\(\int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [1077]

Optimal result

Integrand size = 43, antiderivative size = 200 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \left (10 a b B+5 b^2 (A-C)+a^2 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (a^2 B+3 b^2 B+2 a b (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a (4 A b+5 a B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (4 A b^2+10 a b B+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \] Output:

-2/5*(10*B*a*b+5*b^2*(A-C)+a^2*(3*A+5*C))*EllipticE(sin(1/2*d*x+1/2*c),2^( 
1/2))/d+2/3*(B*a^2+3*B*b^2+2*a*b*(A+3*C))*InverseJacobiAM(1/2*d*x+1/2*c,2^ 
(1/2))/d+2/15*a*(4*A*b+5*B*a)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*(4*A*b^2+1 
0*B*a*b+a^2*(3*A+5*C))*sin(d*x+c)/d/cos(d*x+c)^(1/2)+2/5*A*(a+b*cos(d*x+c) 
)^2*sin(d*x+c)/d/cos(d*x+c)^(5/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 3.92 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {-6 \left (10 a b B+5 b^2 (A-C)+a^2 (3 A+5 C)\right ) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (a^2 B+3 b^2 B+2 a b (A+3 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+20 a A b \sin (c+d x)+10 a^2 B \sin (c+d x)+9 a^2 A \sin (2 (c+d x))+15 A b^2 \sin (2 (c+d x))+30 a b B \sin (2 (c+d x))+15 a^2 C \sin (2 (c+d x))+6 a^2 A \tan (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \] Input:

Integrate[((a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) 
/Cos[c + d*x]^(7/2),x]
 

Output:

(-6*(10*a*b*B + 5*b^2*(A - C) + a^2*(3*A + 5*C))*Cos[c + d*x]^(3/2)*Ellipt 
icE[(c + d*x)/2, 2] + 10*(a^2*B + 3*b^2*B + 2*a*b*(A + 3*C))*Cos[c + d*x]^ 
(3/2)*EllipticF[(c + d*x)/2, 2] + 20*a*A*b*Sin[c + d*x] + 10*a^2*B*Sin[c + 
 d*x] + 9*a^2*A*Sin[2*(c + d*x)] + 15*A*b^2*Sin[2*(c + d*x)] + 30*a*b*B*Si 
n[2*(c + d*x)] + 15*a^2*C*Sin[2*(c + d*x)] + 6*a^2*A*Tan[c + d*x])/(15*d*C 
os[c + d*x]^(3/2))
 

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 3526, 27, 3042, 3510, 27, 3042, 3500, 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.

\(\displaystyle \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

\(\displaystyle \frac {2}{5} \int \frac {(a+b \cos (c+d x)) \left (-b (A-5 C) \cos ^2(c+d x)+(3 a A+5 b B+5 a C) \cos (c+d x)+4 A b+5 a B\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \frac {(a+b \cos (c+d x)) \left (-b (A-5 C) \cos ^2(c+d x)+(3 a A+5 b B+5 a C) \cos (c+d x)+4 A b+5 a B\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3510

\(\displaystyle \frac {1}{5} \left (\frac {2 a (5 a B+4 A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2}{3} \int -\frac {-3 b^2 (A-5 C) \cos ^2(c+d x)+5 \left (B a^2+2 b (A+3 C) a+3 b^2 B\right ) \cos (c+d x)+3 \left ((3 A+5 C) a^2+10 b B a+4 A b^2\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {-3 b^2 (A-5 C) \cos ^2(c+d x)+5 \left (B a^2+2 b (A+3 C) a+3 b^2 B\right ) \cos (c+d x)+3 \left ((3 A+5 C) a^2+10 b B a+4 A b^2\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a (5 a B+4 A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {-3 b^2 (A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+5 \left (B a^2+2 b (A+3 C) a+3 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left ((3 A+5 C) a^2+10 b B a+4 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a (5 a B+4 A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (2 \int \frac {5 \left (B a^2+2 b (A+3 C) a+3 b^2 B\right )-3 \left ((3 A+5 C) a^2+10 b B a+5 b^2 (A-C)\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+10 a b B+4 A b^2\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a (5 a B+4 A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {5 \left (B a^2+2 b (A+3 C) a+3 b^2 B\right )-3 \left ((3 A+5 C) a^2+10 b B a+5 b^2 (A-C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+10 a b B+4 A b^2\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a (5 a B+4 A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {5 \left (B a^2+2 b (A+3 C) a+3 b^2 B\right )-3 \left ((3 A+5 C) a^2+10 b B a+5 b^2 (A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+10 a b B+4 A b^2\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a (5 a B+4 A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right ) \int \sqrt {\cos (c+d x)}dx+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+10 a b B+4 A b^2\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a (5 a B+4 A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+10 a b B+4 A b^2\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a (5 a B+4 A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right )}{d}+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)+10 a b B+4 A b^2\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 a (5 a B+4 A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

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

Input:

Int[((a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c 
 + d*x]^(7/2),x]
 

Output:

(2*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((2*a 
*(4*A*b + 5*a*B)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((-6*(10*a*b*B + 
 5*b^2*(A - C) + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/d + (10*(a^2* 
B + 3*b^2*B + 2*a*b*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/d + (6*(4*A*b^2 
+ 10*a*b*B + a^2*(3*A + 5*C))*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))/3)/5
 

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 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 3510

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

rule 3526

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(946\) vs. \(2(187)=374\).

Time = 5.02 (sec) , antiderivative size = 947, normalized size of antiderivative = 4.74

method	result	size
parts	\(\text {Expression too large to display}\)	\(947\)
default	\(\text {Expression too large to display}\)	\(973\)

Input:

int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, 
method=_RETURNVERBOSE)
 

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.60 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, B a^{2} + 2 i \, {\left (A + 3 \, C\right )} a b + 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, B a^{2} - 2 i \, {\left (A + 3 \, C\right )} a b - 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (i \, {\left (3 \, A + 5 \, C\right )} a^{2} + 10 i \, B a b + 5 i \, {\left (A - C\right )} b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-i \, {\left (3 \, A + 5 \, C\right )} a^{2} - 10 i \, B a b - 5 i \, {\left (A - C\right )} b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, A a^{2} + 3 \, {\left ({\left (3 \, A + 5 \, C\right )} a^{2} + 10 \, B a b + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7 
/2),x, algorithm="fricas")
 

Output:

-1/15*(5*sqrt(2)*(I*B*a^2 + 2*I*(A + 3*C)*a*b + 3*I*B*b^2)*cos(d*x + c)^3* 
weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I* 
B*a^2 - 2*I*(A + 3*C)*a*b - 3*I*B*b^2)*cos(d*x + c)^3*weierstrassPInverse( 
-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(I*(3*A + 5*C)*a^2 + 10* 
I*B*a*b + 5*I*(A - C)*b^2)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstra 
ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-I*(3*A + 5 
*C)*a^2 - 10*I*B*a*b - 5*I*(A - C)*b^2)*cos(d*x + c)^3*weierstrassZeta(-4, 
 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*A*a^ 
2 + 3*((3*A + 5*C)*a^2 + 10*B*a*b + 5*A*b^2)*cos(d*x + c)^2 + 5*(B*a^2 + 2 
*A*a*b)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)* 
*(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7 
/2),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2/c 
os(d*x + c)^(7/2), x)
 

Giac [F]

\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7 
/2),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2/c 
os(d*x + c)^(7/2), x)
 

Mupad [B] (verification not implemented)

Time = 2.58 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.55 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {6\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,A\,b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,A\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,B\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,B\,a\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int(((a + b*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x)^(7/2),x)
 

Output:

(6*A*a^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 30*A* 
b^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2 
) + 20*A*a*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + 
 d*x)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (2*B*b^2* 
ellipticF(c/2 + (d*x)/2, 2))/d + (2*C*b^2*ellipticE(c/2 + (d*x)/2, 2))/d + 
 (4*C*a*b*ellipticF(c/2 + (d*x)/2, 2))/d + (2*B*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)) + (2*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)) + (4*B*a*b*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))
 

Reduce [F]

\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a b c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) b^{3}+\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) a^{3}+3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a^{2} b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a^{2} c +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a \,b^{2}+\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) b^{2} c \] Input:

int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)
 

Output:

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