\(\int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx\) [168]

Optimal result

Integrand size = 35, antiderivative size = 184 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx=\frac {(9 A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {2 (3 A-2 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(9 A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:

1/10*(9*A-C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d+1/6*(3*A+C)*Inver 
seJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^3/d-1/5*(A+C)*cos(d*x+c)^(1/2)*sin(d*x 
+c)/d/(a+a*cos(d*x+c))^3-2/15*(3*A-2*C)*cos(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a 
+a*cos(d*x+c))^2-1/10*(9*A-C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a^3+a^3*cos(d 
*x+c))
 

Mathematica [C] (warning: unable to verify)

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

Time = 8.51 (sec) , antiderivative size = 1029, normalized size of antiderivative = 5.59 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(-2*A*Cos[c/2 + (d*x)/2]^6*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]]]])/(d*(a + a*Cos[c + d*x])^3 
*Sqrt[1 + Cot[c]^2]) - (2*C*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPF 
Q[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTa 
n[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*S 
in[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3* 
d*(a + a*Cos[c + d*x])^3*Sqrt[1 + Cot[c]^2]) + (Cos[c/2 + (d*x)/2]^6*Sqrt[ 
Cos[c + d*x]]*((-4*(9*A - C)*Csc[c])/(5*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2 
]^3*(3*A*Sin[(d*x)/2] - 2*C*Sin[(d*x)/2]))/(15*d) - (4*Sec[c/2]*Sec[c/2 + 
(d*x)/2]*(9*A*Sin[(d*x)/2] - C*Sin[(d*x)/2]))/(5*d) - (2*Sec[c/2]*Sec[c/2 
+ (d*x)/2]^5*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(5*d) - (8*(3*A - 2*C)*Sec 
[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) - (2*(A + C)*Sec[c/2 + (d*x)/2]^4*Tan[c 
/2])/(5*d)))/(a + a*Cos[c + d*x])^3 - (9*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*S 
ec[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 + ArcTan[Tan[c]]]*Sqrt[1 ...
 

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 3521, 27, 3042, 3457, 3042, 3457, 27, 3042, 3227, 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.

\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\)

\(\Big \downarrow \) 3521

\(\displaystyle \frac {\int \frac {a (9 A-C)-a (3 A-7 C) \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (9 A-C)-a (3 A-7 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (9 A-C)-a (3 A-7 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {a^2 (21 A+C)-2 a^2 (3 A-2 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {4 a (3 A-2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {a^2 (21 A+C)-2 a^2 (3 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {4 a (3 A-2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\frac {\int \frac {5 (3 A+C) a^3+3 (9 A-C) \cos (c+d x) a^3}{2 \sqrt {\cos (c+d x)}}dx}{a^2}-\frac {3 a^2 (9 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {4 a (3 A-2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {5 (3 A+C) a^3+3 (9 A-C) \cos (c+d x) a^3}{\sqrt {\cos (c+d x)}}dx}{2 a^2}-\frac {3 a^2 (9 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {4 a (3 A-2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {5 (3 A+C) a^3+3 (9 A-C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {3 a^2 (9 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {4 a (3 A-2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {\frac {5 a^3 (3 A+C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^3 (9 A-C) \int \sqrt {\cos (c+d x)}dx}{2 a^2}-\frac {3 a^2 (9 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {4 a (3 A-2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {5 a^3 (3 A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^3 (9 A-C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a^2}-\frac {3 a^2 (9 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {4 a (3 A-2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {5 a^3 (3 A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^3 (9 A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}-\frac {3 a^2 (9 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {4 a (3 A-2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {\frac {10 a^3 (3 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^3 (9 A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}-\frac {3 a^2 (9 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {4 a (3 A-2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

Input:

Int[(A + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3),x]
 

Output:

-1/5*((A + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^3) 
+ ((-4*a*(3*A - 2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Cos[c + 
d*x])^2) + (((6*a^3*(9*A - C)*EllipticE[(c + d*x)/2, 2])/d + (10*a^3*(3*A 
+ C)*EllipticF[(c + d*x)/2, 2])/d)/(2*a^2) - (3*a^2*(9*A - C)*Sqrt[Cos[c + 
 d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])))/(3*a^2))/(10*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 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 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])
 

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. \(2(171)=342\).

Time = 3.36 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.45

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 (108 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-30 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 )^{5}+54 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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 )-12 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-10 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 )^{5}-6 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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 )-138 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+24 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+3 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-17 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 A +3 C \right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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((A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^3,x,method=_RETUR 
NVERBOSE)
 

Output:

1/60*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(108*A*cos(1 
/2*d*x+1/2*c)^8-30*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)^5+54*A* 
cos(1/2*d*x+1/2*c)^5*(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))-12*C*cos(1/2*d*x+1/2*c)^8- 
10*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)*Ellipt 
icF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5-6*C*cos(1/2*d*x+1/2*c 
)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Ellipti 
cE(cos(1/2*d*x+1/2*c),2^(1/2))-138*A*cos(1/2*d*x+1/2*c)^6+2*C*cos(1/2*d*x+ 
1/2*c)^6+24*A*cos(1/2*d*x+1/2*c)^4+24*C*cos(1/2*d*x+1/2*c)^4+3*A*cos(1/2*d 
*x+1/2*c)^2-17*C*cos(1/2*d*x+1/2*c)^2+3*A+3*C)/a^3/cos(1/2*d*x+1/2*c)^5/(- 
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 = 0.13 (sec) , antiderivative size = 467, normalized size of antiderivative = 2.54 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^3,x, algori 
thm="fricas")
 

Output:

-1/60*(2*(3*(9*A - C)*cos(d*x + c)^2 + 2*(33*A - 7*C)*cos(d*x + c) + 45*A 
- 5*C)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2)*(3*I*A + I*C)*cos(d*x 
+ c)^3 + 3*sqrt(2)*(3*I*A + I*C)*cos(d*x + c)^2 + 3*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)) + 5*(sqrt(2)*(-3*I*A - I*C)*cos(d*x + c)^3 + 3*sqrt( 
2)*(-3*I*A - I*C)*cos(d*x + c)^2 + 3*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)) + 3*(sqrt(2)*(-9*I*A + I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-9*I*A + 
I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-9*I*A + I*C)*cos(d*x + c) + sqrt(2)*(-9* 
I*A + I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) 
 + I*sin(d*x + c))) + 3*(sqrt(2)*(9*I*A - I*C)*cos(d*x + c)^3 + 3*sqrt(2)* 
(9*I*A - I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(9*I*A - I*C)*cos(d*x + c) + sqrt 
(2)*(9*I*A - I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d 
*x + c) - I*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. \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)**2)/cos(d*x+c)**(1/2)/(a+a*cos(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^3,x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^3,x, algori 
thm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)^3*sqrt(cos(d*x + c) 
)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \] Input:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + a*cos(c + d*x))^3),x)
                                                                                    
                                                                                    
 

Output:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + a*cos(c + d*x))^3), x)
 

Reduce [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}+3 \cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) c}{a^{3}} \] Input:

int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^3,x)
 

Output:

(int(sqrt(cos(c + d*x))/(cos(c + d*x)**4 + 3*cos(c + d*x)**3 + 3*cos(c + d 
*x)**2 + cos(c + d*x)),x)*a + int((sqrt(cos(c + d*x))*cos(c + d*x))/(cos(c 
 + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*c)/a**3