\(\int \frac {(a+b \cos (c+d x))^2 (B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\) [867]

Optimal result

Integrand size = 42, antiderivative size = 214 \[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=-\frac {2 \left (6 a b B+3 a^2 C+5 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (5 a^2 B+7 b^2 B+14 a b C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 a^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 a (2 b B+a C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (5 a^2 B+7 b^2 B+14 a b C\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (6 a b B+3 a^2 C+5 b^2 C\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \] Output:

-2/5*(6*B*a*b+3*C*a^2+5*C*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/2 
1*(5*B*a^2+7*B*b^2+14*C*a*b)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/7* 
a^2*B*sin(d*x+c)/d/cos(d*x+c)^(7/2)+2/5*a*(2*B*b+C*a)*sin(d*x+c)/d/cos(d*x 
+c)^(5/2)+2/21*(5*B*a^2+7*B*b^2+14*C*a*b)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/ 
5*(6*B*a*b+3*C*a^2+5*C*b^2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 6.07 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 \left (-21 \left (6 a b B+3 a^2 C+5 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 \left (5 a^2 B+7 b^2 B+14 a b C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {15 a^2 B \sin (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)}+\frac {21 a (2 b B+a C) \sin (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}+\frac {5 \left (5 a^2 B+7 b^2 B+14 a b C\right ) \sin (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {21 \left (6 a b B+3 a^2 C+5 b^2 C\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{105 d} \] Input:

Integrate[((a + b*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos 
[c + d*x]^(11/2),x]
 

Output:

(2*(-21*(6*a*b*B + 3*a^2*C + 5*b^2*C)*EllipticE[(c + d*x)/2, 2] + 5*(5*a^2 
*B + 7*b^2*B + 14*a*b*C)*EllipticF[(c + d*x)/2, 2] + (15*a^2*B*Sin[c + d*x 
])/Cos[c + d*x]^(7/2) + (21*a*(2*b*B + a*C)*Sin[c + d*x])/Cos[c + d*x]^(5/ 
2) + (5*(5*a^2*B + 7*b^2*B + 14*a*b*C)*Sin[c + d*x])/Cos[c + d*x]^(3/2) + 
(21*(6*a*b*B + 3*a^2*C + 5*b^2*C)*Sin[c + d*x])/Sqrt[Cos[c + d*x]]))/(105* 
d)
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.89, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3508, 3042, 3467, 27, 3042, 3500, 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.

\(\displaystyle \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx\)

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3467

\(\displaystyle \frac {2 a^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {2}{7} \int -\frac {7 b^2 C \cos ^2(c+d x)+\left (5 B a^2+14 b C a+7 b^2 B\right ) \cos (c+d x)+7 a (2 b B+a C)}{2 \cos ^{\frac {7}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \int \frac {7 b^2 C \cos ^2(c+d x)+\left (5 B a^2+14 b C a+7 b^2 B\right ) \cos (c+d x)+7 a (2 b B+a C)}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \int \frac {7 b^2 C \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (5 B a^2+14 b C a+7 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+7 a (2 b B+a C)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (5 \left (5 a^2 B+14 a b C+7 b^2 B\right ) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 \left (3 a^2 C+6 a b B+5 b^2 C\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {14 a (a C+2 b B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

Int[((a + b*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d 
*x]^(11/2),x]
 

Output:

(2*a^2*B*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + ((14*a*(2*b*B + a*C)*Sin 
[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (5*(5*a^2*B + 7*b^2*B + 14*a*b*C)*(( 
2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3 
/2))) + 7*(6*a*b*B + 3*a^2*C + 5*b^2*C)*((-2*EllipticE[(c + d*x)/2, 2])/d 
+ (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/5)/7
 

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 3467

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

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 3508

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(831\) vs. \(2(197)=394\).

Time = 11.26 (sec) , antiderivative size = 832, normalized size of antiderivative = 3.89

method	result	size
default	\(\text {Expression too large to display}\)	\(832\)
parts	\(\text {Expression too large to display}\)	\(1024\)

Input:

int((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x,m 
ethod=_RETURNVERBOSE)
 

Output:

-(-(1-2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B*a^2*(-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)/(c 
os(1/2*d*x+1/2*c)^2-1/2)^4-5/42*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)^2+5/21*(sin(1/2*d 
*x+1/2*c)^2)^(1/2)*(1-2*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*C* 
b^2/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)*EllipticE(co 
s(1/2*d*x+1/2*c),2^(1/2)))+2/5*a*(2*B*b+C*a)/sin(1/2*d*x+1/2*c)^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)*(24*cos( 
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2 
))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d 
*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*EllipticE(cos(1/ 
2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(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(c 
os(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)+2*b*(B*b+2*C*a)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+s 
in(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.47 \[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (5 i \, B a^{2} + 14 i \, C a b + 7 i \, B b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-5 i \, B a^{2} - 14 i \, C a b - 7 i \, B b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (3 i \, C a^{2} + 6 i \, B a b + 5 i \, C b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (-3 i \, C a^{2} - 6 i \, B a b - 5 i \, C b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (21 \, {\left (3 \, C a^{2} + 6 \, B a b + 5 \, C b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \, B a^{2} + 5 \, {\left (5 \, B a^{2} + 14 \, C a b + 7 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 21 \, {\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{4}} \] Input:

integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/ 
2),x, algorithm="fricas")
 

Output:

-1/105*(5*sqrt(2)*(5*I*B*a^2 + 14*I*C*a*b + 7*I*B*b^2)*cos(d*x + c)^4*weie 
rstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-5*I*B* 
a^2 - 14*I*C*a*b - 7*I*B*b^2)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, co 
s(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(3*I*C*a^2 + 6*I*B*a*b + 5*I*C*b 
^2)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d 
*x + c) + I*sin(d*x + c))) + 21*sqrt(2)*(-3*I*C*a^2 - 6*I*B*a*b - 5*I*C*b^ 
2)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* 
x + c) - I*sin(d*x + c))) - 2*(21*(3*C*a^2 + 6*B*a*b + 5*C*b^2)*cos(d*x + 
c)^3 + 15*B*a^2 + 5*(5*B*a^2 + 14*C*a*b + 7*B*b^2)*cos(d*x + c)^2 + 21*(C* 
a^2 + 2*B*a*b)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + 
 c)^4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))**2*(B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**( 
11/2),x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {11}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/ 
2),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^2/cos(d 
*x + c)^(11/2), x)
 

Giac [F]

\[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {11}{2}}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/ 
2),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^2/cos(d 
*x + c)^(11/2), x)
 

Mupad [B] (verification not implemented)

Time = 2.05 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {30\,B\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+70\,B\,b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+84\,B\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{105\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {6\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,C\,b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,C\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \] Input:

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^2)/cos(c + d 
*x)^(11/2),x)
 

Output:

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

Reduce [F]

\[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5}}d x \right ) a^{2} b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) a^{2} c +2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) a \,b^{2}+2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a b c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) b^{3}+\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) b^{2} c \] Input:

int((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x)
 

Output:

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