\(\int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx\) [292]

Optimal result

Integrand size = 37, antiderivative size = 199 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=-\frac {5 (15 A-C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(13 A-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(49 A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:

-5/32*(15*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))*2^(1/2)/a^(5/2)/d-1/4*(A+C)*sec(d*x+c)^(1/2)*sin(d*x+ 
c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(13*A-3*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/ 
d/(a+a*sec(d*x+c))^(3/2)+1/16*(49*A+C)*sec(d*x+c)^(1/2)*sin(d*x+c)/a^2/d/( 
a+a*sec(d*x+c))^(1/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(479\) vs. \(2(199)=398\).

Time = 6.64 (sec) , antiderivative size = 479, normalized size of antiderivative = 2.41 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=-\frac {A \sqrt {1+\sec (c+d x)} \left (\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{d (1+\sec (c+d x))^{5/2}}+\frac {26 \sqrt {\sec (c+d x)} \sin (c+d x)}{d (1+\sec (c+d x))^{3/2}}-\frac {98 \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {1+\sec (c+d x)}}-\frac {75 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )}{32 a^2 \sqrt {a (1+\sec (c+d x))}}-\frac {C \sqrt {1+\sec (c+d x)} \left (\frac {8 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (1+\sec (c+d x))^{5/2}}+5 \left (\frac {2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (1+\sec (c+d x))^{3/2}}-\frac {\left (2 \arcsin \left (\sqrt {\sec (c+d x)}\right )-\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {2 \left (\arcsin \left (\sqrt {1-\sec (c+d x)}\right )+\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )\right )}{32 a^2 \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

-1/32*(A*Sqrt[1 + Sec[c + d*x]]*((8*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(1 
 + Sec[c + d*x])^(5/2)) + (26*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(1 + Sec 
[c + d*x])^(3/2)) - (98*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[1 + Sec[c 
 + d*x]]) - (75*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c 
 + d*x]]]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]])) 
)/(a^2*Sqrt[a*(1 + Sec[c + d*x])]) - (C*Sqrt[1 + Sec[c + d*x]]*((8*Sec[c + 
 d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(5/2)) + 5*((2*Sec[c + d*x 
]^(5/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(3/2)) - ((2*ArcSin[Sqrt[Sec[c 
 + d*x]]] - Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d 
*x]]])*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]) - ( 
2*(ArcSin[Sqrt[1 - Sec[c + d*x]]] + Sqrt[1 - Sec[c + d*x]]*Sqrt[Sec[c + d* 
x]])*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]))))/(3 
2*a^2*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {3042, 4573, 27, 3042, 4508, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

\(\displaystyle -\frac {\int -\frac {a (9 A+C)-4 a (A-C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (9 A+C)-4 a (A-C) \sec (c+d x)}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\int \frac {a^2 (49 A+C)-2 a^2 (13 A-3 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {a^2 (49 A+C)-2 a^2 (13 A-3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {a^2 (49 A+C)-2 a^2 (13 A-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4501

\(\displaystyle \frac {\frac {\frac {2 a^2 (49 A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-5 a^2 (15 A-C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (49 A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-5 a^2 (15 A-C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {\frac {\frac {10 a^2 (15 A-C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (49 A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {2 a^2 (49 A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {5 \sqrt {2} a^{3/2} (15 A-C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*((A + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/ 
2)) + (-1/2*(a*(13*A - 3*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec 
[c + d*x])^(3/2)) + ((-5*Sqrt[2]*a^(3/2)*(15*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*(49*A + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) 
)/(4*a^2))/(8*a^2)
 

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 4508

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

rule 4573

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) 
*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m 
+ 1))), x] + Simp[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*C 
sc[e + f*x])^n*Simp[b*C*n + A*b*(2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - 
n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && EqQ[ 
a^2 - b^2, 0] && LtQ[m, -2^(-1)]
 

Maple [A] (warning: unable to verify)

Time = 2.08 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.31

method	result	size
default	\(\frac {\left (-\frac {\left (2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-3 \csc \left (d x +c \right )+3 \cot \left (d x +c \right )\right ) C}{32}-\frac {\left (2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-83 \csc \left (d x +c \right )+83 \cot \left (d x +c \right )\right ) A}{32}+\frac {75 A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}}{32}-\frac {5 C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )}{32}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{3} \sqrt {\sec \left (d x +c \right )}}\)	\(261\)
parts	\(\frac {A \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}}{16}+\frac {17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}}{32}+\frac {75 \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}}{32}+\frac {83 \csc \left (d x +c \right )}{32}-\frac {83 \cot \left (d x +c \right )}{32}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{3} \sqrt {\sec \left (d x +c \right )}}+\frac {C \left (\left (5 \cos \left (d x +c \right )^{2}+10 \cos \left (d x +c \right )+5\right ) \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (5 \cos \left (d x +c \right )+1\right ) \sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \cos \left (d x +c \right )^{2} \sqrt {2}\, \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(314\)

Input:

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

Output:

1/d*(-1/32*(2*(1-cos(d*x+c))^5*csc(d*x+c)^5-(1-cos(d*x+c))^3*csc(d*x+c)^3- 
3*csc(d*x+c)+3*cot(d*x+c))*C-1/32*(2*(1-cos(d*x+c))^5*csc(d*x+c)^5-17*(1-c 
os(d*x+c))^3*csc(d*x+c)^3-83*csc(d*x+c)+83*cot(d*x+c))*A+75/32*A*arctan(1/ 
2*2^(1/2)/(-1/(cos(d*x+c)+1))^(1/2)*(-csc(d*x+c)+cot(d*x+c)))*(-2/(cos(d*x 
+c)+1))^(1/2)-5/32*C*(-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))))/a^3*(a*(1+sec(d*x+c)))^(1/2)/ 
sec(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 520, normalized size of antiderivative = 2.61 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {5 \, \sqrt {2} {\left ({\left (15 \, A - C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right ) + 15 \, A - C\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 (32 \, A \cos \left (d x + c\right )^{3} + 5 \, {\left (17 \, A + C\right )} \cos \left (d x + c\right )^{2} + {\left (49 \, A + C\right )} \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 {5 \, \sqrt {2} {\left ({\left (15 \, A - C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right ) + 15 \, A - C\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 (32 \, A \cos \left (d x + c\right )^{3} + 5 \, {\left (17 \, A + C\right )} \cos \left (d x + c\right )^{2} + {\left (49 \, A + C\right )} \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((A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, al 
gorithm="fricas")
 

Output:

[-1/64*(5*sqrt(2)*((15*A - C)*cos(d*x + c)^3 + 3*(15*A - C)*cos(d*x + c)^2 
 + 3*(15*A - C)*cos(d*x + c) + 15*A - C)*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*(32*A*cos(d*x + c)^3 + 5*(17*A + C)*cos(d*x + c)^2 + (49*A + C) 
*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(c 
os(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*(5*sqrt(2)*((15*A - C)*cos(d*x + c)^3 + 3*(15*A - 
 C)*cos(d*x + c)^2 + 3*(15*A - C)*cos(d*x + c) + 15*A - C)*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*(32*A*cos(d*x + c)^3 + 5*(17*A + C)*cos(d*x + c)^ 
2 + (49*A + C)*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)]
 

Sympy [F(-1)]

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

integrate((A+C*sec(d*x+c)**2)/sec(d*x+c)**(1/2)/(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. 261332 vs. \(2 (168) = 336\).

Time = 2.95 (sec) , antiderivative size = 261332, normalized size of antiderivative = 1313.23 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/32*((576*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*si 
n(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/ 
2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*cos(5/2*d* 
x + 5/2*c)^6 + 14400*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) 
^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2 
*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*c 
os(3/2*d*x + 3/2*c)^6 + 187500*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 
 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin 
(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(1/2*d*x + 1/2*c)^6 
+ 576*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2* 
d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 
 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/ 
2*c)^6 + 5184*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2* 
sin(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 
1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*sin(3/2* 
d*x + 3/2*c)^6 + 262500*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) 
^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d* 
x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(1/2*d*x + 1/2*c)^4*sin(1/2 
*d*x + 1/2*c)^2 + 77700*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) 
^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2...
 

Giac [A] (verification not implemented)

Time = 3.03 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.94 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\frac {{\left ({\left (\frac {2 \, {\left (\sqrt {2} A a^{6} + \sqrt {2} C a^{6}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8}} - \frac {17 \, \sqrt {2} A a^{6} + \sqrt {2} C a^{6}}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {83 \, \sqrt {2} A a^{6} + 3 \, \sqrt {2} C a^{6}}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {5 \, {\left (15 \, \sqrt {2} A - \sqrt {2} C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {5}{2}}}}{32 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

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

Output:

-1/32*(((2*(sqrt(2)*A*a^6 + sqrt(2)*C*a^6)*tan(1/2*d*x + 1/2*c)^2/a^8 - (1 
7*sqrt(2)*A*a^6 + sqrt(2)*C*a^6)/a^8)*tan(1/2*d*x + 1/2*c)^2 - (83*sqrt(2) 
*A*a^6 + 3*sqrt(2)*C*a^6)/a^8)*tan(1/2*d*x + 1/2*c)/sqrt(a*tan(1/2*d*x + 1 
/2*c)^2 + a) - 5*(15*sqrt(2)*A - sqrt(2)*C)*log(abs(-sqrt(a)*tan(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. \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

int((A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(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))/(se 
c(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*c + int((sqrt(s 
ec(c + d*x))*sqrt(sec(c + d*x) + 1))/(sec(c + d*x)**4 + 3*sec(c + d*x)**3 
+ 3*sec(c + d*x)**2 + sec(c + d*x)),x)*a))/a**3