\(\int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\) [260]

Optimal result

Integrand size = 25, antiderivative size = 137 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \] Output:

3/32*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))*2^(1/2)/a^(5/2)/d+1/4*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d* 
x+c))^(5/2)+3/16*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(308\) vs. \(2(137)=274\).

Time = 0.47 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.25 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {6 \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+14 \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)-3 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-6 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)-3 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^2(c+d x) \tan (c+d x)+6 \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) (1+\sec (c+d x))^2 \tan (c+d x)+6 \arcsin \left (\sqrt {\sec (c+d x)}\right ) (1+\sec (c+d x))^2 \tan (c+d x)}{32 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \] Input:

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

Output:

(6*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] + 14*Sqrt[1 - Se 
c[c + d*x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x] - 3*Sqrt[2]*ArcTan[(Sqrt[2]*Sq 
rt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - 6*Sqrt[2]*ArcTan[ 
(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Tan[c + 
d*x] - 3*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x] 
]]*Sec[c + d*x]^2*Tan[c + d*x] + 6*ArcSin[Sqrt[1 - Sec[c + d*x]]]*(1 + Sec 
[c + d*x])^2*Tan[c + d*x] + 6*ArcSin[Sqrt[Sec[c + d*x]]]*(1 + Sec[c + d*x] 
)^2*Tan[c + d*x])/(32*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2 
))
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4297, 3042, 4297, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4297

\(\displaystyle \frac {3 \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a}+\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a}+\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4297

\(\displaystyle \frac {3 \left (\frac {\int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{4 a}+\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}+\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {3 \left (\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {\int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{2 a d}\right )}{8 a}+\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

Input:

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

Output:

(Sec[c + d*x]^(5/2)*Sin[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + (3*(A 
rcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c 
 + d*x]])]/(2*Sqrt[2]*a^(3/2)*d) + (Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*d* 
(a + a*Sec[c + d*x])^(3/2))))/(8*a)
 

Defintions of rubi rules used

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 4297

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

Maple [A] (verified)

Time = 1.88 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.26

method	result	size
default	\(\frac {\left (\left (3 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+3\right ) \arctan \left (\frac {\sqrt {2}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (3 \cos \left (d x +c \right )+7\right ) \sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \cos \left (d x +c \right )^{3} \sqrt {2}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(172\)

Input:

int(sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/32/d/a^3*((3*cos(d*x+c)^2+6*cos(d*x+c)+3)*arctan(1/2*2^(1/2)*(cot(d*x+c) 
-csc(d*x+c))/(-1/(cos(d*x+c)+1))^(1/2))+(3*cos(d*x+c)+7)*sin(d*x+c)*(-2/(c 
os(d*x+c)+1))^(1/2))*cos(d*x+c)^3*2^(1/2)*(a*(1+sec(d*x+c)))^(1/2)*sec(d*x 
+c)^(5/2)/(cos(d*x+c)^3+3*cos(d*x+c)^2+3*cos(d*x+c)+1)/(-1/(cos(d*x+c)+1)) 
^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 426, normalized size of antiderivative = 3.11 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + \frac {4 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 7 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) - \frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 7 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \] Input:

integrate(sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

[1/64*(3*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)* 
sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + 
a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a) 
/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(3*cos(d*x + c)^2 + 7*cos(d*x 
+ c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + 
c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) 
 + a^3*d), -1/32*(3*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x 
 + c) + 1)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos( 
d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) - 2*(3*cos(d*x + c)^2 + 7*c 
os(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos 
(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d 
*x + c) + a^3*d)]
 

Sympy [F(-1)]

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

integrate(sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 84332 vs. \(2 (112) = 224\).

Time = 14.41 (sec) , antiderivative size = 84332, normalized size of antiderivative = 615.56 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

1/32*(512*((2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d*x + 5/2*c) + cos( 
5/2*d*x + 5/2*c)*sin(4*d*x + 4*c) + 2*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c 
) + (2*cos(2*d*x + 2*c) + cos(d*x + c))*sin(5/2*d*x + 5/2*c) + cos(4*d*x + 
 4*c)*sin(5/2*d*x + 5/2*c) + 2*cos(3*d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos( 
5*d*x + 5*c)^2 + 2560*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d*x + 
 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2*c)*s 
in(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*cos(2*d*x 
 + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x + 5*c)*sin( 
5/2*d*x + 5/2*c) - 5*cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) - 10*cos(3*d*x 
+ 3*c)*sin(5/2*d*x + 5/2*c))*cos(8/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2 
*d*x + 5/2*c)))^2 + 10240*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d 
*x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2* 
c)*sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*cos(2 
*d*x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x + 5*c)* 
sin(5/2*d*x + 5/2*c) - 5*cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) - 10*cos(3* 
d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos(6/5*arctan2(sin(5/2*d*x + 5/2*c), cos 
(5/2*d*x + 5/2*c)))^2 + 10240*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5 
/2*d*x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 
5/2*c)*sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*c 
os(2*d*x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x ...
 

Giac [A] (verification not implemented)

Time = 1.30 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.82 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3}} + \frac {5 \, \sqrt {2}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {3 \, \sqrt {2} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {5}{2}}}}{32 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

integrate(sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

1/32*(sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*tan(1/2*d*x + 1/2*c)^2 
/a^3 + 5*sqrt(2)/a^3)*tan(1/2*d*x + 1/2*c) - 3*sqrt(2)*log(abs(-sqrt(a)*ta 
n(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/a^(5/2))/(d*sgn( 
cos(d*x + c)))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

int(sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(5/2),x)
 

Output:

(sqrt(a)*int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2)/( 
sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x))/a**3