3.574 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\)
Optimal. Leaf size=182 \[ \frac {2 b^3 (A b-a B) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^4 d (a+b)}-\frac {2 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a^2 d}-\frac {2 \left (a^2+3 b^2\right ) (A b-a B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^4 d}+\frac {2 \left (3 a^2 A-5 a b B+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d} \]
[Out]
2/5*(3*A*a^2+5*A*b^2-5*B*a*b)*(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))/a^3/d-2/3*(a^2+3*b^2)*(A*b-B*a)*(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))/a^4/d+2*b^3*(A*b-B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2
*c),2*a/(a+b),2^(1/2))/a^4/(a+b)/d+2/5*A*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d-2/3*(A*b-B*a)*sin(d*x+c)*cos(d*x+c)^(
1/2)/a^2/d
________________________________________________________________________________________
Rubi [A] time = 0.86, antiderivative size = 182, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used =
{2954, 2990, 3049, 3059, 2639, 3002, 2641, 2805} \[ -\frac {2 \left (a^2+3 b^2\right ) (A b-a B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^4 d}+\frac {2 \left (3 a^2 A-5 a b B+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d}+\frac {2 b^3 (A b-a B) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^4 d (a+b)}-\frac {2 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a^2 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^(5/2)*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x]),x]
[Out]
(2*(3*a^2*A + 5*A*b^2 - 5*a*b*B)*EllipticE[(c + d*x)/2, 2])/(5*a^3*d) - (2*(a^2 + 3*b^2)*(A*b - a*B)*EllipticF
[(c + d*x)/2, 2])/(3*a^4*d) + (2*b^3*(A*b - a*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a^4*(a + b)*d) -
(2*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a^2*d) + (2*A*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*a*d)
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rule 2641
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]
Rule 2805
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Rule 2954
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]
Rule 2990
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x
])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c -
b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m,
1] && !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Rule 3002
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
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])))
Rule 3059
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]
Rubi steps
\begin {align*} \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx &=\int \frac {\cos ^{\frac {5}{2}}(c+d x) (B+A \cos (c+d x))}{b+a \cos (c+d x)} \, dx\\ &=\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {2 \int \frac {\sqrt {\cos (c+d x)} \left (\frac {3 A b}{2}+\frac {3}{2} a A \cos (c+d x)-\frac {5}{2} (A b-a B) \cos ^2(c+d x)\right )}{b+a \cos (c+d x)} \, dx}{5 a}\\ &=-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {4 \int \frac {-\frac {5}{4} b (A b-a B)+\frac {1}{4} a (4 A b+5 a B) \cos (c+d x)+\frac {3}{4} \left (3 a^2 A+5 A b^2-5 a b B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{15 a^2}\\ &=-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {4 \int \frac {\frac {5}{4} a b (A b-a B)+\frac {5}{4} \left (a^2+3 b^2\right ) (A b-a B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{15 a^3}+\frac {\left (3 a^2 A+5 A b^2-5 a b B\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 a^3}\\ &=\frac {2 \left (3 a^2 A+5 A b^2-5 a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {\left (b^3 (A b-a B)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a^4}-\frac {\left (\left (a^2+3 b^2\right ) (A b-a B)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 a^4}\\ &=\frac {2 \left (3 a^2 A+5 A b^2-5 a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac {2 \left (a^2+3 b^2\right ) (A b-a B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^4 d}+\frac {2 b^3 (A b-a B) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^4 (a+b) d}-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 2.80, size = 260, normalized size = 1.43 \[ \frac {\frac {2 a^2 \left (9 a^2 A-5 a b B+5 A b^2\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {6 \left (3 a^2 A-5 a b B+5 A b^2\right ) \sin (c+d x) \left (\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{b \sqrt {\sin ^2(c+d x)}}+4 a^2 \sin (c+d x) \sqrt {\cos (c+d x)} (3 a A \cos (c+d x)+5 a B-5 A b)+2 a^2 (5 a B+4 A b) \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )}{30 a^4 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^(5/2)*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x]),x]
[Out]
((2*a^2*(9*a^2*A + 5*A*b^2 - 5*a*b*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + 2*a^2*(4*A*b + 5*a*
B)*(2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)) + 4*a^2*Sqrt[Cos[c
+ d*x]]*(-5*A*b + 5*a*B + 3*a*A*Cos[c + d*x])*Sin[c + d*x] + (6*(3*a^2*A + 5*A*b^2 - 5*a*b*B)*(-2*a*b*Elliptic
E[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*Elli
pticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(b*Sqrt[Sin[c + d*x]^2]))/(30*a^4*d)
________________________________________________________________________________________
fricas [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)^(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="fricas")
[Out]
Timed out
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \sec \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*cos(d*x + c)^(5/2)/(b*sec(d*x + c) + a), x)
________________________________________________________________________________________
maple [B] time = 5.94, size = 1074, normalized size = 5.90 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x)
[Out]
-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((-24*A*a^4+24*A*a^3*b)*cos(1/2*d*x+1/2*c)*sin(1
/2*d*x+1/2*c)^6+(24*A*a^4-44*A*a^3*b+20*A*a^2*b^2+20*B*a^4-20*B*a^3*b)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)
+(-6*A*a^4+16*A*a^3*b-10*A*a^2*b^2-10*B*a^4+10*B*a^3*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-9*A*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^4+9*A*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b-15*A*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2+15*A*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3-15*A*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*b^4-5*
A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b+5*
A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2-
15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3
+15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^4+
15*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b
-15*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*
b^2+15*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b)
,2^(1/2))*a*b^3+5*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c)
,2^(1/2))*a^4-5*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2
^(1/2))*a^3*b+15*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),
2^(1/2))*a^2*b^2-15*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*
c),2^(1/2))*a*b^3)/a^4/(a-b)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/
2*d*x+1/2*c)^2-1)^(1/2)/d
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \sec \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="maxima")
[Out]
integrate((B*sec(d*x + c) + A)*cos(d*x + c)^(5/2)/(b*sec(d*x + c) + a), x)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^(5/2)*(A + B/cos(c + d*x)))/(a + b/cos(c + d*x)),x)
[Out]
int((cos(c + d*x)^(5/2)*(A + B/cos(c + d*x)))/(a + b/cos(c + d*x)), x)
________________________________________________________________________________________
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)**(5/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x)
[Out]
Timed out
________________________________________________________________________________________