sin(4/11*arctan2(sin(11/2*d*
x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 385*a^2*cos(11/2*d*x + 11/2*c)*sin(2/11*arctan2(sin(11/2*d*x + 11/2*c)
, cos(11/2*d*x + 11/2*c))) + 126*a^2*sin(11/2*d*x + 11/2*c) + 385*a^2*sin(9/11*arctan2(sin(11/2*d*x + 11/2*c),
cos(11/2*d*x + 11/2*c))) + 1287*a^2*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 3465*
a^2*sin(5/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 8778*a^2*sin(3/11*arctan2(sin(11/2*d*x
+ 11/2*c), cos(11/2*d*x + 11/2*c))) + 31878*a^2*sin(1/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*
c))))*sqrt(a)/d

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mupad [B]  time = 6.93, size = 356, normalized size = 1.48 \[ \frac {\sqrt {a-\frac {a}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {23\,a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{4\,d}+\frac {19\,a^2\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{12\,d}+\frac {5\,a^2\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{8\,d}+\frac {13\,a^2\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{56\,d}+\frac {5\,a^2\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{72\,d}+\frac {a^2\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{88\,d}\right )}{2\,\sqrt {-\frac {1}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {c}{4}+\frac {d\,x}{4}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(5/2)/(1/cos(c + d*x))^(11/2),x)

[Out]

((a - a/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(sin((11*c)/2 + (11*d*x)/2)*1i + 2*sin((11*c)/4 + (11*d*x)/4)^2 -
1)*((23*a^2*sin(c/2 + (d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(4*d) + (
19*a^2*sin((3*c)/2 + (3*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(12*d) +
(5*a^2*sin((5*c)/2 + (5*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(8*d) +
(13*a^2*sin((7*c)/2 + (7*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(56*d)
+ (5*a^2*sin((9*c)/2 + (9*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(72*d
) + (a^2*sin((11*c)/2 + (11*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(88*
d)))/(2*(-1/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(2*sin(c/4 + (d*x)/4)^2 - 1))

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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((a+a*sec(d*x+c))**(5/2)/sec(d*x+c)**(11/2),x)

[Out]

Timed out

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