\(\int \sec ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\) [62]

Optimal result

Integrand size = 31, antiderivative size = 210 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {a^3 (26 A+23 B) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {a^3 (19 A+17 B) \tan (c+d x)}{5 d}+\frac {a^3 (26 A+23 B) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a^3 (22 A+21 B) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {a B \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {(3 A+4 B) \sec ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac {a^3 (19 A+17 B) \tan ^3(c+d x)}{15 d} \] Output:

1/16*a^3*(26*A+23*B)*arctanh(sin(d*x+c))/d+1/5*a^3*(19*A+17*B)*tan(d*x+c)/ 
d+1/16*a^3*(26*A+23*B)*sec(d*x+c)*tan(d*x+c)/d+1/40*a^3*(22*A+21*B)*sec(d* 
x+c)^3*tan(d*x+c)/d+1/6*a*B*sec(d*x+c)^3*(a+a*sec(d*x+c))^2*tan(d*x+c)/d+1 
/15*(3*A+4*B)*sec(d*x+c)^3*(a^3+a^3*sec(d*x+c))*tan(d*x+c)/d+1/15*a^3*(19* 
A+17*B)*tan(d*x+c)^3/d
 

Mathematica [A] (verified)

Time = 3.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.55 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {a^3 \left (15 (26 A+23 B) \text {arctanh}(\sin (c+d x))+\sec (c+d x) \left (390 A+345 B+16 (19 A+17 B) (2+\cos (2 (c+d x))) \sec (c+d x)+10 (18 A+23 B) \sec ^2(c+d x)+48 (A+3 B) \sec ^3(c+d x)+40 B \sec ^4(c+d x)\right ) \tan (c+d x)\right )}{240 d} \] Input:

Integrate[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]
 

Output:

(a^3*(15*(26*A + 23*B)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(390*A + 345*B 
 + 16*(19*A + 17*B)*(2 + Cos[2*(c + d*x)])*Sec[c + d*x] + 10*(18*A + 23*B) 
*Sec[c + d*x]^2 + 48*(A + 3*B)*Sec[c + d*x]^3 + 40*B*Sec[c + d*x]^4)*Tan[c 
 + d*x]))/(240*d)
 

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.98, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4506, 3042, 4506, 27, 3042, 4485, 3042, 4274, 3042, 4254, 2009, 4255, 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.

\(\displaystyle \int \sec ^3(c+d x) (a \sec (c+d x)+a)^3 (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 4506

\(\displaystyle \frac {1}{6} \int \sec ^3(c+d x) (\sec (c+d x) a+a)^2 (3 a (2 A+B)+2 a (3 A+4 B) \sec (c+d x))dx+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (3 a (2 A+B)+2 a (3 A+4 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

\(\Big \downarrow \) 4506

\(\displaystyle \frac {1}{6} \left (\frac {1}{5} \int 3 \sec ^3(c+d x) (\sec (c+d x) a+a) \left ((16 A+13 B) a^2+(22 A+21 B) \sec (c+d x) a^2\right )dx+\frac {2 (3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (\frac {3}{5} \int \sec ^3(c+d x) (\sec (c+d x) a+a) \left ((16 A+13 B) a^2+(22 A+21 B) \sec (c+d x) a^2\right )dx+\frac {2 (3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((16 A+13 B) a^2+(22 A+21 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {2 (3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

\(\Big \downarrow \) 4485

\(\displaystyle \frac {1}{6} \left (\frac {3}{5} \left (\frac {1}{4} \int \sec ^3(c+d x) \left (5 (26 A+23 B) a^3+8 (19 A+17 B) \sec (c+d x) a^3\right )dx+\frac {a^3 (22 A+21 B) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {3}{5} \left (\frac {1}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (5 (26 A+23 B) a^3+8 (19 A+17 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx+\frac {a^3 (22 A+21 B) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {1}{6} \left (\frac {3}{5} \left (\frac {1}{4} \left (8 a^3 (19 A+17 B) \int \sec ^4(c+d x)dx+5 a^3 (26 A+23 B) \int \sec ^3(c+d x)dx\right )+\frac {a^3 (22 A+21 B) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (26 A+23 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+8 a^3 (19 A+17 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {a^3 (22 A+21 B) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {1}{6} \left (\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (26 A+23 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {8 a^3 (19 A+17 B) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {a^3 (22 A+21 B) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{6} \left (\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (26 A+23 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {8 a^3 (19 A+17 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^3 (22 A+21 B) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {1}{6} \left (\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (26 A+23 B) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {8 a^3 (19 A+17 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^3 (22 A+21 B) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (26 A+23 B) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {8 a^3 (19 A+17 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^3 (22 A+21 B) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{6} \left (\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (26 A+23 B) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {8 a^3 (19 A+17 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^3 (22 A+21 B) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 A+4 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\)

Input:

Int[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]
 

Output:

(a*B*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(6*d) + ((2*(3*A 
+ 4*B)*Sec[c + d*x]^3*(a^3 + a^3*Sec[c + d*x])*Tan[c + d*x])/(5*d) + (3*(( 
a^3*(22*A + 21*B)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*a^3*(26*A + 23*B 
)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) - (8*a 
^3*(19*A + 17*B)*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d)/4))/5)/6
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4254

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]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4485

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ 
e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1)   Int[(d*Csc 
[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x 
], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[ 
n, -1]
 

rule 4506

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* 
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), 
 x] + Simp[1/(d*(m + n))   Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] 
)^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* 
Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - 
 a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]
 

Maple [A] (verified)

Time = 1.29 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.08

method	result	size
norman	\(\frac {-\frac {33 a^{3} \left (26 A +23 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 d}+\frac {17 a^{3} \left (26 A +23 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}-\frac {a^{3} \left (26 A +23 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}+\frac {3 a^{3} \left (34 A +35 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {a^{3} \left (838 A +633 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}+\frac {a^{3} \left (998 A +969 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{6}}-\frac {a^{3} \left (26 A +23 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 d}+\frac {a^{3} \left (26 A +23 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d}\)	\(227\)
parts	\(-\frac {\left (a^{3} A +3 B \,a^{3}\right ) \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}-\frac {\left (3 a^{3} A +B \,a^{3}\right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 a^{3} A +3 B \,a^{3}\right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {B \,a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}+\frac {a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(244\)
parallelrisch	\(\frac {25 a^{3} \left (-\frac {39 \left (\frac {2}{3}+\frac {\cos \left (6 d x +6 c \right )}{15}+\frac {2 \cos \left (4 d x +4 c \right )}{5}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {23 B}{26}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{20}+\frac {39 \left (\frac {2}{3}+\frac {\cos \left (6 d x +6 c \right )}{15}+\frac {2 \cos \left (4 d x +4 c \right )}{5}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {23 B}{26}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{20}+\frac {4 \left (\frac {13 A}{5}+3 B \right ) \sin \left (2 d x +2 c \right )}{5}+\frac {\left (\frac {391 B}{6}+63 A \right ) \sin \left (3 d x +3 c \right )}{50}+\frac {8 \left (19 A +17 B \right ) \sin \left (4 d x +4 c \right )}{125}+\frac {\left (13 A +\frac {23 B}{2}\right ) \sin \left (5 d x +5 c \right )}{50}+\frac {4 \left (19 A +17 B \right ) \sin \left (6 d x +6 c \right )}{375}+\sin \left (d x +c \right ) \left (A +\frac {3 B}{2}\right )\right )}{2 d \left (10+\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )\right )}\)	\(251\)
derivativedivides	\(\frac {a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-3 a^{3} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 B \,a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 B \,a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-a^{3} A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\)	\(317\)
default	\(\frac {a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-3 a^{3} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 B \,a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 B \,a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-a^{3} A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\)	\(317\)
risch	\(-\frac {i a^{3} \left (390 A \,{\mathrm e}^{11 i \left (d x +c \right )}+345 B \,{\mathrm e}^{11 i \left (d x +c \right )}+1890 A \,{\mathrm e}^{9 i \left (d x +c \right )}+1955 B \,{\mathrm e}^{9 i \left (d x +c \right )}-1440 A \,{\mathrm e}^{8 i \left (d x +c \right )}-480 B \,{\mathrm e}^{8 i \left (d x +c \right )}+1500 A \,{\mathrm e}^{7 i \left (d x +c \right )}+2250 B \,{\mathrm e}^{7 i \left (d x +c \right )}-6080 A \,{\mathrm e}^{6 i \left (d x +c \right )}-5440 B \,{\mathrm e}^{6 i \left (d x +c \right )}-1500 A \,{\mathrm e}^{5 i \left (d x +c \right )}-2250 B \,{\mathrm e}^{5 i \left (d x +c \right )}-7680 A \,{\mathrm e}^{4 i \left (d x +c \right )}-7680 B \,{\mathrm e}^{4 i \left (d x +c \right )}-1890 A \,{\mathrm e}^{3 i \left (d x +c \right )}-1955 B \,{\mathrm e}^{3 i \left (d x +c \right )}-3648 A \,{\mathrm e}^{2 i \left (d x +c \right )}-3264 B \,{\mathrm e}^{2 i \left (d x +c \right )}-390 \,{\mathrm e}^{i \left (d x +c \right )} A -345 B \,{\mathrm e}^{i \left (d x +c \right )}-608 A -544 B \right )}{120 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}+\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}+\frac {23 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{16 d}-\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}-\frac {23 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{16 d}\)	\(359\)

Input:

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

(-33/20*a^3*(26*A+23*B)/d*tan(1/2*d*x+1/2*c)^7+17/24*a^3*(26*A+23*B)/d*tan 
(1/2*d*x+1/2*c)^9-1/8*a^3*(26*A+23*B)/d*tan(1/2*d*x+1/2*c)^11+3/8*a^3*(34* 
A+35*B)/d*tan(1/2*d*x+1/2*c)-1/24*a^3*(838*A+633*B)/d*tan(1/2*d*x+1/2*c)^3 
+1/20*a^3*(998*A+969*B)/d*tan(1/2*d*x+1/2*c)^5)/(tan(1/2*d*x+1/2*c)^2-1)^6 
-1/16*a^3*(26*A+23*B)/d*ln(tan(1/2*d*x+1/2*c)-1)+1/16*a^3*(26*A+23*B)/d*ln 
(tan(1/2*d*x+1/2*c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.88 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {15 \, {\left (26 \, A + 23 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (26 \, A + 23 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (32 \, {\left (19 \, A + 17 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} + 15 \, {\left (26 \, A + 23 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 16 \, {\left (19 \, A + 17 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 10 \, {\left (18 \, A + 23 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 48 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 40 \, B a^{3}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \] Input:

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="f 
ricas")
 

Output:

1/480*(15*(26*A + 23*B)*a^3*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(26* 
A + 23*B)*a^3*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(32*(19*A + 17*B)* 
a^3*cos(d*x + c)^5 + 15*(26*A + 23*B)*a^3*cos(d*x + c)^4 + 16*(19*A + 17*B 
)*a^3*cos(d*x + c)^3 + 10*(18*A + 23*B)*a^3*cos(d*x + c)^2 + 48*(A + 3*B)* 
a^3*cos(d*x + c) + 40*B*a^3)*sin(d*x + c))/(d*cos(d*x + c)^6)
 

Sympy [F]

\[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=a^{3} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 A \sec ^{4}{\left (c + d x \right )}\, dx + \int 3 A \sec ^{5}{\left (c + d x \right )}\, dx + \int A \sec ^{6}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{5}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{6}{\left (c + d x \right )}\, dx + \int B \sec ^{7}{\left (c + d x \right )}\, dx\right ) \] Input:

integrate(sec(d*x+c)**3*(a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)),x)
 

Output:

a**3*(Integral(A*sec(c + d*x)**3, x) + Integral(3*A*sec(c + d*x)**4, x) + 
Integral(3*A*sec(c + d*x)**5, x) + Integral(A*sec(c + d*x)**6, x) + Integr 
al(B*sec(c + d*x)**4, x) + Integral(3*B*sec(c + d*x)**5, x) + Integral(3*B 
*sec(c + d*x)**6, x) + Integral(B*sec(c + d*x)**7, x))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 405 vs. \(2 (196) = 392\).

Time = 0.04 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.93 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {32 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 96 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a^{3} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} - 5 \, B a^{3} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 90 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 90 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{480 \, d} \] Input:

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="m 
axima")
 

Output:

1/480*(32*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^3 + 
 480*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 96*(3*tan(d*x + c)^5 + 10*t 
an(d*x + c)^3 + 15*tan(d*x + c))*B*a^3 + 160*(tan(d*x + c)^3 + 3*tan(d*x + 
 c))*B*a^3 - 5*B*a^3*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d* 
x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 15*lo 
g(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 90*A*a^3*(2*(3*sin(d*x + 
 c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(si 
n(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 90*B*a^3*(2*(3*sin(d*x + c)^3 
 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x 
 + c) + 1) + 3*log(sin(d*x + c) - 1)) - 120*A*a^3*(2*sin(d*x + c)/(sin(d*x 
 + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.33 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {15 \, {\left (26 \, A a^{3} + 23 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (26 \, A a^{3} + 23 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (390 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 345 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 2210 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1955 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 5148 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4554 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5988 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5814 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4190 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3165 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1530 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \] Input:

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="g 
iac")
 

Output:

1/240*(15*(26*A*a^3 + 23*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(2 
6*A*a^3 + 23*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(390*A*a^3*tan( 
1/2*d*x + 1/2*c)^11 + 345*B*a^3*tan(1/2*d*x + 1/2*c)^11 - 2210*A*a^3*tan(1 
/2*d*x + 1/2*c)^9 - 1955*B*a^3*tan(1/2*d*x + 1/2*c)^9 + 5148*A*a^3*tan(1/2 
*d*x + 1/2*c)^7 + 4554*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 5988*A*a^3*tan(1/2*d 
*x + 1/2*c)^5 - 5814*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 4190*A*a^3*tan(1/2*d*x 
 + 1/2*c)^3 + 3165*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 1530*A*a^3*tan(1/2*d*x + 
 1/2*c) - 1575*B*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6) 
/d
 

Mupad [B] (verification not implemented)

Time = 13.64 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.25 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {\left (-\frac {13\,A\,a^3}{4}-\frac {23\,B\,a^3}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {221\,A\,a^3}{12}+\frac {391\,B\,a^3}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {429\,A\,a^3}{10}-\frac {759\,B\,a^3}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {499\,A\,a^3}{10}+\frac {969\,B\,a^3}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {419\,A\,a^3}{12}-\frac {211\,B\,a^3}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {51\,A\,a^3}{4}+\frac {105\,B\,a^3}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (26\,A+23\,B\right )}{8\,d} \] Input:

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^3)/cos(c + d*x)^3,x)
 

Output:

(tan(c/2 + (d*x)/2)*((51*A*a^3)/4 + (105*B*a^3)/8) - tan(c/2 + (d*x)/2)^11 
*((13*A*a^3)/4 + (23*B*a^3)/8) - tan(c/2 + (d*x)/2)^3*((419*A*a^3)/12 + (2 
11*B*a^3)/8) + tan(c/2 + (d*x)/2)^9*((221*A*a^3)/12 + (391*B*a^3)/24) - ta 
n(c/2 + (d*x)/2)^7*((429*A*a^3)/10 + (759*B*a^3)/20) + tan(c/2 + (d*x)/2)^ 
5*((499*A*a^3)/10 + (969*B*a^3)/20))/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c 
/2 + (d*x)/2)^2 - 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*ta 
n(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) + (a^3*atanh(tan(c/2 + ( 
d*x)/2))*(26*A + 23*B))/(8*d)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 537, normalized size of antiderivative = 2.56 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx =\text {Too large to display} \] Input:

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x)
 

Output:

(a**3*( - 608*cos(c + d*x)*sin(c + d*x)**5*a - 544*cos(c + d*x)*sin(c + d* 
x)**5*b + 1520*cos(c + d*x)*sin(c + d*x)**3*a + 1360*cos(c + d*x)*sin(c + 
d*x)**3*b - 960*cos(c + d*x)*sin(c + d*x)*a - 960*cos(c + d*x)*sin(c + d*x 
)*b - 390*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6*a - 345*log(tan((c + d 
*x)/2) - 1)*sin(c + d*x)**6*b + 1170*log(tan((c + d*x)/2) - 1)*sin(c + d*x 
)**4*a + 1035*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b - 1170*log(tan(( 
c + d*x)/2) - 1)*sin(c + d*x)**2*a - 1035*log(tan((c + d*x)/2) - 1)*sin(c 
+ d*x)**2*b + 390*log(tan((c + d*x)/2) - 1)*a + 345*log(tan((c + d*x)/2) - 
 1)*b + 390*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6*a + 345*log(tan((c + 
 d*x)/2) + 1)*sin(c + d*x)**6*b - 1170*log(tan((c + d*x)/2) + 1)*sin(c + d 
*x)**4*a - 1035*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b + 1170*log(tan 
((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 1035*log(tan((c + d*x)/2) + 1)*sin( 
c + d*x)**2*b - 390*log(tan((c + d*x)/2) + 1)*a - 345*log(tan((c + d*x)/2) 
 + 1)*b - 390*sin(c + d*x)**5*a - 345*sin(c + d*x)**5*b + 960*sin(c + d*x) 
**3*a + 920*sin(c + d*x)**3*b - 570*sin(c + d*x)*a - 615*sin(c + d*x)*b))/ 
(240*d*(sin(c + d*x)**6 - 3*sin(c + d*x)**4 + 3*sin(c + d*x)**2 - 1))