3.14.58 \(\int \frac {(d+e x)^{7/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=145 \[ -\frac {35 e^3 \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{9/2}}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 (a+b x)}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x)^2}-\frac {(d+e x)^{7/2}}{3 b (a+b x)^3}+\frac {35 e^3 \sqrt {d+e x}}{8 b^4} \]

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Rubi [A]  time = 0.07, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 47, 50, 63, 208} \begin {gather*} -\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 (a+b x)}-\frac {35 e^3 \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{9/2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x)^2}-\frac {(d+e x)^{7/2}}{3 b (a+b x)^3}+\frac {35 e^3 \sqrt {d+e x}}{8 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(7/2)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(35*e^3*Sqrt[d + e*x])/(8*b^4) - (35*e^2*(d + e*x)^(3/2))/(24*b^3*(a + b*x)) - (7*e*(d + e*x)^(5/2))/(12*b^2*(
a + b*x)^2) - (d + e*x)^(7/2)/(3*b*(a + b*x)^3) - (35*e^3*Sqrt[b*d - a*e]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt
[b*d - a*e]])/(8*b^(9/2))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {(d+e x)^{7/2}}{(a+b x)^4} \, dx\\ &=-\frac {(d+e x)^{7/2}}{3 b (a+b x)^3}+\frac {(7 e) \int \frac {(d+e x)^{5/2}}{(a+b x)^3} \, dx}{6 b}\\ &=-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x)^2}-\frac {(d+e x)^{7/2}}{3 b (a+b x)^3}+\frac {\left (35 e^2\right ) \int \frac {(d+e x)^{3/2}}{(a+b x)^2} \, dx}{24 b^2}\\ &=-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 (a+b x)}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x)^2}-\frac {(d+e x)^{7/2}}{3 b (a+b x)^3}+\frac {\left (35 e^3\right ) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{16 b^3}\\ &=\frac {35 e^3 \sqrt {d+e x}}{8 b^4}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 (a+b x)}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x)^2}-\frac {(d+e x)^{7/2}}{3 b (a+b x)^3}+\frac {\left (35 e^3 (b d-a e)\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{16 b^4}\\ &=\frac {35 e^3 \sqrt {d+e x}}{8 b^4}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 (a+b x)}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x)^2}-\frac {(d+e x)^{7/2}}{3 b (a+b x)^3}+\frac {\left (35 e^2 (b d-a e)\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{8 b^4}\\ &=\frac {35 e^3 \sqrt {d+e x}}{8 b^4}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 (a+b x)}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x)^2}-\frac {(d+e x)^{7/2}}{3 b (a+b x)^3}-\frac {35 e^3 \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.36 \begin {gather*} \frac {2 e^3 (d+e x)^{9/2} \, _2F_1\left (4,\frac {9}{2};\frac {11}{2};-\frac {b (d+e x)}{a e-b d}\right )}{9 (a e-b d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(7/2)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(2*e^3*(d + e*x)^(9/2)*Hypergeometric2F1[4, 9/2, 11/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(9*(-(b*d) + a*e)^4)

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IntegrateAlgebraic [A]  time = 0.89, size = 215, normalized size = 1.48 \begin {gather*} \frac {e^3 \sqrt {d+e x} \left (105 a^3 e^3+280 a^2 b e^2 (d+e x)-315 a^2 b d e^2+315 a b^2 d^2 e+231 a b^2 e (d+e x)^2-560 a b^2 d e (d+e x)-105 b^3 d^3+280 b^3 d^2 (d+e x)+48 b^3 (d+e x)^3-231 b^3 d (d+e x)^2\right )}{24 b^4 (a e+b (d+e x)-b d)^3}+\frac {35 e^3 \sqrt {a e-b d} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{8 b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^(7/2)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(e^3*Sqrt[d + e*x]*(-105*b^3*d^3 + 315*a*b^2*d^2*e - 315*a^2*b*d*e^2 + 105*a^3*e^3 + 280*b^3*d^2*(d + e*x) - 5
60*a*b^2*d*e*(d + e*x) + 280*a^2*b*e^2*(d + e*x) - 231*b^3*d*(d + e*x)^2 + 231*a*b^2*e*(d + e*x)^2 + 48*b^3*(d
+ e*x)^3))/(24*b^4*(-(b*d) + a*e + b*(d + e*x))^3) + (35*e^3*Sqrt[-(b*d) + a*e]*ArcTan[(Sqrt[b]*Sqrt[-(b*d) +
a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(8*b^(9/2))

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fricas [B]  time = 0.43, size = 498, normalized size = 3.43 \begin {gather*} \left [\frac {105 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (48 \, b^{3} e^{3} x^{3} - 8 \, b^{3} d^{3} - 14 \, a b^{2} d^{2} e - 35 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} - 3 \, {\left (29 \, b^{3} d e^{2} - 77 \, a b^{2} e^{3}\right )} x^{2} - 2 \, {\left (19 \, b^{3} d^{2} e + 49 \, a b^{2} d e^{2} - 140 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}, -\frac {105 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (48 \, b^{3} e^{3} x^{3} - 8 \, b^{3} d^{3} - 14 \, a b^{2} d^{2} e - 35 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} - 3 \, {\left (29 \, b^{3} d e^{2} - 77 \, a b^{2} e^{3}\right )} x^{2} - 2 \, {\left (19 \, b^{3} d^{2} e + 49 \, a b^{2} d e^{2} - 140 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

[1/48*(105*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d -
a*e - 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) + 2*(48*b^3*e^3*x^3 - 8*b^3*d^3 - 14*a*b^2*d^2*e - 35*
a^2*b*d*e^2 + 105*a^3*e^3 - 3*(29*b^3*d*e^2 - 77*a*b^2*e^3)*x^2 - 2*(19*b^3*d^2*e + 49*a*b^2*d*e^2 - 140*a^2*b
*e^3)*x)*sqrt(e*x + d))/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^5*x + a^3*b^4), -1/24*(105*(b^3*e^3*x^3 + 3*a*b^2*e^3
*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e))
- (48*b^3*e^3*x^3 - 8*b^3*d^3 - 14*a*b^2*d^2*e - 35*a^2*b*d*e^2 + 105*a^3*e^3 - 3*(29*b^3*d*e^2 - 77*a*b^2*e^
3)*x^2 - 2*(19*b^3*d^2*e + 49*a*b^2*d*e^2 - 140*a^2*b*e^3)*x)*sqrt(e*x + d))/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^
5*x + a^3*b^4)]

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giac [B]  time = 0.22, size = 248, normalized size = 1.71 \begin {gather*} \frac {35 \, {\left (b d e^{3} - a e^{4}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, \sqrt {-b^{2} d + a b e} b^{4}} + \frac {2 \, \sqrt {x e + d} e^{3}}{b^{4}} - \frac {87 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{3} d e^{3} - 136 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{3} d^{2} e^{3} + 57 \, \sqrt {x e + d} b^{3} d^{3} e^{3} - 87 \, {\left (x e + d\right )}^{\frac {5}{2}} a b^{2} e^{4} + 272 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{2} d e^{4} - 171 \, \sqrt {x e + d} a b^{2} d^{2} e^{4} - 136 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b e^{5} + 171 \, \sqrt {x e + d} a^{2} b d e^{5} - 57 \, \sqrt {x e + d} a^{3} e^{6}}{24 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{3} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

35/8*(b*d*e^3 - a*e^4)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^4) + 2*sqrt(x*e +
d)*e^3/b^4 - 1/24*(87*(x*e + d)^(5/2)*b^3*d*e^3 - 136*(x*e + d)^(3/2)*b^3*d^2*e^3 + 57*sqrt(x*e + d)*b^3*d^3*e
^3 - 87*(x*e + d)^(5/2)*a*b^2*e^4 + 272*(x*e + d)^(3/2)*a*b^2*d*e^4 - 171*sqrt(x*e + d)*a*b^2*d^2*e^4 - 136*(x
*e + d)^(3/2)*a^2*b*e^5 + 171*sqrt(x*e + d)*a^2*b*d*e^5 - 57*sqrt(x*e + d)*a^3*e^6)/(((x*e + d)*b - b*d + a*e)
^3*b^4)

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maple [B]  time = 0.07, size = 352, normalized size = 2.43 \begin {gather*} \frac {19 \sqrt {e x +d}\, a^{3} e^{6}}{8 \left (b e x +a e \right )^{3} b^{4}}-\frac {57 \sqrt {e x +d}\, a^{2} d \,e^{5}}{8 \left (b e x +a e \right )^{3} b^{3}}+\frac {57 \sqrt {e x +d}\, a \,d^{2} e^{4}}{8 \left (b e x +a e \right )^{3} b^{2}}-\frac {19 \sqrt {e x +d}\, d^{3} e^{3}}{8 \left (b e x +a e \right )^{3} b}+\frac {17 \left (e x +d \right )^{\frac {3}{2}} a^{2} e^{5}}{3 \left (b e x +a e \right )^{3} b^{3}}-\frac {34 \left (e x +d \right )^{\frac {3}{2}} a d \,e^{4}}{3 \left (b e x +a e \right )^{3} b^{2}}+\frac {17 \left (e x +d \right )^{\frac {3}{2}} d^{2} e^{3}}{3 \left (b e x +a e \right )^{3} b}+\frac {29 \left (e x +d \right )^{\frac {5}{2}} a \,e^{4}}{8 \left (b e x +a e \right )^{3} b^{2}}-\frac {29 \left (e x +d \right )^{\frac {5}{2}} d \,e^{3}}{8 \left (b e x +a e \right )^{3} b}-\frac {35 a \,e^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}\, b^{4}}+\frac {35 d \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}\, b^{3}}+\frac {2 \sqrt {e x +d}\, e^{3}}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e^3*(e*x+d)^(1/2)/b^4+29/8*e^4/b^2/(b*e*x+a*e)^3*(e*x+d)^(5/2)*a-29/8*e^3/b/(b*e*x+a*e)^3*(e*x+d)^(5/2)*d+17
/3*e^5/b^3/(b*e*x+a*e)^3*(e*x+d)^(3/2)*a^2-34/3*e^4/b^2/(b*e*x+a*e)^3*(e*x+d)^(3/2)*a*d+17/3*e^3/b/(b*e*x+a*e)
^3*(e*x+d)^(3/2)*d^2+19/8*e^6/b^4/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a^3-57/8*e^5/b^3/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a^2
*d+57/8*e^4/b^2/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a*d^2-19/8*e^3/b/(b*e*x+a*e)^3*(e*x+d)^(1/2)*d^3-35/8*e^4/b^4/((a*
e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a+35/8*e^3/b^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^
(1/2)/((a*e-b*d)*b)^(1/2)*b)*d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^2,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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [B]  time = 0.69, size = 303, normalized size = 2.09 \begin {gather*} \frac {\sqrt {d+e\,x}\,\left (\frac {19\,a^3\,e^6}{8}-\frac {57\,a^2\,b\,d\,e^5}{8}+\frac {57\,a\,b^2\,d^2\,e^4}{8}-\frac {19\,b^3\,d^3\,e^3}{8}\right )+\left (\frac {29\,a\,b^2\,e^4}{8}-\frac {29\,b^3\,d\,e^3}{8}\right )\,{\left (d+e\,x\right )}^{5/2}+{\left (d+e\,x\right )}^{3/2}\,\left (\frac {17\,a^2\,b\,e^5}{3}-\frac {34\,a\,b^2\,d\,e^4}{3}+\frac {17\,b^3\,d^2\,e^3}{3}\right )}{b^7\,{\left (d+e\,x\right )}^3-\left (3\,b^7\,d-3\,a\,b^6\,e\right )\,{\left (d+e\,x\right )}^2+\left (d+e\,x\right )\,\left (3\,a^2\,b^5\,e^2-6\,a\,b^6\,d\,e+3\,b^7\,d^2\right )-b^7\,d^3+a^3\,b^4\,e^3-3\,a^2\,b^5\,d\,e^2+3\,a\,b^6\,d^2\,e}+\frac {2\,e^3\,\sqrt {d+e\,x}}{b^4}-\frac {35\,e^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^3\,\sqrt {a\,e-b\,d}\,\sqrt {d+e\,x}}{a\,e^4-b\,d\,e^3}\right )\,\sqrt {a\,e-b\,d}}{8\,b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d + e*x)^(1/2)*((19*a^3*e^6)/8 - (19*b^3*d^3*e^3)/8 + (57*a*b^2*d^2*e^4)/8 - (57*a^2*b*d*e^5)/8) + ((29*a*b^
2*e^4)/8 - (29*b^3*d*e^3)/8)*(d + e*x)^(5/2) + (d + e*x)^(3/2)*((17*a^2*b*e^5)/3 + (17*b^3*d^2*e^3)/3 - (34*a*
b^2*d*e^4)/3))/(b^7*(d + e*x)^3 - (3*b^7*d - 3*a*b^6*e)*(d + e*x)^2 + (d + e*x)*(3*b^7*d^2 + 3*a^2*b^5*e^2 - 6
*a*b^6*d*e) - b^7*d^3 + a^3*b^4*e^3 - 3*a^2*b^5*d*e^2 + 3*a*b^6*d^2*e) + (2*e^3*(d + e*x)^(1/2))/b^4 - (35*e^3
*atan((b^(1/2)*e^3*(a*e - b*d)^(1/2)*(d + e*x)^(1/2))/(a*e^4 - b*d*e^3))*(a*e - b*d)^(1/2))/(8*b^(9/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(7/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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