\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [1114]

Optimal result

Integrand size = 35, antiderivative size = 196 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {4 (14 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac {5 (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}-\frac {5 (3 A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {4 (14 A+5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a^2 d}-\frac {(3 A+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A+C) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:

4/5*(14*A+5*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d-5/3*(3*A+C)*Inv 
erseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^2/d-5/3*(3*A+C)*cos(d*x+c)^(1/2)*sin 
(d*x+c)/a^2/d+4/15*(14*A+5*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a^2/d-(3*A+C)*co 
s(d*x+c)^(5/2)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A+C)*cos(d*x+c)^(7/2)* 
sin(d*x+c)/d/(a+a*cos(d*x+c))^2
 

Mathematica [C] (warning: unable to verify)

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

Time = 8.24 (sec) , antiderivative size = 1162, normalized size of antiderivative = 5.93 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(20*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S 
in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcT 
an[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]]]])/(d 
*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^2) 
 + (20*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4} 
, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*Sec[d*x - A 
rcTan[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*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x 
])^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*(A + C*Sec[c + d*x]^2)*(( 
-16*(10*A + 5*C + 18*A*Cos[c] + 5*C*Cos[c])*Csc[c])/(5*d) - (32*A*Cos[d*x] 
*Sin[c])/(3*d) + (8*A*Cos[2*d*x]*Sin[2*c])/(5*d) + (4*Sec[c/2]*Sec[c/2 + ( 
d*x)/2]^3*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(3*d) - (16*Sec[c/2]*Sec[c/2 
+ (d*x)/2]*(2*A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/d - (32*A*Cos[c]*Sin[d*x]) 
/(3*d) + (8*A*Cos[2*c]*Sin[2*d*x])/(5*d) + (4*(A + C)*Sec[c/2 + (d*x)/2]^2 
*Tan[c/2])/(3*d)))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) 
 - (112*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*(( 
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...
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.99, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4602, 3042, 3521, 27, 3042, 3456, 3042, 3227, 3042, 3115, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4602

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

\(\displaystyle \frac {\int -\frac {\cos ^{\frac {5}{2}}(c+d x) (a (7 A+C)-a (11 A+5 C) \cos (c+d x))}{2 (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\cos ^{\frac {5}{2}}(c+d x) (a (7 A+C)-a (11 A+5 C) \cos (c+d x))}{\cos (c+d x) a+a}dx}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a (7 A+C)-a (11 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3456

\(\displaystyle -\frac {\frac {\int \cos ^{\frac {3}{2}}(c+d x) \left (15 a^2 (3 A+C)-4 a^2 (14 A+5 C) \cos (c+d x)\right )dx}{a^2}+\frac {6 (3 A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (15 a^2 (3 A+C)-4 a^2 (14 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {6 (3 A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3227

\(\displaystyle -\frac {\frac {15 a^2 (3 A+C) \int \cos ^{\frac {3}{2}}(c+d x)dx-4 a^2 (14 A+5 C) \int \cos ^{\frac {5}{2}}(c+d x)dx}{a^2}+\frac {6 (3 A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {15 a^2 (3 A+C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx-4 a^2 (14 A+5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx}{a^2}+\frac {6 (3 A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {\frac {15 a^2 (3 A+C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-4 a^2 (14 A+5 C) \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (3 A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {15 a^2 (3 A+C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-4 a^2 (14 A+5 C) \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (3 A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {\frac {15 a^2 (3 A+C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-4 a^2 (14 A+5 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (3 A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {\frac {15 a^2 (3 A+C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-4 a^2 (14 A+5 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (3 A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

Input:

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

Output:

-1/3*((A + C)*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) 
- ((6*(3*A + C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Cos[c + d*x])) + 
(15*a^2*(3*A + C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d 
*x]]*Sin[c + d*x])/(3*d)) - 4*a^2*(14*A + 5*C)*((6*EllipticE[(c + d*x)/2, 
2])/(5*d) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)))/a^2)/(6*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 3042

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

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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 3456

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[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( 
a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + 
 b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B}, 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] && (In 
tegerQ[2*n] || EqQ[c, 0])
 

rule 3521

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n 
 + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Simp[1/(b*(b*c - a*d)*(2*m + 1))   I 
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) 
- b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2) + C*(b* 
c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
 - d^2, 0] && LtQ[m, -2^(-1)]
 

rule 4602

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x 
_)])^(m_.)*((A_.) + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[d^( 
m + 2)   Int[(b + a*Cos[e + f*x])^m*(d*Cos[e + f*x])^(n - m - 2)*(C + A*Cos 
[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] &&  !IntegerQ[n] 
 && IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(450\) vs. \(2(181)=362\).

Time = 8.78 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.30

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 (96 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-352 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+120 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-150 A \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 ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-336 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-120 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-50 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \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 )-120 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+266 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+190 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-135 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-75 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+5 A +5 C \right )}{30 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{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}}\, \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}\)	\(451\)

Input:

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

Output:

-1/30*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(96*A*cos(1/ 
2*d*x+1/2*c)^10-352*A*cos(1/2*d*x+1/2*c)^8+120*A*cos(1/2*d*x+1/2*c)^6-150* 
A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF 
(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3-336*A*cos(1/2*d*x+1/2*c) 
^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)*Elliptic 
E(cos(1/2*d*x+1/2*c),2^(1/2))-120*C*cos(1/2*d*x+1/2*c)^6-50*C*cos(1/2*d*x+ 
1/2*c)^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)*El 
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))-120*C*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+ 
1/2*c),2^(1/2))+266*A*cos(1/2*d*x+1/2*c)^4+190*C*cos(1/2*d*x+1/2*c)^4-135* 
A*cos(1/2*d*x+1/2*c)^2-75*C*cos(1/2*d*x+1/2*c)^2+5*A+5*C)/a^2/cos(1/2*d*x+ 
1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+ 
1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.93 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {2 \, {\left (6 \, A \cos \left (d x + c\right )^{3} - 8 \, A \cos \left (d x + c\right )^{2} - 2 \, {\left (47 \, A + 15 \, C\right )} \cos \left (d x + c\right ) - 75 \, A - 25 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 25 \, {\left (\sqrt {2} {\left (-3 i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-3 i \, A - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A - i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 25 \, {\left (\sqrt {2} {\left (3 i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (3 i \, A + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A + i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 12 \, {\left (\sqrt {2} {\left (-14 i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-14 i \, A - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-14 i \, A - 5 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 \, {\left (\sqrt {2} {\left (14 i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (14 i \, A + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (14 i \, A + 5 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{30 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

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

Output:

1/30*(2*(6*A*cos(d*x + c)^3 - 8*A*cos(d*x + c)^2 - 2*(47*A + 15*C)*cos(d*x 
 + c) - 75*A - 25*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 25*(sqrt(2)*(-3*I*A 
 - I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(-3*I*A - I*C)*cos(d*x + c) + sqrt(2)*( 
-3*I*A - I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 
 25*(sqrt(2)*(3*I*A + I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(3*I*A + I*C)*cos(d* 
x + c) + sqrt(2)*(3*I*A + I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - 
I*sin(d*x + c)) - 12*(sqrt(2)*(-14*I*A - 5*I*C)*cos(d*x + c)^2 + 2*sqrt(2) 
*(-14*I*A - 5*I*C)*cos(d*x + c) + sqrt(2)*(-14*I*A - 5*I*C))*weierstrassZe 
ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 12* 
(sqrt(2)*(14*I*A + 5*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(14*I*A + 5*I*C)*cos( 
d*x + c) + sqrt(2)*(14*I*A + 5*I*C))*weierstrassZeta(-4, 0, weierstrassPIn 
verse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x + c)^2 + 2*a^ 
2*d*cos(d*x + c) + a^2*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

integrate(cos(d*x+c)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algori 
thm="maxima")
 

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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