3.1041 \(\int \frac {\cos (c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx\)
Optimal. Leaf size=258 \[ \frac {2 \left (8 a^2 C-10 a b B+15 A b^2+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (-8 a^3 C+10 a^2 b B-a b^2 (15 A+7 C)+5 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 b B-4 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 b^2 d}+\frac {2 C \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d} \]
[Out]
2/15*(5*B*b-4*C*a)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^2/d+2/5*C*cos(d*x+c)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/
b/d+2/15*(15*A*b^2-10*B*a*b+8*C*a^2+9*C*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2
*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/b^3/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2/15*(10*a^2*
b*B+5*b^3*B-8*a^3*C-a*b^2*(15*A+7*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/
2*c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/b^3/d/(a+b*cos(d*x+c))^(1/2)
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Rubi [A] time = 0.42, antiderivative size = 258, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used =
{3049, 3023, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 \left (10 a^2 b B-8 a^3 C-a b^2 (15 A+7 C)+5 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^2 C-10 a b B+15 A b^2+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (5 b B-4 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 b^2 d}+\frac {2 C \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Sqrt[a + b*Cos[c + d*x]],x]
[Out]
(2*(15*A*b^2 - 10*a*b*B + 8*a^2*C + 9*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(
15*b^3*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(10*a^2*b*B + 5*b^3*B - 8*a^3*C - a*b^2*(15*A + 7*C))*Sqrt[(
a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(15*b^3*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(
5*b*B - 4*a*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*b^2*d) + (2*C*Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*
Sin[c + d*x])/(5*b*d)
Rule 2653
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 2655
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] && !GtQ[a + b, 0]
Rule 2661
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 2663
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && !GtQ[a + b, 0]
Rule 2752
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 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])))
Rubi steps
\begin {align*} \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx &=\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d}+\frac {2 \int \frac {a C+\frac {1}{2} b (5 A+3 C) \cos (c+d x)+\frac {1}{2} (5 b B-4 a C) \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{5 b}\\ &=\frac {2 (5 b B-4 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d}+\frac {4 \int \frac {\frac {1}{4} b (5 b B+2 a C)+\frac {1}{4} \left (15 A b^2-10 a b B+8 a^2 C+9 b^2 C\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^2}\\ &=\frac {2 (5 b B-4 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d}+\frac {\left (15 A b^2-10 a b B+8 a^2 C+9 b^2 C\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{15 b^3}+\frac {\left (10 a^2 b B+5 b^3 B-8 a^3 C-a b^2 (15 A+7 C)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^3}\\ &=\frac {2 (5 b B-4 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d}+\frac {\left (\left (15 A b^2-10 a b B+8 a^2 C+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{15 b^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (\left (10 a^2 b B+5 b^3 B-8 a^3 C-a b^2 (15 A+7 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{15 b^3 \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 \left (15 A b^2-10 a b B+8 a^2 C+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (10 a^2 b B+5 b^3 B-8 a^3 C-a b^2 (15 A+7 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 b B-4 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d}\\ \end {align*}
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Mathematica [A] time = 1.09, size = 186, normalized size = 0.72 \[ \frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (\left (8 a^2 C-10 a b B+15 A b^2+9 b^2 C\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )+b^2 (2 a C+5 b B) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )+2 b \sin (c+d x) (a+b \cos (c+d x)) (-4 a C+5 b B+3 b C \cos (c+d x))}{15 b^3 d \sqrt {a+b \cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Sqrt[a + b*Cos[c + d*x]],x]
[Out]
(2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(5*b*B + 2*a*C)*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + (15*A*b^2 -
10*a*b*B + 8*a^2*C + 9*b^2*C)*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)
/(a + b)])) + 2*b*(a + b*Cos[c + d*x])*(5*b*B - 4*a*C + 3*b*C*Cos[c + d*x])*Sin[c + d*x])/(15*b^3*d*Sqrt[a + b
*Cos[c + d*x]])
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fricas [F] time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C \cos \left (d x + c\right )^{3} + B \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )}{\sqrt {b \cos \left (d x + c\right ) + a}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
integral((C*cos(d*x + c)^3 + B*cos(d*x + c)^2 + A*cos(d*x + c))/sqrt(b*cos(d*x + c) + a), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{\sqrt {b \cos \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)/sqrt(b*cos(d*x + c) + a), x)
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maple [B] time = 3.51, size = 1258, normalized size = 4.88 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x)
[Out]
2/15*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-10*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1
/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b+15*A*a*b^2*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1
/2))-6*C*cos(1/2*d*x+1/2*c)^3*a*b^2+8*C*cos(1/2*d*x+1/2*c)^3*a^2*b-10*B*cos(1/2*d*x+1/2*c)^3*a*b^2+10*B*cos(1/
2*d*x+1/2*c)*a*b^2+4*C*cos(1/2*d*x+1/2*c)^5*a*b^2-8*C*cos(1/2*d*x+1/2*c)*a^2*b+2*C*cos(1/2*d*x+1/2*c)*a*b^2+6*
C*cos(1/2*d*x+1/2*c)*b^3+48*C*cos(1/2*d*x+1/2*c)^5*b^3-30*C*cos(1/2*d*x+1/2*c)^3*b^3-24*C*cos(1/2*d*x+1/2*c)^7
*b^3-20*B*cos(1/2*d*x+1/2*c)^5*b^3+30*B*cos(1/2*d*x+1/2*c)^3*b^3-10*B*cos(1/2*d*x+1/2*c)*b^3+8*C*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*
a^3-8*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c)
,(-2*b/(a-b))^(1/2))*a^3+15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*Ellipt
icE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-5*B*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b
+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+9*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1
/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+10*B*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2
*b-10*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c)
,(-2*b/(a-b))^(1/2))*a*b^2+7*a*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*Ell
ipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^2+8*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+
a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b-9*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*
cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-15*A*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/
2))*a*b^2)/b^3/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x
+1/2*c)^2*b+a+b)^(1/2)/d
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{\sqrt {b \cos \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)/sqrt(b*cos(d*x + c) + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\cos \left (c+d\,x\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^(1/2),x)
[Out]
int((cos(c + d*x)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^(1/2), 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(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(1/2),x)
[Out]
Timed out
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