\(\int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [531]

Optimal result

Integrand size = 35, antiderivative size = 247 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a^{3/2} (88 A+75 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{64 d}+\frac {a^2 (8 A+9 B) \sin (c+d x)}{24 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (88 A+75 B) \sin (c+d x)}{96 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (88 A+75 B) \sin (c+d x)}{64 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a B \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x)} \] Output:

1/64*a^(3/2)*(88*A+75*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2) 
)*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+1/24*a^2*(8*A+9*B)*sin(d*x+c)/d/cos( 
d*x+c)^(7/2)/(a+a*sec(d*x+c))^(1/2)+1/96*a^2*(88*A+75*B)*sin(d*x+c)/d/cos( 
d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+1/64*a^2*(88*A+75*B)*sin(d*x+c)/d/cos( 
d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+1/4*a*B*(a+a*sec(d*x+c))^(1/2)*sin(d*x 
+c)/d/cos(d*x+c)^(7/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 1.04 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.85 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a^2 \sqrt {\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \left (3 (88 A+75 B) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+2 (88 A+75 B) \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+8 (8 A+15 B) \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+48 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {7}{2}}(c+d x)+3 (88 A+75 B) \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \sin (c+d x)}{192 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a^2*Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*(3*(88*A + 75*B)*ArcSin[Sqrt[1 
- Sec[c + d*x]]] + 2*(88*A + 75*B)*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/ 
2) + 8*(8*A + 15*B)*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2) + 48*B*Sqrt[ 
1 - Sec[c + d*x]]*Sec[c + d*x]^(7/2) + 3*(88*A + 75*B)*Sqrt[-((-1 + Sec[c 
+ d*x])*Sec[c + d*x])])*Sin[c + d*x])/(192*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a 
*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.34 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.97, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3434, 3042, 4506, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3434

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \int \frac {1}{2} \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (a (8 A+5 B)+a (8 A+9 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (a (8 A+5 B)+a (8 A+9 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (8 A+5 B)+a (8 A+9 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 4504

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 (8 A+9 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 (8 A+9 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 4290

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \left (\frac {3}{4} \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 A+9 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 A+9 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 4290

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 A+9 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 A+9 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 4288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {1}{6} a (88 A+75 B) \left (\frac {3}{4} \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 A+9 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

\(\Big \downarrow \) 222

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {a^2 (8 A+9 B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {1}{6} a (88 A+75 B) \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\right )+\frac {a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((a*B*Sec[c + d*x]^(7/2)*Sqrt[a + a* 
Sec[c + d*x]]*Sin[c + d*x])/(4*d) + ((a^2*(8*A + 9*B)*Sec[c + d*x]^(7/2)*S 
in[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(88*A + 75*B)*((a*Sec[c + 
 d*x]^(5/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (3*((Sqrt[a]*Ar 
cSinh[(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)/8)
 

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 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 3434

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* 
(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim 
p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p   Int[(a + b*Csc[e + f*x])^m*((c + 
d*Csc[e + f*x])^n/(g*Csc[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, n, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && I 
ntegerQ[n])
 

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]
 

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 = 2.58 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.35

method	result	size
default	\(\frac {\left (264 A \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+225 B \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+264 A \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+225 B \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (264 \cos \left (d x +c \right )^{2}+176 \cos \left (d x +c \right )+64\right ) \sqrt {2}\, A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+\sin \left (d x +c \right ) \left (225 \cos \left (d x +c \right )^{3}+150 \cos \left (d x +c \right )^{2}+120 \cos \left (d x +c \right )+48\right ) \sqrt {2}\, B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, a}{384 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )^{\frac {7}{2}}}\)	\(334\)

Input:

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

Output:

1/384/d*(264*A*cos(d*x+c)^4*arctan(1/2*(cot(d*x+c)-csc(d*x+c)-1)/(-1/(1+co 
s(d*x+c)))^(1/2))+225*B*cos(d*x+c)^4*arctan(1/2*(cot(d*x+c)-csc(d*x+c)-1)/ 
(-1/(1+cos(d*x+c)))^(1/2))+264*A*cos(d*x+c)^4*arctan(1/2*(cot(d*x+c)-csc(d 
*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))+225*B*cos(d*x+c)^4*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)*(264*co 
s(d*x+c)^2+176*cos(d*x+c)+64)*2^(1/2)*A*(-2/(1+cos(d*x+c)))^(1/2)+sin(d*x+ 
c)*(225*cos(d*x+c)^3+150*cos(d*x+c)^2+120*cos(d*x+c)+48)*2^(1/2)*B*(-2/(1+ 
cos(d*x+c)))^(1/2))*(a*(1+sec(d*x+c)))^(1/2)*a/(1+cos(d*x+c))/(-1/(1+cos(d 
*x+c)))^(1/2)/cos(d*x+c)^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 485, normalized size of antiderivative = 1.96 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\left [\frac {4 \, {\left (3 \, {\left (88 \, A + 75 \, B\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (88 \, A + 75 \, B\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, A + 15 \, B\right )} a \cos \left (d x + c\right ) + 48 \, B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (88 \, A + 75 \, B\right )} a \cos \left (d x + c\right )^{5} + {\left (88 \, A + 75 \, B\right )} a \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{768 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}}, \frac {2 \, {\left (3 \, {\left (88 \, A + 75 \, B\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (88 \, A + 75 \, B\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, A + 15 \, B\right )} a \cos \left (d x + c\right ) + 48 \, B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (88 \, A + 75 \, B\right )} a \cos \left (d x + c\right )^{5} + {\left (88 \, A + 75 \, B\right )} a \cos \left (d x + c\right )^{4}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{384 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}}\right ] \] Input:

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

Output:

[1/768*(4*(3*(88*A + 75*B)*a*cos(d*x + c)^3 + 2*(88*A + 75*B)*a*cos(d*x + 
c)^2 + 8*(8*A + 15*B)*a*cos(d*x + c) + 48*B*a)*sqrt((a*cos(d*x + c) + a)/c 
os(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((88*A + 75*B)*a*cos(d*x 
+ c)^5 + (88*A + 75*B)*a*cos(d*x + c)^4)*sqrt(a)*log((a*cos(d*x + c)^3 - 4 
*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(c 
os(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + co 
s(d*x + c)^2)))/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4), 1/384*(2*(3*(88*A + 
 75*B)*a*cos(d*x + c)^3 + 2*(88*A + 75*B)*a*cos(d*x + c)^2 + 8*(8*A + 15*B 
)*a*cos(d*x + c) + 48*B*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(co 
s(d*x + c))*sin(d*x + c) + 3*((88*A + 75*B)*a*cos(d*x + c)^5 + (88*A + 75* 
B)*a*cos(d*x + c)^4)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/ 
cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d* 
x + c) - 2*a)))/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5879 vs. \(2 (211) = 422\).

Time = 0.59 (sec) , antiderivative size = 5879, normalized size of antiderivative = 23.80 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/768*(8*(132*(sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d*x + 4*c) 
+ 3*sqrt(2)*a*sin(2*d*x + 2*c))*cos(11/4*arctan2(sin(2*d*x + 2*c), cos(2*d 
*x + 2*c))) + 44*(sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d*x + 4*c 
) + 3*sqrt(2)*a*sin(2*d*x + 2*c))*cos(9/4*arctan2(sin(2*d*x + 2*c), cos(2* 
d*x + 2*c))) + 216*(sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d*x + 4 
*c) + 3*sqrt(2)*a*sin(2*d*x + 2*c))*cos(7/4*arctan2(sin(2*d*x + 2*c), cos( 
2*d*x + 2*c))) - 216*(sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d*x + 
 4*c) + 3*sqrt(2)*a*sin(2*d*x + 2*c))*cos(5/4*arctan2(sin(2*d*x + 2*c), co 
s(2*d*x + 2*c))) - 44*(sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d*x 
+ 4*c) + 3*sqrt(2)*a*sin(2*d*x + 2*c))*cos(3/4*arctan2(sin(2*d*x + 2*c), c 
os(2*d*x + 2*c))) - 132*(sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d* 
x + 4*c) + 3*sqrt(2)*a*sin(2*d*x + 2*c))*cos(1/4*arctan2(sin(2*d*x + 2*c), 
 cos(2*d*x + 2*c))) - 33*(a*cos(6*d*x + 6*c)^2 + 9*a*cos(4*d*x + 4*c)^2 + 
9*a*cos(2*d*x + 2*c)^2 + a*sin(6*d*x + 6*c)^2 + 9*a*sin(4*d*x + 4*c)^2 + 1 
8*a*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*a*sin(2*d*x + 2*c)^2 + 2*(3*a*co 
s(4*d*x + 4*c) + 3*a*cos(2*d*x + 2*c) + a)*cos(6*d*x + 6*c) + 6*(3*a*cos(2 
*d*x + 2*c) + a)*cos(4*d*x + 4*c) + 6*a*cos(2*d*x + 2*c) + 6*(a*sin(4*d*x 
+ 4*c) + a*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + a)*log(2*cos(1/4*arctan2(s 
in(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c) 
, cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), co...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 942 vs. \(2 (211) = 422\).

Time = 0.72 (sec) , antiderivative size = 942, normalized size of antiderivative = 3.81 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/384*(3*(88*A*a^(3/2)*sgn(cos(d*x + c)) + 75*B*a^(3/2)*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*(88*A*a^(3/2)*sgn(cos(d*x + c)) + 75*B*a^(3 
/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*sqrt(2)*(264*(sqrt(a) 
*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^14*A*a^(5/2)*s 
gn(cos(d*x + c)) + 225*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x 
+ 1/2*c)^2 + a))^14*B*a^(5/2)*sgn(cos(d*x + c)) - 4008*(sqrt(a)*tan(1/2*d* 
x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^12*A*a^(7/2)*sgn(cos(d*x 
+ c)) - 6261*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 
 + a))^12*B*a^(7/2)*sgn(cos(d*x + c)) + 33960*(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^(9/2)*sgn(cos(d*x + c)) + 3 
5925*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^1 
0*B*a^(9/2)*sgn(cos(d*x + c)) - 131784*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqr 
t(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*A*a^(11/2)*sgn(cos(d*x + c)) - 127449*( 
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^( 
11/2)*sgn(cos(d*x + c)) + 108312*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*ta 
n(1/2*d*x + 1/2*c)^2 + a))^6*A*a^(13/2)*sgn(cos(d*x + c)) + 101667*(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^(13/2)* 
sgn(cos(d*x + c)) - 29432*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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