\(\int \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx\) [224]

Optimal result

Integrand size = 35, antiderivative size = 176 \[ \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {\sqrt {a} (6 A+5 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a (6 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {a (6 A+5 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a B \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}} \] Output:

1/8*a^(1/2)*(6*A+5*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d 
+1/8*a*(6*A+5*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/12 
*a*(6*A+5*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/3*a*B* 
sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 0.57 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.89 \[ \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {a \left (3 (6 A+5 B) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+2 (6 A+5 B) \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+8 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+3 (6 A+5 B) \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \tan (c+d x)}{24 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a*(3*(6*A + 5*B)*ArcSin[Sqrt[1 - Sec[c + d*x]]] + 2*(6*A + 5*B)*Sqrt[1 - 
Sec[c + d*x]]*Sec[c + d*x]^(3/2) + 8*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x] 
^(5/2) + 3*(6*A + 5*B)*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])*Tan[c + 
d*x])/(24*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 4504, 3042, 4290, 3042, 4290, 3042, 4288, 222}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4504

\(\displaystyle \frac {1}{6} (6 A+5 B) \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} (6 A+5 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4290

\(\displaystyle \frac {1}{6} (6 A+5 B) \left (\frac {3}{4} \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} (6 A+5 B) \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4290

\(\displaystyle \frac {1}{6} (6 A+5 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} (6 A+5 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {1}{6} (6 A+5 B) \left (\frac {3}{4} \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {1}{6} (6 A+5 B) \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

Input:

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

Output:

(a*B*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + ((6 
*A + 5*B)*((a*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x 
]]) + (3*((Sqrt[a]*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]] 
])/d + (a*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/ 
4))/6
 

Defintions of rubi rules used

rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4288

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

rule 4290

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)], x_Symbol] :> Simp[-2*b*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/( 
f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*a*d*((n - 1)/(b*(2*n - 
1)))   Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; Fre 
eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
 

rule 4504

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C 
ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] 
 + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1))   Int[Sqrt[a + b*Csc[e + f* 
x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(312\) vs. \(2(150)=300\).

Time = 1.98 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.78

method	result	size
default	\(-\frac {\left (18 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+15 B \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+18 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+15 B \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-18 \cos \left (d x +c \right )-12\right ) \sqrt {2}\, A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+\sin \left (d x +c \right ) \left (-15 \cos \left (d x +c \right )^{2}-10 \cos \left (d x +c \right )-8\right ) \sqrt {2}\, B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \sec \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{48 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(313\)
parts	\(-\frac {A \sec \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (3 \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+3 \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\sin \left (d x +c \right ) \left (-3 \cos \left (d x +c \right )-2\right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )\right )}{8 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}-\frac {B \sec \left (d x +c \right )^{\frac {7}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (15 \cos \left (d x +c \right )^{4} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+15 \cos \left (d x +c \right )^{4} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\sin \left (d x +c \right ) \left (-15 \cos \left (d x +c \right )^{2}-10 \cos \left (d x +c \right )-8\right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )\right )}{48 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(366\)

Input:

int(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x,method=_RET 
URNVERBOSE)
 

Output:

-1/48/d*(18*A*cos(d*x+c)^3*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(-1/(1+co 
s(d*x+c)))^(1/2))+15*B*cos(d*x+c)^3*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/ 
(-1/(1+cos(d*x+c)))^(1/2))+18*A*cos(d*x+c)^3*arctan(1/2*(-cot(d*x+c)+csc(d 
*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))+15*B*cos(d*x+c)^3*arctan(1/2*(-cot(d*x 
+c)+csc(d*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))+sin(d*x+c)*cos(d*x+c)*(-18*co 
s(d*x+c)-12)*2^(1/2)*A*(-2/(1+cos(d*x+c)))^(1/2)+sin(d*x+c)*(-15*cos(d*x+c 
)^2-10*cos(d*x+c)-8)*2^(1/2)*B*(-2/(1+cos(d*x+c)))^(1/2))*sec(d*x+c)^(5/2) 
*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 445, normalized size of antiderivative = 2.53 \[ \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\left [\frac {3 \, {\left ({\left (6 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (3 \, {\left (6 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 8 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{96 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac {3 \, {\left ({\left (6 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (3 \, {\left (6 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 8 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{48 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \] Input:

integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algo 
rithm="fricas")
 

Output:

[1/96*(3*((6*A + 5*B)*cos(d*x + c)^3 + (6*A + 5*B)*cos(d*x + c)^2)*sqrt(a) 
*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d* 
x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt( 
cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(3*(6*A + 5*B) 
*cos(d*x + c)^2 + 2*(6*A + 5*B)*cos(d*x + c) + 8*B)*sqrt((a*cos(d*x + c) + 
 a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^3 + d*c 
os(d*x + c)^2), 1/48*(3*((6*A + 5*B)*cos(d*x + c)^3 + (6*A + 5*B)*cos(d*x 
+ c)^2)*sqrt(-a)*arctan(1/2*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(-a)*sqr 
t((a*cos(d*x + c) + a)/cos(d*x + c))/(a*sqrt(cos(d*x + c))*sin(d*x + c))) 
+ 2*(3*(6*A + 5*B)*cos(d*x + c)^2 + 2*(6*A + 5*B)*cos(d*x + c) + 8*B)*sqrt 
((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*co 
s(d*x + c)^3 + d*cos(d*x + c)^2)]
 

Sympy [F(-1)]

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

integrate(sec(d*x+c)**(5/2)*(a+a*sec(d*x+c))**(1/2)*(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. 3342 vs. \(2 (150) = 300\).

Time = 0.41 (sec) , antiderivative size = 3342, normalized size of antiderivative = 18.99 \[ \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algo 
rithm="maxima")
 

Output:

-1/96*(6*(12*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(7 
/2*arctan2(sin(d*x + c), cos(d*x + c))) + 4*(sqrt(2)*sin(4*d*x + 4*c) + 2* 
sqrt(2)*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(d*x + c), cos(d*x + c))) - 4 
*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(3/2*arctan2(s 
in(d*x + c), cos(d*x + c))) - 12*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin 
(2*d*x + 2*c))*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 3*(2*(2*cos( 
2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2* 
c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2* 
d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), 
 cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*s 
qrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*ar 
ctan2(sin(d*x + c), cos(d*x + c))) + 2) + 3*(2*(2*cos(2*d*x + 2*c) + 1)*co 
s(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4 
*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos 
(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 
2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arcta 
n2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), 
cos(d*x + c))) + 2) - 3*(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos 
(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x 
+ 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 783 vs. \(2 (150) = 300\).

Time = 2.04 (sec) , antiderivative size = 783, normalized size of antiderivative = 4.45 \[ \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algo 
rithm="giac")
 

Output:

1/48*(3*(6*A*sqrt(a)*sgn(cos(d*x + c)) + 5*B*sqrt(a)*sgn(cos(d*x + c)))*lo 
g(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*(6*A*sqrt(a)*sgn(cos(d*x + c)) + 5*B*sqrt(a)*s 
gn(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*(30*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*a^(3/2)*sgn(cos 
(d*x + c)) - 63*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*a^(3/2)*sgn(cos(d*x + c)) - 66*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*a^(5/2)*sgn(co 
s(d*x + c)) + 369*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*a^(5/2)*sgn(cos(d*x + c)) - 756*sqrt(2)*(sqrt(a)*t 
an(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*a^(7/2)*sgn( 
cos(d*x + c)) - 1638*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*a^(7/2)*sgn(cos(d*x + c)) + 732*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*A*a^(9/2)*s 
gn(cos(d*x + c)) + 1074*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*a^(9/2)*sgn(cos(d*x + c)) - 138*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))^2*A*a^(11/ 
2)*sgn(cos(d*x + c)) - 171*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*a^(11/2)*sgn(cos(d*x + c)) + 6*sqrt(2)...
 

Mupad [F(-1)]

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

int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(5/2) 
,x)
 

Output:

int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(5/2) 
, x)
 

Reduce [F]

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

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

Output:

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