3.1245 \(\int x^{-7-2 p} (d+e x^2)^p (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=466 \[ -\frac {e^2 x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (p+1) (p+2) (p+3)}+\frac {e x^{-2 (p+2)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (p+2) (p+3)}-\frac {x^{-2 (p+3)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{2 d (p+3)}-\frac {b x^{-2 p-5} \left (c^4 d^2 \left (p^2+3 p+2\right )+2 c^2 d e (p+1)+2 e^2\right ) \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} F_1\left (\frac {1}{2} (-2 p-5);1,-p-1;\frac {1}{2} (-2 p-3);-c^2 x^2,-\frac {e x^2}{d}\right )}{2 c^3 d^2 (p+1) (p+2) (p+3) (2 p+5)}+\frac {b e x^{-2 p-5} \left (c^2 d (p+1)+e\right ) \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} \, _2F_1\left (\frac {1}{2} (-2 p-5),-p-1;\frac {1}{2} (-2 p-3);-\frac {e x^2}{d}\right )}{c^3 d^2 (p+1) (p+2) (p+3) (2 p+5)}-\frac {b e^2 x^{-2 p-3} \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} \, _2F_1\left (\frac {1}{2} (-2 p-3),-p-1;\frac {1}{2} (-2 p-1);-\frac {e x^2}{d}\right )}{c d^2 (p+1) (p+2) (p+3) (2 p+3)} \]

[Out]

-1/2*b*(2*e^2+2*c^2*d*e*(1+p)+c^4*d^2*(p^2+3*p+2))*x^(-5-2*p)*(e*x^2+d)^p*AppellF1(-5/2-p,1,-1-p,-3/2-p,-c^2*x
^2,-e*x^2/d)/c^3/d^2/(3+p)/(5+2*p)/(p^2+3*p+2)/((1+e*x^2/d)^p)-e^2*(e*x^2+d)^(1+p)*(a+b*arctan(c*x))/d^3/(2+p)
/(p^2+4*p+3)/(x^(2+2*p))+e*(e*x^2+d)^(1+p)*(a+b*arctan(c*x))/d^2/(2+p)/(3+p)/(x^(4+2*p))-1/2*(e*x^2+d)^(1+p)*(
a+b*arctan(c*x))/d/(3+p)/(x^(6+2*p))+b*e*(e+c^2*d*(1+p))*x^(-5-2*p)*(e*x^2+d)^p*hypergeom([-1-p, -5/2-p],[-3/2
-p],-e*x^2/d)/c^3/d^2/(3+p)/(5+2*p)/(p^2+3*p+2)/((1+e*x^2/d)^p)-b*e^2*x^(-3-2*p)*(e*x^2+d)^p*hypergeom([-1-p,
-3/2-p],[-1/2-p],-e*x^2/d)/c/d^2/(p^2+3*p+2)/(2*p^2+9*p+9)/((1+e*x^2/d)^p)

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Rubi [A]  time = 1.43, antiderivative size = 466, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {271, 264, 4976, 12, 6725, 365, 364, 511, 510} \[ -\frac {e^2 x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (p+1) (p+2) (p+3)}+\frac {e x^{-2 (p+2)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (p+2) (p+3)}-\frac {x^{-2 (p+3)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{2 d (p+3)}-\frac {b x^{-2 p-5} \left (c^4 d^2 \left (p^2+3 p+2\right )+2 c^2 d e (p+1)+2 e^2\right ) \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} F_1\left (\frac {1}{2} (-2 p-5);1,-p-1;\frac {1}{2} (-2 p-3);-c^2 x^2,-\frac {e x^2}{d}\right )}{2 c^3 d^2 (p+1) (p+2) (p+3) (2 p+5)}+\frac {b e x^{-2 p-5} \left (c^2 d (p+1)+e\right ) \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} \, _2F_1\left (\frac {1}{2} (-2 p-5),-p-1;\frac {1}{2} (-2 p-3);-\frac {e x^2}{d}\right )}{c^3 d^2 (p+1) (p+2) (p+3) (2 p+5)}-\frac {b e^2 x^{-2 p-3} \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} \, _2F_1\left (\frac {1}{2} (-2 p-3),-p-1;\frac {1}{2} (-2 p-1);-\frac {e x^2}{d}\right )}{c d^2 (p+1) (p+2) (p+3) (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Int[x^(-7 - 2*p)*(d + e*x^2)^p*(a + b*ArcTan[c*x]),x]

[Out]

-(b*(2*e^2 + 2*c^2*d*e*(1 + p) + c^4*d^2*(2 + 3*p + p^2))*x^(-5 - 2*p)*(d + e*x^2)^p*AppellF1[(-5 - 2*p)/2, 1,
-1 - p, (-3 - 2*p)/2, -(c^2*x^2), -((e*x^2)/d)])/(2*c^3*d^2*(1 + p)*(2 + p)*(3 + p)*(5 + 2*p)*(1 + (e*x^2)/d)
^p) - (e^2*(d + e*x^2)^(1 + p)*(a + b*ArcTan[c*x]))/(d^3*(1 + p)*(2 + p)*(3 + p)*x^(2*(1 + p))) + (e*(d + e*x^
2)^(1 + p)*(a + b*ArcTan[c*x]))/(d^2*(2 + p)*(3 + p)*x^(2*(2 + p))) - ((d + e*x^2)^(1 + p)*(a + b*ArcTan[c*x])
)/(2*d*(3 + p)*x^(2*(3 + p))) + (b*e*(e + c^2*d*(1 + p))*x^(-5 - 2*p)*(d + e*x^2)^p*Hypergeometric2F1[(-5 - 2*
p)/2, -1 - p, (-3 - 2*p)/2, -((e*x^2)/d)])/(c^3*d^2*(1 + p)*(2 + p)*(3 + p)*(5 + 2*p)*(1 + (e*x^2)/d)^p) - (b*
e^2*x^(-3 - 2*p)*(d + e*x^2)^p*Hypergeometric2F1[(-3 - 2*p)/2, -1 - p, (-1 - 2*p)/2, -((e*x^2)/d)])/(c*d^2*(1
+ p)*(2 + p)*(3 + p)*(3 + 2*p)*(1 + (e*x^2)/d)^p)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 4976

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^
2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m +
2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] &&
!ILtQ[(m - 1)/2, 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 x^{-7-2 p} \left (d+e x^2\right )^p \left (a+b \tan ^{-1}(c x)\right ) \, dx &=-\frac {e^2 x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (1+p) (2+p) (3+p)}+\frac {e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (2+p) (3+p)}-\frac {x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (3+p)}-(b c) \int \frac {x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (-d^2 \left (2+3 p+p^2\right )+2 d e (1+p) x^2-2 e^2 x^4\right )}{2 d^3 (1+p) (2+p) (3+p) \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {e^2 x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (1+p) (2+p) (3+p)}+\frac {e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (2+p) (3+p)}-\frac {x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (3+p)}-\frac {(b c) \int \frac {x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (-d^2 \left (2+3 p+p^2\right )+2 d e (1+p) x^2-2 e^2 x^4\right )}{1+c^2 x^2} \, dx}{2 d^3 (1+p) (2+p) (3+p)}\\ &=-\frac {e^2 x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (1+p) (2+p) (3+p)}+\frac {e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (2+p) (3+p)}-\frac {x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (3+p)}-\frac {(b c) \int \left (\frac {2 e \left (e+c^2 d (1+p)\right ) x^{-2 (3+p)} \left (d+e x^2\right )^{1+p}}{c^4}-\frac {2 e^2 x^{2-2 (3+p)} \left (d+e x^2\right )^{1+p}}{c^2}+\frac {\left (-2 c^4 d^2-2 c^2 d e-2 e^2-3 c^4 d^2 p-2 c^2 d e p-c^4 d^2 p^2\right ) x^{-2 (3+p)} \left (d+e x^2\right )^{1+p}}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx}{2 d^3 (1+p) (2+p) (3+p)}\\ &=-\frac {e^2 x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (1+p) (2+p) (3+p)}+\frac {e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (2+p) (3+p)}-\frac {x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (3+p)}+\frac {\left (b e^2\right ) \int x^{2-2 (3+p)} \left (d+e x^2\right )^{1+p} \, dx}{c d^3 (1+p) (2+p) (3+p)}-\frac {\left (b e \left (e+c^2 d (1+p)\right )\right ) \int x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \, dx}{c^3 d^3 (1+p) (2+p) (3+p)}+\frac {\left (b \left (2 e^2+2 c^2 d e (1+p)+c^4 d^2 \left (2+3 p+p^2\right )\right )\right ) \int \frac {x^{-2 (3+p)} \left (d+e x^2\right )^{1+p}}{1+c^2 x^2} \, dx}{2 c^3 d^3 (1+p) (2+p) (3+p)}\\ &=-\frac {e^2 x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (1+p) (2+p) (3+p)}+\frac {e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (2+p) (3+p)}-\frac {x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (3+p)}+\frac {\left (b e^2 \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p}\right ) \int x^{2-2 (3+p)} \left (1+\frac {e x^2}{d}\right )^{1+p} \, dx}{c d^2 (1+p) (2+p) (3+p)}-\frac {\left (b e \left (e+c^2 d (1+p)\right ) \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p}\right ) \int x^{-2 (3+p)} \left (1+\frac {e x^2}{d}\right )^{1+p} \, dx}{c^3 d^2 (1+p) (2+p) (3+p)}+\frac {\left (b \left (2 e^2+2 c^2 d e (1+p)+c^4 d^2 \left (2+3 p+p^2\right )\right ) \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p}\right ) \int \frac {x^{-2 (3+p)} \left (1+\frac {e x^2}{d}\right )^{1+p}}{1+c^2 x^2} \, dx}{2 c^3 d^2 (1+p) (2+p) (3+p)}\\ &=-\frac {b \left (2 e^2+2 c^2 d e (1+p)+c^4 d^2 \left (2+3 p+p^2\right )\right ) x^{-5-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} F_1\left (\frac {1}{2} (-5-2 p);1,-1-p;\frac {1}{2} (-3-2 p);-c^2 x^2,-\frac {e x^2}{d}\right )}{2 c^3 d^2 (1+p) (2+p) (3+p) (5+2 p)}-\frac {e^2 x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (1+p) (2+p) (3+p)}+\frac {e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (2+p) (3+p)}-\frac {x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (3+p)}+\frac {b e \left (e+c^2 d (1+p)\right ) x^{-5-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} \, _2F_1\left (\frac {1}{2} (-5-2 p),-1-p;\frac {1}{2} (-3-2 p);-\frac {e x^2}{d}\right )}{c^3 d^2 (1+p) (2+p) (3+p) (5+2 p)}-\frac {b e^2 x^{-3-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} \, _2F_1\left (\frac {1}{2} (-3-2 p),-1-p;\frac {1}{2} (-1-2 p);-\frac {e x^2}{d}\right )}{c d^2 (1+p) (2+p) (3+p) (3+2 p)}\\ \end {align*}

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Mathematica [F]  time = 5.32, size = 0, normalized size = 0.00 \[ \int x^{-7-2 p} \left (d+e x^2\right )^p \left (a+b \tan ^{-1}(c x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[x^(-7 - 2*p)*(d + e*x^2)^p*(a + b*ArcTan[c*x]),x]

[Out]

Integrate[x^(-7 - 2*p)*(d + e*x^2)^p*(a + b*ArcTan[c*x]), x]

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \arctan \left (c x\right ) + a\right )} {\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 7}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-7-2*p)*(e*x^2+d)^p*(a+b*arctan(c*x)),x, algorithm="fricas")

[Out]

integral((b*arctan(c*x) + a)*(e*x^2 + d)^p*x^(-2*p - 7), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \arctan \left (c x\right ) + a\right )} {\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 7}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-7-2*p)*(e*x^2+d)^p*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

integrate((b*arctan(c*x) + a)*(e*x^2 + d)^p*x^(-2*p - 7), x)

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maple [F]  time = 1.46, size = 0, normalized size = 0.00 \[ \int x^{-7-2 p} \left (e \,x^{2}+d \right )^{p} \left (a +b \arctan \left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-7-2*p)*(e*x^2+d)^p*(a+b*arctan(c*x)),x)

[Out]

int(x^(-7-2*p)*(e*x^2+d)^p*(a+b*arctan(c*x)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {\arctan \left (c x\right ) e^{\left (p \log \left (e x^{2} + d\right ) - 2 \, p \log \relax (x)\right )}}{x^{7}}\,{d x} - \frac {{\left (2 \, e^{3} x^{6} - 2 \, d e^{2} p x^{4} + {\left (p^{2} + p\right )} d^{2} e x^{2} + {\left (p^{2} + 3 \, p + 2\right )} d^{3}\right )} a e^{\left (p \log \left (e x^{2} + d\right ) - 2 \, p \log \relax (x)\right )}}{2 \, {\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} d^{3} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-7-2*p)*(e*x^2+d)^p*(a+b*arctan(c*x)),x, algorithm="maxima")

[Out]

b*integrate(arctan(c*x)*e^(p*log(e*x^2 + d) - 2*p*log(x))/x^7, x) - 1/2*(2*e^3*x^6 - 2*d*e^2*p*x^4 + (p^2 + p)
*d^2*e*x^2 + (p^2 + 3*p + 2)*d^3)*a*e^(p*log(e*x^2 + d) - 2*p*log(x))/((p^3 + 6*p^2 + 11*p + 6)*d^3*x^6)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^p}{x^{2\,p+7}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + e*x^2)^p)/x^(2*p + 7),x)

[Out]

int(((a + b*atan(c*x))*(d + e*x^2)^p)/x^(2*p + 7), x)

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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(x**(-7-2*p)*(e*x**2+d)**p*(a+b*atan(c*x)),x)

[Out]

Timed out

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