\(\int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) [445]

Optimal result

Integrand size = 43, antiderivative size = 215 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {4 a^2 (3 A+4 B+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (6 A+7 B+14 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 a^2 (33 A+49 B+35 C) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (3 A+4 B+5 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (4 A+7 B) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)} \] Output:

-4/5*a^2*(3*A+4*B+5*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/21*a^2*(6 
*A+7*B+14*C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/105*a^2*(33*A+49*B 
+35*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+4/5*a^2*(3*A+4*B+5*C)*sin(d*x+c)/d/co 
s(d*x+c)^(1/2)+2/7*A*(a+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(7/2)+2/35 
*(4*A+7*B)*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(5/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 10.31 (sec) , antiderivative size = 1310, normalized size of antiderivative = 6.09 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

Integrate[((a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) 
/Cos[c + d*x]^(9/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(((3*A + 4* 
B + 5*C)*Csc[c]*Sec[c])/(5*d) + (A*Sec[c]*Sec[c + d*x]^4*Sin[d*x])/(14*d) 
+ (Sec[c]*Sec[c + d*x]^3*(5*A*Sin[c] + 14*A*Sin[d*x] + 7*B*Sin[d*x]))/(70* 
d) + (Sec[c]*Sec[c + d*x]^2*(42*A*Sin[c] + 21*B*Sin[c] + 60*A*Sin[d*x] + 7 
0*B*Sin[d*x] + 35*C*Sin[d*x]))/(210*d) + (Sec[c]*Sec[c + d*x]*(60*A*Sin[c] 
 + 70*B*Sin[c] + 35*C*Sin[c] + 126*A*Sin[d*x] + 168*B*Sin[d*x] + 210*C*Sin 
[d*x]))/(210*d)) - (2*A*(a + a*Cos[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1 
/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*Sec[d*x 
 - ArcTan[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] 
]]])/(7*d*Sqrt[1 + Cot[c]^2]) - (B*(a + a*Cos[c + d*x])^2*Csc[c]*Hypergeom 
etricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2 
]^4*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(S 
qrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - Ar 
cTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) - (2*C*(a + a*Cos[c + d*x])^2*Csc 
[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[ 
c/2 + (d*x)/2]^4*Sec[d*x - ArcTan[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*Sqrt[1 + Cot[c]^2]) + (3*A*(a + a*Cos[c 
+ d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((HypergeometricPFQ[{-1/2, -1/4},...
 

Rubi [A] (verified)

Time = 1.39 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.01, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 3522, 27, 3042, 3454, 27, 3042, 3447, 3042, 3500, 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.

\(\displaystyle \int \frac {(a \cos (c+d x)+a)^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\)

\(\Big \downarrow \) 3522

\(\displaystyle \frac {2 \int \frac {(\cos (c+d x) a+a)^2 (a (4 A+7 B)+a (A+7 C) \cos (c+d x))}{2 \cos ^{\frac {7}{2}}(c+d x)}dx}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^2 (a (4 A+7 B)+a (A+7 C) \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)}dx}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (4 A+7 B)+a (A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {2}{5} \int \frac {(\cos (c+d x) a+a) \left ((33 A+49 B+35 C) a^2+(9 A+7 B+35 C) \cos (c+d x) a^2\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 (4 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \int \frac {(\cos (c+d x) a+a) \left ((33 A+49 B+35 C) a^2+(9 A+7 B+35 C) \cos (c+d x) a^2\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 (4 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((33 A+49 B+35 C) a^2+(9 A+7 B+35 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 (4 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\frac {1}{5} \int \frac {(9 A+7 B+35 C) \cos ^2(c+d x) a^3+(33 A+49 B+35 C) a^3+\left ((9 A+7 B+35 C) a^3+(33 A+49 B+35 C) a^3\right ) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 (4 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \int \frac {(9 A+7 B+35 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3+(33 A+49 B+35 C) a^3+\left ((9 A+7 B+35 C) a^3+(33 A+49 B+35 C) a^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 (4 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {21 (3 A+4 B+5 C) a^3+5 (6 A+7 B+14 C) \cos (c+d x) a^3}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^3 (33 A+49 B+35 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (4 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {21 (3 A+4 B+5 C) a^3+5 (6 A+7 B+14 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^3 (33 A+49 B+35 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (4 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (21 a^3 (3 A+4 B+5 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx+5 a^3 (6 A+7 B+14 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )+\frac {2 a^3 (33 A+49 B+35 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (4 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (21 a^3 (3 A+4 B+5 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+5 a^3 (6 A+7 B+14 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 a^3 (33 A+49 B+35 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (4 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (6 A+7 B+14 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+21 a^3 (3 A+4 B+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {2 a^3 (33 A+49 B+35 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (4 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (6 A+7 B+14 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+21 a^3 (3 A+4 B+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {2 a^3 (33 A+49 B+35 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (4 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (6 A+7 B+14 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+21 a^3 (3 A+4 B+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a^3 (33 A+49 B+35 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (4 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2 a^3 (33 A+49 B+35 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \left (\frac {10 a^3 (6 A+7 B+14 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+21 a^3 (3 A+4 B+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\right )+\frac {2 (4 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

Input:

Int[((a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c 
 + d*x]^(9/2),x]
 

Output:

(2*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + ((2*( 
4*A + 7*B)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) 
 + ((2*a^3*(33*A + 49*B + 35*C)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ( 
2*((10*a^3*(6*A + 7*B + 14*C)*EllipticF[(c + d*x)/2, 2])/d + 21*a^3*(3*A + 
 4*B + 5*C)*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[C 
os[c + d*x]]))))/3)/5)/(7*a)
 

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 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3454

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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ 
e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + 
 a*d))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp 
[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B 
*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{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] 
&& GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 
])
 

rule 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

rule 3522

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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - 
d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^m* 
(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( 
a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* 
(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, 
 x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ 
[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(904\) vs. \(2(198)=396\).

Time = 11.23 (sec) , antiderivative size = 905, normalized size of antiderivative = 4.21

method	result	size
default	\(\text {Expression too large to display}\)	\(905\)
parts	\(\text {Expression too large to display}\)	\(1060\)

Input:

int((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, 
method=_RETURNVERBOSE)
 

Output:

-8*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(1/4*C*(s 
in(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/ 
2))+1/4*A*(-1/56*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1 
/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2*d*x+1/2*c)*(-2*si 
n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^ 
2+5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)*EllipticF(cos(1/2*d*x+1/2 
*c),2^(1/2)))+1/5*(1/2*A+1/4*B)/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2 
*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(24*cos(1/2*d*x+1/2*c 
)*sin(1/2*d*x+1/2*c)^6-12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/ 
2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-2 
4*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1 
/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin 
(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x 
+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/ 
2*c),2^(1/2)))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(1/4*B 
+1/2*C)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/ 
2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2 
*c)-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellip...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.31 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (6 \, A + 7 \, B + 14 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (6 \, A + 7 \, B + 14 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (3 \, A + 4 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (3 \, A + 4 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (42 \, {\left (3 \, A + 4 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (12 \, A + 14 \, B + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 21 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 15 \, A a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{105 \, d \cos \left (d x + c\right )^{4}} \] Input:

integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9 
/2),x, algorithm="fricas")
 

Output:

-2/105*(5*I*sqrt(2)*(6*A + 7*B + 14*C)*a^2*cos(d*x + c)^4*weierstrassPInve 
rse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(6*A + 7*B + 14*C) 
*a^2*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + 
c)) + 21*I*sqrt(2)*(3*A + 4*B + 5*C)*a^2*cos(d*x + c)^4*weierstrassZeta(-4 
, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*I*sqr 
t(2)*(3*A + 4*B + 5*C)*a^2*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstra 
ssPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (42*(3*A + 4*B + 5*C)* 
a^2*cos(d*x + c)^3 + 5*(12*A + 14*B + 7*C)*a^2*cos(d*x + c)^2 + 21*(2*A + 
B)*a^2*cos(d*x + c) + 15*A*a^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d* 
x + c)^4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)* 
*(9/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \] Input:

integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9 
/2),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^2/c 
os(d*x + c)^(9/2), x)
 

Giac [F]

\[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \] Input:

integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9 
/2),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^2/c 
os(d*x + c)^(9/2), x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 1.83 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.61 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {30\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+84\,A\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+70\,A\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{105\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {6\,B\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,B\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,B\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int(((a + a*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x)^(9/2),x)
 

Output:

(30*A*a^2*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2) + 84*A 
*a^2*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2 
) + 70*A*a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c 
 + d*x)^2))/(105*d*cos(c + d*x)^(7/2)*(1 - cos(c + d*x)^2)^(1/2)) + (6*B*a 
^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 20*B*a^2*co 
s(c + d*x)*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2) + 30*B 
*a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^ 
2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (2*C*a^2*ellipt 
icF(c/2 + (d*x)/2, 2))/d + (4*C*a^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/ 
4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (2*C*a 
^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + 
d*x)^(3/2)*(sin(c + d*x)^2)^(1/2))
 

Reduce [F]

\[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=a^{2} \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5}}d x \right ) a +2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a +2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) b +2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) c \right ) \] Input:

int((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x)
 

Output:

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