\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [1155]

Optimal result

Integrand size = 37, antiderivative size = 201 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {2 (13 A+15 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {2 A \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}} \] Output:

-2^(1/2)*(A+C)*arctanh(1/2*a^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)*2^(1/2)/(a+ 
a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(1/2)/d+2/15*(13* 
A+15*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)-2/15*A*cos(d* 
x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/5*A*cos(d*x+c)^(3/2)*sin( 
d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\cos ^{\frac {3}{2}}(c+d x) \left ((29 A+30 C-2 A \cos (c+d x)+3 A \cos (2 (c+d x))) \sqrt {1-\sec (c+d x)} \sec ^2(c+d x)+15 \sqrt {2} (A+C) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^{\frac {5}{2}}(c+d x)\right ) \sin (c+d x)}{15 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(Cos[c + d*x]^(3/2)*((29*A + 30*C - 2*A*Cos[c + d*x] + 3*A*Cos[2*(c + d*x) 
])*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^2 + 15*Sqrt[2]*(A + C)*ArcTan[(Sqrt 
[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]^(5/2))*Sin[c 
+ d*x])/(15*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.25 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.09, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.351, Rules used = {3042, 4753, 3042, 4575, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4575

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int -\frac {a A-a (4 A+5 C) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{5 a}+\frac {2 A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a A-a (4 A+5 C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{5 a}\right )\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 \int -\frac {a^2 (13 A+15 C)-2 a^2 A \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {2 a A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{5 a}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (13 A+15 C)-2 a^2 A \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (13 A+15 C)-2 a^2 A \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 4501

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-15 a^2 (A+C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-15 a^2 (A+C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 4295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {30 a^2 (A+C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (13 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {15 \sqrt {2} a^{3/2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}}{5 a}\right )\)

Input:

Int[(Cos[c + d*x]^(5/2)*(A + C*Sec[c + d*x]^2))/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])/(5*d*Sec[c + d*x 
]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) - ((2*a*A*Sin[c + d*x])/(3*d*Sqrt[Sec[c 
+ d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((-15*Sqrt[2]*a^(3/2)*(A + C)*ArcTanh[ 
(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x] 
])])/d + (2*a^2*(13*A + 15*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + 
 a*Sec[c + d*x]]))/(3*a))/(5*a))
 

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 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]
 

rule 4575

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

rule 4753

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a 
+ b*x])^m*(c*Sec[a + b*x])^m   Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[u, x 
]
 

Maple [A] (verified)

Time = 1.68 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {\left (\left (15 \cos \left (d x +c \right )+15\right ) A \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (15 \cos \left (d x +c \right )+15\right ) C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (6 \cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )+26\right ) \sin \left (d x +c \right ) A +30 C \sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{15 d a \left (\cos \left (d x +c \right )+1\right )}\)	\(199\)

Input:

int(cos(d*x+c)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x,method=_R 
ETURNVERBOSE)
 

Output:

1/15/d*((15*cos(d*x+c)+15)*A*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)/ 
(-1/(cos(d*x+c)+1))^(1/2)*(-csc(d*x+c)+cot(d*x+c)))+(15*cos(d*x+c)+15)*C*( 
-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)/(-1/(cos(d*x+c)+1))^(1/2)*(-cs 
c(d*x+c)+cot(d*x+c)))+(6*cos(d*x+c)^2-2*cos(d*x+c)+26)*sin(d*x+c)*A+30*C*s 
in(d*x+c))*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2)/a/(cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.80 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {4 \, {\left (3 \, A \cos \left (d x + c\right )^{2} - A \cos \left (d x + c\right ) + 13 \, A + 15 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \frac {15 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right ) + {\left (A + C\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac {15 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right ) + {\left (A + C\right )} a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right ) + 1\right )}}\right ) + 2 \, {\left (3 \, A \cos \left (d x + c\right )^{2} - A \cos \left (d x + c\right ) + 13 \, A + 15 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \] Input:

integrate(cos(d*x+c)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, al 
gorithm="fricas")
 

Output:

[1/30*(4*(3*A*cos(d*x + c)^2 - A*cos(d*x + c) + 13*A + 15*C)*sqrt((a*cos(d 
*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 15*sqrt(2)*(( 
A + C)*a*cos(d*x + c) + (A + C)*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), 1/15*(15*sqrt(2)*((A + C)*a*cos(d*x + c) + (A + C)*a 
)*sqrt(-1/a)*arctan(1/2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sq 
rt(-1/a)*sqrt(cos(d*x + c))*sin(d*x + c)/(cos(d*x + c) + 1)) + 2*(3*A*cos( 
d*x + c)^2 - A*cos(d*x + c) + 13*A + 15*C)*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)]
 

Sympy [F(-1)]

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

integrate(cos(d*x+c)**(5/2)*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 553 vs. \(2 (168) = 336\).

Time = 0.27 (sec) , antiderivative size = 553, normalized size of antiderivative = 2.75 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, al 
gorithm="maxima")
 

Output:

1/60*(sqrt(2)*(60*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2* 
c)))*sin(5/2*d*x + 5/2*c) - 5*cos(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/ 
2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 60*cos(5/2*d*x + 5/2*c)*sin(4/5*ar 
ctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 5*cos(5/2*d*x + 5/2*c 
)*sin(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 30*log(co 
s(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + sin(1/5*arc 
tan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + 2*sin(1/5*arctan2(si 
n(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 1) + 30*log(cos(1/5*arctan2(s 
in(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + sin(1/5*arctan2(sin(5/2*d* 
x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 - 2*sin(1/5*arctan2(sin(5/2*d*x + 5/2 
*c), cos(5/2*d*x + 5/2*c))) + 1) + 6*sin(5/2*d*x + 5/2*c) - 5*sin(3/5*arct 
an2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 60*sin(1/5*arctan2(sin( 
5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*A/sqrt(a) - 30*(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), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), co 
s(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), cos(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))))*C/sqrt(a))/d
 

Giac [A] (verification not implemented)

Time = 173.65 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\frac {15 \, {\left (\sqrt {2} A + \sqrt {2} C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {2 \, {\left (15 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (20 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 30 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (17 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {5}{2}}}}{15 \, d} \] Input:

integrate(cos(d*x+c)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, al 
gorithm="giac")
 

Output:

1/15*(15*(sqrt(2)*A + sqrt(2)*C)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + s 
qrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(a)*sgn(cos(d*x + c))) + 2*(15*sq 
rt(2)*A*a^2*sgn(cos(d*x + c)) + 15*sqrt(2)*C*a^2*sgn(cos(d*x + c)) + (20*s 
qrt(2)*A*a^2*sgn(cos(d*x + c)) + 30*sqrt(2)*C*a^2*sgn(cos(d*x + c)) + (17* 
sqrt(2)*A*a^2*sgn(cos(d*x + c)) + 15*sqrt(2)*C*a^2*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)/(a*tan(1/ 
2*d*x + 1/2*c)^2 + a)^(5/2))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**2*s 
ec(c + d*x)**2)/(sec(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x) + 1)*sqrt 
(cos(c + d*x))*cos(c + d*x)**2)/(sec(c + d*x) + 1),x)*a))/a