\(\int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [350]

Optimal result

Integrand size = 31, antiderivative size = 103 \[ \int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 (A b+a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 (a A+3 b B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b+a B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \] Output:

-2*(A*b+B*a)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*(A*a+3*B*b)*Inver 
seJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/3*a*A*sin(d*x+c)/d/cos(d*x+c)^(3/2)+ 
2*(A*b+B*a)*sin(d*x+c)/d/cos(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \left (-3 (A b+a B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+(a A+3 b B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+3 A b \sin (c+d x)+3 a B \sin (c+d x)+a A \tan (c+d x)\right )}{3 d \sqrt {\cos (c+d x)}} \] Input:

Integrate[((a + b*Cos[c + d*x])*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(5/2),x 
]
 

Output:

(2*(-3*(A*b + a*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + (a*A + 3 
*b*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 3*A*b*Sin[c + d*x] + 
3*a*B*Sin[c + d*x] + a*A*Tan[c + d*x]))/(3*d*Sqrt[Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 3447, 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.

\(\displaystyle \int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

\(\displaystyle \int \frac {(a B+A b) \cos (c+d x)+a A+b B \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {3 (A b+a B)+(a A+3 b B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

(2*a*A*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((2*(a*A + 3*b*B)*Elliptic 
F[(c + d*x)/2, 2])/d + 3*(A*b + a*B)*((-2*EllipticE[(c + d*x)/2, 2])/d + ( 
2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/3
 

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 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(400\) vs. \(2(98)=196\).

Time = 6.78 (sec) , antiderivative size = 401, normalized size of antiderivative = 3.89

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {2 B b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {2 \left (A b +B a \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+2 A a \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{6 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(401\)
parts	\(-\frac {2 \left (A b +B a \right ) \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}-\frac {2 A a \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d}+\frac {2 B b \,\operatorname {InverseJacobiAM}\left (\frac {d x}{2}+\frac {c}{2}, \sqrt {2}\right )}{d}\)	\(423\)

Input:

int((a+cos(d*x+c)*b)*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x,method=_RETURNVER 
BOSE)
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B*b*(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))+2* 
(A*b+B*a)/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)*Ellipt 
icE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*A*a*(-1/6*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+1/ 
3*(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))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.07 \[ \int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\sqrt {2} {\left (-i \, A a - 3 i \, B b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A a + 3 i \, B b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (i \, B a + i \, A b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (-i \, B a - i \, A b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (A a + 3 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm= 
"fricas")
 

Output:

1/3*(sqrt(2)*(-I*A*a - 3*I*B*b)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, 
cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(I*A*a + 3*I*B*b)*cos(d*x + c)^2* 
weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*(I*B 
*a + I*A*b)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 
0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(-I*B*a - I*A*b)*cos(d*x + 
c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*si 
n(d*x + c))) + 2*(A*a + 3*(B*a + A*b)*cos(d*x + c))*sqrt(cos(d*x + c))*sin 
(d*x + c))/(d*cos(d*x + c)^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))/cos(d*x+c)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm= 
"maxima")
 

Output:

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

Giac [F]

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

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm= 
"giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 25.62 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.46 \[ \int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2\,B\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

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

Output:

(2*B*b*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*a*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*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)) + (2*B*a*sin(c + d*x)*hypergeo 
m([-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]

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

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

Output:

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