3.2.48 \(\int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx\) [148]

Optimal. Leaf size=110 \[ -\frac {2 (-1)^{3/4} a \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{7/2} f}-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac {2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {2 a}{d^3 f \sqrt {d \tan (e+f x)}} \]

[Out]

-2*(-1)^(3/4)*a*arctanh((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/d^(7/2)/f+2*a/d^3/f/(d*tan(f*x+e))^(1/2)-2/5*
a/d/f/(d*tan(f*x+e))^(5/2)+2/3*I*a/d^2/f/(d*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.11, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3610, 3614, 214} \begin {gather*} -\frac {2 (-1)^{3/4} a \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{7/2} f}+\frac {2 a}{d^3 f \sqrt {d \tan (e+f x)}}+\frac {2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - I*a*Tan[e + f*x])/(d*Tan[e + f*x])^(7/2),x]

[Out]

(-2*(-1)^(3/4)*a*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(7/2)*f) - (2*a)/(5*d*f*(d*Tan[e + f*x
])^(5/2)) + (((2*I)/3)*a)/(d^2*f*(d*Tan[e + f*x])^(3/2)) + (2*a)/(d^3*f*Sqrt[d*Tan[e + f*x]])

Rule 214

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

Rule 3610

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b
*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3614

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2*(c^2/f), S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rubi steps

\begin {align*} \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx &=-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac {\int \frac {-i a d-a d \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx}{d^2}\\ &=-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac {2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {\int \frac {-a d^2+i a d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{d^4}\\ &=-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac {2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {2 a}{d^3 f \sqrt {d \tan (e+f x)}}+\frac {\int \frac {i a d^3+a d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{d^6}\\ &=-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac {2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {2 a}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{i a d^4-a d^3 x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=-\frac {2 (-1)^{3/4} a \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{7/2} f}-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac {2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {2 a}{d^3 f \sqrt {d \tan (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.17, size = 41, normalized size = 0.37 \begin {gather*} -\frac {2 a \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-i \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a - I*a*Tan[e + f*x])/(d*Tan[e + f*x])^(7/2),x]

[Out]

(-2*a*Hypergeometric2F1[-5/2, 1, -3/2, (-I)*Tan[e + f*x]])/(5*d*f*(d*Tan[e + f*x])^(5/2))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (89 ) = 178\).
time = 0.10, size = 326, normalized size = 2.96

method	result	size
derivativedivides	\(-\frac {a \left (\frac {-\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{3}}-\frac {2}{d^{3} \sqrt {d \tan \left (f x +e \right )}}-\frac {2 i}{3 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {2}{5 d \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f}\)	\(326\)
default	\(-\frac {a \left (\frac {-\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{3}}-\frac {2}{d^{3} \sqrt {d \tan \left (f x +e \right )}}-\frac {2 i}{3 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {2}{5 d \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f}\)	\(326\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a-I*a*tan(f*x+e))/(d*tan(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/f*a*(2/d^3*(-1/8*I/d*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(
1/2))/(d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan
(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))-1/8/(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*
x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1
/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*
x+e))^(1/2)+1)))-2/d^3/(d*tan(f*x+e))^(1/2)-2/3*I/d^2/(d*tan(f*x+e))^(3/2)+2/5/d/(d*tan(f*x+e))^(5/2))

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (92) = 184\).
time = 0.51, size = 220, normalized size = 2.00 \begin {gather*} -\frac {\frac {15 \, a {\left (-\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d^{2}} - \frac {8 \, {\left (15 \, a d^{2} \tan \left (f x + e\right )^{2} + 5 i \, a d^{2} \tan \left (f x + e\right ) - 3 \, a d^{2}\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} d^{2}}}{60 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*tan(f*x+e))/(d*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

-1/60*(15*a*(-(2*I + 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d)
- (2*I + 2)*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - (I - 1)
*sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) + (I - 1)*sqrt(2)*log(d*tan(f*
x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d))/d^2 - 8*(15*a*d^2*tan(f*x + e)^2 + 5*I*a*d^2*tan(f
*x + e) - 3*a*d^2)/((d*tan(f*x + e))^(5/2)*d^2))/(d*f)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 459 vs. \(2 (92) = 184\).
time = 0.40, size = 459, normalized size = 4.17 \begin {gather*} -\frac {15 \, {\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )} \sqrt {-\frac {4 i \, a^{2}}{d^{7} f^{2}}} \log \left (-\frac {{\left ({\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 i \, a^{2}}{d^{7} f^{2}}} + 2 \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3} f}\right ) - 15 \, {\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )} \sqrt {-\frac {4 i \, a^{2}}{d^{7} f^{2}}} \log \left (\frac {{\left ({\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 i \, a^{2}}{d^{7} f^{2}}} - 2 \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3} f}\right ) + 8 \, {\left (-13 i \, a e^{\left (6 i \, f x + 6 i \, e\right )} + 11 i \, a e^{\left (4 i \, f x + 4 i \, e\right )} + i \, a e^{\left (2 i \, f x + 2 i \, e\right )} - 23 i \, a\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, {\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*tan(f*x+e))/(d*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-1/60*(15*(d^4*f*e^(6*I*f*x + 6*I*e) - 3*d^4*f*e^(4*I*f*x + 4*I*e) + 3*d^4*f*e^(2*I*f*x + 2*I*e) - d^4*f)*sqrt
(-4*I*a^2/(d^7*f^2))*log(-((d^3*f*e^(2*I*f*x + 2*I*e) + d^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f
*x + 2*I*e) + 1))*sqrt(-4*I*a^2/(d^7*f^2)) + 2*a)*e^(-2*I*f*x - 2*I*e)/(d^3*f)) - 15*(d^4*f*e^(6*I*f*x + 6*I*e
) - 3*d^4*f*e^(4*I*f*x + 4*I*e) + 3*d^4*f*e^(2*I*f*x + 2*I*e) - d^4*f)*sqrt(-4*I*a^2/(d^7*f^2))*log(((d^3*f*e^
(2*I*f*x + 2*I*e) + d^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-4*I*a^2/(d^7
*f^2)) - 2*a)*e^(-2*I*f*x - 2*I*e)/(d^3*f)) + 8*(-13*I*a*e^(6*I*f*x + 6*I*e) + 11*I*a*e^(4*I*f*x + 4*I*e) + I*
a*e^(2*I*f*x + 2*I*e) - 23*I*a)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d^4*f*e^(6*
I*f*x + 6*I*e) - 3*d^4*f*e^(4*I*f*x + 4*I*e) + 3*d^4*f*e^(2*I*f*x + 2*I*e) - d^4*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i a \left (\int \frac {i}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {\tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*tan(f*x+e))/(d*tan(f*x+e))**(7/2),x)

[Out]

-I*a*(Integral(I/(d*tan(e + f*x))**(7/2), x) + Integral(tan(e + f*x)/(d*tan(e + f*x))**(7/2), x))

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Giac [A]
time = 0.66, size = 130, normalized size = 1.18 \begin {gather*} -\frac {2}{15} \, a {\left (-\frac {15 i \, \sqrt {2} \arctan \left (\frac {8 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{d^{\frac {7}{2}} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {15 \, d^{2} \tan \left (f x + e\right )^{2} + 5 i \, d^{2} \tan \left (f x + e\right ) - 3 \, d^{2}}{\sqrt {d \tan \left (f x + e\right )} d^{5} f \tan \left (f x + e\right )^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*tan(f*x+e))/(d*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

-2/15*a*(-15*I*sqrt(2)*arctan(8*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sq
rt(d)))/(d^(7/2)*f*(I*d/sqrt(d^2) + 1)) - (15*d^2*tan(f*x + e)^2 + 5*I*d^2*tan(f*x + e) - 3*d^2)/(sqrt(d*tan(f
*x + e))*d^5*f*tan(f*x + e)^2))

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Mupad [B]
time = 4.91, size = 87, normalized size = 0.79 \begin {gather*} \frac {2\,{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{d^{7/2}\,f}-\frac {\frac {2\,a}{5\,d}-\frac {2\,a\,{\mathrm {tan}\left (e+f\,x\right )}^2}{d}}{f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}+\frac {a\,2{}\mathrm {i}}{3\,d^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - a*tan(e + f*x)*1i)/(d*tan(e + f*x))^(7/2),x)

[Out]

(a*2i)/(3*d^2*f*(d*tan(e + f*x))^(3/2)) - ((2*a)/(5*d) - (2*a*tan(e + f*x)^2)/d)/(f*(d*tan(e + f*x))^(5/2)) +
(2*(-1)^(1/4)*a*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2)))/(d^(7/2)*f)

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