\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\) [161]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 33, antiderivative size = 180 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=-\frac {(A+9 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(A+3 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(A-B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-6 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(A+9 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:

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

Mathematica [C] (verified)

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

Time = 6.78 (sec) , antiderivative size = 845, normalized size of antiderivative = 4.69 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=-\frac {(A-B) \sqrt {\cos (c+d x)} \csc (c+d x)}{15 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}+\frac {4 B \sqrt {\cos (c+d x)} \csc (c+d x)}{3 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}+\frac {4 B \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x)}{3 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {2 B \cos ^{\frac {5}{2}}(c+d x) \csc (c+d x)}{a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}+\frac {2 (A-B) \sqrt {\cos (c+d x)} \csc ^3(c+d x)}{5 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}+\frac {4 (A-B) \cos ^{\frac {3}{2}}(c+d x) \csc ^3(c+d x)}{21 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {6 (A-B) \cos ^{\frac {5}{2}}(c+d x) \csc ^3(c+d x)}{5 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}+\frac {2 (A-B) \cos ^{\frac {7}{2}}(c+d x) \csc ^3(c+d x)}{3 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {2 B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {(A-B) \sqrt {\cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cos ^2(c+d x)\right ) \sin (c+d x) \sqrt {\sin ^2(c+d x)}}{6 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {2 B \sqrt {\cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cos ^2(c+d x)\right ) \sin (c+d x) \sqrt {\sin ^2(c+d x)}}{3 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {4 B \cos ^{\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sin (c+d x) \sqrt {\sin ^2(c+d x)}}{3 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))}-\frac {4 (A-B) \cos ^{\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sin (c+d x) \sqrt {\sin ^2(c+d x)}}{21 a^3 d (1-\cos (c+d x)) (1+\cos (c+d x))} \] Input:

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

Output:

-1/15*((A - B)*Sqrt[Cos[c + d*x]]*Csc[c + d*x])/(a^3*d*(1 - Cos[c + d*x])* 
(1 + Cos[c + d*x])) + (4*B*Sqrt[Cos[c + d*x]]*Csc[c + d*x])/(3*a^3*d*(1 - 
Cos[c + d*x])*(1 + Cos[c + d*x])) + (4*B*Cos[c + d*x]^(3/2)*Csc[c + d*x])/ 
(3*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + d*x])) - (2*B*Cos[c + d*x]^(5/2)* 
Csc[c + d*x])/(a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + d*x])) + (2*(A - B)*S 
qrt[Cos[c + d*x]]*Csc[c + d*x]^3)/(5*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + 
 d*x])) + (4*(A - B)*Cos[c + d*x]^(3/2)*Csc[c + d*x]^3)/(21*a^3*d*(1 - Cos 
[c + d*x])*(1 + Cos[c + d*x])) - (6*(A - B)*Cos[c + d*x]^(5/2)*Csc[c + d*x 
]^3)/(5*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + d*x])) + (2*(A - B)*Cos[c + 
d*x]^(7/2)*Csc[c + d*x]^3)/(3*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + d*x])) 
 - ((A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(6*a^3*d*(1 - Cos[c + d*x])*( 
1 + Cos[c + d*x])) - (2*B*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a^3*d*(1 - C 
os[c + d*x])*(1 + Cos[c + d*x])) - ((A - B)*Sqrt[Cos[c + d*x]]*Hypergeomet 
ric2F1[1/4, 1/2, 5/4, Cos[c + d*x]^2]*Sin[c + d*x]*Sqrt[Sin[c + d*x]^2])/( 
6*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + d*x])) - (2*B*Sqrt[Cos[c + d*x]]*H 
ypergeometric2F1[1/4, 1/2, 5/4, Cos[c + d*x]^2]*Sin[c + d*x]*Sqrt[Sin[c + 
d*x]^2])/(3*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + d*x])) - (4*B*Cos[c + d* 
x]^(3/2)*Hypergeometric2F1[3/4, 5/2, 7/4, Cos[c + d*x]^2]*Sin[c + d*x]*Sqr 
t[Sin[c + d*x]^2])/(3*a^3*d*(1 - Cos[c + d*x])*(1 + Cos[c + d*x])) - (4*(A 
 - B)*Cos[c + d*x]^(3/2)*Hypergeometric2F1[3/4, 7/2, 7/4, Cos[c + d*x]^...
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 3456, 27, 3042, 3456, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3456

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\frac {\int \frac {(A-6 B) a^2+(4 A+21 B) \cos (c+d x) a^2}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)}dx}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {(A-6 B) a^2+(4 A+21 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\frac {\int \frac {5 a^3 (A+3 B)-3 a^3 (A+9 B) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx}{a^2}+\frac {3 a^2 (A+9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {5 a^3 (A+3 B)-3 a^3 (A+9 B) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx}{2 a^2}+\frac {3 a^2 (A+9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^3 (A+9 B) \int \sqrt {\cos (c+d x)}dx}{2 a^2}+\frac {3 a^2 (A+9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

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

Leaf count of result is larger than twice the leaf count of optimal. \(450\) vs. \(2(167)=334\).

Time = 6.32 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.51

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 (12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+10 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}+6 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 )+108 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+30 B \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 B \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 )-2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-198 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+114 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+17 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-27 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 A +3 B \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 {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(451\)

Input:

int(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x,method=_RETURNV 
ERBOSE)
 

Output:

-1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*A*cos(1/ 
2*d*x+1/2*c)^8+10*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+6*A*co 
s(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))+108*B*cos(1/2*d*x+1/2*c)^8+3 
0*B*(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 
cF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+54*B*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))-2*A*cos(1/2*d*x+1/2*c)^6-198*B*cos(1/2*d*x+ 
1/2*c)^6-24*A*cos(1/2*d*x+1/2*c)^4+114*B*cos(1/2*d*x+1/2*c)^4+17*A*cos(1/2 
*d*x+1/2*c)^2-27*B*cos(1/2*d*x+1/2*c)^2-3*A+3*B)/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)/(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 = 0.11 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.58 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorith 
m="fricas")
 

Output:

1/60*(2*(3*(A + 9*B)*cos(d*x + c)^2 + 2*(7*A + 18*B)*cos(d*x + c) + 5*A + 
15*B)*sqrt(cos(d*x + c))*sin(d*x + c) - 5*(sqrt(2)*(I*A + 3*I*B)*cos(d*x + 
 c)^3 + 3*sqrt(2)*(I*A + 3*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A + 3*I*B)*c 
os(d*x + c) + sqrt(2)*(I*A + 3*I*B))*weierstrassPInverse(-4, 0, cos(d*x + 
c) + I*sin(d*x + c)) - 5*(sqrt(2)*(-I*A - 3*I*B)*cos(d*x + c)^3 + 3*sqrt(2 
)*(-I*A - 3*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A - 3*I*B)*cos(d*x + c) + 
sqrt(2)*(-I*A - 3*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* 
x + c)) - 3*(sqrt(2)*(I*A + 9*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A + 9*I*B 
)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A + 9*I*B)*cos(d*x + c) + sqrt(2)*(I*A + 9 
*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I* 
sin(d*x + c))) - 3*(sqrt(2)*(-I*A - 9*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-I* 
A - 9*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A - 9*I*B)*cos(d*x + c) + sqrt(2 
)*(-I*A - 9*I*B))*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. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorith 
m="maxima")
 

Output:

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

Giac [F]

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

integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorith 
m="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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