\(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [135]

Optimal result

Integrand size = 33, antiderivative size = 237 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^3 (803 A+710 B) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (209 A+194 B) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a^2 (803 A+710 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a^2 (11 A+14 B) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a (803 A+710 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d} \] Output:

2/495*a^3*(803*A+710*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/693*a^3*(209 
*A+194*B)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/3465*a^2*(803 
*A+710*B)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/99*a^2*(11*A+14*B)*sec(d*x 
+c)^3*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/1155*a*(803*A+710*B)*(a+a*sec( 
d*x+c))^(3/2)*tan(d*x+c)/d+2/11*a*B*sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*ta 
n(d*x+c)/d
 

Mathematica [A] (verified)

Time = 4.35 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.49 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^3 \left (8 (803 A+710 B)+4 (803 A+710 B) \sec (c+d x)+3 (803 A+710 B) \sec ^2(c+d x)+5 (286 A+355 B) \sec ^3(c+d x)+35 (11 A+32 B) \sec ^4(c+d x)+315 B \sec ^5(c+d x)\right ) \tan (c+d x)}{3465 d \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(2*a^3*(8*(803*A + 710*B) + 4*(803*A + 710*B)*Sec[c + d*x] + 3*(803*A + 71 
0*B)*Sec[c + d*x]^2 + 5*(286*A + 355*B)*Sec[c + d*x]^3 + 35*(11*A + 32*B)* 
Sec[c + d*x]^4 + 315*B*Sec[c + d*x]^5)*Tan[c + d*x])/(3465*d*Sqrt[a*(1 + S 
ec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.51 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4506, 27, 3042, 4506, 27, 3042, 4504, 3042, 4287, 27, 3042, 4489, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

\(\displaystyle \frac {2}{11} \int \frac {1}{2} \sec ^3(c+d x) (\sec (c+d x) a+a)^{3/2} (a (11 A+6 B)+a (11 A+14 B) \sec (c+d x))dx+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{11} \int \sec ^3(c+d x) (\sec (c+d x) a+a)^{3/2} (a (11 A+6 B)+a (11 A+14 B) \sec (c+d x))dx+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{11} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (11 A+6 B)+a (11 A+14 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

\(\Big \downarrow \) 4506

\(\displaystyle \frac {1}{11} \left (\frac {2}{9} \int \frac {1}{2} \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (55 A+46 B) a^2+(209 A+194 B) \sec (c+d x) a^2\right )dx+\frac {2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (55 A+46 B) a^2+(209 A+194 B) \sec (c+d x) a^2\right )dx+\frac {2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 (55 A+46 B) a^2+(209 A+194 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

\(\Big \downarrow \) 4504

\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 A+710 B) \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^3 (209 A+194 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 A+710 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^3 (209 A+194 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

\(\Big \downarrow \) 4287

\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 A+710 B) \left (\frac {2 \int \frac {1}{2} \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^3 (209 A+194 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 A+710 B) \left (\frac {\int \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^3 (209 A+194 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 A+710 B) \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^3 (209 A+194 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 A+710 B) \left (\frac {\frac {7}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^3 (209 A+194 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {3}{7} a^2 (803 A+710 B) \left (\frac {\frac {7}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^3 (209 A+194 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

\(\Big \downarrow \) 4279

\(\displaystyle \frac {1}{11} \left (\frac {2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}+\frac {1}{9} \left (\frac {2 a^3 (209 A+194 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}+\frac {3}{7} a^2 (803 A+710 B) \left (\frac {\frac {14 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )\right )\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}\)

Input:

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

Output:

(2*a*B*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(11*d) + (( 
2*a^2*(11*A + 14*B)*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/ 
(9*d) + ((2*a^3*(209*A + 194*B)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + 
 a*Sec[c + d*x]]) + (3*a^2*(803*A + 710*B)*((2*(a + a*Sec[c + d*x])^(3/2)* 
Tan[c + d*x])/(5*a*d) + ((14*a^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x 
]]) - (4*a*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/(5*a)))/7)/9)/11
 

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 3042

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

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 4287

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), 
x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2 
))), x] + Simp[1/(b*(m + 2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b*( 
m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - 
b^2, 0] &&  !LtQ[m, -2^(-1)]
 

rule 4489

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

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 = 237.90 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.62

method	result	size
default	\(\frac {2 a^{2} \left (\cos \left (d x +c \right ) \left (6424 \cos \left (d x +c \right )^{4}+3212 \cos \left (d x +c \right )^{3}+2409 \cos \left (d x +c \right )^{2}+1430 \cos \left (d x +c \right )+385\right ) A +\left (5680 \cos \left (d x +c \right )^{5}+2840 \cos \left (d x +c \right )^{4}+2130 \cos \left (d x +c \right )^{3}+1775 \cos \left (d x +c \right )^{2}+1120 \cos \left (d x +c \right )+315\right ) B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{3465 d \left (1+\cos \left (d x +c \right )\right )}\)	\(146\)
parts	\(\frac {2 A \,a^{2} \left (584 \cos \left (d x +c \right )^{4}+292 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+130 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (1+\cos \left (d x +c \right )\right )}+\frac {2 B \,a^{2} \left (1136 \cos \left (d x +c \right )^{5}+568 \cos \left (d x +c \right )^{4}+426 \cos \left (d x +c \right )^{3}+355 \cos \left (d x +c \right )^{2}+224 \cos \left (d x +c \right )+63\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{693 d \left (1+\cos \left (d x +c \right )\right )}\)	\(182\)

Input:

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

Output:

2/3465/d*a^2*(cos(d*x+c)*(6424*cos(d*x+c)^4+3212*cos(d*x+c)^3+2409*cos(d*x 
+c)^2+1430*cos(d*x+c)+385)*A+(5680*cos(d*x+c)^5+2840*cos(d*x+c)^4+2130*cos 
(d*x+c)^3+1775*cos(d*x+c)^2+1120*cos(d*x+c)+315)*B)*(a*(1+sec(d*x+c)))^(1/ 
2)/(1+cos(d*x+c))*tan(d*x+c)*sec(d*x+c)^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.66 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (8 \, {\left (803 \, A + 710 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 4 \, {\left (803 \, A + 710 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (803 \, A + 710 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (286 \, A + 355 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, A + 32 \, B\right )} a^{2} \cos \left (d x + c\right ) + 315 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \] Input:

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

Output:

2/3465*(8*(803*A + 710*B)*a^2*cos(d*x + c)^5 + 4*(803*A + 710*B)*a^2*cos(d 
*x + c)^4 + 3*(803*A + 710*B)*a^2*cos(d*x + c)^3 + 5*(286*A + 355*B)*a^2*c 
os(d*x + c)^2 + 35*(11*A + 32*B)*a^2*cos(d*x + c) + 315*B*a^2)*sqrt((a*cos 
(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x + 
c)^5)
 

Sympy [F(-1)]

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

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

integrate(sec(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 [A] (verification not implemented)

Time = 0.82 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.29 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {8 \, {\left ({\left ({\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (143 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, \sqrt {2} {\left (143 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 99 \, \sqrt {2} {\left (143 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 231 \, \sqrt {2} {\left (69 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 65 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1155 \, \sqrt {2} {\left (9 \, A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3465 \, \sqrt {2} {\left (A a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{3465 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \] Input:

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

Output:

8/3465*((((4*(2*sqrt(2)*(143*A*a^8*sgn(cos(d*x + c)) + 125*B*a^8*sgn(cos(d 
*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 11*sqrt(2)*(143*A*a^8*sgn(cos(d*x + c)) 
 + 125*B*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 99*sqrt(2)*(143* 
A*a^8*sgn(cos(d*x + c)) + 125*B*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2* 
c)^2 - 231*sqrt(2)*(69*A*a^8*sgn(cos(d*x + c)) + 65*B*a^8*sgn(cos(d*x + c) 
)))*tan(1/2*d*x + 1/2*c)^2 + 1155*sqrt(2)*(9*A*a^8*sgn(cos(d*x + c)) + 7*B 
*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 3465*sqrt(2)*(A*a^8*sgn( 
cos(d*x + c)) + B*a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2 
*d*x + 1/2*c)^2 - a)^5*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)
 

Mupad [B] (verification not implemented)

Time = 21.91 (sec) , antiderivative size = 856, normalized size of antiderivative = 3.61 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((A*a^2*8i) 
/(3*d) - (a^2*exp(c*1i + d*x*1i)*(803*A + 710*B)*8i)/(3465*d)))/((exp(c*1i 
 + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) - ((a + a/(exp(- c*1i - d*x*1i)/ 
2 + exp(c*1i + d*x*1i)/2))^(1/2)*((A*a^2*24i)/(7*d) - exp(c*1i + d*x*1i)*( 
(a^2*(5*A + 16*B)*8i)/(7*d) - (a^2*(5*A + 2*B)*8i)/(7*d) + (a^2*(11*A + 50 
*B)*32i)/(693*d)) + (a^2*(9*A + 10*B)*8i)/(7*d)))/((exp(c*1i + d*x*1i) + 1 
)*(exp(c*2i + d*x*2i) + 1)^3) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i 
 + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2*8i)/(11*d) + (a^2*(3*A + 
4*B)*40i)/(11*d) - (a^2*(5*A + 2*B)*8i)/(11*d) - (a^2*(11*A + 10*B)*8i)/(1 
1*d)) + (A*a^2*8i)/(11*d) + (a^2*(3*A + 4*B)*40i)/(11*d) - (a^2*(5*A + 2*B 
)*8i)/(11*d) - (a^2*(11*A + 10*B)*8i)/(11*d)))/((exp(c*1i + d*x*1i) + 1)*( 
exp(c*2i + d*x*2i) + 1)^5) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + 
d*x*1i)/2))^(1/2)*((A*a^2*8i)/(9*d) - exp(c*1i + d*x*1i)*((a^2*(A - 8*B)*8 
i)/(9*d) - (B*a^2*64i)/(99*d) + (a^2*(5*A + 2*B)*8i)/(9*d) - (a^2*(5*A + 9 
*B)*16i)/(9*d)) + (a^2*(A + 2*B)*40i)/(9*d) + (B*a^2*64i)/(9*d) - (a^2*(A 
+ B)*80i)/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) + 
((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + 
 d*x*1i)*((a^2*(5*A + 2*B)*8i)/(5*d) + (a^2*(44*A - 31*B)*16i)/(1155*d)) - 
 (A*a^2*8i)/(5*d) + (a^2*(4*A + 5*B)*16i)/(5*d)))/((exp(c*1i + d*x*1i) + 1 
)*(exp(c*2i + d*x*2i) + 1)^2) - (a^2*exp(c*1i + d*x*1i)*(a + a/(exp(- c...
 

Reduce [F]

\[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{6}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}d x \right ) b +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) a \right ) \] Input:

int(sec(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)*sec(c + d*x)**6,x)*b + int(sqrt(s 
ec(c + d*x) + 1)*sec(c + d*x)**5,x)*a + 2*int(sqrt(sec(c + d*x) + 1)*sec(c 
 + d*x)**5,x)*b + 2*int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**4,x)*a + int( 
sqrt(sec(c + d*x) + 1)*sec(c + d*x)**4,x)*b + int(sqrt(sec(c + d*x) + 1)*s 
ec(c + d*x)**3,x)*a)