3.535 \(\int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=179 \[ -\frac {a^{3/2} (c+7 d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{4 d^{3/2} f (c+d)^{5/2}}-\frac {a^2 (c+7 d) \cos (e+f x)}{4 d f (c+d)^2 \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}+\frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2} \]

[Out]

-1/4*a^(3/2)*(c+7*d)*arctanh(cos(f*x+e)*a^(1/2)*d^(1/2)/(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))/d^(3/2)/(c+d)^(5/2
)/f+1/2*a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2)-1/4*a^2*(c+7*d)*cos(f*x+e)/d/
(c+d)^2/f/(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2762, 21, 2772, 2773, 208} \[ -\frac {a^{3/2} (c+7 d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{4 d^{3/2} f (c+d)^{5/2}}-\frac {a^2 (c+7 d) \cos (e+f x)}{4 d f (c+d)^2 \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}+\frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(3/2)/(c + d*Sin[e + f*x])^3,x]

[Out]

-(a^(3/2)*(c + 7*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(4*d^(3/2)
*(c + d)^(5/2)*f) + (a^2*(c - d)*Cos[e + f*x])/(2*d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^2)
- (a^2*(c + 7*d)*Cos[e + f*x])/(4*d*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x]))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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]

Rule 2762

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c
+ a*d)), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*
Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
&& (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 2773

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

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^3} \, dx &=\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {a \int \frac {-\frac {1}{2} a (c+7 d)-\frac {1}{2} a (c+7 d) \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx}{2 d (c+d)}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}+\frac {(a (c+7 d)) \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^2} \, dx}{4 d (c+d)}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {a^2 (c+7 d) \cos (e+f x)}{4 d (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac {(a (c+7 d)) \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{8 d (c+d)^2}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {a^2 (c+7 d) \cos (e+f x)}{4 d (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac {\left (a^2 (c+7 d)\right ) \operatorname {Subst}\left (\int \frac {1}{a c+a d-d x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{4 d (c+d)^2 f}\\ &=-\frac {a^{3/2} (c+7 d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 d^{3/2} (c+d)^{5/2} f}+\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {a^2 (c+7 d) \cos (e+f x)}{4 d (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 4.06, size = 313, normalized size = 1.75 \[ \frac {(a (\sin (e+f x)+1))^{3/2} \left (\frac {4 \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right ) \left (-c^2+d (c+7 d) \sin (e+f x)+7 c d+2 d^2\right )}{(c+d)^2 (c+d \sin (e+f x))^2}-\frac {4 \sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right ) \left (-c^2+d (c+7 d) \sin (e+f x)+7 c d+2 d^2\right )}{(c+d)^2 (c+d \sin (e+f x))^2}-\frac {2 (c+7 d) \left (\log \left (-\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (-\sqrt {d} \sqrt {c+d} \sin \left (\frac {1}{2} (e+f x)\right )+\sqrt {d} \sqrt {c+d} \cos \left (\frac {1}{2} (e+f x)\right )+c+d\right )\right )-\log \left (\sqrt {d} \sqrt {c+d} \left (\tan ^2\left (\frac {1}{4} (e+f x)\right )+2 \tan \left (\frac {1}{4} (e+f x)\right )-1\right )+(c+d) \sec ^2\left (\frac {1}{4} (e+f x)\right )\right )\right )}{(c+d)^{5/2}}\right )}{16 d^{3/2} f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(3/2)/(c + d*Sin[e + f*x])^3,x]

[Out]

((a*(1 + Sin[e + f*x]))^(3/2)*((-2*(c + 7*d)*(Log[-(Sec[(e + f*x)/4]^2*(c + d + Sqrt[d]*Sqrt[c + d]*Cos[(e + f
*x)/2] - Sqrt[d]*Sqrt[c + d]*Sin[(e + f*x)/2]))] - Log[(c + d)*Sec[(e + f*x)/4]^2 + Sqrt[d]*Sqrt[c + d]*(-1 +
2*Tan[(e + f*x)/4] + Tan[(e + f*x)/4]^2)]))/(c + d)^(5/2) - (4*Sqrt[d]*Cos[(e + f*x)/2]*(-c^2 + 7*c*d + 2*d^2
+ d*(c + 7*d)*Sin[e + f*x]))/((c + d)^2*(c + d*Sin[e + f*x])^2) + (4*Sqrt[d]*Sin[(e + f*x)/2]*(-c^2 + 7*c*d +
2*d^2 + d*(c + 7*d)*Sin[e + f*x]))/((c + d)^2*(c + d*Sin[e + f*x])^2)))/(16*d^(3/2)*f*(Cos[(e + f*x)/2] + Sin[
(e + f*x)/2])^3)

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fricas [B]  time = 0.66, size = 1558, normalized size = 8.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

[-1/16*((a*c^3 + 9*a*c^2*d + 15*a*c*d^2 + 7*a*d^3 - (a*c*d^2 + 7*a*d^3)*cos(f*x + e)^3 - (2*a*c^2*d + 15*a*c*d
^2 + 7*a*d^3)*cos(f*x + e)^2 + (a*c^3 + 7*a*c^2*d + a*c*d^2 + 7*a*d^3)*cos(f*x + e) + (a*c^3 + 9*a*c^2*d + 15*
a*c*d^2 + 7*a*d^3 - (a*c*d^2 + 7*a*d^3)*cos(f*x + e)^2 + 2*(a*c^2*d + 7*a*c*d^2)*cos(f*x + e))*sin(f*x + e))*s
qrt(a/(c*d + d^2))*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 +
4*(c^2*d + 4*c*d^2 + 3*d^3 - (c*d^2 + d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e) - (c^2*d +
4*c*d^2 + 3*d^3 + (c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(a/(c*d + d^2)) - (a*
c^2 + 8*a*c*d + 9*a*d^2)*cos(f*x + e) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2
)*cos(f*x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 +
d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4*(a*c^2 -
6*a*c*d + 5*a*d^2 - (a*c*d + 7*a*d^2)*cos(f*x + e)^2 + (a*c^2 - 7*a*c*d - 2*a*d^2)*cos(f*x + e) - (a*c^2 - 6*a
*c*d + 5*a*d^2 + (a*c*d + 7*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c^2*d^3 + 2*c*d^4 +
d^5)*f*cos(f*x + e)^3 + (2*c^3*d^2 + 5*c^2*d^3 + 4*c*d^4 + d^5)*f*cos(f*x + e)^2 - (c^4*d + 2*c^3*d^2 + 2*c^2
*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e) - (c^4*d + 4*c^3*d^2 + 6*c^2*d^3 + 4*c*d^4 + d^5)*f + ((c^2*d^3 + 2*c*d^4
+ d^5)*f*cos(f*x + e)^2 - 2*(c^3*d^2 + 2*c^2*d^3 + c*d^4)*f*cos(f*x + e) - (c^4*d + 4*c^3*d^2 + 6*c^2*d^3 + 4
*c*d^4 + d^5)*f)*sin(f*x + e)), 1/8*((a*c^3 + 9*a*c^2*d + 15*a*c*d^2 + 7*a*d^3 - (a*c*d^2 + 7*a*d^3)*cos(f*x +
e)^3 - (2*a*c^2*d + 15*a*c*d^2 + 7*a*d^3)*cos(f*x + e)^2 + (a*c^3 + 7*a*c^2*d + a*c*d^2 + 7*a*d^3)*cos(f*x +
e) + (a*c^3 + 9*a*c^2*d + 15*a*c*d^2 + 7*a*d^3 - (a*c*d^2 + 7*a*d^3)*cos(f*x + e)^2 + 2*(a*c^2*d + 7*a*c*d^2)*
cos(f*x + e))*sin(f*x + e))*sqrt(-a/(c*d + d^2))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d
)*sqrt(-a/(c*d + d^2))/(a*cos(f*x + e))) - 2*(a*c^2 - 6*a*c*d + 5*a*d^2 - (a*c*d + 7*a*d^2)*cos(f*x + e)^2 + (
a*c^2 - 7*a*c*d - 2*a*d^2)*cos(f*x + e) - (a*c^2 - 6*a*c*d + 5*a*d^2 + (a*c*d + 7*a*d^2)*cos(f*x + e))*sin(f*x
+ e))*sqrt(a*sin(f*x + e) + a))/((c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e)^3 + (2*c^3*d^2 + 5*c^2*d^3 + 4*c*d^
4 + d^5)*f*cos(f*x + e)^2 - (c^4*d + 2*c^3*d^2 + 2*c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e) - (c^4*d + 4*c^3*d^
2 + 6*c^2*d^3 + 4*c*d^4 + d^5)*f + ((c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e)^2 - 2*(c^3*d^2 + 2*c^2*d^3 + c*d^
4)*f*cos(f*x + e) - (c^4*d + 4*c^3*d^2 + 6*c^2*d^3 + 4*c*d^4 + d^5)*f)*sin(f*x + e))]

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giac [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*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^3,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 1.50, size = 429, normalized size = 2.40 \[ \frac {\left (-\arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) a^{2} c \,d^{2}-7 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) a^{2} d^{3}-2 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sin \left (f x +e \right ) a^{2} c^{2} d -14 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sin \left (f x +e \right ) a^{2} c \,d^{2}+\left (-a \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, c d +7 \left (-a \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, d^{2}-\arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{2} c^{3}-7 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{2} c^{2} d +\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\, a \,c^{2}-8 \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\, a c d -9 \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\, a \,d^{2}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (1+\sin \left (f x +e \right )\right )}{4 \sqrt {a \left (c +d \right ) d}\, \left (c +d \sin \left (f x +e \right )\right )^{2} \left (c +d \right )^{2} d \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^3,x)

[Out]

1/4*(-arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)^2*a^2*c*d^2-7*arctanh((-a*(sin(f*x+e)-
1))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)^2*a^2*d^3-2*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*s
in(f*x+e)*a^2*c^2*d-14*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)*a^2*c*d^2+(-a*(sin(f*
x+e)-1))^(3/2)*(a*(c+d)*d)^(1/2)*c*d+7*(-a*(sin(f*x+e)-1))^(3/2)*(a*(c+d)*d)^(1/2)*d^2-arctanh((-a*(sin(f*x+e)
-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*c^3-7*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*c^2*d+(-a
*(sin(f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*a*c^2-8*(-a*(sin(f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*a*c*d-9*(-a*(sin(
f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*a*d^2)*(-a*(sin(f*x+e)-1))^(1/2)*(1+sin(f*x+e))/(a*(c+d)*d)^(1/2)/(c+d*sin(
f*x+e))^2/(c+d)^2/d/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)/(d*sin(f*x + e) + c)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^(3/2)/(c + d*sin(e + f*x))^3,x)

[Out]

int((a + a*sin(e + f*x))^(3/2)/(c + d*sin(e + f*x))^3, 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((a+a*sin(f*x+e))**(3/2)/(c+d*sin(f*x+e))**3,x)

[Out]

Timed out

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