\(\int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [141]

Optimal result

Integrand size = 33, antiderivative size = 164 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^{5/2} (25 A+38 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a^3 (49 A+54 B) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (3 A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d} \] Output:

1/8*a^(5/2)*(25*A+38*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/ 
d+1/24*a^3*(49*A+54*B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/4*a^2*(3*A+2* 
B)*cos(d*x+c)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d+1/3*a*A*cos(d*x+c)^2*(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 = 0.73 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.71 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=-\frac {2 a^2 \left (\cos (c+d x) (A+3 B+(A+4 B) \cos (c+d x)+A \cos (2 (c+d x)))-19 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right )-15 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4,\frac {3}{2},1-\sec (c+d x)\right )\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{3 d (1+\cos (c+d x))} \] Input:

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

Output:

(-2*a^2*(Cos[c + d*x]*(A + 3*B + (A + 4*B)*Cos[c + d*x] + A*Cos[2*(c + d*x 
)]) - 19*B*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]] - 15*A*Hyperge 
ometric2F1[1/2, 4, 3/2, 1 - Sec[c + d*x]])*Sqrt[a*(1 + Sec[c + d*x])]*Sin[ 
c + d*x])/(3*d*(1 + Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4505, 27, 3042, 4505, 27, 3042, 4503, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4505

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4505

\(\displaystyle \frac {1}{6} \left (\frac {1}{2} \int \frac {1}{2} \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left ((49 A+54 B) a^2+(13 A+30 B) \sec (c+d x) a^2\right )dx+\frac {3 a^2 (3 A+2 B) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (\frac {1}{4} \int \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left ((49 A+54 B) a^2+(13 A+30 B) \sec (c+d x) a^2\right )dx+\frac {3 a^2 (3 A+2 B) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {1}{4} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((49 A+54 B) a^2+(13 A+30 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {3 a^2 (3 A+2 B) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^2 (25 A+38 B) \int \sqrt {\sec (c+d x) a+a}dx+\frac {a^3 (49 A+54 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {3 a^2 (3 A+2 B) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^2 (25 A+38 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^3 (49 A+54 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {3 a^2 (3 A+2 B) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {a^3 (49 A+54 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 a^3 (25 A+38 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {3 a^2 (3 A+2 B) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{6} \left (\frac {3 a^2 (3 A+2 B) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}+\frac {1}{4} \left (\frac {3 a^{5/2} (25 A+38 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^3 (49 A+54 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

Input:

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

Output:

(a*A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + ((3*a 
^2*(3*A + 2*B)*Cos[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + 
 ((3*a^(5/2)*(25*A + 38*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c 
+ d*x]]])/d + (a^3*(49*A + 54*B)*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 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]
 

Maple [A] (verified)

Time = 244.39 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.60

method	result	size
default	\(\frac {a^{2} \left (\left (-75 \cos \left (d x +c \right )-75\right ) A \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+\left (-114 \cos \left (d x +c \right )-114\right ) B \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (8 \cos \left (d x +c \right )^{2}+34 \cos \left (d x +c \right )+75\right ) A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (12 \cos \left (d x +c \right )+66\right ) B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{24 d \left (1+\cos \left (d x +c \right )\right )}\)	\(262\)

Input:

int(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

Output:

1/24/d*a^2*((-75*cos(d*x+c)-75)*A*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcta 
nh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(cot(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+ 
csc(d*x+c)^2-1)^(1/2))+(-114*cos(d*x+c)-114)*B*(-cos(d*x+c)/(1+cos(d*x+c)) 
)^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(cot(d*x+c)^2-2*csc(d*x+c 
)*cot(d*x+c)+csc(d*x+c)^2-1)^(1/2))+sin(d*x+c)*cos(d*x+c)*(8*cos(d*x+c)^2+ 
34*cos(d*x+c)+75)*A+sin(d*x+c)*cos(d*x+c)*(12*cos(d*x+c)+66)*B)*(a*(1+sec( 
d*x+c)))^(1/2)/(1+cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.32 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\left [\frac {3 \, {\left ({\left (25 \, A + 38 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (25 \, A + 38 \, B\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (17 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (25 \, A + 22 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (25 \, A + 38 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (25 \, A + 38 \, B\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (17 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (25 \, A + 22 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:

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

Output:

[1/48*(3*((25*A + 38*B)*a^2*cos(d*x + c) + (25*A + 38*B)*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*(8 
*A*a^2*cos(d*x + c)^3 + 2*(17*A + 6*B)*a^2*cos(d*x + c)^2 + 3*(25*A + 22*B 
)*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), -1/24*(3*((25*A + 38*B)*a^2*cos(d*x + c) + (25*A + 3 
8*B)*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))) - (8*A*a^2*cos(d*x + c)^3 + 2*(17*A + 6*B)*a^2 
*cos(d*x + c)^2 + 3*(25*A + 22*B)*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. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 905 vs. \(2 (144) = 288\).

Time = 0.96 (sec) , antiderivative size = 905, normalized size of antiderivative = 5.52 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/48*(3*(25*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 38*B*sqrt(-a)*a^2*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*(25*A*sqrt(-a)*a^2*sgn(cos(d*x 
+ c)) + 38*B*sqrt(-a)*a^2*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*(75*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*A*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 114*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*B*sqrt(-a)*a^3 
*sgn(cos(d*x + c)) - 1125*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*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 1710*sq 
rt(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^4*sgn(cos(d*x + c)) + 6174*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*A*sqrt(-a)*a^5*sgn(cos(d 
*x + c)) + 6804*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^5*sgn(cos(d*x + c)) - 4314*sqrt(2)*(sqr 
t(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt 
(-a)*a^6*sgn(cos(d*x + c)) - 4284*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^6*sgn(cos(d*x + c)) + 
 807*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*A*sqrt(-a)*a^7*sgn(cos(d*x + c)) + 858*sqrt(2)*(sqrt(-a)*tan...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a**2*(int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**3,x 
)*b + int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**2,x)*a + 2* 
int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**2,x)*b + 2*int(sq 
rt(sec(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x),x)*a + int(sqrt(sec(c + 
d*x) + 1)*cos(c + d*x)**3*sec(c + d*x),x)*b + int(sqrt(sec(c + d*x) + 1)*c 
os(c + d*x)**3,x)*a)