3.3.5 \(\int \frac {\cos (c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [205]

3.3.5.1 Optimal result
3.3.5.2 Mathematica [C] (verified)
3.3.5.3 Rubi [A] (verified)
3.3.5.4 Maple [B] (warning: unable to verify)
3.3.5.5 Fricas [A] (verification not implemented)
3.3.5.6 Sympy [F]
3.3.5.7 Maxima [F]
3.3.5.8 Giac [B] (verification not implemented)
3.3.5.9 Mupad [F(-1)]

3.3.5.1 Optimal result

Integrand size = 33, antiderivative size = 199 \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {5 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {(115 A+3 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(15 A-C) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(35 A+3 C) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}} \]

output
-5*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-1/4*(A+C) 
*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(15*A-C)*sin(d*x+c)/a/d/(a+a*sec 
(d*x+c))^(3/2)+1/32*(115*A+3*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a 
*sec(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+1/16*(35*A+3*C)*sin(d*x+c)/a^2/d/(a+ 
a*sec(d*x+c))^(1/2)
 
3.3.5.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 3.17 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.10 \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\left (2 A \left (-640 \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+460 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+(55+94 \cos (c+d x)+55 \cos (2 (c+d x))+8 \cos (3 (c+d x))) \sqrt {1-\sec (c+d x)}\right )+64 C \cos ^4\left (\frac {1}{2} (c+d x)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right ) \sqrt {1-\sec (c+d x)}\right ) \sec ^2(c+d x) \tan (c+d x)}{64 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \]

input
Integrate[(Cos[c + d*x]*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2) 
,x]
 
output
((2*A*(-640*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^4 + 460*Sqrt[ 
2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^4 + (55 + 94*C 
os[c + d*x] + 55*Cos[2*(c + d*x)] + 8*Cos[3*(c + d*x)])*Sqrt[1 - Sec[c + d 
*x]]) + 64*C*Cos[(c + d*x)/2]^4*Hypergeometric2F1[1/2, 3, 3/2, (1 - Sec[c 
+ d*x])/2]*Sqrt[1 - Sec[c + d*x]])*Sec[c + d*x]^2*Tan[c + d*x])/(64*d*Sqrt 
[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2))
 
3.3.5.3 Rubi [A] (verified)

Time = 1.29 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.08, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 4573, 27, 3042, 4508, 27, 3042, 4510, 25, 3042, 4408, 3042, 4261, 216, 4282, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (2 a^2 (35 A+3 C)-3 a^2 (15 A-C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (15 A-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (2 a^2 (35 A+3 C)-3 a^2 (15 A-C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (15 A-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 a^2 (35 A+3 C)-3 a^2 (15 A-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (15 A-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\frac {\frac {\int -\frac {80 a^3 A-a^3 (35 A+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {2 a^2 (35 A+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (15 A-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {80 a^3 A-a^3 (35 A+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {80 a^3 A-a^3 (35 A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4408

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {80 a^2 A \int \sqrt {\sec (c+d x) a+a}dx-a^3 (115 A+3 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {80 a^2 A \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-a^3 (115 A+3 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-\left (a^3 (115 A+3 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )-\frac {160 a^3 A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}}{4 a^2}-\frac {a (15 A-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {160 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-a^3 (115 A+3 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (115 A+3 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {160 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {a (15 A-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {160 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{5/2} (115 A+3 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {a (15 A-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

input
Int[(Cos[c + d*x]*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]
 
output
-1/4*((A + C)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) + (-1/2*(a*(15* 
A - C)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) + (-(((160*a^(5/2)*A*A 
rcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (Sqrt[2]*a^(5/ 
2)*(115*A + 3*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + 
 d*x]])])/d)/a) + (2*a^2*(35*A + 3*C)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + 
d*x]]))/(4*a^2))/(8*a^2)
 

3.3.5.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4261
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 4282
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2/f   Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ 
a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4408
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c/a   Int[Sqrt[a + b*Csc[e + f*x]], x], x] - 
Simp[(b*c - a*d)/a   Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; F 
reeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
 

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 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 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)]
 
3.3.5.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(726\) vs. \(2(170)=340\).

Time = 1.07 (sec) , antiderivative size = 727, normalized size of antiderivative = 3.65

method result size
default \(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (-2 A \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}-2 C \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+80 A \sqrt {2}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right ) \left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+21 A \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+5 C \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-115 A \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-3 C \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+80 A \,\operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right ) \sqrt {2}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}+34 A \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+2 C \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-115 A \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}-3 C \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}-53 A \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-5 C \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{32 a^{3} d \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1\right )}\) \(727\)

input
int(cos(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x,method=_RETURNV 
ERBOSE)
 
output
-1/32/a^3/d*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(-2*A*(1-cos(d* 
x+c))^7*csc(d*x+c)^7-2*C*(1-cos(d*x+c))^7*csc(d*x+c)^7+80*A*2^(1/2)*((1-co 
s(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d* 
x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*(1-cos(d*x+c))^2*csc(d*x+c)^2+21 
*A*(1-cos(d*x+c))^5*csc(d*x+c)^5+5*C*(1-cos(d*x+c))^5*csc(d*x+c)^5-115*A*( 
(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d* 
x+c))^2*csc(d*x+c)^2-1)^(1/2))*(1-cos(d*x+c))^2*csc(d*x+c)^2-3*C*((1-cos(d 
*x+c))^2*csc(d*x+c)^2-1)^(1/2)*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2* 
csc(d*x+c)^2-1)^(1/2))*(1-cos(d*x+c))^2*csc(d*x+c)^2+80*A*arctanh(2^(1/2)/ 
((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*2^(1/2)* 
((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)+34*A*(1-cos(d*x+c))^3*csc(d*x+c)^3 
+2*C*(1-cos(d*x+c))^3*csc(d*x+c)^3-115*A*ln(csc(d*x+c)-cot(d*x+c)+((1-cos( 
d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)-3 
*C*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*((1-c 
os(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)-53*A*(-cot(d*x+c)+csc(d*x+c))-5*C*(-cot 
(d*x+c)+csc(d*x+c)))/((1-cos(d*x+c))^2*csc(d*x+c)^2+1)
 
3.3.5.5 Fricas [A] (verification not implemented)

Time = 3.56 (sec) , antiderivative size = 692, normalized size of antiderivative = 3.48 \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {\sqrt {2} {\left ({\left (115 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (115 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (115 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 115 \, A + 3 \, C\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 160 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (16 \, A \cos \left (d x + c\right )^{3} + {\left (55 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (35 \, A + 3 \, 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 )}{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 {\sqrt {2} {\left ({\left (115 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (115 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (115 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 115 \, A + 3 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 160 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, {\left (16 \, A \cos \left (d x + c\right )^{3} + {\left (55 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (35 \, A + 3 \, 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 )}{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(cos(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorith 
m="fricas")
 
output
[-1/64*(sqrt(2)*((115*A + 3*C)*cos(d*x + c)^3 + 3*(115*A + 3*C)*cos(d*x + 
c)^2 + 3*(115*A + 3*C)*cos(d*x + c) + 115*A + 3*C)*sqrt(-a)*log((2*sqrt(2) 
*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c 
) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x 
 + c) + 1)) + 160*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos(d*x + c 
) + A)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) 
+ a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d* 
x + c) + 1)) - 4*(16*A*cos(d*x + c)^3 + (55*A + 7*C)*cos(d*x + c)^2 + (35* 
A + 3*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/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)*((115*A + 3*C)*cos(d*x + c)^3 + 3*(115*A + 3*C)*co 
s(d*x + c)^2 + 3*(115*A + 3*C)*cos(d*x + c) + 115*A + 3*C)*sqrt(a)*arctan( 
sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin( 
d*x + c))) - 160*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos(d*x + c) 
 + A)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/ 
(sqrt(a)*sin(d*x + c))) - 2*(16*A*cos(d*x + c)^3 + (55*A + 7*C)*cos(d*x + 
c)^2 + (35*A + 3*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/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)]
 
3.3.5.6 Sympy [F]

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

input
integrate(cos(d*x+c)*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)
 
output
Integral((A + C*sec(c + d*x)**2)*cos(c + d*x)/(a*(sec(c + d*x) + 1))**(5/2 
), x)
 
3.3.5.7 Maxima [F]

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

input
integrate(cos(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorith 
m="maxima")
 
output
integrate((C*sec(d*x + c)^2 + A)*cos(d*x + c)/(a*sec(d*x + c) + a)^(5/2), 
x)
 
3.3.5.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 457 vs. \(2 (170) = 340\).

Time = 1.59 (sec) , antiderivative size = 457, normalized size of antiderivative = 2.30 \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {2 \, \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} {\left (A a^{5} + C a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\sqrt {2} {\left (21 \, A a^{5} + 5 \, C a^{5}\right )}}{a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {\sqrt {2} {\left (115 \, A + 3 \, 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 )}^{2}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {160 \, A \log \left (\frac {{\left | -2 \, {\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 )}^{2} - 4 \, \sqrt {2} {\left | a \right |} + 6 \, a \right |}}{{\left | -2 \, {\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 )}^{2} + 4 \, \sqrt {2} {\left | a \right |} + 6 \, a \right |}}\right )}{\sqrt {-a} a {\left | a \right |} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {128 \, \sqrt {2} {\left (3 \, {\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 )}^{2} A - A a\right )}}{{\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 )}^{4} - 6 \, {\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 )}^{2} a + a^{2}\right )} \sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{64 \, d} \]

input
integrate(cos(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorith 
m="giac")
 
output
-1/64*(2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(A*a^5 + C*a^5)*ta 
n(1/2*d*x + 1/2*c)^2/(a^8*sgn(cos(d*x + c))) - sqrt(2)*(21*A*a^5 + 5*C*a^5 
)/(a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c) + sqrt(2)*(115*A + 3*C)*lo 
g((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2) 
/(sqrt(-a)*a^2*sgn(cos(d*x + c))) - 160*A*log(abs(-2*(sqrt(-a)*tan(1/2*d*x 
 + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 4*sqrt(2)*abs(a) + 6* 
a)/abs(-2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 
+ a))^2 + 4*sqrt(2)*abs(a) + 6*a))/(sqrt(-a)*a*abs(a)*sgn(cos(d*x + c))) - 
 128*sqrt(2)*(3*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2 
*c)^2 + a))^2*A - A*a)/(((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2* 
d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/ 
2*d*x + 1/2*c)^2 + a))^2*a + a^2)*sqrt(-a)*a*sgn(cos(d*x + c))))/d
 
3.3.5.9 Mupad [F(-1)]

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

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