\(\int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [532]

Optimal result

Integrand size = 35, antiderivative size = 275 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {16 a^3 (710 A+803 B) \sin (c+d x)}{3465 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {8 a^3 (710 A+803 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (710 A+803 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (194 A+209 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (14 A+11 B) \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{99 d}+\frac {2 a A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d} \]

[Out]

2/11*a*A*cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/1155*a^3*(710*A+803*B)*cos(d*x+c)^(3/2)*sin(d*
x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/693*a^3*(194*A+209*B)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/
3465*a^3*(710*A+803*B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+8/3465*a^3*(710*A+803*B)*sin(d*x+c
)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/99*a^2*(14*A+11*B)*cos(d*x+c)^(7/2)*sin(d*x+c)*(a+a*sec(d*x+c))^
(1/2)/d

Rubi [A] (verified)

Time = 0.99 (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} \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^3 (194 A+209 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (710 A+803 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{1155 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a^3 (710 A+803 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3465 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^3 (710 A+803 B) \sin (c+d x)}{3465 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (14 A+11 B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{99 d}+\frac {2 a A \sin (c+d x) \cos ^{\frac {9}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d} \]

[In]

Int[Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(16*a^3*(710*A + 803*B)*Sin[c + d*x])/(3465*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a^3*(710*A + 8
03*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3465*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(710*A + 803*B)*Cos[c + d*x]
^(3/2)*Sin[c + d*x])/(1155*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(194*A + 209*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x
])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(14*A + 11*B)*Cos[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c +
d*x])/(99*d) + (2*a*A*Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*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))^{5/2} (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) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac {1}{11} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{3/2} \left (\frac {1}{2} a (14 A+11 B)+\frac {1}{2} a (6 A+11 B) \sec (c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (14 A+11 B) \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{99 d}+\frac {2 a A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac {1}{99} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (194 A+209 B)+\frac {3}{4} a^2 (46 A+55 B) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 a^3 (194 A+209 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 (14 A+11 B) \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{99 d}+\frac {2 a A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac {1}{231} \left (a^2 (710 A+803 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 {2 a^3 (710 A+803 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (194 A+209 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 (14 A+11 B) \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{99 d}+\frac {2 a A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac {\left (4 a^2 (710 A+803 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 {8 a^3 (710 A+803 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (710 A+803 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (194 A+209 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 (14 A+11 B) \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{99 d}+\frac {2 a A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac {\left (8 a^2 (710 A+803 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 {16 a^3 (710 A+803 B) \sin (c+d x)}{3465 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {8 a^3 (710 A+803 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (710 A+803 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (194 A+209 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 (14 A+11 B) \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{99 d}+\frac {2 a A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.50 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^2 \sqrt {\cos (c+d x)} \left (8 (710 A+803 B)+4 (710 A+803 B) \cos (c+d x)+3 (710 A+803 B) \cos ^2(c+d x)+5 (355 A+286 B) \cos ^3(c+d x)+35 (32 A+11 B) \cos ^4(c+d x)+315 A \cos ^5(c+d x)\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{3465 d (1+\cos (c+d x))} \]

[In]

Integrate[Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(2*a^2*Sqrt[Cos[c + d*x]]*(8*(710*A + 803*B) + 4*(710*A + 803*B)*Cos[c + d*x] + 3*(710*A + 803*B)*Cos[c + d*x]
^2 + 5*(355*A + 286*B)*Cos[c + d*x]^3 + 35*(32*A + 11*B)*Cos[c + d*x]^4 + 315*A*Cos[c + d*x]^5)*Sqrt[a*(1 + Se
c[c + d*x])]*Sin[c + d*x])/(3465*d*(1 + Cos[c + d*x]))

Maple [A] (verified)

Time = 3.86 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.51

\[-\frac {2 a^{2} \left (\left (315 \cos \left (d x +c \right )^{5}+1120 \cos \left (d x +c \right )^{4}+1775 \cos \left (d x +c \right )^{3}+2130 \cos \left (d x +c \right )^{2}+2840 \cos \left (d x +c \right )+5680\right ) A +\left (385 \cos \left (d x +c \right )^{4}+1430 \cos \left (d x +c \right )^{3}+2409 \cos \left (d x +c \right )^{2}+3212 \cos \left (d x +c \right )+6424\right ) B \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{3465 d}\]

[In]

int(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)

[Out]

-2/3465*a^2/d*((315*cos(d*x+c)^5+1120*cos(d*x+c)^4+1775*cos(d*x+c)^3+2130*cos(d*x+c)^2+2840*cos(d*x+c)+5680)*A
+(385*cos(d*x+c)^4+1430*cos(d*x+c)^3+2409*cos(d*x+c)^2+3212*cos(d*x+c)+6424)*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 = 0.26 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.56 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (315 \, A a^{2} \cos \left (d x + c\right )^{5} + 35 \, {\left (32 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (355 \, A + 286 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 4 \, {\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, {\left (710 \, A + 803 \, B\right )} a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

2/3465*(315*A*a^2*cos(d*x + c)^5 + 35*(32*A + 11*B)*a^2*cos(d*x + c)^4 + 5*(355*A + 286*B)*a^2*cos(d*x + c)^3
+ 3*(710*A + 803*B)*a^2*cos(d*x + c)^2 + 4*(710*A + 803*B)*a^2*cos(d*x + c) + 8*(710*A + 803*B)*a^2)*sqrt((a*c
os(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. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(11/2)*(a+a*sec(d*x+c))**(5/2)*(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. 754 vs. \(2 (239) = 478\).

Time = 0.49 (sec) , antiderivative size = 754, normalized size of antiderivative = 2.74 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/110880*(5*sqrt(2)*(31878*a^2*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) + 8778*a^2*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)
+ 3465*a^2*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) + 1287*a^
2*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) + 385*a^2*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) - 31878*a^2*cos(11/2*d*x + 11/
2*c)*sin(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 8778*a^2*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))) - 3465*a^2*cos(11/2*d*x + 11/2*c)*sin(6/11*arcta
n2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 1287*a^2*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))) - 385*a^2*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))) + 126*a^2*sin(11/2*d*x + 11/2*c) + 385*a^2*sin(9/11*arctan2(sin(11/2*d*x + 11/2
*c), cos(11/2*d*x + 11/2*c))) + 1287*a^2*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 3
465*a^2*sin(5/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 8778*a^2*sin(3/11*arctan2(sin(11/2
*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 31878*a^2*sin(1/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 1
1/2*c))))*A*sqrt(a) + 44*sqrt(2)*(225*a^2*sin(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 378*a^2*sin(5
/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2100*a^2*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))
) + 4095*a^2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 63*(65*a^2*sin(4*d*x + 4*c) + 6*a^2*sin(2*
d*x + 2*c))*cos(9/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 7*(585*a^2*cos(4*d*x + 4*c) + 54*a^2*cos(2*
d*x + 2*c) + 5*a^2)*sin(9/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*B*sqrt(a))/d

Giac [F]

\[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {11}{2}} \,d x } \]

[In]

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^{11/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

[In]

int(cos(c + d*x)^(11/2)*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^(11/2)*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2), x)