\(\int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\) [180]

Optimal result

Integrand size = 35, antiderivative size = 200 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (163 A+304 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^3 (299 A+432 C) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (17 A+16 C) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {5 a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d} \]

[Out]

1/64*a^(5/2)*(163*A+304*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+5/24*a*A*cos(d*x+c)^2*(a+a*sec(
d*x+c))^(3/2)*sin(d*x+c)/d+1/4*A*cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+1/192*a^3*(299*A+432*C)*sin(
d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/32*a^2*(17*A+16*C)*cos(d*x+c)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4172, 4102, 4100, 3859, 209} \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (163 A+304 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{64 d}+\frac {a^3 (299 A+432 C) \sin (c+d x)}{192 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (17 A+16 C) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{32 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}+\frac {5 a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{24 d} \]

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^(5/2)*(163*A + 304*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a^3*(299*A + 432*C
)*Sin[c + d*x])/(192*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(17*A + 16*C)*Cos[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*Si
n[c + d*x])/(32*d) + (5*a*A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(24*d) + (A*Cos[c + d*x]^3
*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(4*d)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3859

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(a + x^2), x], x, b*(C
ot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

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])

Rubi steps \begin{align*} \text {integral}& = \frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {5 a A}{2}+\frac {1}{2} a (A+8 C) \sec (c+d x)\right ) \, dx}{4 a} \\ & = \frac {5 a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {\int \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {3}{4} a^2 (17 A+16 C)+\frac {1}{4} a^2 (11 A+48 C) \sec (c+d x)\right ) \, dx}{12 a} \\ & = \frac {a^2 (17 A+16 C) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {5 a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {\int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{8} a^3 (299 A+432 C)+\frac {5}{8} a^3 (19 A+48 C) \sec (c+d x)\right ) \, dx}{24 a} \\ & = \frac {a^3 (299 A+432 C) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (17 A+16 C) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {5 a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {1}{128} \left (a^2 (163 A+304 C)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {a^3 (299 A+432 C) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (17 A+16 C) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {5 a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}-\frac {\left (a^3 (163 A+304 C)\right ) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d} \\ & = \frac {a^{5/2} (163 A+304 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^3 (299 A+432 C) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (17 A+16 C) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {5 a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 3.16 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.68 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \left (675 A \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+560 C \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+231 A \sqrt {1-\sec (c+d x)}-240 C \sqrt {1-\sec (c+d x)}+849 A \cos (c+d x) \sqrt {1-\sec (c+d x)}-80 C \cos (c+d x) \sqrt {1-\sec (c+d x)}+233 A \cos (2 (c+d x)) \sqrt {1-\sec (c+d x)}-240 C \cos (2 (c+d x)) \sqrt {1-\sec (c+d x)}+58 A \cos (3 (c+d x)) \sqrt {1-\sec (c+d x)}+2 A \cos (4 (c+d x)) \sqrt {1-\sec (c+d x)}+2560 C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}+512 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},5,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{320 d \sqrt {1-\sec (c+d x)}} \]

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(675*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 560*C*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 231*A*Sqrt[1 - Sec[c + d
*x]] - 240*C*Sqrt[1 - Sec[c + d*x]] + 849*A*Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]] - 80*C*Cos[c + d*x]*Sqrt[1 - S
ec[c + d*x]] + 233*A*Cos[2*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] - 240*C*Cos[2*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] +
58*A*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 2*A*Cos[4*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 2560*C*Hypergeom
etric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]] + 512*A*Hypergeometric2F1[1/2, 5, 3/2, 1 - Sec[
c + d*x]]*Sqrt[1 - Sec[c + d*x]])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(320*d*Sqrt[1 - Sec[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(385\) vs. \(2(176)=352\).

Time = 0.32 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.93

\[\frac {a^{2} \left (48 A \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+184 A \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )+489 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+326 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+912 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+96 C \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+489 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+489 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+912 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+528 C \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{192 d \left (\cos \left (d x +c \right )+1\right )}\]

[In]

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)

[Out]

1/192*a^2/d*(48*A*cos(d*x+c)^4*sin(d*x+c)+184*A*cos(d*x+c)^3*sin(d*x+c)+489*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/
2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+326*A*cos(d*x+c)^2*sin(d*x
+c)+912*C*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1
/2))*cos(d*x+c)+96*C*cos(d*x+c)^2*sin(d*x+c)+489*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(
d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+489*A*cos(d*x+c)*sin(d*x+c)+912*C*(-cos(d*x+c)/(cos(d*x+c)+1))^(
1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+528*C*cos(d*x+c)*sin(d*x+c))*(a*(1+
sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.04 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {3 \, {\left ({\left (163 \, A + 304 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (163 \, A + 304 \, C\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (48 \, A a^{2} \cos \left (d x + c\right )^{4} + 184 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (163 \, A + 48 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (163 \, A + 176 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{384 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (163 \, A + 304 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (163 \, A + 304 \, C\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (48 \, A a^{2} \cos \left (d x + c\right )^{4} + 184 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (163 \, A + 48 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (163 \, A + 176 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/384*(3*((163*A + 304*C)*a^2*cos(d*x + c) + (163*A + 304*C)*a^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-
a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1))
+ 2*(48*A*a^2*cos(d*x + c)^4 + 184*A*a^2*cos(d*x + c)^3 + 2*(163*A + 48*C)*a^2*cos(d*x + c)^2 + 3*(163*A + 17
6*C)*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/192*(3*(
(163*A + 304*C)*a^2*cos(d*x + c) + (163*A + 304*C)*a^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))
*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (48*A*a^2*cos(d*x + c)^4 + 184*A*a^2*cos(d*x + c)^3 + 2*(163*A + 48*C)
*a^2*cos(d*x + c)^2 + 3*(163*A + 176*C)*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)
)/(d*cos(d*x + c) + d)]

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*(a+a*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cos(d*x+c)^4*(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. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^4\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

[In]

int(cos(c + d*x)^4*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^4*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2), x)