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

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

[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)

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Rubi [A]  time = 0.831167, antiderivative size = 275, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2955, 4017, 4015, 3805, 3804} \[ \frac{2 a^2 (14 A+11 B) \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{99 d}+\frac{2 a^3 (194 A+209 B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^3 (710 A+803 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{1155 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a^3 (710 A+803 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3465 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^3 (710 A+803 B) \sin (c+d x)}{3465 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 a A \sin (c+d x) \cos ^{\frac{9}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d} \]

Antiderivative was successfully verified.

[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 2955

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 4017

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]

Rule 4015

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 3805

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 3804

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]

Rubi steps

\begin{align*} \int \cos ^{\frac{11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx &=\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]  time = 0.583902, size = 137, normalized size = 0.5 \[ \frac{2 a^2 \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a (\sec (c+d x)+1)} \left (35 (32 A+11 B) \cos ^4(c+d x)+5 (355 A+286 B) \cos ^3(c+d x)+3 (710 A+803 B) \cos ^2(c+d x)+4 (710 A+803 B) \cos (c+d x)+8 (710 A+803 B)+315 A \cos ^5(c+d x)\right )}{3465 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[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]))

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Maple [A]  time = 0.34, size = 155, normalized size = 0.6 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 315\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1120\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+385\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1775\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+1430\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+2130\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2409\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2840\,A\cos \left ( dx+c \right ) +3212\,B\cos \left ( dx+c \right ) +5680\,A+6424\,B \right ) }{3465\,d\sin \left ( dx+c \right ) }\sqrt{\cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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/d*a^2*(-1+cos(d*x+c))*(315*A*cos(d*x+c)^5+1120*A*cos(d*x+c)^4+385*B*cos(d*x+c)^4+1775*A*cos(d*x+c)^3+1
430*B*cos(d*x+c)^3+2130*A*cos(d*x+c)^2+2409*B*cos(d*x+c)^2+2840*A*cos(d*x+c)+3212*B*cos(d*x+c)+5680*A+6424*B)*
cos(d*x+c)^(1/2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)

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Maxima [B]  time = 2.16263, size = 1018, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 0.494805, size = 412, normalized size = 1.5 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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