\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^3} \, dx\) [471]

Optimal result

Integrand size = 43, antiderivative size = 195 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=-\frac {(A+9 B-49 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(A+3 B-13 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(2 A+3 B-8 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(A+3 B-13 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:

-1/10*(A+9*B-49*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d+1/6*(A+3*B- 
13*C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^3/d-1/5*(A-B+C)*cos(d*x+c)^ 
(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+1/15*(2*A+3*B-8*C)*cos(d*x+c)^(3/2)* 
sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2+1/6*(A+3*B-13*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 = 9.15 (sec) , antiderivative size = 1455, normalized size of antiderivative = 7.46 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

Integrate[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(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]]]])/(3*d*(a + a*Cos[c + d*x]) 
^3*Sqrt[1 + Cot[c]^2]) - (2*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Hypergeometric 
PFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - Arc 
Tan[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])^3*Sqrt[1 + Cot[c]^2]) + (26*C*Cos[c/2 + (d*x)/2]^6* 
Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2] 
*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sq 
rt[-(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])^3*Sqrt[1 + Cot[c]^2]) + (C 
os[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*((-4*(-A - 9*B + 29*C + 20*C*Cos[c] 
)*Csc[c])/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + 9*B*Sin 
[(d*x)/2] - 29*C*Sin[(d*x)/2]))/(5*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*( 
A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(5*d) + (4*Sec[c/2]*Sec 
[c/2 + (d*x)/2]^3*(4*A*Sin[(d*x)/2] - 9*B*Sin[(d*x)/2] + 14*C*Sin[(d*x)/2] 
))/(15*d) + (4*(4*A - 9*B + 14*C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) - 
(2*(A - B + C)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/(a + a*Cos[c + d*...
 

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.302, Rules used = {3042, 3520, 27, 3042, 3456, 3042, 3456, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3520

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) (5 a (A+B-C)+a (A-B+11 C) \cos (c+d x))}{(\cos (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (5 a (A+B-C)+a (A-B+11 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\cos (c+d x)} \left (3 (2 A+3 B-8 C) a^2+(A-6 B+41 C) \cos (c+d x) a^2\right )}{\cos (c+d x) a+a}dx}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (3 (2 A+3 B-8 C) a^2+(A-6 B+41 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\frac {\frac {\int \frac {5 a^3 (A+3 B-13 C)-3 a^3 (A+9 B-49 C) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx}{a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {5 a^3 (A+3 B-13 C)-3 a^3 (A+9 B-49 C) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {5 a^3 (A+3 B-13 C)-3 a^3 (A+9 B-49 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B-13 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^3 (A+9 B-49 C) \int \sqrt {\cos (c+d x)}dx}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B-13 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^3 (A+9 B-49 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B-13 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a^3 (A+9 B-49 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}+\frac {\frac {10 a^3 (A+3 B-13 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^3 (A+9 B-49 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

Input:

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

Output:

-1/5*((A - B + C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]) 
^3) + ((2*a*(2*A + 3*B - 8*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(3*d*(a + a 
*Cos[c + d*x])^2) + (((-6*a^3*(A + 9*B - 49*C)*EllipticE[(c + d*x)/2, 2])/ 
d + (10*a^3*(A + 3*B - 13*C)*EllipticF[(c + d*x)/2, 2])/d)/(2*a^2) + (5*a^ 
2*(A + 3*B - 13*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 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 3520

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + 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))   Int[(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)) + B*(b*c*m + a 
*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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. \(623\) vs. \(2(182)=364\).

Time = 8.75 (sec) , antiderivative size = 624, normalized size of antiderivative = 3.20

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 (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 )-348 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-130 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}-294 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 )-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}+578 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}+114 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-264 C \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}+37 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 A +3 B -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}\)	\(624\)

Input:

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

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)*(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*c 
os(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+ 
30*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)*Ellipt 
icF(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)*Ellipt 
icE(cos(1/2*d*x+1/2*c),2^(1/2))-348*C*cos(1/2*d*x+1/2*c)^8-130*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)^5-294*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)*EllipticE(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+578 
*C*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) 
^4-264*C*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+37*C*cos(1/2*d*x+1/2*c)^2-3*A+3*B-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 = 0.15 (sec) , antiderivative size = 522, normalized size of antiderivative = 2.68 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

1/60*(2*(3*(A + 9*B - 29*C)*cos(d*x + c)^2 + 2*(7*A + 18*B - 73*C)*cos(d*x 
 + c) + 5*A + 15*B - 65*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 5*(sqrt(2)*(I 
*A + 3*I*B - 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A + 3*I*B - 13*I*C)*cos 
(d*x + c)^2 + 3*sqrt(2)*(I*A + 3*I*B - 13*I*C)*cos(d*x + c) + sqrt(2)*(I*A 
 + 3*I*B - 13*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + 
c)) - 5*(sqrt(2)*(-I*A - 3*I*B + 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A 
- 3*I*B + 13*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A - 3*I*B + 13*I*C)*cos(d 
*x + c) + sqrt(2)*(-I*A - 3*I*B + 13*I*C))*weierstrassPInverse(-4, 0, cos( 
d*x + c) - I*sin(d*x + c)) - 3*(sqrt(2)*(I*A + 9*I*B - 49*I*C)*cos(d*x + c 
)^3 + 3*sqrt(2)*(I*A + 9*I*B - 49*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A + 9 
*I*B - 49*I*C)*cos(d*x + c) + sqrt(2)*(I*A + 9*I*B - 49*I*C))*weierstrassZ 
eta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3* 
(sqrt(2)*(-I*A - 9*I*B + 49*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A - 9*I*B 
+ 49*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A - 9*I*B + 49*I*C)*cos(d*x + c) 
+ sqrt(2)*(-I*A - 9*I*B + 49*I*C))*weierstrassZeta(-4, 0, weierstrassPInve 
rse(-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. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(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)**3)/(c 
os(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*c + int((sqrt( 
cos(c + d*x))*cos(c + d*x)**2)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*co 
s(c + d*x) + 1),x)*b)/a**3