\(\int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx\) [489]

Optimal result

Integrand size = 33, antiderivative size = 216 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx=\frac {(A-B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {(A+B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(4 A+B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(A+B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:

1/10*(A-B)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+ 
c)^(1/2)/a^3/d+1/6*(A+B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^ 
(1/2))*sec(d*x+c)^(1/2)/a^3/d-1/5*(A-B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a 
*sec(d*x+c))^3-1/15*(4*A+B)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c 
))^2+1/6*(A+B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a^3+a^3*sec(d*x+c))
 

Mathematica [C] (warning: unable to verify)

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

Time = 8.57 (sec) , antiderivative size = 792, normalized size of antiderivative = 3.67 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

Integrate[(A + B*Cos[c + d*x])/((a + a*Cos[c + d*x])^3*Sqrt[Sec[c + d*x]]) 
,x]
 

Output:

-1/15*(Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + 
E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*( 
c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 
7/4, -E^((2*I)*(c + d*x))])*Sec[c/2])/(d*E^(I*d*x)*(a + a*Cos[c + d*x])^3) 
 + (Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^( 
(2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + 
 d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4 
, -E^((2*I)*(c + d*x))])*Sec[c/2])/(15*d*E^(I*d*x)*(a + a*Cos[c + d*x])^3) 
 + (2*A*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d* 
x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(3*d*(a + a*Cos[c + d*x])^3) 
+ (2*B*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x 
)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(3*d*(a + a*Cos[c + d*x])^3) + 
 (Cos[c/2 + (d*x)/2]^6*Sqrt[Sec[c + d*x]]*((-2*(A - B)*Cos[d*x]*Csc[c/2]*S 
ec[c/2])/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(2*A*Sin[(d*x)/2] - 7*B* 
Sin[(d*x)/2]))/(15*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] - 
 B*Sin[(d*x)/2]))/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + 
 B*Sin[(d*x)/2]))/(3*d) + (4*(A + B)*Tan[c/2])/(3*d) + (4*(2*A - 7*B)*Sec[ 
c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) - (2*(A - B)*Sec[c/2 + (d*x)/2]^4*Tan[c/ 
2])/(5*d)))/(a + a*Cos[c + d*x])^3
 

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.07, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 3439, 3042, 4507, 27, 3042, 4507, 25, 3042, 4508, 27, 3042, 4274, 3042, 4258, 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.

\(\displaystyle \int \frac {A+B \cos (c+d x)}{\sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3439

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\int -\frac {\sqrt {\sec (c+d x)} (a (A-B)-a (7 A+3 B) \sec (c+d x))}{2 (\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\sqrt {\sec (c+d x)} (a (A-B)-a (7 A+3 B) \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a (A-B)-a (7 A+3 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4507

\(\displaystyle -\frac {\frac {\int -\frac {(4 A+B) a^2+3 (3 A+2 B) \sec (c+d x) a^2}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)}dx}{3 a^2}+\frac {2 a (4 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {2 a (4 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\int \frac {(4 A+B) a^2+3 (3 A+2 B) \sec (c+d x) a^2}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)}dx}{3 a^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

\(\displaystyle -\frac {\frac {2 a (4 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\frac {\int \frac {3 (A-B) a^3+5 (A+B) \sec (c+d x) a^3}{2 \sqrt {\sec (c+d x)}}dx}{a^2}+\frac {5 a^2 (A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {2 a (4 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\frac {\int \frac {3 (A-B) a^3+5 (A+B) \sec (c+d x) a^3}{\sqrt {\sec (c+d x)}}dx}{2 a^2}+\frac {5 a^2 (A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

\(\displaystyle -\frac {\frac {2 a (4 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\frac {3 a^3 (A-B) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+5 a^3 (A+B) \int \sqrt {\sec (c+d x)}dx}{2 a^2}+\frac {5 a^2 (A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

Int[(A + B*Cos[c + d*x])/((a + a*Cos[c + d*x])^3*Sqrt[Sec[c + d*x]]),x]
 

Output:

-1/5*((A - B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^3) 
- ((2*a*(4*A + B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x 
])^2) - (((6*a^3*(A - B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt 
[Sec[c + d*x]])/d + (10*a^3*(A + B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x) 
/2, 2]*Sqrt[Sec[c + d*x]])/d)/(2*a^2) + (5*a^2*(A + B)*Sqrt[Sec[c + d*x]]* 
Sin[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2))/(10*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3439

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[g^(m + n)   Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(d + 
c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c 
- a*d, 0] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4507

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b 
- a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 
2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)* 
(d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m 
 - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, 
A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G 
tQ[n, 0]
 

rule 4508

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b 
- a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 
 1))), x] - Simp[1/(a^2*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs 
c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ 
e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B 
, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(450\) vs. \(2(195)=390\).

Time = 6.39 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.09

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 (12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-10 A \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 ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+6 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-10 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 ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-6 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-22 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+6 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+7 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-17 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 A +3 B \right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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}\)	\(451\)

Input:

int((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3/sec(d*x+c)^(1/2),x,method=_RETURNV 
ERBOSE)
                                                                                    
                                                                                    
 

Output:

1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*A*cos(1/2 
*d*x+1/2*c)^8-10*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1 
)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+6*A*cos 
(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1) 
^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-12*B*cos(1/2*d*x+1/2*c)^8-10* 
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)*EllipticF 
(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5-6*B*cos(1/2*d*x+1/2*c)^5 
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE( 
cos(1/2*d*x+1/2*c),2^(1/2))-22*A*cos(1/2*d*x+1/2*c)^6+2*B*cos(1/2*d*x+1/2* 
c)^6+6*A*cos(1/2*d*x+1/2*c)^4+24*B*cos(1/2*d*x+1/2*c)^4+7*A*cos(1/2*d*x+1/ 
2*c)^2-17*B*cos(1/2*d*x+1/2*c)^2-3*A+3*B)/a^3/cos(1/2*d*x+1/2*c)^5/(-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*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 = 472, normalized size of antiderivative = 2.19 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3/sec(d*x+c)^(1/2),x, algorith 
m="fricas")
 

Output:

-1/60*(5*(sqrt(2)*(I*A + I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A + I*B)*cos(d 
*x + c)^2 + 3*sqrt(2)*(I*A + I*B)*cos(d*x + c) + sqrt(2)*(I*A + I*B))*weie 
rstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-I*A - 
 I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A - I*B)*cos(d*x + c)^2 + 3*sqrt(2)*( 
-I*A - I*B)*cos(d*x + c) + sqrt(2)*(-I*A - I*B))*weierstrassPInverse(-4, 0 
, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(-I*A + I*B)*cos(d*x + c)^3 
+ 3*sqrt(2)*(-I*A + I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A + I*B)*cos(d*x + 
 c) + sqrt(2)*(-I*A + I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 
 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(I*A - I*B)*cos(d*x + c)^ 
3 + 3*sqrt(2)*(I*A - I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A - I*B)*cos(d*x + 
 c) + sqrt(2)*(I*A - I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 
0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*(A - B)*cos(d*x + c)^3 + 2*(2*A 
- 7*B)*cos(d*x + c)^2 - 5*(A + B)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(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(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))**3/sec(d*x+c)**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3/sec(d*x+c)^(1/2),x, algorith 
m="maxima")
 

Output:

integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^3*sqrt(sec(d*x + c))) 
, x)
 

Giac [F]

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

integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3/sec(d*x+c)^(1/2),x, algorith 
m="giac")
 

Output:

integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^3*sqrt(sec(d*x + c))) 
, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^3),x 
)
 

Output:

int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^3), 
x)
 

Reduce [F]

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

int((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3/sec(d*x+c)^(1/2),x)
 

Output:

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