\(\int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [120]

Optimal result

Integrand size = 31, antiderivative size = 164 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {2 A \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}-\frac {(43 A-3 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(11 A-3 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \] Output:

2*A*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d-1/32*(43* 
A-3*B)*arctanh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))*2^(1 
/2)/a^(5/2)/d-1/4*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-1/16*(11*A-3*B 
)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)
 

Mathematica [A] (verified)

Time = 1.68 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.77 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {-2 (43 A-3 B) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )+64 \sqrt {2} A \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )+(-15 A+7 B+(-11 A+3 B) \cos (c+d x)) \tan \left (\frac {1}{2} (c+d x)\right )}{16 a d (a (1+\cos (c+d x)))^{3/2}} \] Input:

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

Output:

(-2*(43*A - 3*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + 64*Sqrt[2] 
*A*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + (-15*A + 7*B + ( 
-11*A + 3*B)*Cos[c + d*x])*Tan[(c + d*x)/2])/(16*a*d*(a*(1 + Cos[c + d*x]) 
)^(3/2))
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 3457, 27, 3042, 3457, 27, 3042, 3464, 3042, 3128, 219, 3252, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\int \frac {(8 a A-3 a (A-B) \cos (c+d x)) \sec (c+d x)}{2 (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(8 a A-3 a (A-B) \cos (c+d x)) \sec (c+d x)}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {8 a A-3 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {\left (32 a^2 A-a^2 (11 A-3 B) \cos (c+d x)\right ) \sec (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {a (11 A-3 B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\left (32 a^2 A-a^2 (11 A-3 B) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {a (11 A-3 B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {32 a^2 A-a^2 (11 A-3 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (11 A-3 B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3464

\(\displaystyle \frac {\frac {32 a A \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-a^2 (43 A-3 B) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {a (11 A-3 B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {32 a A \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a^2 (43 A-3 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (11 A-3 B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {\frac {2 a^2 (43 A-3 B) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}+32 a A \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{4 a^2}-\frac {a (11 A-3 B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {32 a A \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {\sqrt {2} a^{3/2} (43 A-3 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (11 A-3 B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {-\frac {64 a^2 A \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {\sqrt {2} a^{3/2} (43 A-3 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (11 A-3 B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {64 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{3/2} (43 A-3 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (11 A-3 B) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*((A - B)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(5/2)) + (((64*a^(3/2) 
*A*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (Sqrt[2]* 
a^(3/2)*(43*A - 3*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Co 
s[c + d*x]])])/d)/(4*a^2) - (a*(11*A - 3*B)*Sin[c + d*x])/(2*d*(a + a*Cos[ 
c + d*x])^(3/2)))/(8*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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3252

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b*c + a*d - d*x^2), 
x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, 
 e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

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])
 

rule 3464

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A 
*b - a*B)/(b*c - a*d)   Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c 
- A*d)/(b*c - a*d)   Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(429\) vs. \(2(139)=278\).

Time = 4.88 (sec) , antiderivative size = 430, normalized size of antiderivative = 2.62

method	result	size
default	\(\frac {\sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (16 A \sqrt {2}\, \ln \left (-\frac {2 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+16 A \sqrt {2}\, \ln \left (\frac {2 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+4 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-43 A \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+3 B \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -11 A \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 B \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 A \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+2 B \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right ) \sqrt {2}}{32 a^{\frac {7}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(430\)
parts	\(-\frac {A \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (43 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-32 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -32 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +11 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \sqrt {2}\, \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right )}{32 a^{\frac {7}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {B \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (3 \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +3 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\right ) \sqrt {2}}{32 a^{\frac {7}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(492\)

Input:

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

Output:

1/32/a^(7/2)/cos(1/2*d*x+1/2*c)^3*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*A*2^( 
1/2)*ln(-2/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^ 
(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a*cos(1/2*d*x+1/2*c)^4+ 
16*A*2^(1/2)*ln(2/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/ 
2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a*cos(1/2*d*x+1/ 
2*c)^4-43*A*ln(2*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/ 
2*c))*a*cos(1/2*d*x+1/2*c)^4+3*B*ln(2*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1 
/2)+a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^4*a-11*A*(a*sin(1/2*d*x+1/2* 
c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^2+3*B*(a*sin(1/2*d*x+1/2*c)^2)^(1/2 
)*a^(1/2)*cos(1/2*d*x+1/2*c)^2-2*A*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+ 
2*B*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/sin(1/2*d*x+1/2*c)*2^(1/2)/(a* 
cos(1/2*d*x+1/2*c)^2)^(1/2)/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 339 vs. \(2 (139) = 278\).

Time = 0.11 (sec) , antiderivative size = 339, normalized size of antiderivative = 2.07 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (43 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (43 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (43 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 43 \, A - 3 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 32 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left ({\left (11 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 15 \, A - 7 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:

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

Output:

-1/64*(sqrt(2)*((43*A - 3*B)*cos(d*x + c)^3 + 3*(43*A - 3*B)*cos(d*x + c)^ 
2 + 3*(43*A - 3*B)*cos(d*x + c) + 43*A - 3*B)*sqrt(a)*log(-(a*cos(d*x + c) 
^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x 
 + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 32*(A*cos(d*x + c)^3 
 + 3*A*cos(d*x + c)^2 + 3*A*cos(d*x + c) + A)*sqrt(a)*log((a*cos(d*x + c)^ 
3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) 
- 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*((11*A - 3 
*B)*cos(d*x + c) + 15*A - 7*B)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^3 
*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
 

Sympy [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

Integral((A + B*cos(c + d*x))*sec(c + d*x)/(a*(cos(c + d*x) + 1))**(5/2), 
x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 84333 vs. \(2 (139) = 278\).

Time = 14.56 (sec) , antiderivative size = 84333, normalized size of antiderivative = 514.23 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/32*(512*((2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d*x + 5/2*c) + cos( 
5/2*d*x + 5/2*c)*sin(4*d*x + 4*c) + 2*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c 
) + (2*cos(2*d*x + 2*c) + cos(d*x + c))*sin(5/2*d*x + 5/2*c) + cos(4*d*x + 
 4*c)*sin(5/2*d*x + 5/2*c) + 2*cos(3*d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos( 
5*d*x + 5*c)^2 + 2560*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d*x + 
 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2*c)*s 
in(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*cos(2*d*x 
 + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x + 5*c)*sin( 
5/2*d*x + 5/2*c) - 5*cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) - 10*cos(3*d*x 
+ 3*c)*sin(5/2*d*x + 5/2*c))*cos(8/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2 
*d*x + 5/2*c)))^2 + 10240*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d 
*x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2* 
c)*sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*cos(2 
*d*x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x + 5*c)* 
sin(5/2*d*x + 5/2*c) - 5*cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) - 10*cos(3* 
d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos(6/5*arctan2(sin(5/2*d*x + 5/2*c), cos 
(5/2*d*x + 5/2*c)))^2 + 10240*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5 
/2*d*x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 
5/2*c)*sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*c 
os(2*d*x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x ...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) a \right )}{a^{3}} \] Input:

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

Output:

(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x))/(cos(c + 
d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*b + int((sqrt(cos(c + 
 d*x) + 1)*sec(c + d*x))/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + 
d*x) + 1),x)*a))/a**3