3.865 \(\int \frac {e^{-3 \tanh ^{-1}(a+b x)}}{x^2} \, dx\)

Optimal. Leaf size=130 \[ \frac {6 (1-a) b \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{(a+1)^2 \sqrt {1-a^2}}-\frac {(-a-b x+1)^{3/2}}{(a+1) x \sqrt {a+b x+1}}-\frac {6 b \sqrt {-a-b x+1}}{(a+1)^2 \sqrt {a+b x+1}} \]

[Out]

6*(1-a)*b*arctanh((1-a)^(1/2)*(b*x+a+1)^(1/2)/(1+a)^(1/2)/(-b*x-a+1)^(1/2))/(1+a)^2/(-a^2+1)^(1/2)-(-b*x-a+1)^
(3/2)/(1+a)/x/(b*x+a+1)^(1/2)-6*b*(-b*x-a+1)^(1/2)/(1+a)^2/(b*x+a+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6163, 94, 93, 208} \[ \frac {6 (1-a) b \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{(a+1)^2 \sqrt {1-a^2}}-\frac {(-a-b x+1)^{3/2}}{(a+1) x \sqrt {a+b x+1}}-\frac {6 b \sqrt {-a-b x+1}}{(a+1)^2 \sqrt {a+b x+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a + b*x])*x^2),x]

[Out]

(-6*b*Sqrt[1 - a - b*x])/((1 + a)^2*Sqrt[1 + a + b*x]) - (1 - a - b*x)^(3/2)/((1 + a)*x*Sqrt[1 + a + b*x]) + (
6*(1 - a)*b*ArcTanh[(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 + a]*Sqrt[1 - a - b*x])])/((1 + a)^2*Sqrt[1 - a^2]
)

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

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 6163

Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(1
+ a*c + b*c*x)^(n/2))/(1 - a*c - b*c*x)^(n/2), x] /; FreeQ[{a, b, c, d, e, m, n}, x]

Rubi steps

\begin {align*} \int \frac {e^{-3 \tanh ^{-1}(a+b x)}}{x^2} \, dx &=\int \frac {(1-a-b x)^{3/2}}{x^2 (1+a+b x)^{3/2}} \, dx\\ &=-\frac {(1-a-b x)^{3/2}}{(1+a) x \sqrt {1+a+b x}}-\frac {(3 b) \int \frac {\sqrt {1-a-b x}}{x (1+a+b x)^{3/2}} \, dx}{1+a}\\ &=-\frac {6 b \sqrt {1-a-b x}}{(1+a)^2 \sqrt {1+a+b x}}-\frac {(1-a-b x)^{3/2}}{(1+a) x \sqrt {1+a+b x}}-\frac {(3 (1-a) b) \int \frac {1}{x \sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx}{(1+a)^2}\\ &=-\frac {6 b \sqrt {1-a-b x}}{(1+a)^2 \sqrt {1+a+b x}}-\frac {(1-a-b x)^{3/2}}{(1+a) x \sqrt {1+a+b x}}-\frac {(6 (1-a) b) \operatorname {Subst}\left (\int \frac {1}{-1-a-(-1+a) x^2} \, dx,x,\frac {\sqrt {1+a+b x}}{\sqrt {1-a-b x}}\right )}{(1+a)^2}\\ &=-\frac {6 b \sqrt {1-a-b x}}{(1+a)^2 \sqrt {1+a+b x}}-\frac {(1-a-b x)^{3/2}}{(1+a) x \sqrt {1+a+b x}}+\frac {6 (1-a) b \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {1+a+b x}}{\sqrt {1+a} \sqrt {1-a-b x}}\right )}{(1+a)^2 \sqrt {1-a^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 106, normalized size = 0.82 \[ \frac {\sqrt {-a-b x+1} \left (a^2+a b x-5 b x-1\right )}{(a+1)^2 x \sqrt {a+b x+1}}+\frac {6 \sqrt {a-1} b \tanh ^{-1}\left (\frac {\sqrt {-a-1} \sqrt {-a-b x+1}}{\sqrt {a-1} \sqrt {a+b x+1}}\right )}{(-a-1)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(3*ArcTanh[a + b*x])*x^2),x]

[Out]

(Sqrt[1 - a - b*x]*(-1 + a^2 - 5*b*x + a*b*x))/((1 + a)^2*x*Sqrt[1 + a + b*x]) + (6*Sqrt[-1 + a]*b*ArcTanh[(Sq
rt[-1 - a]*Sqrt[1 - a - b*x])/(Sqrt[-1 + a]*Sqrt[1 + a + b*x])])/(-1 - a)^(5/2)

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 369, normalized size = 2.84 \[ \left [\frac {3 \, {\left (b^{2} x^{2} + {\left (a + 1\right )} b x\right )} \sqrt {-\frac {a - 1}{a + 1}} \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 4 \, a^{2} - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{3} + {\left (a^{2} + a\right )} b x + a^{2} - a - 1\right )} \sqrt {-\frac {a - 1}{a + 1}} + 2}{x^{2}}\right ) + 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left ({\left (a - 5\right )} b x + a^{2} - 1\right )}}{2 \, {\left ({\left (a^{2} + 2 \, a + 1\right )} b x^{2} + {\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} x\right )}}, \frac {3 \, {\left (b^{2} x^{2} + {\left (a + 1\right )} b x\right )} \sqrt {\frac {a - 1}{a + 1}} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {\frac {a - 1}{a + 1}}}{{\left (a - 1\right )} b^{2} x^{2} + a^{3} + 2 \, {\left (a^{2} - a\right )} b x - a^{2} - a + 1}\right ) + \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left ({\left (a - 5\right )} b x + a^{2} - 1\right )}}{{\left (a^{2} + 2 \, a + 1\right )} b x^{2} + {\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a+1)^3*(1-(b*x+a)^2)^(3/2)/x^2,x, algorithm="fricas")

[Out]

[1/2*(3*(b^2*x^2 + (a + 1)*b*x)*sqrt(-(a - 1)/(a + 1))*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^3 - a)*b*x - 4*
a^2 - 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a^3 + (a^2 + a)*b*x + a^2 - a - 1)*sqrt(-(a - 1)/(a + 1)) + 2)/x^2
) + 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*((a - 5)*b*x + a^2 - 1))/((a^2 + 2*a + 1)*b*x^2 + (a^3 + 3*a^2 + 3*a
+ 1)*x), (3*(b^2*x^2 + (a + 1)*b*x)*sqrt((a - 1)/(a + 1))*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a
^2 - 1)*sqrt((a - 1)/(a + 1))/((a - 1)*b^2*x^2 + a^3 + 2*(a^2 - a)*b*x - a^2 - a + 1)) + sqrt(-b^2*x^2 - 2*a*b
*x - a^2 + 1)*((a - 5)*b*x + a^2 - 1))/((a^2 + 2*a + 1)*b*x^2 + (a^3 + 3*a^2 + 3*a + 1)*x)]

________________________________________________________________________________________

giac [B]  time = 0.21, size = 604, normalized size = 4.65 \[ \frac {6 \, {\left (a b^{2} - b^{2}\right )} \arctan \left (\frac {\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt {a^{2} - 1}}\right )}{{\left (a^{2} {\left | b \right |} + 2 \, a {\left | b \right |} + {\left | b \right |}\right )} \sqrt {a^{2} - 1}} + \frac {2 \, {\left (\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a^{2} b^{2}}{b^{2} x + a b} + \frac {4 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2} a^{2} b^{2}}{{\left (b^{2} x + a b\right )}^{2}} + 5 \, a^{2} b^{2} - \frac {10 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a b^{2}}{b^{2} x + a b} - \frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2} a b^{2}}{{\left (b^{2} x + a b\right )}^{2}} - a b^{2} + \frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} b^{2}}{b^{2} x + a b} + \frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2} b^{2}}{{\left (b^{2} x + a b\right )}^{2}}\right )}}{{\left (a^{3} {\left | b \right |} + 2 \, a^{2} {\left | b \right |} + a {\left | b \right |}\right )} {\left (\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} + \frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2} a}{{\left (b^{2} x + a b\right )}^{2}} + \frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{3} a}{{\left (b^{2} x + a b\right )}^{3}} + a - \frac {2 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}}{b^{2} x + a b} - \frac {2 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2}}{{\left (b^{2} x + a b\right )}^{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a+1)^3*(1-(b*x+a)^2)^(3/2)/x^2,x, algorithm="giac")

[Out]

6*(a*b^2 - b^2)*arctan(((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a/(b^2*x + a*b) - 1)/sqrt(a^2 - 1))/((
a^2*abs(b) + 2*a*abs(b) + abs(b))*sqrt(a^2 - 1)) + 2*((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a^2*b^2/
(b^2*x + a*b) + 4*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a^2*b^2/(b^2*x + a*b)^2 + 5*a^2*b^2 - 10*(
sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a*b^2/(b^2*x + a*b) - (sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b
) + b)^2*a*b^2/(b^2*x + a*b)^2 - a*b^2 + (sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*b^2/(b^2*x + a*b) + (
sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*b^2/(b^2*x + a*b)^2)/((a^3*abs(b) + 2*a^2*abs(b) + a*abs(b))*
((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a/(b^2*x + a*b) + (sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b)
+ b)^2*a/(b^2*x + a*b)^2 + (sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a/(b^2*x + a*b)^3 + a - 2*(sqrt(-
b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)/(b^2*x + a*b) - 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2/(
b^2*x + a*b)^2))

________________________________________________________________________________________

maple [B]  time = 0.06, size = 1710, normalized size = 13.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a+1)^3*(1-(b*x+a)^2)^(3/2)/x^2,x)

[Out]

-3/(1+a)^3*a^4*b^2/(-a^2+1)/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))+9/2/(1+a)^3
*a^2*b^2/(-a^2+1)/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))+3/2/(1+a)^3*a^2*b^2/(
-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x-3/(1+a)^2*b^2/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1+a)/b-1/b)/(-(x+(1+
a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2))+2/(1+a)^3/b/(x+1/b+a/b)^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)+3/(1+
a)^3*b^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*x+3/(1+a)^3*b*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a
-1/(1+a)^2/b^2/(x+1/b+a/b)^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)-2/(1+a)^2/b/(x+1/b+a/b)^2*(-(x+(1+a)/b
)^2*b^2+2*b*(x+(1+a)/b))^(5/2)-3/(1+a)^2*b^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*x-3/(1+a)^2*b*(-(x+(1+
a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a+3/2/(1+a)^4*b^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*x+3/2/(1+a)^4*
b*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a+3/2/(1+a)^4*b^2/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1+a)/b-1/b)/
(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2))-1/(1+a)^3/(-a^2+1)/x*(-b^2*x^2-2*a*b*x-a^2+1)^(5/2)+9/2/(1+a)^4*b*
a^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+3/(1+a)^4*b*(-a^2+1)^(1/2)*ln((-2*a^2+2-2*a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-
2*a*b*x-a^2+1)^(1/2))/x)+3/(1+a)^3*b^2/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1+a)/b-1/b)/(-(x+(1+a)/b)^2*b^2+2*b*
(x+(1+a)/b))^(1/2))+1/(1+a)^4*b*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/2)-1/(1+a)^4*b*(-b^2*x^2-2*a*b*x-a^2+1
)^(3/2)-3/(1+a)^4*b*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-2/(1+a)^2*b*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/2)+2/(1
+a)^3*b*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/2)+3/2/(1+a)^4*b^2*a*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x+9/2/(1+a
)^4*b^2*a/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))-3/(1+a)^4*b^2*a^3/(b^2)^(1/2)
*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))-3/(1+a)^4*b*(-a^2+1)^(1/2)*ln((-2*a^2+2-2*a*b*x+2*
(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)*a^2-2/(1+a)^3*a*b/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(3/2)+9/
2/(1+a)^3*a^3*b/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-9/2/(1+a)^3*a*b/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2
)-3/(1+a)^3*a^3*b/(-a^2+1)^(1/2)*ln((-2*a^2+2-2*a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)+3/(1
+a)^3*a*b/(-a^2+1)^(1/2)*ln((-2*a^2+2-2*a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)-1/(1+a)^3*b^
2/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(3/2)*x-3/2/(1+a)^3*b^2/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x-3/2/(1+a
)^3*b^2/(-a^2+1)/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}}}{{\left (b x + a + 1\right )}^{3} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a+1)^3*(1-(b*x+a)^2)^(3/2)/x^2,x, algorithm="maxima")

[Out]

integrate((-(b*x + a)^2 + 1)^(3/2)/((b*x + a + 1)^3*x^2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - (a + b*x)^2)^(3/2)/(x^2*(a + b*x + 1)^3),x)

[Out]

int((1 - (a + b*x)^2)^(3/2)/(x^2*(a + b*x + 1)^3), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac {3}{2}}}{x^{2} \left (a + b x + 1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a+1)**3*(1-(b*x+a)**2)**(3/2)/x**2,x)

[Out]

Integral((-(a + b*x - 1)*(a + b*x + 1))**(3/2)/(x**2*(a + b*x + 1)**3), x)

________________________________________________________________________________________