3.6.43 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [543]

3.6.43.1 Optimal result
3.6.43.2 Mathematica [A] (verified)
3.6.43.3 Rubi [A] (verified)
3.6.43.4 Maple [B] (verified)
3.6.43.5 Fricas [A] (verification not implemented)
3.6.43.6 Sympy [F(-1)]
3.6.43.7 Maxima [B] (verification not implemented)
3.6.43.8 Giac [F]
3.6.43.9 Mupad [F(-1)]

3.6.43.1 Optimal result

Integrand size = 35, antiderivative size = 162 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}-\frac {2 (A-3 B) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 A \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}} \]

output
(A-B)*arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x 
+c))^(1/2))*2^(1/2)*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d/a^(1/2)-2/3*(A-3*B 
)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+2/3*A*sin(d*x+c)*co 
s(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)
 
3.6.43.2 Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (2 (-A+3 B+A \cos (c+d x)) \sqrt {1-\sec (c+d x)}-3 \sqrt {2} (A-B) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sqrt {\sec (c+d x)}\right ) \sin (c+d x)}{3 d \sqrt {-1+\cos (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

input
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/Sqrt[a + a*Sec[c + d*x 
]],x]
 
output
((2*(-A + 3*B + A*Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]] - 3*Sqrt[2]*(A - B) 
*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sqrt[Sec[c + 
d*x]])*Sin[c + d*x])/(3*d*Sqrt[-1 + Cos[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x] 
)])
 
3.6.43.3 Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3434, 3042, 4510, 27, 3042, 4501, 3042, 4295, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3434

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int -\frac {a (A-3 B)-2 a A \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a (A-3 B)-2 a A \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}\right )\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4501

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a (A-B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}\right )\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4295

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

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 \sqrt {2} \sqrt {a} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}\right )\)

input
Int[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/Sqrt[a + a*Sec[c + d*x]],x]
 
output
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*Sin[c + d*x])/(3*d*Sqrt[Sec[c 
+ d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((-3*Sqrt[2]*Sqrt[a]*(A - B)*ArcTanh[( 
Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]] 
)])/d + (2*a*(A - 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[ 
c + d*x]]))/(3*a))
 

3.6.43.3.1 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 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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 4295
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[-2*b*(d/(a*f))   Subst[Int[1/(2*b - d*x^2), x], 
x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]))], x] /; 
 FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4501
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[(a*A*m 
 - b*B*n)/(b*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] 
, x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a 
^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]
 

rule 4510
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d 
*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B* 
n - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 
 m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
 
3.6.43.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(308\) vs. \(2(135)=270\).

Time = 4.61 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.91

method result size
default \(\frac {\left (3 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {2}\, \cos \left (d x +c \right )-3 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {2}\, \cos \left (d x +c \right )+3 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {2}-3 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {2}+2 A \cos \left (d x +c \right ) \sin \left (d x +c \right )-2 A \sin \left (d x +c \right )+6 B \sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{3 d \left (\cos \left (d x +c \right )+1\right ) a}\) \(309\)

input
int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x,method=_RET 
URNVERBOSE)
 
output
1/3/d*(3*A*arctan(1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1) 
)^(1/2))*(-1/(cos(d*x+c)+1))^(1/2)*2^(1/2)*cos(d*x+c)-3*B*arctan(1/2*sin(d 
*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*(-1/(cos(d*x+c)+1) 
)^(1/2)*2^(1/2)*cos(d*x+c)+3*A*arctan(1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1 
)/(-1/(cos(d*x+c)+1))^(1/2))*(-1/(cos(d*x+c)+1))^(1/2)*2^(1/2)-3*B*arctan( 
1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*(-1/(cos( 
d*x+c)+1))^(1/2)*2^(1/2)+2*A*cos(d*x+c)*sin(d*x+c)-2*A*sin(d*x+c)+6*B*sin( 
d*x+c))*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/a
 
3.6.43.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.07 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {4 \, {\left (A \cos \left (d x + c\right ) - A + 3 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \frac {3 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac {3 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - 2 \, {\left (A \cos \left (d x + c\right ) - A + 3 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \]

input
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algo 
rithm="fricas")
 
output
[1/6*(4*(A*cos(d*x + c) - A + 3*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)) 
*sqrt(cos(d*x + c))*sin(d*x + c) - 3*sqrt(2)*((A - B)*a*cos(d*x + c) + (A 
- B)*a)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x 
 + c))*sqrt(cos(d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos( 
d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c) + a*d), -1/3* 
(3*sqrt(2)*((A - B)*a*cos(d*x + c) + (A - B)*a)*sqrt(-1/a)*arctan(sqrt(2)* 
sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*sqrt(cos(d*x + c))/sin( 
d*x + c)) - 2*(A*cos(d*x + c) - A + 3*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x 
 + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)]
 
3.6.43.6 Sympy [F(-1)]

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

input
integrate(cos(d*x+c)**(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(1/2),x)
 
output
Timed out
 
3.6.43.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 478 vs. \(2 (135) = 270\).

Time = 0.45 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.95 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\frac {{\left (3 \, \sqrt {2} \cos \left (\frac {2}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 \, \sqrt {2} \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \sin \left (\frac {2}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) - 3 \, \sqrt {2} \log \left (\cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 1\right ) + 3 \, \sqrt {2} \log \left (\cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 1\right ) - 2 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sqrt {2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )} A}{\sqrt {a}} + \frac {3 \, {\left (\sqrt {2} \log \left (\cos \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 1\right ) - \sqrt {2} \log \left (\cos \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 1\right ) - 4 \, \sqrt {2} \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )\right )} B}{\sqrt {a}}}{6 \, d} \]

input
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algo 
rithm="maxima")
 
output
-1/6*((3*sqrt(2)*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c 
)))*sin(3/2*d*x + 3/2*c) - 3*sqrt(2)*cos(3/2*d*x + 3/2*c)*sin(2/3*arctan2( 
sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 3*sqrt(2)*log(cos(1/3*arcta 
n2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(1/3*arctan2(sin(3/ 
2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 
 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + 3*sqrt(2)*log(cos(1/3*arctan2(sin(3 
/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(1/3*arctan2(sin(3/2*d*x + 
3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), 
 cos(3/2*d*x + 3/2*c))) + 1) - 2*sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sqrt(2)* 
sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*A/sqrt(a) + 
3*(sqrt(2)*log(cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + si 
n(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(s 
in(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - sqrt(2)*log(cos(1/4*arctan2(sin 
(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/4*arctan2(sin(2*d*x + 2*c), co 
s(2*d*x + 2*c)))^2 - 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)) 
) + 1) - 4*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*B 
/sqrt(a))/d
 
3.6.43.8 Giac [F]

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

input
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algo 
rithm="giac")
 
output
integrate((B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/sqrt(a*sec(d*x + c) + a) 
, x)
 
3.6.43.9 Mupad [F(-1)]

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

input
int((cos(c + d*x)^(3/2)*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x))^(1/2),x 
)
 
output
int((cos(c + d*x)^(3/2)*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x))^(1/2), 
x)