\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx\) [474]

Optimal result

Integrand size = 43, antiderivative size = 237 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=-\frac {(49 A-9 B-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(13 A-3 B-C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(49 A-9 B-C) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}-\frac {(13 A-3 B-C) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )} \] Output:

-1/10*(49*A-9*B-C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d-1/6*(13*A-3 
*B-C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^3/d+1/10*(49*A-9*B-C)*sin(d 
*x+c)/a^3/d/cos(d*x+c)^(1/2)-1/5*(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+ 
a*cos(d*x+c))^3-1/15*(8*A-3*B-2*C)*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)/(a+a*co 
s(d*x+c))^2-1/6*(13*A-3*B-C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(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 = 9.05 (sec) , antiderivative size = 1487, normalized size of antiderivative = 6.27 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(26*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]) - (2*C*Cos[c/2 + (d*x)/2]^6*C 
sc[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]]]]*Sqr 
t[-(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]) + (Co 
s[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*((2*(20*A + 29*A*Cos[c] - 9*B*Cos[c] 
 - C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x) 
/2]*(29*A*Sin[(d*x)/2] - 9*B*Sin[(d*x)/2] - C*Sin[(d*x)/2]))/(5*d) + (4*Se 
c[c/2]*Sec[c/2 + (d*x)/2]^3*(11*A*Sin[(d*x)/2] - 6*B*Sin[(d*x)/2] + C*Sin[ 
(d*x)/2]))/(15*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] - B*S 
in[(d*x)/2] + C*Sin[(d*x)/2]))/(5*d) + (16*A*Sec[c]*Sec[c + d*x]*Sin[d*x]) 
/d + (4*(11*A - 6*B + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) + (2*(A ...
 

Rubi [A] (verified)

Time = 1.40 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.01, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.349, Rules used = {3042, 3520, 27, 3042, 3457, 3042, 3457, 27, 3042, 3227, 3042, 3116, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3520

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

-1/5*((A - B + C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x]) 
^3) + ((-2*a*(8*A - 3*B - 2*C)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*(a + 
a*Cos[c + d*x])^2) + ((-5*a^2*(13*A - 3*B - C)*Sin[c + d*x])/(d*Sqrt[Cos[c 
 + d*x]]*(a + a*Cos[c + d*x])) + ((-10*a^3*(13*A - 3*B - C)*EllipticF[(c + 
 d*x)/2, 2])/d + 3*a^3*(49*A - 9*B - C)*((-2*EllipticE[(c + d*x)/2, 2])/d 
+ (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/(2*a^2))/(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 3116

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 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 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] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(792\) vs. \(2(220)=440\).

Time = 7.40 (sec) , antiderivative size = 793, normalized size of antiderivative = 3.35

method	result	size
default	\(\text {Expression too large to display}\)	\(793\)

Input:

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

Output:

-1/60*(-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(- 
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2 
*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*Elli 
pticF(cos(1/2*d*x+1/2*c),2^(1/2))+27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2 
))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(cos(1/2*d*x+1/2 
*c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+4*(sin(1/2*d*x+1/2*c 
)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*E 
llipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*EllipticF(cos(1/2*d*x+1/2*c),2^( 
1/2))+27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*C*EllipticF(cos(1/2*d*x 
+1/2*c),2^(1/2))+3*C*EllipticE(cos(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*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/ 
2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(65*A 
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c), 
2^(1/2))-15*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+27*B*EllipticE(cos(1/2 
*d*x+1/2*c),2^(1/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*Ellipti 
cE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)+12*(-2*sin(1/2*d*x+1/2* 
c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(49*A-9*B-C)*sin(1/2*d*x+1/2*c)^8-2*(-2*s 
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(817*A-147*B-13*C)*sin(1/2 
*d*x+1/2*c)^6+6*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 576, normalized size of antiderivative = 2.43 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

1/60*(2*(3*(49*A - 9*B - C)*cos(d*x + c)^3 + 2*(188*A - 33*B - 2*C)*cos(d* 
x + c)^2 + 5*(59*A - 9*B + C)*cos(d*x + c) + 60*A)*sqrt(cos(d*x + c))*sin( 
d*x + c) - 5*(sqrt(2)*(-13*I*A + 3*I*B + I*C)*cos(d*x + c)^4 + 3*sqrt(2)*( 
-13*I*A + 3*I*B + I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-13*I*A + 3*I*B + I*C)* 
cos(d*x + c)^2 + sqrt(2)*(-13*I*A + 3*I*B + I*C)*cos(d*x + c))*weierstrass 
PInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*(sqrt(2)*(13*I*A - 3*I* 
B - I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(13*I*A - 3*I*B - I*C)*cos(d*x + c)^3 
+ 3*sqrt(2)*(13*I*A - 3*I*B - I*C)*cos(d*x + c)^2 + sqrt(2)*(13*I*A - 3*I* 
B - I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x 
 + c)) - 3*(sqrt(2)*(49*I*A - 9*I*B - I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(49* 
I*A - 9*I*B - I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(49*I*A - 9*I*B - I*C)*cos(d 
*x + c)^2 + sqrt(2)*(49*I*A - 9*I*B - I*C)*cos(d*x + c))*weierstrassZeta(- 
4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt 
(2)*(-49*I*A + 9*I*B + I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(-49*I*A + 9*I*B + 
I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-49*I*A + 9*I*B + I*C)*cos(d*x + c)^2 + s 
qrt(2)*(-49*I*A + 9*I*B + I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weiers 
trassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^3*d*cos(d*x + c)^ 
4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(int(sqrt(cos(c + d*x))/(cos(c + d*x)**5 + 3*cos(c + d*x)**4 + 3*cos(c + d 
*x)**3 + cos(c + d*x)**2),x)*a + 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)*b + int(sqrt(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