\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [542]

Optimal result

Integrand size = 35, antiderivative size = 207 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {2 (13 A-5 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {2 (A-5 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}} \] Output:

-2^(1/2)*(A-B)*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-5*B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)-2/15*(A-5*B)*c 
os(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 = 0.49 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.74 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\cos ^{\frac {3}{2}}(c+d x) \left (15 \sqrt {2} (A-B) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^{\frac {5}{2}}(c+d x)+2 \sqrt {1-\sec (c+d x)} \left (3 A-(A-5 B) \sec (c+d x)+(13 A-5 B) \sec ^2(c+d x)\right )\right ) \sin (c+d x)}{15 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(Cos[c + d*x]^(3/2)*(15*Sqrt[2]*(A - B)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]] 
)/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]^(5/2) + 2*Sqrt[1 - Sec[c + d*x]]*(3 
*A - (A - 5*B)*Sec[c + d*x] + (13*A - 5*B)*Sec[c + d*x]^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 = 226, normalized size of antiderivative = 1.09, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 3434, 3042, 4510, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3434

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4501

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4295

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

\(\Big \downarrow \) 219

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

Input:

Int[(Cos[c + d*x]^(5/2)*(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])/(5*d*Sec[c + d*x 
]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) - ((2*a*(A - 5*B)*Sin[c + d*x])/(3*d*Sqr 
t[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((-15*Sqrt[2]*a^(3/2)*(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^2*(13*A - 5*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*S 
qrt[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 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]
 

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {\left (\left (15 \cos \left (d x +c \right )+15\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (-15 \cos \left (d x +c \right )-15\right ) B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (6 \cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )+26\right ) \sin \left (d x +c \right ) A +\left (10 \cos \left (d x +c \right )-10\right ) \sin \left (d x +c \right ) B \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{15 d a \left (1+\cos \left (d x +c \right )\right )}\)	\(208\)

Input:

int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x,method=_RET 
URNVERBOSE)
 

Output:

1/15/d*((15*cos(d*x+c)+15)*(-2/(1+cos(d*x+c)))^(1/2)*A*arctan(1/2*2^(1/2)* 
(-csc(d*x+c)+cot(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+(-15*cos(d*x+c)-15)*B* 
(-2/(1+cos(d*x+c)))^(1/2)*arctan(1/2*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(-1/ 
(1+cos(d*x+c)))^(1/2))+(6*cos(d*x+c)^2-2*cos(d*x+c)+26)*sin(d*x+c)*A+(10*c 
os(d*x+c)-10)*sin(d*x+c)*B)*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2)/a/(1 
+cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.82 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {4 \, {\left (3 \, A \cos \left (d x + c\right )^{2} - {\left (A - 5 \, B\right )} \cos \left (d x + c\right ) + 13 \, A - 5 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \frac {15 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac {15 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right ) + 1\right )}}\right ) + 2 \, {\left (3 \, A \cos \left (d x + c\right )^{2} - {\left (A - 5 \, B\right )} \cos \left (d x + c\right ) + 13 \, A - 5 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \] Input:

integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algo 
rithm="fricas")
 

Output:

[1/30*(4*(3*A*cos(d*x + c)^2 - (A - 5*B)*cos(d*x + c) + 13*A - 5*B)*sqrt(( 
a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 15*sqr 
t(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)/s 
qrt(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 - B)*a*cos(d*x + c) + (A 
 - B)*a)*sqrt(-1/a)*arctan(1/2*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)/(cos(d*x + c) + 1)) + 2*(3 
*A*cos(d*x + c)^2 - (A - 5*B)*cos(d*x + c) + 13*A - 5*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)]
 

Sympy [F(-1)]

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

integrate(cos(d*x+c)**(5/2)*(A+B*sec(d*x+c))/(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. 581 vs. \(2 (174) = 348\).

Time = 0.26 (sec) , antiderivative size = 581, normalized size of antiderivative = 2.81 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algo 
rithm="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) + 10*(3*sqrt(2)*log(co 
s(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) - 3*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) + 2*sqrt 
(2)*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 6*sqrt(2)*sin(1 
/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*B/sqrt(a))/d
 

Giac [A] (verification not implemented)

Time = 171.59 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\frac {15 \, {\left (\sqrt {2} A - \sqrt {2} B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {2 \, {\left ({\left (20 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 10 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (17 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 10 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {15 \, \sqrt {2} A a^{2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {5}{2}}}}{15 \, d} \] Input:

integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algo 
rithm="giac")
 

Output:

1/15*(15*(sqrt(2)*A - sqrt(2)*B)*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*((20*s 
qrt(2)*A*a^2*sgn(cos(d*x + c)) - 10*sqrt(2)*B*a^2*sgn(cos(d*x + c)) + (17* 
sqrt(2)*A*a^2*sgn(cos(d*x + c)) - 10*sqrt(2)*B*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 + 15*sqrt(2)*A*a^2/sgn(cos(d*x 
+ c)))*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5/2))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(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))/(sec(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x) + 1)*sqrt(co 
s(c + d*x))*cos(c + d*x)**2)/(sec(c + d*x) + 1),x)*a))/a