\(\int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\) [523]

Optimal result

Integrand size = 35, antiderivative size = 275 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {32 a^2 (168 A+187 B) \sin (c+d x)}{3465 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (168 A+187 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {4 a^2 (168 A+187 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (168 A+187 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (12 A+11 B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d} \]

[Out]

4/1155*a^2*(168*A+187*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/693*a^2*(168*A+187*B)*cos(d*x+
c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/99*a^2*(12*A+11*B)*cos(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c
))^(1/2)+32/3465*a^2*(168*A+187*B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+16/3465*a^2*(168*A+187
*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/11*a*A*cos(d*x+c)^(9/2)*sin(d*x+c)*(a+a*sec(d*x+c))
^(1/2)/d

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3034, 4102, 4100, 3890, 3889} \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^2 (12 A+11 B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{99 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (168 A+187 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {4 a^2 (168 A+187 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{1155 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (168 A+187 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3465 d \sqrt {a \sec (c+d x)+a}}+\frac {32 a^2 (168 A+187 B) \sin (c+d x)}{3465 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a A \sin (c+d x) \cos ^{\frac {9}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d} \]

[In]

Int[Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]),x]

[Out]

(32*a^2*(168*A + 187*B)*Sin[c + d*x])/(3465*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(168*A +
187*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3465*d*Sqrt[a + a*Sec[c + d*x]]) + (4*a^2*(168*A + 187*B)*Cos[c + d*x
]^(3/2)*Sin[c + d*x])/(1155*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(168*A + 187*B)*Cos[c + d*x]^(5/2)*Sin[c + d*
x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(12*A + 11*B)*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(99*d*Sqrt[a + a*
Sec[c + d*x]]) + (2*a*A*Cos[c + d*x]^(9/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(11*d)

Rule 3034

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*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[(a + b*Csc[e + f*x])^m*((
c + d*Csc[e + f*x])^n/(g*Csc[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 3889

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[-2*a*(Co
t[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^
2, 0]

Rule 3890

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*Cot[e
+ f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Dist[a*((2*n + 1)/(2*b*d*n)), Int[Sqrt[a + b
*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2
^(-1)] && IntegerQ[2*n]

Rule 4100

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

Rule 4102

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

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx \\ & = \frac {2 a A \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d}+\frac {1}{11} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{2} a (12 A+11 B)+\frac {1}{2} a (8 A+11 B) \sec (c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (12 A+11 B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d}+\frac {1}{99} \left (a (168 A+187 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (168 A+187 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (12 A+11 B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d}+\frac {1}{231} \left (2 a (168 A+187 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {4 a^2 (168 A+187 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (168 A+187 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (12 A+11 B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d}+\frac {\left (8 a (168 A+187 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{1155} \\ & = \frac {16 a^2 (168 A+187 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {4 a^2 (168 A+187 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (168 A+187 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (12 A+11 B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d}+\frac {\left (16 a (168 A+187 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{3465} \\ & = \frac {32 a^2 (168 A+187 B) \sin (c+d x)}{3465 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (168 A+187 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {4 a^2 (168 A+187 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (168 A+187 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (12 A+11 B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{11 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.85 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.48 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {2 a \sqrt {\cos (c+d x)} \left (2688 A+2992 B+8 (168 A+187 B) \cos (c+d x)+6 (168 A+187 B) \cos ^2(c+d x)+(840 A+935 B) \cos ^3(c+d x)+35 (21 A+11 B) \cos ^4(c+d x)+315 A \cos ^5(c+d x)\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{3465 d (1+\cos (c+d x))} \]

[In]

Integrate[Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]),x]

[Out]

(2*a*Sqrt[Cos[c + d*x]]*(2688*A + 2992*B + 8*(168*A + 187*B)*Cos[c + d*x] + 6*(168*A + 187*B)*Cos[c + d*x]^2 +
(840*A + 935*B)*Cos[c + d*x]^3 + 35*(21*A + 11*B)*Cos[c + d*x]^4 + 315*A*Cos[c + d*x]^5)*Sqrt[a*(1 + Sec[c +
d*x])]*Sin[c + d*x])/(3465*d*(1 + Cos[c + d*x]))

Maple [A] (verified)

Time = 5.38 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.50

method	result	size
default	\(-\frac {2 a \left (\left (315 \cos \left (d x +c \right )^{5}+735 \cos \left (d x +c \right )^{4}+840 \cos \left (d x +c \right )^{3}+1008 \cos \left (d x +c \right )^{2}+1344 \cos \left (d x +c \right )+2688\right ) A +\left (385 \cos \left (d x +c \right )^{4}+935 \cos \left (d x +c \right )^{3}+1122 \cos \left (d x +c \right )^{2}+1496 \cos \left (d x +c \right )+2992\right ) B \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{3465 d}\)	\(137\)

[In]

int(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-2/3465*a/d*((315*cos(d*x+c)^5+735*cos(d*x+c)^4+840*cos(d*x+c)^3+1008*cos(d*x+c)^2+1344*cos(d*x+c)+2688)*A+(38
5*cos(d*x+c)^4+935*cos(d*x+c)^3+1122*cos(d*x+c)^2+1496*cos(d*x+c)+2992)*B)*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))
^(1/2)*(cot(d*x+c)-csc(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.52 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (315 \, A a \cos \left (d x + c\right )^{5} + 35 \, {\left (21 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{4} + 5 \, {\left (168 \, A + 187 \, B\right )} a \cos \left (d x + c\right )^{3} + 6 \, {\left (168 \, A + 187 \, B\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (168 \, A + 187 \, B\right )} a \cos \left (d x + c\right ) + 16 \, {\left (168 \, A + 187 \, B\right )} a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

2/3465*(315*A*a*cos(d*x + c)^5 + 35*(21*A + 11*B)*a*cos(d*x + c)^4 + 5*(168*A + 187*B)*a*cos(d*x + c)^3 + 6*(1
68*A + 187*B)*a*cos(d*x + c)^2 + 8*(168*A + 187*B)*a*cos(d*x + c) + 16*(168*A + 187*B)*a)*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c) + d)

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(11/2)*(a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 703 vs. \(2 (239) = 478\).

Time = 0.47 (sec) , antiderivative size = 703, normalized size of antiderivative = 2.56 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/110880*(21*sqrt(2)*(3630*a*cos(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x +
11/2*c) + 990*a*cos(8/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 42
9*a*cos(6/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 165*a*cos(4/11*
arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 55*a*cos(2/11*arctan2(sin(11
/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) - 3630*a*cos(11/2*d*x + 11/2*c)*sin(10/11*ar
ctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 990*a*cos(11/2*d*x + 11/2*c)*sin(8/11*arctan2(sin(11/
2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 429*a*cos(11/2*d*x + 11/2*c)*sin(6/11*arctan2(sin(11/2*d*x + 11/2*
c), cos(11/2*d*x + 11/2*c))) - 165*a*cos(11/2*d*x + 11/2*c)*sin(4/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*
d*x + 11/2*c))) - 55*a*cos(11/2*d*x + 11/2*c)*sin(2/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))
) + 30*a*sin(11/2*d*x + 11/2*c) + 55*a*sin(9/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 165
*a*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 429*a*sin(5/11*arctan2(sin(11/2*d*x + 1
1/2*c), cos(11/2*d*x + 11/2*c))) + 990*a*sin(3/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 3
630*a*sin(1/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))))*A*sqrt(a) - 44*sqrt(2)*(189*(10*a*sin
(4*d*x + 4*c) + a*sin(2*d*x + 2*c))*cos(9/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 7*(270*a*cos(4*d*x
+ 4*c) + 27*a*cos(2*d*x + 2*c) + 5*a)*sin(9/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 135*a*sin(7/4*arc
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 189*a*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1050*
a*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1890*a*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c))))*B*sqrt(a))/d

Giac [F]

\[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {11}{2}} \,d x } \]

[In]

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^{11/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]

[In]

int(cos(c + d*x)^(11/2)*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^(11/2)*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(3/2), x)