Integrand size = 37, antiderivative size = 92 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a (2 A+B+C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (3 A+3 B+2 C) \tan (c+d x)}{3 d}+\frac {a (B+C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a C \sec ^2(c+d x) \tan (c+d x)}{3 d} \]
1/2*a*(2*A+B+C)*arctanh(sin(d*x+c))/d+1/3*a*(3*A+3*B+2*C)*tan(d*x+c)/d+1/2 *a*(B+C)*sec(d*x+c)*tan(d*x+c)/d+1/3*a*C*sec(d*x+c)^2*tan(d*x+c)/d
Time = 1.48 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.68 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (3 (2 A+B+C) \text {arctanh}(\sin (c+d x))+\left (6 A+6 B+4 C+3 (B+C) \sec (c+d x)+2 C \sec ^2(c+d x)\right ) \tan (c+d x)\right )}{6 d} \]
(a*(3*(2*A + B + C)*ArcTanh[Sin[c + d*x]] + (6*A + 6*B + 4*C + 3*(B + C)*S ec[c + d*x] + 2*C*Sec[c + d*x]^2)*Tan[c + d*x]))/(6*d)
Time = 0.60 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.270, Rules used = {3042, 4564, 3042, 4535, 3042, 4254, 24, 4534, 3042, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sec (c+d x) (a \sec (c+d x)+a) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4564 |
| \(\displaystyle \frac {1}{3} \int \sec (c+d x) \left (3 a (B+C) \sec ^2(c+d x)+a (3 A+3 B+2 C) \sec (c+d x)+3 a A\right )dx+\frac {a C \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a (B+C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (3 A+3 B+2 C) \csc \left (c+d x+\frac {\pi }{2}\right )+3 a A\right )dx+\frac {a C \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{3} \left (a (3 A+3 B+2 C) \int \sec ^2(c+d x)dx+\int \sec (c+d x) \left (3 a (B+C) \sec ^2(c+d x)+3 a A\right )dx\right )+\frac {a C \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (a (3 A+3 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a (B+C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 a A\right )dx\right )+\frac {a C \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{3} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a (B+C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 a A\right )dx-\frac {a (3 A+3 B+2 C) \int 1d(-\tan (c+d x))}{d}\right )+\frac {a C \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{3} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a (B+C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 a A\right )dx+\frac {a (3 A+3 B+2 C) \tan (c+d x)}{d}\right )+\frac {a C \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} a (2 A+B+C) \int \sec (c+d x)dx+\frac {a (3 A+3 B+2 C) \tan (c+d x)}{d}+\frac {3 a (B+C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a C \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} a (2 A+B+C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {a (3 A+3 B+2 C) \tan (c+d x)}{d}+\frac {3 a (B+C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a C \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 a (2 A+B+C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (3 A+3 B+2 C) \tan (c+d x)}{d}+\frac {3 a (B+C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a C \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
(a*C*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((3*a*(2*A + B + C)*ArcTanh[Sin[ c + d*x]])/(2*d) + (a*(3*A + 3*B + 2*C)*Tan[c + d*x])/d + (3*a*(B + C)*Sec [c + d*x]*Tan[c + d*x])/(2*d))/3
3.5.10.3.1 Defintions of rubi rules used
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f*x])^ n/(f*(n + 2))), x] + Simp[1/(n + 2) Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1) + A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*( n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && !LtQ[n, -1]
Time = 0.50 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.15
| method | result | size |
| parts | \(\frac {\left (a A +a B \right ) \tan \left (d x +c \right )}{d}+\frac {\left (a B +C a \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(106\) |
| derivativedivides | \(\frac {a A \tan \left (d x +c \right )+a B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a B \tan \left (d x +c \right )+C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(131\) |
| default | \(\frac {a A \tan \left (d x +c \right )+a B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a B \tan \left (d x +c \right )+C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(131\) |
| norman | \(\frac {-\frac {a \left (2 A +B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a \left (2 A +3 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a \left (3 A +3 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {a \left (2 A +B +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (2 A +B +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(140\) |
| parallelrisch | \(\frac {\left (-3 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {B}{2}+\frac {C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {B}{2}+\frac {C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (A +B +\frac {2 C}{3}\right ) \sin \left (3 d x +3 c \right )+\left (B +C \right ) \sin \left (2 d x +2 c \right )+\sin \left (d x +c \right ) \left (A +B +2 C \right )\right ) a}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(149\) |
| risch | \(-\frac {i a \left (3 B \,{\mathrm e}^{5 i \left (d x +c \right )}+3 C \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A \,{\mathrm e}^{4 i \left (d x +c \right )}-6 B \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 B \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-3 C \,{\mathrm e}^{i \left (d x +c \right )}-6 A -6 B -4 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\) | \(259\) |
(A*a+B*a)/d*tan(d*x+c)+(B*a+C*a)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d *x+c)+tan(d*x+c)))-C*a/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a*A/d*ln(sec(d *x+c)+tan(d*x+c))
Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.24 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, A + B + C\right )} a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A + B + C\right )} a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, A + 3 \, B + 2 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (B + C\right )} a \cos \left (d x + c\right ) + 2 \, C a\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
1/12*(3*(2*A + B + C)*a*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*A + B + C)*a*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(3*A + 3*B + 2*C)*a*co s(d*x + c)^2 + 3*(B + C)*a*cos(d*x + c) + 2*C*a)*sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a \left (\int A \sec {\left (c + d x \right )}\, dx + \int A \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]
a*(Integral(A*sec(c + d*x), x) + Integral(A*sec(c + d*x)**2, x) + Integral (B*sec(c + d*x)**2, x) + Integral(B*sec(c + d*x)**3, x) + Integral(C*sec(c + d*x)**3, x) + Integral(C*sec(c + d*x)**4, x))
Time = 0.23 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.68 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a - 3 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, A a \tan \left (d x + c\right ) + 12 \, B a \tan \left (d x + c\right )}{12 \, d} \]
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a - 3*B*a*(2*sin(d*x + c)/(sin (d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 3*C*a* (2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 12*A*a*log(sec(d*x + c) + tan(d*x + c)) + 12*A*a*tan(d*x + c ) + 12*B*a*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (84) = 168\).
Time = 0.31 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.23 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, A a + B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a + B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
1/6*(3*(2*A*a + B*a + C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a + B*a + C*a)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*a*tan(1/2*d*x + 1/ 2*c)^5 + 3*B*a*tan(1/2*d*x + 1/2*c)^5 + 3*C*a*tan(1/2*d*x + 1/2*c)^5 - 12* A*a*tan(1/2*d*x + 1/2*c)^3 - 12*B*a*tan(1/2*d*x + 1/2*c)^3 - 4*C*a*tan(1/2 *d*x + 1/2*c)^3 + 6*A*a*tan(1/2*d*x + 1/2*c) + 9*B*a*tan(1/2*d*x + 1/2*c) + 9*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 20.56 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.79 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a\,\mathrm {atanh}\left (\frac {2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A+B+C\right )}{4\,A\,a+2\,B\,a+2\,C\,a}\right )\,\left (2\,A+B+C\right )}{d}-\frac {\left (2\,A\,a+B\,a+C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,A\,a-4\,B\,a-\frac {4\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+3\,B\,a+3\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
(a*atanh((2*a*tan(c/2 + (d*x)/2)*(2*A + B + C))/(4*A*a + 2*B*a + 2*C*a))*( 2*A + B + C))/d - (tan(c/2 + (d*x)/2)*(2*A*a + 3*B*a + 3*C*a) + tan(c/2 + (d*x)/2)^5*(2*A*a + B*a + C*a) - tan(c/2 + (d*x)/2)^3*(4*A*a + 4*B*a + (4* C*a)/3))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + ( d*x)/2)^6 - 1))