\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^2} \, dx\) [158]

Optimal result

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

4/5*(5*A+14*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d-5/3*(A+3*C)*Inv 
erseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^2/d-5/3*(A+3*C)*cos(d*x+c)^(1/2)*sin 
(d*x+c)/a^2/d+4/15*(5*A+14*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a^2/d-(A+3*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 = 9.04 (sec) , antiderivative size = 1012, normalized size of antiderivative = 5.16 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(10*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]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Si 
n[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 + a*Cos[c + d*x]) 
^2*Sqrt[1 + Cot[c]^2]) + (10*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Hypergeometri 
cPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - Ar 
cTan[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 + a*Cos[c + d*x])^2*Sqrt[1 + Cot[c]^2]) + (Cos[c/2 + (d*x)/2]^4*Sqrt 
[Cos[c + d*x]]*((-8*(5*A + 10*C + 5*A*Cos[c] + 18*C*Cos[c])*Csc[c])/(5*d) 
- (16*C*Cos[d*x]*Sin[c])/(3*d) + (4*C*Cos[2*d*x]*Sin[2*c])/(5*d) + (2*Sec[ 
c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(3*d) - (8*Se 
c[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + 2*C*Sin[(d*x)/2]))/d - (16*C*C 
os[c]*Sin[d*x])/(3*d) + (4*C*Cos[2*c]*Sin[2*d*x])/(5*d) + (2*(A + C)*Sec[c 
/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/(a + a*Cos[c + d*x])^2 - (4*A*Cos[c/2 + 
(d*x)/2]^4*Csc[c/2]*Sec[c/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 + 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 + Ar...
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3456

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

Int[(Cos[c + d*x]^(5/2)*(A + C*Cos[c + d*x]^2))/(a + a*Cos[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*(A + 3*C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Cos[c + d*x])) + 
(15*a^2*(A + 3*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*(5*A + 14*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)]
 

Maple [B] (verified)

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

Time = 8.40 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.30

method	result	size
default	\(\frac {\sqrt {\left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-96 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+352 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+120 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+50 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}+120 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}+150 C \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 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 )-190 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-266 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+75 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+135 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 {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(451\)

Input:

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

Output:

1/30*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-96*C*cos(1 
/2*d*x+1/2*c)^10+352*C*cos(1/2*d*x+1/2*c)^8+120*A*cos(1/2*d*x+1/2*c)^6+50* 
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+120*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+150*C*(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*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))-190*A*cos(1/2*d*x+1/2*c)^4-266*C*cos(1/2*d*x+1/2*c)^4+75* 
A*cos(1/2*d*x+1/2*c)^2+135*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)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.93 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {2 \, {\left (6 \, C \cos \left (d x + c\right )^{3} - 8 \, C \cos \left (d x + c\right )^{2} - 2 \, {\left (15 \, A + 47 \, C\right )} \cos \left (d x + c\right ) - 25 \, A - 75 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 25 \, {\left (\sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 25 \, {\left (\sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 12 \, {\left (\sqrt {2} {\left (-5 i \, A - 14 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-5 i \, A - 14 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A - 14 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 \, {\left (\sqrt {2} {\left (5 i \, A + 14 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (5 i \, A + 14 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A + 14 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{30 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

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

Output:

1/30*(2*(6*C*cos(d*x + c)^3 - 8*C*cos(d*x + c)^2 - 2*(15*A + 47*C)*cos(d*x 
 + c) - 25*A - 75*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 25*(sqrt(2)*(-I*A - 
 3*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c) + sqrt(2)*( 
-I*A - 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 
 25*(sqrt(2)*(I*A + 3*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(I*A + 3*I*C)*cos(d* 
x + c) + sqrt(2)*(I*A + 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - 
I*sin(d*x + c)) - 12*(sqrt(2)*(-5*I*A - 14*I*C)*cos(d*x + c)^2 + 2*sqrt(2) 
*(-5*I*A - 14*I*C)*cos(d*x + c) + sqrt(2)*(-5*I*A - 14*I*C))*weierstrassZe 
ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 12* 
(sqrt(2)*(5*I*A + 14*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(5*I*A + 14*I*C)*cos( 
d*x + c) + sqrt(2)*(5*I*A + 14*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. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

(int((sqrt(cos(c + d*x))*cos(c + d*x)**4)/(cos(c + d*x)**2 + 2*cos(c + d*x 
) + 1),x)*c + int((sqrt(cos(c + d*x))*cos(c + d*x)**2)/(cos(c + d*x)**2 + 
2*cos(c + d*x) + 1),x)*a)/a**2