\(\int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\) [1145]

Optimal result

Integrand size = 37, antiderivative size = 266 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 a^3 (568 A+759 C) \sin (c+d x)}{693 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (568 A+759 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (232 A+297 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+33 C) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d}+\frac {10 a A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d} \]

[Out]

10/99*a*A*cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/11*A*cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*
sin(d*x+c)/d+2/693*a^3*(232*A+297*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+4/693*a^3*(568*A+759
*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+2/693*a^3*(568*A+759*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d
/(a+a*sec(d*x+c))^(1/2)+2/231*a^2*(32*A+33*C)*cos(d*x+c)^(5/2)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4350, 4172, 4102, 4100, 3890, 3889} \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^3 (232 A+297 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (568 A+759 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {4 a^3 (568 A+759 C) \sin (c+d x)}{693 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (32 A+33 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{231 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {9}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}+\frac {10 a A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{99 d} \]

[In]

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

[Out]

(4*a^3*(568*A + 759*C)*Sin[c + d*x])/(693*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(568*A + 759
*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(232*A + 297*C)*Cos[c + d*x]^(3
/2)*Sin[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(32*A + 33*C)*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c
+ d*x]]*Sin[c + d*x])/(231*d) + (10*a*A*Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(99*d) + (
2*A*Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(11*d)

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]

Rule 4172

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

Rule 4350

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

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} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx \\ & = \frac {2 A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{5/2} \left (\frac {5 a A}{2}+\frac {1}{2} a (4 A+11 C) \sec (c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx}{11 a} \\ & = \frac {10 a A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{3/2} \left (\frac {3}{4} a^2 (32 A+33 C)+\frac {1}{4} a^2 (56 A+99 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{99 a} \\ & = \frac {2 a^2 (32 A+33 C) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d}+\frac {10 a A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {5}{8} a^3 (232 A+297 C)+\frac {1}{8} a^3 (776 A+1089 C) \sec (c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{693 a} \\ & = \frac {2 a^3 (232 A+297 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+33 C) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d}+\frac {10 a A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d}+\frac {1}{231} \left (a^2 (568 A+759 C) \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 \\ & = \frac {2 a^3 (568 A+759 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (232 A+297 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+33 C) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d}+\frac {10 a A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d}+\frac {1}{693} \left (2 a^2 (568 A+759 C) \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 \\ & = \frac {4 a^3 (568 A+759 C) \sin (c+d x)}{693 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (568 A+759 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (232 A+297 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+33 C) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d}+\frac {10 a A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.69 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.50 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \sqrt {\cos (c+d x)} (22928 A+27456 C+2 (6989 A+6666 C) \cos (c+d x)+16 (325 A+198 C) \cos (2 (c+d x))+1735 A \cos (3 (c+d x))+396 C \cos (3 (c+d x))+448 A \cos (4 (c+d x))+63 A \cos (5 (c+d x))) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{5544 d (1+\cos (c+d x))} \]

[In]

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

[Out]

(a^2*Sqrt[Cos[c + d*x]]*(22928*A + 27456*C + 2*(6989*A + 6666*C)*Cos[c + d*x] + 16*(325*A + 198*C)*Cos[2*(c +
d*x)] + 1735*A*Cos[3*(c + d*x)] + 396*C*Cos[3*(c + d*x)] + 448*A*Cos[4*(c + d*x)] + 63*A*Cos[5*(c + d*x)])*Sqr
t[a*(1 + Sec[c + d*x])]*Sin[c + d*x])/(5544*d*(1 + Cos[c + d*x]))

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.51

\[\frac {2 a^{2} \sin \left (d x +c \right ) \left (63 A \cos \left (d x +c \right )^{5}+224 A \cos \left (d x +c \right )^{4}+355 A \cos \left (d x +c \right )^{3}+99 C \cos \left (d x +c \right )^{3}+426 A \cos \left (d x +c \right )^{2}+396 C \cos \left (d x +c \right )^{2}+568 A \cos \left (d x +c \right )+759 C \cos \left (d x +c \right )+1136 A +1518 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\cos \left (d x +c \right )}}{693 d \left (1+\cos \left (d x +c \right )\right )}\]

[In]

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

[Out]

2/693*a^2/d*sin(d*x+c)*(63*A*cos(d*x+c)^5+224*A*cos(d*x+c)^4+355*A*cos(d*x+c)^3+99*C*cos(d*x+c)^3+426*A*cos(d*
x+c)^2+396*C*cos(d*x+c)^2+568*A*cos(d*x+c)+759*C*cos(d*x+c)+1136*A+1518*C)*(a*(1+sec(d*x+c)))^(1/2)*cos(d*x+c)
^(1/2)/(1+cos(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.55 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (63 \, A a^{2} \cos \left (d x + c\right )^{5} + 224 \, A a^{2} \cos \left (d x + c\right )^{4} + {\left (355 \, A + 99 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 6 \, {\left (71 \, A + 66 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (568 \, A + 759 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (568 \, A + 759 \, C\right )} a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{693 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

2/693*(63*A*a^2*cos(d*x + c)^5 + 224*A*a^2*cos(d*x + c)^4 + (355*A + 99*C)*a^2*cos(d*x + c)^3 + 6*(71*A + 66*C
)*a^2*cos(d*x + c)^2 + (568*A + 759*C)*a^2*cos(d*x + c) + 2*(568*A + 759*C)*a^2)*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. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 695 vs. \(2 (230) = 460\).

Time = 0.50 (sec) , antiderivative size = 695, normalized size of antiderivative = 2.61 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/22176*(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*c
os(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*arc
tan2(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(8/1
1*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*arctan2(
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))) + 3465
*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 + 11/2
*c))))*A*sqrt(a) - 132*sqrt(2)*(77*a^2*cos(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) -
42*a^2*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 77*a^2*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*
d*x + 2*c))) - 630*a^2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (77*a^2*cos(2*d*x + 2*c) + 6*a^2
)*sin(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*C*sqrt(a))/d

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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