\(\int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [138]

Optimal result

Integrand size = 25, antiderivative size = 142 \[ \int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^3 (35 A+32 B) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (5 A+8 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 a B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \] Output:

2*a^(5/2)*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2/15*a^3*( 
35*A+32*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/15*a^2*(5*A+8*B)*(a+a*sec 
(d*x+c))^(1/2)*tan(d*x+c)/d+2/5*a*B*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.90 \[ \int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt {a (1+\sec (c+d x))} \left (30 \sqrt {2} A \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {5}{2}}(c+d x)+2 (40 A+49 B+2 (5 A+14 B) \cos (c+d x)+(40 A+43 B) \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{30 d} \] Input:

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

Output:

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]^2*Sqrt[a*(1 + Sec[c + d*x])]*(30*Sqrt[2 
]*A*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(5/2) + 2*(40*A + 49*B + 
 2*(5*A + 14*B)*Cos[c + d*x] + (40*A + 43*B)*Cos[2*(c + d*x)])*Sin[(c + d* 
x)/2]))/(30*d)
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4405, 27, 3042, 4405, 27, 3042, 4403, 3042, 4261, 216, 4279}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4405

\(\displaystyle \frac {2}{5} \int \frac {1}{2} (\sec (c+d x) a+a)^{3/2} (5 a A+a (5 A+8 B) \sec (c+d x))dx+\frac {2 a B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int (\sec (c+d x) a+a)^{3/2} (5 a A+a (5 A+8 B) \sec (c+d x))dx+\frac {2 a B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (5 a A+a (5 A+8 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 a B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4405

\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {1}{2} \sqrt {\sec (c+d x) a+a} \left (15 A a^2+(35 A+32 B) \sec (c+d x) a^2\right )dx+\frac {2 a^2 (5 A+8 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \sqrt {\sec (c+d x) a+a} \left (15 A a^2+(35 A+32 B) \sec (c+d x) a^2\right )dx+\frac {2 a^2 (5 A+8 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (15 A a^2+(35 A+32 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {2 a^2 (5 A+8 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4403

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (a^2 (35 A+32 B) \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+15 a^2 A \int \sqrt {\sec (c+d x) a+a}dx\right )+\frac {2 a^2 (5 A+8 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (a^2 (35 A+32 B) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+15 a^2 A \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\right )+\frac {2 a^2 (5 A+8 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (a^2 (35 A+32 B) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {30 a^3 A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {2 a^2 (5 A+8 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (a^2 (35 A+32 B) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {30 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}\right )+\frac {2 a^2 (5 A+8 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

\(\Big \downarrow \) 4279

\(\displaystyle \frac {1}{5} \left (\frac {2 a^2 (5 A+8 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}+\frac {1}{3} \left (\frac {30 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (35 A+32 B) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )+\frac {2 a B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\)

Input:

Int[(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)*Tan[c + d*x])/(5*d) + ((2*a^2*(5*A + 8*B 
)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) + ((30*a^(5/2)*A*ArcTan[(Sq 
rt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^3*(35*A + 32*B)*Ta 
n[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/3)/5
 

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 4279

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free 
Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4403

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_ 
.) + (c_)), x_Symbol] :> Simp[c   Int[Sqrt[a + b*Csc[e + f*x]], x], x] + Si 
mp[d   Int[Sqrt[a + b*Csc[e + f*x]]*Csc[e + f*x], x], x] /; FreeQ[{a, b, c, 
 d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
 

rule 4405

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[(-b)*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m 
 - 1)/(f*m)), x] + Simp[1/m   Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m + 
 (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, 
f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2 
*m]
 

Maple [A] (verified)

Time = 7.74 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.19

method	result	size
default	\(\frac {2 a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (A \left (40 \sin \left (d x +c \right )+5 \tan \left (d x +c \right )\right )+B \left (43 \sin \left (d x +c \right )+14 \tan \left (d x +c \right )+3 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )-15 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) A \right )}{15 d \left (1+\cos \left (d x +c \right )\right )}\)	\(169\)
parts	\(\frac {A \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (16 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )+3 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right ) \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )\right )}{3 d \left (1+\cos \left (d x +c \right )\right )}+\frac {B \left (86 \sin \left (d x +c \right )+28 \tan \left (d x +c \right )+6 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (15 \cos \left (d x +c \right )+15\right )}\)	\(200\)

Input:

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

Output:

2/15/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*(A*(40*sin(d*x+c)+5*tan 
(d*x+c))+B*(43*sin(d*x+c)+14*tan(d*x+c)+3*sec(d*x+c)*tan(d*x+c))-15*(1+cos 
(d*x+c))*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+c 
ot(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))*A)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.66 \[ \int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\left [\frac {15 \, {\left (A a^{2} \cos \left (d x + c\right )^{3} + A a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left ({\left (40 \, A + 43 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (5 \, A + 14 \, B\right )} a^{2} \cos \left (d x + c\right ) + 3 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, -\frac {2 \, {\left (15 \, {\left (A a^{2} \cos \left (d x + c\right )^{3} + A a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left ({\left (40 \, A + 43 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (5 \, A + 14 \, B\right )} a^{2} \cos \left (d x + c\right ) + 3 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \] Input:

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

Output:

[1/15*(15*(A*a^2*cos(d*x + c)^3 + A*a^2*cos(d*x + c)^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*((40*A + 
 43*B)*a^2*cos(d*x + c)^2 + (5*A + 14*B)*a^2*cos(d*x + c) + 3*B*a^2)*sqrt( 
(a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3 + d*cos 
(d*x + c)^2), -2/15*(15*(A*a^2*cos(d*x + c)^3 + A*a^2*cos(d*x + c)^2)*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))) - ((40*A + 43*B)*a^2*cos(d*x + c)^2 + (5*A + 14*B)*a^2*cos(d 
*x + c) + 3*B*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/( 
d*cos(d*x + c)^3 + d*cos(d*x + c)^2)]
 

Sympy [F]

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

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

Output:

Integral((a*(sec(c + d*x) + 1))**(5/2)*(A + B*sec(c + d*x)), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1396 vs. \(2 (124) = 248\).

Time = 0.21 (sec) , antiderivative size = 1396, normalized size of antiderivative = 9.83 \[ \int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

1/6*(30*(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) 
*((12*a^2*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 
 2*c) - 3*a^2*sin(2*d*x + 2*c) - 4*(3*a^2*cos(2*d*x + 2*c) + 4*a^2)*sin(3/ 
2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(3/2*arctan2(sin(2*d*x 
+ 2*c), cos(2*d*x + 2*c) + 1)) + (12*a^2*sin(2*d*x + 2*c)*sin(3/2*arctan2( 
sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 3*a^2*cos(2*d*x + 2*c) - a^2 + 4*(3 
*a^2*cos(2*d*x + 2*c) + 4*a^2)*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x 
 + 2*c))))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt( 
a) + 3*((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 + sin(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)) 
)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arcta 
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2* 
c), cos(2*d*x + 2*c)))), (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(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c 
) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arc 
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 
2*c), cos(2*d*x + 2*c)))) + 1) - (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (124) = 248\).

Time = 0.75 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.18 \[ \int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=-\frac {\frac {15 \, A \sqrt {-a} a^{3} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} - \frac {2 \, {\left (45 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 60 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (80 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 80 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (35 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 32 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{15 \, d} \] Input:

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

Output:

-1/15*(15*A*sqrt(-a)*a^3*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(- 
a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a) 
*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqrt(2) 
*abs(a) - 6*a))*sgn(cos(d*x + c))/abs(a) - 2*(45*sqrt(2)*A*a^5*sgn(cos(d*x 
 + c)) + 60*sqrt(2)*B*a^5*sgn(cos(d*x + c)) - (80*sqrt(2)*A*a^5*sgn(cos(d* 
x + c)) + 80*sqrt(2)*B*a^5*sgn(cos(d*x + c)) - (35*sqrt(2)*A*a^5*sgn(cos(d 
*x + c)) + 32*sqrt(2)*B*a^5*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*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)^ 
2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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