\(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [1288]

Optimal result

Integrand size = 45, antiderivative size = 231 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {(75 A-19 B-5 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {(13 A-5 B-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(49 A-9 B+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \] Output:

-1/32*(75*A-19*B-5*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)*2^(1/2)/a^(5 
/2)/d-1/4*(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2)-1/1 
6*(13*A-5*B-3*C)*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(3/2)+1/ 
16*(49*A-9*B+C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 3.42 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (-8 (75 A-19 B-5 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+4 (65 A-9 B+C+(85 A-13 B+5 C) \cos (c+d x)+16 A \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 a d \cos ^{\frac {3}{2}}(c+d x) (a (1+\sec (c+d x)))^{3/2}} \] Input:

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

Output:

(Sec[(c + d*x)/2]*(-8*(75*A - 19*B - 5*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c 
 + d*x)/2]^4 + 4*(65*A - 9*B + C + (85*A - 13*B + 5*C)*Cos[c + d*x] + 16*A 
*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(64*a*d*Cos[c + d*x]^(3/2)*(a*(1 + S 
ec[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 4753, 3042, 4572, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4572

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (49 A-9 B+C)-2 a^2 (13 A-5 B-3 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (49 A-9 B+C)-2 a^2 (13 A-5 B-3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (49 A-9 B+C)-2 a^2 (13 A-5 B-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4501

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (49 A-9 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-a^2 (75 A-19 B-5 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (49 A-9 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-a^2 (75 A-19 B-5 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (75 A-19 B-5 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (49 A-9 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (49 A-9 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {\sqrt {2} a^{3/2} (75 A-19 B-5 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/4*((A - B + C)*Sqrt[Sec[c + d*x] 
]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) + (-1/2*(a*(13*A - 5*B - 3* 
C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) + (-((S 
qrt[2]*a^(3/2)*(75*A - 19*B - 5*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 - 9*B + 
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 4572

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + 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*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - 
 A*b*(2*m + n + 1) - (b*B*(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, B, C, n}, x] && EqQ[a^2 - 
b^2, 0] && LtQ[m, -2^(-1)]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(506\) vs. \(2(196)=392\).

Time = 4.77 (sec) , antiderivative size = 507, normalized size of antiderivative = 2.19

method	result	size
default	\(\frac {\sqrt {2}\, \left (-\frac {A \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-64 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )+75 \ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-75 \ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}}\, a^{2}}+\frac {C \left (5 \ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-5 \ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{4 d \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}}\, a^{2}}+\frac {B \left (19 \ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-19 \ln \left (-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\csc \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{4 d \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}}\, a^{2}}\right )}{8}\)	\(507\)

Input:

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

Output:

1/8*2^(1/2)*(-1/4*A/d/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/(a/(2*cos(1/2*d*x+1 
/2*c)^2-1)*cos(1/2*d*x+1/2*c)^2)^(1/2)/a^2*(2*tan(1/2*d*x+1/2*c)*sec(1/2*d 
*x+1/2*c)^2-64*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)+75*ln(-cot(1/2*d*x+1/ 
2*c)+csc(1/2*d*x+1/2*c)+1)*cos(1/2*d*x+1/2*c)-75*ln(-cot(1/2*d*x+1/2*c)+cs 
c(1/2*d*x+1/2*c)-1)*cos(1/2*d*x+1/2*c)-21*tan(1/2*d*x+1/2*c))+1/4*C/d/(2*c 
os(1/2*d*x+1/2*c)^2-1)^(1/2)/(a/(2*cos(1/2*d*x+1/2*c)^2-1)*cos(1/2*d*x+1/2 
*c)^2)^(1/2)/a^2*(5*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)+1)*cos(1/2*d 
*x+1/2*c)-5*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)-1)*cos(1/2*d*x+1/2*c 
)+5*tan(1/2*d*x+1/2*c)-2*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^2)+1/4*B/d/ 
(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/(a/(2*cos(1/2*d*x+1/2*c)^2-1)*cos(1/2*d*x 
+1/2*c)^2)^(1/2)/a^2*(19*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)+1)*cos( 
1/2*d*x+1/2*c)-19*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)-1)*cos(1/2*d*x 
+1/2*c)-13*tan(1/2*d*x+1/2*c)+2*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 544, normalized size of antiderivative = 2.35 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/64*(sqrt(2)*((75*A - 19*B - 5*C)*cos(d*x + c)^3 + 3*(75*A - 19*B - 5*C 
)*cos(d*x + c)^2 + 3*(75*A - 19*B - 5*C)*cos(d*x + c) + 75*A - 19*B - 5*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)^2 + (85*A 
- 13*B + 5*C)*cos(d*x + c) + 49*A - 9*B + C)*sqrt((a*cos(d*x + c) + a)/cos 
(d*x + c))*sqrt(cos(d*x + c))*sin(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*(sqrt(2)*((75*A - 1 
9*B - 5*C)*cos(d*x + c)^3 + 3*(75*A - 19*B - 5*C)*cos(d*x + c)^2 + 3*(75*A 
 - 19*B - 5*C)*cos(d*x + c) + 75*A - 19*B - 5*C)*sqrt(-a)*arctan(1/2*sqrt( 
2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin 
(d*x + c)/(a*cos(d*x + c) + a)) + 2*(32*A*cos(d*x + c)^2 + (85*A - 13*B + 
5*C)*cos(d*x + c) + 49*A - 9*B + C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c) 
)*sqrt(cos(d*x + c))*sin(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. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**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. 264379 vs. \(2 (196) = 392\).

Time = 3.17 (sec) , antiderivative size = 264379, normalized size of antiderivative = 1144.50 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) 
)^(5/2),x, algorithm="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 = 166.33 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.23 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\frac {{\left ({\left (\frac {2 \, {\left (\sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8}} - \frac {17 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 9 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {83 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 11 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {{\left (75 \, \sqrt {2} A - 19 \, \sqrt {2} B - 5 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{32 \, d} \] Input:

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

Output:

-1/32*(((2*(sqrt(2)*A*a^6*sgn(cos(d*x + c)) - sqrt(2)*B*a^6*sgn(cos(d*x + 
c)) + sqrt(2)*C*a^6*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2/a^8 - (17*sq 
rt(2)*A*a^6*sgn(cos(d*x + c)) - 9*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + sqrt(2 
)*C*a^6*sgn(cos(d*x + c)))/a^8)*tan(1/2*d*x + 1/2*c)^2 - (83*sqrt(2)*A*a^6 
*sgn(cos(d*x + c)) - 11*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 3*sqrt(2)*C*a^6* 
sgn(cos(d*x + c)))/a^8)*tan(1/2*d*x + 1/2*c)/sqrt(a*tan(1/2*d*x + 1/2*c)^2 
 + a) - (75*sqrt(2)*A - 19*sqrt(2)*B - 5*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)*sgn(cos(d* 
x + c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)**2)/ 
(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*c + int((sqr 
t(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x))/(sec(c + d*x)**3 + 3* 
sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x) + 1)*s 
qrt(cos(c + d*x)))/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 
 1),x)*a))/a**3