\(\int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [373]

Optimal result

Integrand size = 42, antiderivative size = 209 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{3/2} (75 B+88 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^2 (75 B+88 C) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (75 B+88 C) \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (9 B+8 C) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a B \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d} \] Output:

1/64*a^(3/2)*(75*B+88*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2)) 
/d+1/64*a^2*(75*B+88*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/96*a^2*(75*B 
+88*C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/24*a^2*(9*B+8*C)*c 
os(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/4*a*B*cos(d*x+c)^3*(a+a* 
sec(d*x+c))^(1/2)*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 9.02 (sec) , antiderivative size = 1026, normalized size of antiderivative = 4.91 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

Integrate[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Se 
c[c + d*x]^2),x]
 

Output:

a*((C*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x 
])*Sec[c/2 + (d*x)/2]^2*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])])/(2*Sq 
rt[2]*d) - (C*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*Sec[(c + d*x)/2]*Sqr 
t[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sqrt[C 
os[c + d*x]] + 2*Cos[c + d*x]*(4*Sin[(c + d*x)/2] - 3*Sin[(3*(c + d*x))/2] 
 - 2*Sin[(5*(c + d*x))/2])))/(96*d) - (C*(1 + Cos[c + d*x])*Sec[c/2 + (d*x 
)/2]^2*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(Sqrt[2]*ArcSin[Sqrt[2] 
*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + Sin[(c + d*x)/2] - 2*Sin[(3*(c + d 
*x))/2] - Sin[(5*(c + d*x))/2]))/(16*d) - (B*(1 + Cos[c + d*x])*Sec[c/2 + 
(d*x)/2]^2*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*ArcSin[S 
qrt[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + 5*Sin[(c + d*x)/2] - 16*Sin[ 
(3*(c + d*x))/2] - 9*Sin[(5*(c + d*x))/2] - 8*Sin[(7*(c + d*x))/2] - 6*Sin 
[(9*(c + d*x))/2]))/(768*d) + (3*B*(1 + Cos[c + d*x])*Hypergeometric2F1[1/ 
2, 3, 3/2, 1 - Sec[c + d*x]]*Sec[c/2 + (d*x)/2]^2*Sqrt[a*(1 + Sec[c + d*x] 
)]*Tan[c + d*x])/(4*d*(1 + Sec[c + d*x])) + (B*(1 + Cos[c + d*x])*Sec[c/2 
+ (d*x)/2]^2*(ArcTanh[Sqrt[1 - Sec[c + d*x]]] + Cos[c + d*x]*Sqrt[1 - Sec[ 
c + d*x]])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[c + d*x])/(4*d*Sqrt[1 + Sec[c + 
d*x]]*Sqrt[-Tan[c + d*x]^2]) + (3*C*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^ 
2*(ArcTanh[Sqrt[1 - Sec[c + d*x]]] + Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]])* 
Sqrt[a*(1 + Sec[c + d*x])]*Tan[c + d*x])/(8*d*Sqrt[1 + Sec[c + d*x]]*Sq...
 

Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.96, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4560, 3042, 4505, 27, 3042, 4503, 3042, 4292, 3042, 4292, 3042, 4261, 216}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \cos ^5(c+d x) (a \sec (c+d x)+a)^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\)

\(\Big \downarrow \) 4560

\(\displaystyle \int \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2} (B+C \sec (c+d x))dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\)

\(\Big \downarrow \) 4505

\(\displaystyle \frac {1}{4} \int \frac {1}{2} \cos ^3(c+d x) \sqrt {\sec (c+d x) a+a} (a (9 B+8 C)+a (5 B+8 C) \sec (c+d x))dx+\frac {a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} \int \cos ^3(c+d x) \sqrt {\sec (c+d x) a+a} (a (9 B+8 C)+a (5 B+8 C) \sec (c+d x))dx+\frac {a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (9 B+8 C)+a (5 B+8 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 B+88 C) \int \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 (9 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 B+88 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 (9 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 4292

\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 B+88 C) \left (\frac {3}{4} \int \cos (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (9 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 B+88 C) \left (\frac {3}{4} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (9 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 4292

\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 B+88 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (9 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 B+88 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (9 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 B+88 C) \left (\frac {3}{4} \left (\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (9 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{8} \left (\frac {a^2 (9 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {1}{6} a (75 B+88 C) \left (\frac {3}{4} \left (\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\right )+\frac {a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

Input:

Int[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + 
d*x]^2),x]
 

Output:

(a*B*Cos[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*d) + ((a^2*( 
9*B + 8*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + ( 
a*(75*B + 88*C)*((a*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x 
]]) + (3*((Sqrt[a]*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]] 
)/d + (a*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4))/6)/8
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4261

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

rule 4292

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] + Simp[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 4503

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*Co 
t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp 
[(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] /; FreeQ[{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 4505

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] - Sim 
p[b/(a*d*n)   Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim 
p[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 4560

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. 
)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) 
*(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2   Int[(a + b*Csc[e + f*x])^(m 
+ 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ 
[{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
 

Maple [A] (verified)

Time = 4.99 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.34

method	result	size
default	\(-\frac {a \left (\left (225 \cos \left (d x +c \right )+225\right ) B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (264 \cos \left (d x +c \right )+264\right ) C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-48 \cos \left (d x +c \right )^{3}-120 \cos \left (d x +c \right )^{2}-150 \cos \left (d x +c \right )-225\right ) B +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-64 \cos \left (d x +c \right )^{2}-176 \cos \left (d x +c \right )-264\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{192 d \left (\cos \left (d x +c \right )+1\right )}\)	\(280\)

Input:

int(cos(d*x+c)^5*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,me 
thod=_RETURNVERBOSE)
 

Output:

-1/192/d*a*((225*cos(d*x+c)+225)*B*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arct 
anh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(csc(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c) 
+cot(d*x+c)^2-1)^(1/2))+(264*cos(d*x+c)+264)*C*(-cos(d*x+c)/(cos(d*x+c)+1) 
)^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(csc(d*x+c)^2-2*csc(d*x+c 
)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2))+sin(d*x+c)*cos(d*x+c)*(-48*cos(d*x+c)^ 
3-120*cos(d*x+c)^2-150*cos(d*x+c)-225)*B+sin(d*x+c)*cos(d*x+c)*(-64*cos(d* 
x+c)^2-176*cos(d*x+c)-264)*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.89 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {3 \, {\left ({\left (75 \, B + 88 \, C\right )} a \cos \left (d x + c\right ) + {\left (75 \, B + 88 \, C\right )} a\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (48 \, B a \cos \left (d x + c\right )^{4} + 8 \, {\left (15 \, B + 8 \, C\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (75 \, B + 88 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (75 \, B + 88 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{384 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (75 \, B + 88 \, C\right )} a \cos \left (d x + c\right ) + {\left (75 \, B + 88 \, C\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (48 \, B a \cos \left (d x + c\right )^{4} + 8 \, {\left (15 \, B + 8 \, C\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (75 \, B + 88 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (75 \, B + 88 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="fricas")
 

Output:

[1/384*(3*((75*B + 88*C)*a*cos(d*x + c) + (75*B + 88*C)*a)*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*B 
*a*cos(d*x + c)^4 + 8*(15*B + 8*C)*a*cos(d*x + c)^3 + 2*(75*B + 88*C)*a*co 
s(d*x + c)^2 + 3*(75*B + 88*C)*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/c 
os(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/192*(3*((75*B + 88*C)* 
a*cos(d*x + c) + (75*B + 88*C)*a)*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*B*a*cos(d*x + c) 
^4 + 8*(15*B + 8*C)*a*cos(d*x + c)^3 + 2*(75*B + 88*C)*a*cos(d*x + c)^2 + 
3*(75*B + 88*C)*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*si 
n(d*x + c))/(d*cos(d*x + c) + d)]
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**5*(a+a*sec(d*x+c))**(3/2)*(B*sec(d*x+c)+C*sec(d*x+c) 
**2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1088 vs. \(2 (185) = 370\).

Time = 1.09 (sec) , antiderivative size = 1088, normalized size of antiderivative = 5.21 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="giac")
 

Output:

-1/384*(3*(75*B*sqrt(-a)*a*sgn(cos(d*x + c)) + 88*C*sqrt(-a)*a*sgn(cos(d*x 
 + c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2 
*c)^2 + a))^2 - a*(2*sqrt(2) + 3))) - 3*(75*B*sqrt(-a)*a*sgn(cos(d*x + c)) 
 + 88*C*sqrt(-a)*a*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2* 
c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*sqrt 
(2)*(225*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + 
 a))^14*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 264*(sqrt(-a)*tan(1/2*d*x + 1/2 
*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*C*sqrt(-a)*a^2*sgn(cos(d*x + 
 c)) - 6261*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^ 
2 + a))^12*B*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 4008*(sqrt(-a)*tan(1/2*d*x + 
 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*C*sqrt(-a)*a^3*sgn(cos(d 
*x + c)) + 35925*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/ 
2*c)^2 + a))^10*B*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 33960*(sqrt(-a)*tan(1/2 
*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*C*sqrt(-a)*a^4*sgn 
(cos(d*x + c)) - 127449*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d 
*x + 1/2*c)^2 + a))^8*B*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 131784*(sqrt(-a)* 
tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*C*sqrt(-a)*a 
^5*sgn(cos(d*x + c)) + 101667*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan 
(1/2*d*x + 1/2*c)^2 + a))^6*B*sqrt(-a)*a^6*sgn(cos(d*x + c)) + 108312*(sqr 
t(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*s...
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^5\,\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:

int(cos(c + d*x)^5*(B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x) 
)^(3/2),x)
 

Output:

int(cos(c + d*x)^5*(B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x) 
)^(3/2), x)
 

Reduce [F]

\[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{5} \sec \left (d x +c \right )^{3}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{5} \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{5} \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{5} \sec \left (d x +c \right )d x \right ) b \right ) \] Input:

int(cos(d*x+c)^5*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

sqrt(a)*a*(int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**5*sec(c + d*x)**3,x)*c 
 + int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**5*sec(c + d*x)**2,x)*b + int(s 
qrt(sec(c + d*x) + 1)*cos(c + d*x)**5*sec(c + d*x)**2,x)*c + int(sqrt(sec( 
c + d*x) + 1)*cos(c + d*x)**5*sec(c + d*x),x)*b)