\(\int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x)) \, dx\) [171]

Optimal result

Integrand size = 35, antiderivative size = 200 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{3/2} (75 A+112 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^2 (75 A+112 C) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (13 A+16 C) \cos (c+d x) \sin (c+d x)}{32 d \sqrt {a+a \sec (c+d x)}}+\frac {a A \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d} \] Output:

1/64*a^(3/2)*(75*A+112*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2) 
)/d+1/64*a^2*(75*A+112*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/32*a^2*(13 
*A+16*C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/8*a*A*cos(d*x+c) 
^2*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d+1/4*A*cos(d*x+c)^3*(a+a*sec(d*x+c)) 
^(3/2)*sin(d*x+c)/d
 

Mathematica [C] (verified)

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

Time = 1.79 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.76 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (49 C \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+\cos (c+d x) \left (49 C+14 C \cos (c+d x)-8 A \cos ^3(c+d x)\right ) \sqrt {1-\sec (c+d x)}+120 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},5,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{28 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \] Input:

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2) 
,x]
 

Output:

(a*(49*C*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + Cos[c + d*x]*(49*C + 14*C*Cos[c 
 + d*x] - 8*A*Cos[c + d*x]^3)*Sqrt[1 - Sec[c + d*x]] + 120*A*Hypergeometri 
c2F1[1/2, 5, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]])*Sqrt[a*(1 + Se 
c[c + d*x])]*Sin[c + d*x])/(28*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]] 
)
 

Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 4575, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4575

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \cos ^3(c+d x) (\sec (c+d x) a+a)^{3/2} (3 a A+a (3 A+8 C) \sec (c+d x))dx}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 a A+a (3 A+8 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 4505

\(\displaystyle \frac {\frac {1}{3} \int \frac {3}{2} \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a} \left ((13 A+16 C) a^2+(9 A+16 C) \sec (c+d x) a^2\right )dx+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{2} \int \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a} \left ((13 A+16 C) a^2+(9 A+16 C) \sec (c+d x) a^2\right )dx+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((13 A+16 C) a^2+(9 A+16 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \int \cos (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^3 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^3 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 4292

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \left (\frac {1}{2} \int \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \left (\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^3 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}+\frac {1}{2} \left (\frac {a^3 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}+\frac {1}{4} a^2 (75 A+112 C) \left (\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\)

Input:

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2),x]
 

Output:

(A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(4*d) + ((a^2*A 
*Cos[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d + ((a^3*(13*A + 1 
6*C)*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(75* 
A + 112*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)/2)/(8*a)
 

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 4575

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*Co 
t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[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])
 

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.35

method	result	size
default	\(-\frac {a \left (\left (75 \cos \left (d x +c \right )+75\right ) A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (112 \cos \left (d x +c \right )+112\right ) C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-16 \cos \left (d x +c \right )^{3}-40 \cos \left (d x +c \right )^{2}-50 \cos \left (d x +c \right )-75\right ) A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-32 \cos \left (d x +c \right )-112\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{64 d \left (\cos \left (d x +c \right )+1\right )}\)	\(270\)

Input:

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x,method=_RETUR 
NVERBOSE)
 

Output:

-1/64/d*a*((75*cos(d*x+c)+75)*A*(-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))+(112*cos(d*x+c)+112)*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)*c 
ot(d*x+c)+cot(d*x+c)^2-1)^(1/2))+sin(d*x+c)*cos(d*x+c)*(-16*cos(d*x+c)^3-4 
0*cos(d*x+c)^2-50*cos(d*x+c)-75)*A+sin(d*x+c)*cos(d*x+c)*(-32*cos(d*x+c)-1 
12)*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.90 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {{\left ({\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right ) + {\left (75 \, A + 112 \, C\right )} a\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (16 \, A a \cos \left (d x + c\right )^{4} + 40 \, A a \cos \left (d x + c\right )^{3} + 2 \, {\left (25 \, A + 16 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{128 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {{\left ({\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right ) + {\left (75 \, A + 112 \, C\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (16 \, A a \cos \left (d x + c\right )^{4} + 40 \, A a \cos \left (d x + c\right )^{3} + 2 \, {\left (25 \, A + 16 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:

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

Output:

[1/128*(((75*A + 112*C)*a*cos(d*x + c) + (75*A + 112*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*(16*A 
*a*cos(d*x + c)^4 + 40*A*a*cos(d*x + c)^3 + 2*(25*A + 16*C)*a*cos(d*x + c) 
^2 + (75*A + 112*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), -1/64*(((75*A + 112*C)*a*cos(d*x + c 
) + (75*A + 112*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))) - (16*A*a*cos(d*x + c)^4 + 40*A*a* 
cos(d*x + c)^3 + 2*(25*A + 16*C)*a*cos(d*x + c)^2 + (75*A + 112*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)]
 

Sympy [F(-1)]

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

integrate(cos(d*x+c)**4*(a+a*sec(d*x+c))**(3/2)*(A+C*sec(d*x+c)**2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

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

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1087 vs. \(2 (176) = 352\).

Time = 1.06 (sec) , antiderivative size = 1087, normalized size of antiderivative = 5.44 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/128*((75*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 112*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))) - (75*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 
112*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 
)*(75*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a) 
)^14*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 112*(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) 
) - 2087*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + 
 a))^12*A*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 2864*(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)) + 11975*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c 
)^2 + a))^10*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 23344*(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(co 
s(d*x + c)) - 42483*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 
 1/2*c)^2 + a))^8*A*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 69360*(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*sg 
n(cos(d*x + c)) + 33889*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d 
*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^6*sgn(cos(d*x + c)) + 51536*(sqrt(-a)*t 
an(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a)...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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