\(\int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [357]

Optimal result

Integrand size = 42, antiderivative size = 187 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 a (9 B+8 C) \tan (c+d x)}{45 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (9 B+8 C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}-\frac {8 (9 B+8 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {4 (9 B+8 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 a d} \] Output:

4/45*a*(9*B+8*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/63*a*(9*B+8*C)*sec( 
d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/9*a*C*sec(d*x+c)^4*tan(d*x+ 
c)/d/(a+a*sec(d*x+c))^(1/2)-8/315*(9*B+8*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x 
+c)/d+4/105*(9*B+8*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.52 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a \left (16 (9 B+8 C)+8 (9 B+8 C) \sec (c+d x)+6 (9 B+8 C) \sec ^2(c+d x)+5 (9 B+8 C) \sec ^3(c+d x)+35 C \sec ^4(c+d x)\right ) \tan (c+d x)}{315 d \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*(B*Sec[c + d*x] + C*Sec[ 
c + d*x]^2),x]
 

Output:

(2*a*(16*(9*B + 8*C) + 8*(9*B + 8*C)*Sec[c + d*x] + 6*(9*B + 8*C)*Sec[c + 
d*x]^2 + 5*(9*B + 8*C)*Sec[c + d*x]^3 + 35*C*Sec[c + d*x]^4)*Tan[c + d*x]) 
/(315*d*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 4560, 3042, 4504, 3042, 4290, 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.

\(\displaystyle \int \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

\(\displaystyle \int \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a} (B+C \sec (c+d x))dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 4504

\(\displaystyle \frac {1}{9} (9 B+8 C) \int \sec ^4(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} (9 B+8 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4290

\(\displaystyle \frac {1}{9} (9 B+8 C) \left (\frac {6}{7} \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} (9 B+8 C) \left (\frac {6}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4287

\(\displaystyle \frac {1}{9} (9 B+8 C) \left (\frac {6}{7} \left (\frac {2 \int \frac {1}{2} \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} (9 B+8 C) \left (\frac {6}{7} \left (\frac {\int \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4489

\(\displaystyle \frac {1}{9} (9 B+8 C) \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4279

\(\displaystyle \frac {1}{9} (9 B+8 C) \left (\frac {6}{7} \left (\frac {\frac {14 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

Input:

Int[Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*(B*Sec[c + d*x] + C*Sec[c + d* 
x]^2),x]
 

Output:

(2*a*C*Sec[c + d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) + ((9*B 
 + 8*C)*((2*a*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) 
+ (6*((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]]*Ta 
n[c + d*x])/(3*d))/(5*a)))/7))/9
 

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 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 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 4560

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

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {2 \left (\cos \left (d x +c \right ) \left (144 \cos \left (d x +c \right )^{3}+72 \cos \left (d x +c \right )^{2}+54 \cos \left (d x +c \right )+45\right ) B +\left (128 \cos \left (d x +c \right )^{4}+64 \cos \left (d x +c \right )^{3}+48 \cos \left (d x +c \right )^{2}+40 \cos \left (d x +c \right )+35\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\)	\(123\)
parts	\(\frac {2 B \left (16 \cos \left (d x +c \right )^{3}+8 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+5\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{35 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \left (128 \cos \left (d x +c \right )^{4}+64 \cos \left (d x +c \right )^{3}+48 \cos \left (d x +c \right )^{2}+40 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\)	\(156\)

Input:

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,me 
thod=_RETURNVERBOSE)
 

Output:

2/315/d*(cos(d*x+c)*(144*cos(d*x+c)^3+72*cos(d*x+c)^2+54*cos(d*x+c)+45)*B+ 
(128*cos(d*x+c)^4+64*cos(d*x+c)^3+48*cos(d*x+c)^2+40*cos(d*x+c)+35)*C)*(a* 
(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^3
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.65 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (16 \, {\left (9 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{4} + 8 \, {\left (9 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (9 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (9 \, B + 8 \, C\right )} \cos \left (d x + c\right ) + 35 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \] Input:

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="fricas")
 

Output:

2/315*(16*(9*B + 8*C)*cos(d*x + c)^4 + 8*(9*B + 8*C)*cos(d*x + c)^3 + 6*(9 
*B + 8*C)*cos(d*x + c)^2 + 5*(9*B + 8*C)*cos(d*x + c) + 35*C)*sqrt((a*cos( 
d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c 
)^4)
 

Sympy [F]

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

integrate(sec(d*x+c)**3*(a+a*sec(d*x+c))**(1/2)*(B*sec(d*x+c)+C*sec(d*x+c) 
**2),x)
 

Output:

Integral(sqrt(a*(sec(c + d*x) + 1))*(B + C*sec(c + d*x))*sec(c + d*x)**4, 
x)
 

Maxima [F]

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

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="maxima")
 

Output:

16/315*(315*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) 
+ 1)^(1/4)*((B*d*cos(2*d*x + 2*c)^4 + B*d*sin(2*d*x + 2*c)^4 + 4*B*d*cos(2 
*d*x + 2*c)^3 + 6*B*d*cos(2*d*x + 2*c)^2 + 4*B*d*cos(2*d*x + 2*c) + 2*(B*d 
*cos(2*d*x + 2*c)^2 + 2*B*d*cos(2*d*x + 2*c) + B*d)*sin(2*d*x + 2*c)^2 + B 
*d)*integrate((((cos(12*d*x + 12*c)*cos(2*d*x + 2*c) + 5*cos(10*d*x + 10*c 
)*cos(2*d*x + 2*c) + 10*cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 10*cos(6*d*x + 
 6*c)*cos(2*d*x + 2*c) + 5*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 
 2*c)^2 + sin(12*d*x + 12*c)*sin(2*d*x + 2*c) + 5*sin(10*d*x + 10*c)*sin(2 
*d*x + 2*c) + 10*sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 10*sin(6*d*x + 6*c)*s 
in(2*d*x + 2*c) + 5*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2 
)*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c) 
*sin(12*d*x + 12*c) + 5*cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 10*cos(2*d*x 
 + 2*c)*sin(8*d*x + 8*c) + 10*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 5*cos(2* 
d*x + 2*c)*sin(4*d*x + 4*c) - cos(12*d*x + 12*c)*sin(2*d*x + 2*c) - 5*cos( 
10*d*x + 10*c)*sin(2*d*x + 2*c) - 10*cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 1 
0*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 5*cos(4*d*x + 4*c)*sin(2*d*x + 2*c)) 
*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin 
(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*sin(12*d*x + 12 
*c) + 5*cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 10*cos(2*d*x + 2*c)*sin(8*d* 
x + 8*c) + 10*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 5*cos(2*d*x + 2*c)*si...
 

Giac [A] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.43 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (315 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 315 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (630 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 420 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (756 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 882 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (522 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 324 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (81 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 107 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \] Input:

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="giac")
 

Output:

2/315*(315*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 315*sqrt(2)*C*a^5*sgn(cos(d*x 
 + c)) - (630*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 420*sqrt(2)*C*a^5*sgn(cos( 
d*x + c)) - (756*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 882*sqrt(2)*C*a^5*sgn(c 
os(d*x + c)) - (522*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 324*sqrt(2)*C*a^5*sg 
n(cos(d*x + c)) - (81*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 107*sqrt(2)*C*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)^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)^4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)
 

Mupad [B] (verification not implemented)

Time = 20.64 (sec) , antiderivative size = 512, normalized size of antiderivative = 2.74 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

int(((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(1/2))/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)*(exp(c*1i + 
 d*x*1i)*((B*16i)/(5*d) + ((48*B - 32*C)*1i)/(105*d)) + ((336*B + 672*C)*1 
i)/(105*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - ((a + 
 a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x* 
1i)*((B*16i)/(7*d) - (C*320i)/(63*d)) + (C*32i)/(7*d) + ((144*B + 288*C)*1 
i)/(63*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*1 
i)*((B*16i)/(9*d) - ((16*B + 32*C)*1i)/(9*d)) - (B*16i)/(9*d) + ((16*B + 3 
2*C)*1i)/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) - ( 
exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)) 
^(1/2)*(288*B + 256*C)*1i)/(315*d*(exp(c*1i + d*x*1i) + 1)) - (exp(c*1i + 
d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(144 
*B + 128*C)*1i)/(315*d*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1))
 

Reduce [F]

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

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

sqrt(a)*(int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**5,x)*c + int(sqrt(sec(c 
+ d*x) + 1)*sec(c + d*x)**4,x)*b)