\(\int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x)) \, dx\) [1135]

Optimal result

Integrand size = 37, antiderivative size = 266 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {16 a^2 (112 A+143 C) \sin (c+d x)}{1155 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {8 a^2 (112 A+143 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (112 A+143 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{385 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (28 A+33 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{33 d}+\frac {2 A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d} \]

[Out]

2/11*A*cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/385*a^2*(112*A+143*C)*cos(d*x+c)^(3/2)*sin(d*x+c
)/d/(a+a*sec(d*x+c))^(1/2)+2/231*a^2*(28*A+33*C)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/1155*
a^2*(112*A+143*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+8/1155*a^2*(112*A+143*C)*sin(d*x+c)*cos
(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/33*a*A*cos(d*x+c)^(7/2)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.98 (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} \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^2 (28 A+33 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{231 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (112 A+143 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{385 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a^2 (112 A+143 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{1155 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (112 A+143 C) \sin (c+d x)}{1155 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 A \sin (c+d x) \cos ^{\frac {9}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}+\frac {2 a A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{33 d} \]

[In]

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

[Out]

(16*a^2*(112*A + 143*C)*Sin[c + d*x])/(1155*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a^2*(112*A + 1
43*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(1155*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(112*A + 143*C)*Cos[c + d*x]
^(3/2)*Sin[c + d*x])/(385*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(28*A + 33*C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/
(231*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*A*Cos[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(33*d) + (
2*A*Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*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))^{3/2} \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))^{3/2} \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))^{3/2} \left (\frac {3 a A}{2}+\frac {1}{2} a (6 A+11 C) \sec (c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx}{11 a} \\ & = \frac {2 a A \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{33 d}+\frac {2 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 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {3}{4} a^2 (28 A+33 C)+\frac {9}{4} a^2 (8 A+11 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{99 a} \\ & = \frac {2 a^2 (28 A+33 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{33 d}+\frac {2 A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac {1}{77} \left (a (112 A+143 C) \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^2 (112 A+143 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{385 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (28 A+33 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{33 d}+\frac {2 A \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac {1}{385} \left (4 a (112 A+143 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 {8 a^2 (112 A+143 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (112 A+143 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{385 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (28 A+33 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{33 d}+\frac {2 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 (112 A+143 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}{1155} \\ & = \frac {16 a^2 (112 A+143 C) \sin (c+d x)}{1155 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {8 a^2 (112 A+143 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (112 A+143 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{385 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (28 A+33 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{33 d}+\frac {2 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 = 7.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.47 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a \sqrt {\cos (c+d x)} (18494 A+21736 C+2 (5789 A+5566 C) \cos (c+d x)+8 (581 A+429 C) \cos (2 (c+d x))+1645 A \cos (3 (c+d x))+660 C \cos (3 (c+d x))+490 A \cos (4 (c+d x))+105 A \cos (5 (c+d x))) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{9240 d} \]

[In]

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

[Out]

(a*Sqrt[Cos[c + d*x]]*(18494*A + 21736*C + 2*(5789*A + 5566*C)*Cos[c + d*x] + 8*(581*A + 429*C)*Cos[2*(c + d*x
)] + 1645*A*Cos[3*(c + d*x)] + 660*C*Cos[3*(c + d*x)] + 490*A*Cos[4*(c + d*x)] + 105*A*Cos[5*(c + d*x)])*Sqrt[
a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(9240*d)

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.48

method	result	size
default	\(-\frac {2 a \left (\left (105 \cos \left (d x +c \right )^{5}+245 \cos \left (d x +c \right )^{4}+280 \cos \left (d x +c \right )^{3}+336 \cos \left (d x +c \right )^{2}+448 \cos \left (d x +c \right )+896\right ) A +\left (165 \cos \left (d x +c \right )^{3}+429 \cos \left (d x +c \right )^{2}+572 \cos \left (d x +c \right )+1144\right ) C \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{1155 d}\)	\(127\)

[In]

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

[Out]

-2/1155*a/d*((105*cos(d*x+c)^5+245*cos(d*x+c)^4+280*cos(d*x+c)^3+336*cos(d*x+c)^2+448*cos(d*x+c)+896)*A+(165*c
os(d*x+c)^3+429*cos(d*x+c)^2+572*cos(d*x+c)+1144)*C)*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 = 0.30 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.51 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (105 \, A a \cos \left (d x + c\right )^{5} + 245 \, A a \cos \left (d x + c\right )^{4} + 5 \, {\left (56 \, A + 33 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (112 \, A + 143 \, C\right )} a \cos \left (d x + c\right )^{2} + 4 \, {\left (112 \, A + 143 \, C\right )} a \cos \left (d x + c\right ) + 8 \, {\left (112 \, A + 143 \, C\right )} a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{1155 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

2/1155*(105*A*a*cos(d*x + c)^5 + 245*A*a*cos(d*x + c)^4 + 5*(56*A + 33*C)*a*cos(d*x + c)^3 + 3*(112*A + 143*C)
*a*cos(d*x + c)^2 + 4*(112*A + 143*C)*a*cos(d*x + c) + 8*(112*A + 143*C)*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. \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(11/2)*(a+a*sec(d*x+c))**(3/2)*(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. 652 vs. \(2 (230) = 460\).

Time = 0.50 (sec) , antiderivative size = 652, normalized size of antiderivative = 2.45 \[ \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/36960*(7*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 + 1
1/2*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) + 429*
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*ar
ctan2(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*arct
an2(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 + 11/
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))) + 363
0*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)*(175*a*cos(7/4*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) - 5*(35*a*cos(2*d*x + 2*c) + 6*a)*sin(7/4*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 126*a*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 175*a*sin
(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1470*a*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))
))*C*sqrt(a))/d

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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