3.6.1 \(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx\) [501]

3.6.1.1 Optimal result
3.6.1.2 Mathematica [C] (warning: unable to verify)
3.6.1.3 Rubi [A] (verified)
3.6.1.4 Maple [B] (verified)
3.6.1.5 Fricas [C] (verification not implemented)
3.6.1.6 Sympy [F(-1)]
3.6.1.7 Maxima [F]
3.6.1.8 Giac [F]
3.6.1.9 Mupad [F(-1)]

3.6.1.1 Optimal result

Integrand size = 33, antiderivative size = 152 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=-\frac {3 (A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {(3 A-5 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}-\frac {(3 A-5 B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {3 (A-B) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}+\frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))} \]

output
-3*(A-B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2 
*d*x+1/2*c),2^(1/2))/a/d-1/3*(3*A-5*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/ 
2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a/d-1/3*(3*A-5*B)*sin(d 
*x+c)/a/d/cos(d*x+c)^(3/2)+(A-B)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d* 
x+c))+3*(A-B)*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)
 
3.6.1.2 Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 8.48 (sec) , antiderivative size = 1041, normalized size of antiderivative = 6.85 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {\cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} (A+B \sec (c+d x)) \left (-\frac {(-A+B) (2+\cos (c)) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c)}{d}-\frac {2 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-A \sin \left (\frac {d x}{2}\right )+B \sin \left (\frac {d x}{2}\right )\right )}{d}+\frac {4 B \sec (c) \sec ^2(c+d x) \sin (d x)}{3 d}+\frac {4 \sec (c) \sec (c+d x) (B \sin (c)+3 A \sin (d x)-3 B \sin (d x))}{3 d}\right )}{(B+A \cos (c+d x)) (a+a \sec (c+d x))}+\frac {A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) (A+B \sec (c+d x)) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{d (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))}-\frac {5 B \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) (A+B \sec (c+d x)) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))}+\frac {3 A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) (A+B \sec (c+d x)) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{2 d (B+A \cos (c+d x)) (a+a \sec (c+d x))}-\frac {3 B \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) (A+B \sec (c+d x)) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{2 d (B+A \cos (c+d x)) (a+a \sec (c+d x))} \]

input
Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])),x 
]
 
output
(Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x])*(-(((-A + B) 
*(2 + Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2 
]*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/d + (4*B*Sec[c]*Sec[c + d*x]^2*Sin 
[d*x])/(3*d) + (4*Sec[c]*Sec[c + d*x]*(B*Sin[c] + 3*A*Sin[d*x] - 3*B*Sin[d 
*x]))/(3*d)))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])) + (A*Cos[c/2 + ( 
d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[C 
ot[c]]]^2]*Sec[c/2]*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 
- Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - Ar 
cTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(B + A*Cos[c + d*x 
])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) - (5*B*Cos[c/2 + (d*x)/2]^2*Cs 
c[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*S 
ec[c/2]*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - 
ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]] 
])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(B + A*Cos[c + d*x])*Sqrt[1 
+ Cot[c]^2]*(a + a*Sec[c + d*x])) + (3*A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Sec 
[c/2]*(A + B*Sec[c + d*x])*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d* 
x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x 
 + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d* 
x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + 
ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcT...
 
3.6.1.3 Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.94, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3433, 3042, 3457, 27, 3042, 3227, 3042, 3116, 3042, 3119, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3433

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle \frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {a (3 A-5 B) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx-3 a (A-B) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {a (3 A-5 B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-3 a (A-B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{2 a^2}\)

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {a (3 A-5 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-3 a (A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {(A-B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {a (3 A-5 B) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-3 a (A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}\)

input
Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])),x]
 
output
((A - B)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])) - (a*(3 
*A - 5*B)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d*Cos 
[c + d*x]^(3/2))) - 3*a*(A - B)*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin 
[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/(2*a^2)
 

3.6.1.3.1 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 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3116
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3227
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3433
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^(m + n)   Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + 
c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c 
- a*d, 0] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
 

rule 3457
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( 
n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ 
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 
3.6.1.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(465\) vs. \(2(194)=388\).

Time = 15.51 (sec) , antiderivative size = 466, normalized size of antiderivative = 3.07

method result size
default \(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {\left (-A +B \right ) \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+2 B \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{6 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )+\frac {\left (2 A -2 B \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\right )}{a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(466\)

input
int((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x,method=_RETURNVER 
BOSE)
 
output
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*((-A+B)*(cos( 
1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/ 
2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^( 
1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-2 
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*B*(-1/6*cos(1/2*d*x+1/ 
2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2 
*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^ 
(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1 
/2*d*x+1/2*c),2^(1/2)))+(2*A-2*B)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2* 
c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d* 
x+1/2*c)*sin(1/2*d*x+1/2*c)^2-(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1 
/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)))/sin(1/2*d*x+1/2* 
c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 
3.6.1.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.09 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {2 \, {\left (9 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right ) + 2 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (3 i \, A - 5 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (3 i \, A - 5 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (-3 i \, A + 5 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-3 i \, A + 5 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \]

input
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm= 
"fricas")
 
output
1/6*(2*(9*(A - B)*cos(d*x + c)^2 + 2*(3*A - 2*B)*cos(d*x + c) + 2*B)*sqrt( 
cos(d*x + c))*sin(d*x + c) + (sqrt(2)*(3*I*A - 5*I*B)*cos(d*x + c)^3 + sqr 
t(2)*(3*I*A - 5*I*B)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + 
c) + I*sin(d*x + c)) + (sqrt(2)*(-3*I*A + 5*I*B)*cos(d*x + c)^3 + sqrt(2)* 
(-3*I*A + 5*I*B)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - 
 I*sin(d*x + c)) - 9*(sqrt(2)*(I*A - I*B)*cos(d*x + c)^3 + sqrt(2)*(I*A - 
I*B)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos 
(d*x + c) + I*sin(d*x + c))) - 9*(sqrt(2)*(-I*A + I*B)*cos(d*x + c)^3 + sq 
rt(2)*(-I*A + I*B)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInve 
rse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*cos(d*x + c)^3 + a*d*cos( 
d*x + c)^2)
 
3.6.1.6 Sympy [F(-1)]

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

input
integrate((A+B*sec(d*x+c))/cos(d*x+c)**(5/2)/(a+a*sec(d*x+c)),x)
 
output
Timed out
 
3.6.1.7 Maxima [F]

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

input
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm= 
"maxima")
 
output
integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)*cos(d*x + c)^(5/2)), 
x)
 
3.6.1.8 Giac [F]

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

input
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm= 
"giac")
 
output
integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)*cos(d*x + c)^(5/2)), 
x)
 
3.6.1.9 Mupad [F(-1)]

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

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