\(\int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [139]

Optimal result

Integrand size = 31, antiderivative size = 143 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^{5/2} (5 A+2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}-\frac {a^3 (3 A+14 B) \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (A+2 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a B (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d} \] Output:

a^(5/2)*(5*A+2*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d-1/3* 
a^3*(3*A+14*B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2*a^2*(A+2*B)*(a+a*sec( 
d*x+c))^(1/2)*sin(d*x+c)/d+2/3*a*B*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 3.44 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.82 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^3 \left (3 (5 A+2 B) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)+\sqrt {1-\sec (c+d x)} (3 A \sin (c+d x)+2 (3 A+8 B+B \sec (c+d x)) \tan (c+d x))\right )}{3 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a^3*(3*(5*A + 2*B)*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] + Sqrt[1 
- Sec[c + d*x]]*(3*A*Sin[c + d*x] + 2*(3*A + 8*B + B*Sec[c + d*x])*Tan[c + 
 d*x])))/(3*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4506, 27, 3042, 4506, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

\(\displaystyle \frac {1}{3} \left (2 \int -\frac {1}{2} \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left (a^2 (3 A+14 B)-a^2 (9 A+10 B) \sec (c+d x)\right )dx+\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}-\int \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left (a^2 (3 A+14 B)-a^2 (9 A+10 B) \sec (c+d x)\right )dx\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}-\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a^2 (3 A+14 B)-a^2 (9 A+10 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {1}{3} \left (\frac {3}{2} a^2 (5 A+2 B) \int \sqrt {\sec (c+d x) a+a}dx-\frac {a^3 (3 A+14 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}+\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {3}{2} a^2 (5 A+2 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {a^3 (3 A+14 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}+\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {1}{3} \left (-\frac {3 a^3 (5 A+2 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {a^3 (3 A+14 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}+\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{3} \left (\frac {3 a^{5/2} (5 A+2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {a^3 (3 A+14 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}+\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

Input:

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

Output:

(2*a*B*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + ((3*a^(5/2)*(5*A + 
 2*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (a^3*(3 
*A + 14*B)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) + (6*a^2*(A + 2*B)*S 
qrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d)/3
 

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 4506

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

Maple [A] (verified)

Time = 23.50 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.73

method	result	size
default	\(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (15 \cos \left (d x +c \right )+15\right ) \sqrt {2}\, A \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+\left (6 \cos \left (d x +c \right )+6\right ) \sqrt {2}\, B \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+\sin \left (d x +c \right ) \left (6 \cos \left (d x +c \right )+12\right ) A +B \left (32 \sin \left (d x +c \right )+4 \tan \left (d x +c \right )\right )\right )}{6 d \left (1+\cos \left (d x +c \right )\right )}\)	\(247\)

Input:

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

Output:

1/6/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*((15*cos(d*x+c)+15)*2^(1 
/2)*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(2^(1/2)/(cot(d*x+c)^2-2 
*csc(d*x+c)*cot(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c)))+(6*c 
os(d*x+c)+6)*2^(1/2)*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(2^(1/2 
)/(cot(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(csc(d*x+c)- 
cot(d*x+c)))+sin(d*x+c)*(6*cos(d*x+c)+12)*A+B*(32*sin(d*x+c)+4*tan(d*x+c)) 
)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.70 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\left [\frac {3 \, {\left ({\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (3 \, A a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, -\frac {3 \, {\left ({\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (3 \, A a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \] Input:

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

Output:

[1/6*(3*((5*A + 2*B)*a^2*cos(d*x + c)^2 + (5*A + 2*B)*a^2*cos(d*x + c))*sq 
rt(-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*(3*A*a^2*cos(d*x + c)^2 + 2*(3*A + 8*B)*a^2*cos(d*x + c) + 2*B*a^2 
)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 
+ d*cos(d*x + c)), -1/3*(3*((5*A + 2*B)*a^2*cos(d*x + c)^2 + (5*A + 2*B)*a 
^2*cos(d*x + c))*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*co 
s(d*x + c)/(sqrt(a)*sin(d*x + c))) - (3*A*a^2*cos(d*x + c)^2 + 2*(3*A + 8* 
B)*a^2*cos(d*x + c) + 2*B*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin 
(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c))]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2780 vs. \(2 (127) = 254\).

Time = 0.32 (sec) , antiderivative size = 2780, normalized size of antiderivative = 19.44 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

1/12*(3*(18*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) 
+ 1)^(3/4)*a^(5/2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1) 
) + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^( 
1/4)*((4*a^2*sin(3*d*x + 3*c) + 5*a^2*sin(2*d*x + 2*c) + 4*a^2*sin(d*x + c 
))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (a^2*cos(2*d 
*x + 2*c)^2*sin(d*x + c) + a^2*sin(2*d*x + 2*c)^2*sin(d*x + c) + 2*a^2*cos 
(2*d*x + 2*c)*sin(d*x + c) + a^2*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 
 2*c), cos(2*d*x + 2*c) + 1)) - (4*a^2*cos(3*d*x + 3*c) + 5*a^2*cos(2*d*x 
+ 2*c) + 4*a^2*cos(d*x + c) + 5*a^2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos 
(2*d*x + 2*c) + 1)) - ((a^2*cos(d*x + c) - a^2)*cos(2*d*x + 2*c)^2 + a^2*c 
os(d*x + c) + (a^2*cos(d*x + c) - a^2)*sin(2*d*x + 2*c)^2 - a^2 + 2*(a^2*c 
os(d*x + c) - a^2)*cos(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos 
(2*d*x + 2*c) + 1)))*sqrt(a) + 5*((a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x 
+ 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2(-(cos(2*d*x + 2*c)^2 + si 
n(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d* 
x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arcta 
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2* 
d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2 
(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(s 
in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) + 1) - (a^2*cos(2*d*x + 2*c)^2...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 480 vs. \(2 (127) = 254\).

Time = 0.81 (sec) , antiderivative size = 480, normalized size of antiderivative = 3.36 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/6*(3*(5*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 2*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*(5*A*sqrt(-a)*a^2*sgn(cos(d*x + c) 
) + 2*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*(3 
*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 9*sqrt(2)*B*a^4*sgn(cos(d*x + c)) - (3* 
sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 7*sqrt(2)*B*a^4*sgn(cos(d*x + c)))*tan(1 
/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)*sq 
rt(-a*tan(1/2*d*x + 1/2*c)^2 + a)) + 12*(3*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^3*sgn(cos(d* 
x + c)) - sqrt(2)*A*sqrt(-a)*a^4*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x 
 + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d 
*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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