\(\int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [372]

Optimal result

Integrand size = 42, antiderivative size = 164 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{3/2} (11 B+14 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a^2 (11 B+14 C) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (7 B+6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a B \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d} \] Output:

1/8*a^(3/2)*(11*B+14*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/ 
d+1/8*a^2*(11*B+14*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/12*a^2*(7*B+6* 
C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/3*a*B*cos(d*x+c)^2*(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 = 8.77 (sec) , antiderivative size = 738, normalized size of antiderivative = 4.50 \[ \int \cos ^4(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=a \left (\frac {C \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)} (1+\cos (c+d x)) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sec (c+d x))}}{2 \sqrt {2} d}-\frac {C (1+\cos (c+d x)) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sec (c+d x))} \left (\sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+\sin \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {3}{2} (c+d x)\right )-\sin \left (\frac {5}{2} (c+d x)\right )\right )}{16 d}+\frac {B (1+\cos (c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a (1+\sec (c+d x))} \tan (c+d x)}{d (1+\sec (c+d x))}+\frac {B (1+\cos (c+d x)) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+\cos (c+d x) \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \tan (c+d x)}{4 d \sqrt {1+\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}}+\frac {C (1+\cos (c+d x)) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+\cos (c+d x) \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \tan (c+d x)}{2 d \sqrt {1+\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}}+\frac {B (1+\cos (c+d x)) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a (1+\sec (c+d x))} \left (-\sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {1+\sec (c+d x)} \left (3 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+2 \cos (c+d x) \left (4 \sin \left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {3}{2} (c+d x)\right )-2 \sin \left (\frac {5}{2} (c+d x)\right )\right )\right )+\frac {12 \left (\text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+\cos (c+d x) \sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{\sqrt {-\tan ^2(c+d x)}}\right )}{96 d \sqrt {1+\sec (c+d x)}}\right ) \] Input:

Integrate[Cos[c + d*x]^4*(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])]*(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])*Hypergeometric2F1[1/2, 3, 3/2, 1 - Se 
c[c + d*x]]*Sec[c/2 + (d*x)/2]^2*Sqrt[a*(1 + Sec[c + d*x])]*Tan[c + d*x])/ 
(d*(1 + Sec[c + d*x])) + (B*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*(ArcTa 
nh[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]) + (C*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*(ArcTanh[Sqrt[1 - S 
ec[c + d*x]]] + Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]])*Sqrt[a*(1 + Sec[c + d 
*x])]*Tan[c + d*x])/(2*d*Sqrt[1 + Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2]) + ( 
B*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*Sqrt[a*(1 + Sec[c + d*x])]*(-(Se 
c[(c + d*x)/2]*Sqrt[1 + Sec[c + d*x]]*(3*Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d 
*x)/2]]*Sqrt[Cos[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]))) + (12*(ArcTanh[Sqrt[1 - Sec[c + 
 d*x]]] + Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]])*Tan[c + d*x])/Sqrt[-Tan[c + 
 d*x]^2]))/(96*d*Sqrt[1 + Sec[c + d*x]]))
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4560, 3042, 4505, 27, 3042, 4503, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4505

\(\displaystyle \frac {1}{3} \int \frac {1}{2} \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a} (a (7 B+6 C)+3 a (B+2 C) \sec (c+d x))dx+\frac {a B \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \int \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a} (a (7 B+6 C)+3 a (B+2 C) \sec (c+d x))dx+\frac {a B \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (7 B+6 C)+3 a (B+2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a B \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (11 B+14 C) \int \cos (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 (7 B+6 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (11 B+14 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 (7 B+6 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 4292

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (11 B+14 C) \left (\frac {1}{2} \int \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (7 B+6 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (11 B+14 C) \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (7 B+6 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (11 B+14 C) \left (\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^2 (7 B+6 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{6} \left (\frac {a^2 (7 B+6 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}+\frac {3}{4} a (11 B+14 C) \left (\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

Input:

Int[Cos[c + d*x]^4*(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]^2*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d) + ((a^2*( 
7*B + 6*C)*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (3* 
a*(11*B + 14*C)*((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
 

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 = 4.43 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.59

method	result	size
default	\(-\frac {a \left (\left (33 \cos \left (d x +c \right )+33\right ) B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (42 \cos \left (d x +c \right )+42\right ) C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-8 \cos \left (d x +c \right )^{2}-22 \cos \left (d x +c \right )-33\right ) B +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-12 \cos \left (d x +c \right )-42\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{24 d \left (\cos \left (d x +c \right )+1\right )}\)	\(260\)

Input:

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,me 
thod=_RETURNVERBOSE)
 

Output:

-1/24/d*a*((33*cos(d*x+c)+33)*B*(-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)+co 
t(d*x+c)^2-1)^(1/2))+(42*cos(d*x+c)+42)*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)*(-8*cos(d*x+c)^2-22*c 
os(d*x+c)-33)*B+sin(d*x+c)*cos(d*x+c)*(-12*cos(d*x+c)-42)*C)*(a*(1+sec(d*x 
+c)))^(1/2)/(cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.20 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {3 \, {\left ({\left (11 \, B + 14 \, C\right )} a \cos \left (d x + c\right ) + {\left (11 \, B + 14 \, C\right )} a\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (8 \, B a \cos \left (d x + c\right )^{3} + 2 \, {\left (11 \, B + 6 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (11 \, B + 14 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (11 \, B + 14 \, C\right )} a \cos \left (d x + c\right ) + {\left (11 \, B + 14 \, C\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (8 \, B a \cos \left (d x + c\right )^{3} + 2 \, {\left (11 \, B + 6 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (11 \, B + 14 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="fricas")
 

Output:

[1/48*(3*((11*B + 14*C)*a*cos(d*x + c) + (11*B + 14*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*(8*B*a 
*cos(d*x + c)^3 + 2*(11*B + 6*C)*a*cos(d*x + c)^2 + 3*(11*B + 14*C)*a*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), -1/24*(3*((11*B + 14*C)*a*cos(d*x + c) + (11*B + 14*C)*a)*sqrt 
(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*s 
in(d*x + c))) - (8*B*a*cos(d*x + c)^3 + 2*(11*B + 6*C)*a*cos(d*x + c)^2 + 
3*(11*B + 14*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. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**4*(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. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^4*(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. 897 vs. \(2 (144) = 288\).

Time = 0.96 (sec) , antiderivative size = 897, normalized size of antiderivative = 5.47 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="giac")
 

Output:

-1/48*(3*(11*B*sqrt(-a)*a*sgn(cos(d*x + c)) + 14*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*(11*B*sqrt(-a)*a*sgn(cos(d*x + c)) 
+ 14*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*(33*s 
qrt(2)*(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^2*sgn(cos(d*x + c)) + 42*sqrt(2)*(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^2*sgn(cos( 
d*x + c)) - 303*sqrt(2)*(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^3*sgn(cos(d*x + c)) - 822*sqrt(2)*(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^3*sgn(cos(d*x + c)) + 2394*sqrt(2)*(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^4*sgn(cos(d*x + c)) + 
3780*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c) 
^2 + a))^6*C*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 1806*sqrt(2)*(sqrt(-a)*tan(1 
/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*sqrt(-a)*a^5*sg 
n(cos(d*x + c)) - 2508*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*ta 
n(1/2*d*x + 1/2*c)^2 + a))^4*C*sqrt(-a)*a^5*sgn(cos(d*x + c)) + 309*sqrt(2 
)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2* 
B*sqrt(-a)*a^6*sgn(cos(d*x + c)) + 498*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + ...
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^4\,\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:

int(cos(c + d*x)^4*(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)^4*(B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x) 
)^(3/2), x)
 

Reduce [F]

\[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{3}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )d x \right ) b \right ) \] Input:

int(cos(d*x+c)^4*(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)**4*sec(c + d*x)**3,x)*c 
 + int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**4*sec(c + d*x)**2,x)*b + int(s 
qrt(sec(c + d*x) + 1)*cos(c + d*x)**4*sec(c + d*x)**2,x)*c + int(sqrt(sec( 
c + d*x) + 1)*cos(c + d*x)**4*sec(c + d*x),x)*b)