\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^3} \, dx\) [165]

Optimal result
Mathematica [C] (warning: unable to verify)
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 = 35, antiderivative size = 209 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=-\frac {(9 A+119 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(A+11 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{2 a^3 d}+\frac {(A+11 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a^3 d}-\frac {(A+C) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {2 C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \cos (c+d x))^2}-\frac {(9 A+119 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:

-1/10*(9*A+119*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d+1/2*(A+11*C) 
*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^3/d+1/2*(A+11*C)*cos(d*x+c)^(1/2 
)*sin(d*x+c)/a^3/d-1/5*(A+C)*cos(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*cos(d*x+c) 
)^3-2/3*C*cos(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2-1/30*(9*A+119 
*C)*cos(d*x+c)^(3/2)*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 = 9.63 (sec) , antiderivative size = 1060, normalized size of antiderivative = 5.07 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

Integrate[(Cos[c + d*x]^(5/2)*(A + 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]]]])/(d*(a + a*Cos[c + d*x])^3 
*Sqrt[1 + Cot[c]^2]) - (22*C*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricP 
FQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcT 
an[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]) + (Cos[c/2 + (d*x)/2]^6*Sqrt[C 
os[c + d*x]]*((4*(9*A + 59*C + 60*C*Cos[c])*Csc[c])/(5*d) + (16*C*Cos[d*x] 
*Sin[c])/(3*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] + C*Sin[ 
(d*x)/2]))/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(9*A*Sin[(d*x)/2] + 19 
*C*Sin[(d*x)/2]))/(15*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(9*A*Sin[(d*x)/2 
] + 59*C*Sin[(d*x)/2]))/(5*d) + (16*C*Cos[c]*Sin[d*x])/(3*d) - (4*(9*A + 1 
9*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*Cs 
c[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[T 
an[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[T...
 

Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.05, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 3521, 27, 3042, 3456, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

\(\displaystyle \frac {\int \frac {\cos ^{\frac {5}{2}}(c+d x) (a (3 A-7 C)+a (3 A+13 C) \cos (c+d x))}{2 (\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\cos ^{\frac {5}{2}}(c+d x) (a (3 A-7 C)+a (3 A+13 C) \cos (c+d x))}{(\cos (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a (3 A-7 C)+a (3 A+13 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\frac {\int -\frac {\cos ^{\frac {3}{2}}(c+d x) \left (50 a^2 C-3 a^2 (3 A+23 C) \cos (c+d x)\right )}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {20 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (50 a^2 C-3 a^2 (3 A+23 C) \cos (c+d x)\right )}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {20 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (50 a^2 C-3 a^2 (3 A+23 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {20 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3456

\(\displaystyle \frac {-\frac {\frac {\int \frac {3}{2} \sqrt {\cos (c+d x)} \left (a^3 (9 A+119 C)-15 a^3 (A+11 C) \cos (c+d x)\right )dx}{a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\frac {3 \int \sqrt {\cos (c+d x)} \left (a^3 (9 A+119 C)-15 a^3 (A+11 C) \cos (c+d x)\right )dx}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a^3 (9 A+119 C)-15 a^3 (A+11 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {-\frac {\frac {3 \left (a^3 (9 A+119 C) \int \sqrt {\cos (c+d x)}dx-15 a^3 (A+11 C) \int \cos ^{\frac {3}{2}}(c+d x)dx\right )}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3 \left (a^3 (9 A+119 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-15 a^3 (A+11 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {-\frac {\frac {3 \left (a^3 (9 A+119 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-15 a^3 (A+11 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3 \left (a^3 (9 A+119 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-15 a^3 (A+11 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {-\frac {\frac {3 \left (\frac {2 a^3 (9 A+119 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-15 a^3 (A+11 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )}{2 a^2}+\frac {a^2 (9 A+119 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {20 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {-\frac {\frac {a^2 (9 A+119 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}+\frac {3 \left (\frac {2 a^3 (9 A+119 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-15 a^3 (A+11 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )}{2 a^2}}{3 a^2}-\frac {20 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

Input:

Int[(Cos[c + d*x]^(5/2)*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^3,x]
 

Output:

-1/5*((A + C)*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^3) 
+ ((-20*a*C*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) 
- ((a^2*(9*A + 119*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d 
*x])) + (3*((2*a^3*(9*A + 119*C)*EllipticE[(c + d*x)/2, 2])/d - 15*a^3*(A 
+ 11*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c 
 + d*x])/(3*d))))/(2*a^2))/(3*a^2))/(10*a^2)
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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. \(464\) vs. \(2(192)=384\).

Time = 16.63 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.22

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 (160 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+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 )+468 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+330 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}+714 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 )-198 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-1058 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+114 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+474 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-27 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-47 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}\) \(465\)

Input:

int(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^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)*(160*C*cos( 
1/2*d*x+1/2*c)^10+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 
))+468*C*cos(1/2*d*x+1/2*c)^8+330*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+714*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))-198*A*cos( 
1/2*d*x+1/2*c)^6-1058*C*cos(1/2*d*x+1/2*c)^6+114*A*cos(1/2*d*x+1/2*c)^4+47 
4*C*cos(1/2*d*x+1/2*c)^4-27*A*cos(1/2*d*x+1/2*c)^2-47*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 = 0.11 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.29 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algori 
thm="fricas")
 

Output:

1/60*(2*(20*C*cos(d*x + c)^3 + 3*(9*A + 79*C)*cos(d*x + c)^2 + 4*(9*A + 94 
*C)*cos(d*x + c) + 15*A + 165*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 15*(sqr 
t(2)*(I*A + 11*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A + 11*I*C)*cos(d*x + c) 
^2 + 3*sqrt(2)*(I*A + 11*I*C)*cos(d*x + c) + sqrt(2)*(I*A + 11*I*C))*weier 
strassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 15*(sqrt(2)*(-I*A - 
 11*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A - 11*I*C)*cos(d*x + c)^2 + 3*sqr 
t(2)*(-I*A - 11*I*C)*cos(d*x + c) + sqrt(2)*(-I*A - 11*I*C))*weierstrassPI 
nverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(sqrt(2)*(9*I*A + 119*I*C 
)*cos(d*x + c)^3 + 3*sqrt(2)*(9*I*A + 119*I*C)*cos(d*x + c)^2 + 3*sqrt(2)* 
(9*I*A + 119*I*C)*cos(d*x + c) + sqrt(2)*(9*I*A + 119*I*C))*weierstrassZet 
a(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(s 
qrt(2)*(-9*I*A - 119*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-9*I*A - 119*I*C)*co 
s(d*x + c)^2 + 3*sqrt(2)*(-9*I*A - 119*I*C)*cos(d*x + c) + sqrt(2)*(-9*I*A 
 - 119*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. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(5/2)*(A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**3,x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,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)^3 
, x)
 

Giac [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,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)^3 
, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \] Input:

int((cos(c + d*x)^(5/2)*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^3,x)
 

Output:

int((cos(c + d*x)^(5/2)*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^3, x)
 

Reduce [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) a}{a^{3}} \] Input:

int(cos(d*x+c)^(5/2)*(A+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)**4)/(cos(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*cos(c + d*x) + 1),x)*a)/a**3