\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^2} \, dx\) [159]

Optimal result

Integrand size = 35, antiderivative size = 161 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {(A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {2 (A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {2 (A+5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}-\frac {(A+7 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:

-(A+7*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d+2/3*(A+5*C)*InverseJa 
cobiAM(1/2*d*x+1/2*c,2^(1/2))/a^2/d+2/3*(A+5*C)*cos(d*x+c)^(1/2)*sin(d*x+c 
)/a^2/d-1/3*(A+7*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*( 
A+C)*cos(d*x+c)^(5/2)*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 = 8.41 (sec) , antiderivative size = 968, normalized size of antiderivative = 6.01 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(-4*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]) - (20*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]]]])/ 
(3*d*(a + a*Cos[c + d*x])^2*Sqrt[1 + Cot[c]^2]) + (Cos[c/2 + (d*x)/2]^4*Sq 
rt[Cos[c + d*x]]*((4*(A + 3*C + 4*C*Cos[c])*Csc[c])/d + (8*C*Cos[d*x]*Sin[ 
c])/(3*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + C*Sin[(d*x) 
/2]))/(3*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + 3*C*Sin[(d* 
x)/2]))/d + (8*C*Cos[c]*Sin[d*x])/(3*d) - (2*(A + C)*Sec[c/2 + (d*x)/2]^2* 
Tan[c/2])/(3*d)))/(a + a*Cos[c + d*x])^2 + (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 + ArcTan[Tan[c]]]*Sqrt[1 
+ Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]...
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.02, 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, 3456, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3456

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

-1/3*((A + C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) 
+ ((-2*(A + 7*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(d*(1 + Cos[c + d*x])) - 
 (3*((2*a^2*(A + 7*C)*EllipticE[(c + d*x)/2, 2])/d - 2*a^2*(A + 5*C)*((2*E 
llipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*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. \(436\) vs. \(2(150)=300\).

Time = 4.22 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.71

method	result	size
default	\(-\frac {\sqrt {\left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (16 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+4 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+6 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+20 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+42 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-20 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-48 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+9 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+21 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A -C \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(437\)

Input:

int(cos(d*x+c)^(3/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x,method=_RETUR 
NVERBOSE)
 

Output:

-1/6*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*C*cos(1/ 
2*d*x+1/2*c)^8+12*A*cos(1/2*d*x+1/2*c)^6+4*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))*co 
s(1/2*d*x+1/2*c)^3+6*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+12* 
C*cos(1/2*d*x+1/2*c)^6+20*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+42*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))-20*A*cos(1/2*d*x+1/ 
2*c)^4-48*C*cos(1/2*d*x+1/2*c)^4+9*A*cos(1/2*d*x+1/2*c)^2+21*C*cos(1/2*d*x 
+1/2*c)^2-A-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 = 366, normalized size of antiderivative = 2.27 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {2 \, {\left (2 \, C \cos \left (d x + c\right )^{2} + {\left (3 \, A + 13 \, C\right )} \cos \left (d x + c\right ) + 2 \, A + 10 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, {\left (\sqrt {2} {\left (i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (i \, A + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (\sqrt {2} {\left (-i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-i \, A - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (i \, A + 7 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (i \, A + 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 7 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-i \, A - 7 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-i \, A - 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 7 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

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

Output:

1/6*(2*(2*C*cos(d*x + c)^2 + (3*A + 13*C)*cos(d*x + c) + 2*A + 10*C)*sqrt( 
cos(d*x + c))*sin(d*x + c) - 2*(sqrt(2)*(I*A + 5*I*C)*cos(d*x + c)^2 + 2*s 
qrt(2)*(I*A + 5*I*C)*cos(d*x + c) + sqrt(2)*(I*A + 5*I*C))*weierstrassPInv 
erse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 2*(sqrt(2)*(-I*A - 5*I*C)*cos 
(d*x + c)^2 + 2*sqrt(2)*(-I*A - 5*I*C)*cos(d*x + c) + sqrt(2)*(-I*A - 5*I* 
C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(sqrt(2) 
*(I*A + 7*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(I*A + 7*I*C)*cos(d*x + c) + sqr 
t(2)*(I*A + 7*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos( 
d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-I*A - 7*I*C)*cos(d*x + c)^2 + 2 
*sqrt(2)*(-I*A - 7*I*C)*cos(d*x + c) + sqrt(2)*(-I*A - 7*I*C))*weierstrass 
Zeta(-4, 0, weierstrassPInverse(-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. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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