\(\int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx\) [157]

Optimal result

Integrand size = 33, antiderivative size = 201 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx=\frac {(7 A-4 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {5 (2 A-B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {5 (2 A-B) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(7 A-4 B) \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {(7 A-4 B) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x) (1+\cos (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \] Output:

(7*A-4*B)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d+5/3*(2*A-B)*InverseJ 
acobiAM(1/2*d*x+1/2*c,2^(1/2))/a^2/d+5/3*(2*A-B)*sin(d*x+c)/a^2/d/cos(d*x+ 
c)^(3/2)-(7*A-4*B)*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)-1/3*(7*A-4*B)*sin(d*x 
+c)/a^2/d/cos(d*x+c)^(3/2)/(1+cos(d*x+c))-1/3*(A-B)*sin(d*x+c)/d/cos(d*x+c 
)^(3/2)/(a+a*cos(d*x+c))^2
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 8.85 (sec) , antiderivative size = 1020, normalized size of antiderivative = 5.07 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(-20*A*Cos[c/2 + (d*x)/2]^4*Csc[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 - S 
in[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTa 
n[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x] 
)^2*Sqrt[1 + Cot[c]^2]) + (10*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Hypergeometr 
icPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - A 
rcTan[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*(a + a*Cos[c + d*x])^2*Sqrt[1 + Cot[c]^2]) + (Cos[c/2 + (d*x)/2]^4*S 
qrt[Cos[c + d*x]]*((-2*(4*A - 2*B + 3*A*Cos[c] - 2*B*Cos[c])*Csc[c/2]*Sec[ 
c/2]*Sec[c])/d - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(3*A*Sin[(d*x)/2] - 2*B*Si 
n[(d*x)/2]))/d - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] - B*Sin[ 
(d*x)/2]))/(3*d) + (8*A*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(3*d) + (8*Sec[c]* 
Sec[c + d*x]*(A*Sin[c] - 6*A*Sin[d*x] + 3*B*Sin[d*x]))/(3*d) - (2*(A - B)* 
Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/(a + a*Cos[c + d*x])^2 - (7*A*Cos[c 
/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, 
 Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - 
Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c] 
*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin 
[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x...
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 3457, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \frac {5 a^2 (2 A-B)-a^2 (7 A-4 B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {2 (7 A-4 B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

-1/3*((A - B)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2) 
+ ((-2*(7*A - 4*B)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(1 + Cos[c + d*x])) 
 + (3*(5*a^2*(2*A - B)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sin[c + d 
*x])/(3*d*Cos[c + d*x]^(3/2))) - a^2*(7*A - 4*B)*((-2*EllipticE[(c + d*x)/ 
2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))))/a^2)/(6*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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(722\) vs. \(2(188)=376\).

Time = 5.44 (sec) , antiderivative size = 723, normalized size of antiderivative = 3.60

method	result	size
default	\(\text {Expression too large to display}\)	\(723\)

Input:

int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^2,x,method=_RETURNV 
ERBOSE)
 

Output:

-1/2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*((-8*A+ 
4*B)/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(c 
os(1/2*d*x+1/2*c),2^(1/2)))+1/3*(A-B)*(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)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*El 
lipticE(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*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*Elli 
pticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))) 
*cos(1/2*d*x+1/2*c)-12*sin(1/2*d*x+1/2*c)^6+20*sin(1/2*d*x+1/2*c)^4-7*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)^4+sin(1/2*d*x+ 
1/2*c)^2)^(1/2)/(sin(1/2*d*x+1/2*c)^2-1)+(4*A-2*B)*(cos(1/2*d*x+1/2*c)*(2* 
sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos( 
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+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 
)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+4*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 = 0.11 (sec) , antiderivative size = 436, normalized size of antiderivative = 2.17 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx=-\frac {2 \, {\left (3 \, {\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (32 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (4 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 2 \, A\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 5 \, {\left (\sqrt {2} {\left (2 i \, A - i \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, \sqrt {2} {\left (2 i \, A - i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (2 i \, A - i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (-2 i \, A + i \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, \sqrt {2} {\left (-2 i \, A + i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-2 i \, A + i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-7 i \, A + 4 i \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, \sqrt {2} {\left (-7 i \, A + 4 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-7 i \, A + 4 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\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 \, {\left (\sqrt {2} {\left (7 i \, A - 4 i \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, \sqrt {2} {\left (7 i \, A - 4 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (7 i \, A - 4 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\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 )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \] Input:

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

Output:

-1/6*(2*(3*(7*A - 4*B)*cos(d*x + c)^3 + (32*A - 19*B)*cos(d*x + c)^2 + 2*( 
4*A - 3*B)*cos(d*x + c) - 2*A)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2 
)*(2*I*A - I*B)*cos(d*x + c)^4 + 2*sqrt(2)*(2*I*A - I*B)*cos(d*x + c)^3 + 
sqrt(2)*(2*I*A - I*B)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + 
 c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-2*I*A + I*B)*cos(d*x + c)^4 + 2*sqrt( 
2)*(-2*I*A + I*B)*cos(d*x + c)^3 + sqrt(2)*(-2*I*A + I*B)*cos(d*x + c)^2)* 
weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(-7 
*I*A + 4*I*B)*cos(d*x + c)^4 + 2*sqrt(2)*(-7*I*A + 4*I*B)*cos(d*x + c)^3 + 
 sqrt(2)*(-7*I*A + 4*I*B)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstra 
ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(7*I*A - 4* 
I*B)*cos(d*x + c)^4 + 2*sqrt(2)*(7*I*A - 4*I*B)*cos(d*x + c)^3 + sqrt(2)*( 
7*I*A - 4*I*B)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInverse( 
-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*co 
s(d*x + c)^3 + a^2*d*cos(d*x + c)^2)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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