\(\int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [270]

Optimal result

Integrand size = 29, antiderivative size = 214 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (6 a^4 A b-5 a^2 A b^3+2 A b^5-2 a^5 B-a^3 b^2 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac {A \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {b (A b-a B) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b \left (5 a^2 A b-2 A b^3-3 a^3 B\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

-(6*A*a^4*b-5*A*a^2*b^3+2*A*b^5-2*B*a^5-B*a^3*b^2)*arctan((a-b)^(1/2)*tan( 
1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/(a-b)^(5/2)/(a+b)^(5/2)/d+A*arctanh(sin(d* 
x+c))/a^3/d+1/2*b*(A*b-B*a)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^2+1/ 
2*b*(5*A*a^2*b-2*A*b^3-3*B*a^3)*sin(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*cos(d*x+ 
c))
 

Mathematica [A] (verified)

Time = 2.37 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.26 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\cos (c+d x) (B+A \sec (c+d x)) \left (-\frac {2 \left (-6 a^4 A b+5 a^2 A b^3-2 A b^5+2 a^5 B+a^3 b^2 B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}-2 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2 b (A b-a B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+\frac {a b \left (5 a^2 A b-2 A b^3-3 a^3 B\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}\right )}{2 a^3 d (A+B \cos (c+d x))} \] Input:

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

Output:

(Cos[c + d*x]*(B + A*Sec[c + d*x])*((-2*(-6*a^4*A*b + 5*a^2*A*b^3 - 2*A*b^ 
5 + 2*a^5*B + a^3*b^2*B)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^ 
2]])/(-a^2 + b^2)^(5/2) - 2*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2 
*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^2*b*(A*b - a*B)*Sin[c + d 
*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + (a*b*(5*a^2*A*b - 2*A*b^3 
- 3*a^3*B)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x]))))/(2*a 
^3*d*(A + B*Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.21, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3479, 3042, 3534, 3042, 3480, 3042, 3138, 218, 4257}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3479

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4257

\(\displaystyle \frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\frac {b \left (-3 a^3 B+5 a^2 A b-2 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {2 A \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{a d}-\frac {2 \left (-2 a^5 B+6 a^4 A b-a^3 b^2 B-5 a^2 A b^3+2 A b^5\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\)

Input:

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

Output:

(b*(A*b - a*B)*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + 
(((-2*(6*a^4*A*b - 5*a^2*A*b^3 + 2*A*b^5 - 2*a^5*B - a^3*b^2*B)*ArcTan[(Sq 
rt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + 
(2*A*(a^2 - b^2)^2*ArcTanh[Sin[c + d*x]])/(a*d))/(a*(a^2 - b^2)) + (b*(5*a 
^2*A*b - 2*A*b^3 - 3*a^3*B)*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + 
d*x])))/(2*a*(a^2 - b^2))
 

Defintions of rubi rules used

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 3479

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin 
[e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 
1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e 
 + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 
2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) 
*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n 
}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat 
ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(I 
ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0]) 
))
 

rule 3480

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((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/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ 
(b*c - a*d)   Int[1/(c + d*Sin[e + f*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]
 

rule 3534

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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 2.47 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.41

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (\frac {-\frac {\left (6 A \,a^{2} b +A a \,b^{2}-2 A \,b^{3}-4 a^{3} B -B \,a^{2} b \right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {b a \left (6 A \,a^{2} b -A a \,b^{2}-2 A \,b^{3}-4 a^{3} B +B \,a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (6 A \,a^{4} b -5 A \,a^{2} b^{3}+2 A \,b^{5}-2 B \,a^{5}-B \,a^{3} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3}}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}}{d}\)	\(302\)
default	\(\frac {-\frac {2 \left (\frac {-\frac {\left (6 A \,a^{2} b +A a \,b^{2}-2 A \,b^{3}-4 a^{3} B -B \,a^{2} b \right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {b a \left (6 A \,a^{2} b -A a \,b^{2}-2 A \,b^{3}-4 a^{3} B +B \,a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (6 A \,a^{4} b -5 A \,a^{2} b^{3}+2 A \,b^{5}-2 B \,a^{5}-B \,a^{3} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3}}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}}{d}\)	\(302\)
risch	\(\text {Expression too large to display}\)	\(1191\)

Input:

int((A+B*cos(d*x+c))*sec(d*x+c)/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBOSE 
)
 

Output:

1/d*(-2/a^3*((-1/2*(6*A*a^2*b+A*a*b^2-2*A*b^3-4*B*a^3-B*a^2*b)*a*b/(a-b)/( 
a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1/2*b*a*(6*A*a^2*b-A*a*b^2-2*A*b^3-4*B 
*a^3+B*a^2*b)/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-ta 
n(1/2*d*x+1/2*c)^2*b+a+b)^2+1/2*(6*A*a^4*b-5*A*a^2*b^3+2*A*b^5-2*B*a^5-B*a 
^3*b^2)/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1 
/2*c)/((a-b)*(a+b))^(1/2)))-A/a^3*ln(tan(1/2*d*x+1/2*c)-1)+A/a^3*ln(tan(1/ 
2*d*x+1/2*c)+1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 665 vs. \(2 (199) = 398\).

Time = 11.39 (sec) , antiderivative size = 1400, normalized size of antiderivative = 6.54 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))^3,x, algorithm="fri 
cas")
 

Output:

[1/4*((2*B*a^7 - 6*A*a^6*b + B*a^5*b^2 + 5*A*a^4*b^3 - 2*A*a^2*b^5 + (2*B* 
a^5*b^2 - 6*A*a^4*b^3 + B*a^3*b^4 + 5*A*a^2*b^5 - 2*A*b^7)*cos(d*x + c)^2 
+ 2*(2*B*a^6*b - 6*A*a^5*b^2 + B*a^4*b^3 + 5*A*a^3*b^4 - 2*A*a*b^6)*cos(d* 
x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + 
 c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2 
)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) + 2*(A*a^8 - 3*A*a^6*b^ 
2 + 3*A*a^4*b^4 - A*a^2*b^6 + (A*a^6*b^2 - 3*A*a^4*b^4 + 3*A*a^2*b^6 - A*b 
^8)*cos(d*x + c)^2 + 2*(A*a^7*b - 3*A*a^5*b^3 + 3*A*a^3*b^5 - A*a*b^7)*cos 
(d*x + c))*log(sin(d*x + c) + 1) - 2*(A*a^8 - 3*A*a^6*b^2 + 3*A*a^4*b^4 - 
A*a^2*b^6 + (A*a^6*b^2 - 3*A*a^4*b^4 + 3*A*a^2*b^6 - A*b^8)*cos(d*x + c)^2 
 + 2*(A*a^7*b - 3*A*a^5*b^3 + 3*A*a^3*b^5 - A*a*b^7)*cos(d*x + c))*log(-si 
n(d*x + c) + 1) - 2*(4*B*a^7*b - 6*A*a^6*b^2 - 5*B*a^5*b^3 + 9*A*a^4*b^4 + 
 B*a^3*b^5 - 3*A*a^2*b^6 + (3*B*a^6*b^2 - 5*A*a^5*b^3 - 3*B*a^4*b^4 + 7*A* 
a^3*b^5 - 2*A*a*b^7)*cos(d*x + c))*sin(d*x + c))/((a^9*b^2 - 3*a^7*b^4 + 3 
*a^5*b^6 - a^3*b^8)*d*cos(d*x + c)^2 + 2*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - 
 a^4*b^7)*d*cos(d*x + c) + (a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d), 1/ 
2*((2*B*a^7 - 6*A*a^6*b + B*a^5*b^2 + 5*A*a^4*b^3 - 2*A*a^2*b^5 + (2*B*a^5 
*b^2 - 6*A*a^4*b^3 + B*a^3*b^4 + 5*A*a^2*b^5 - 2*A*b^7)*cos(d*x + c)^2 + 2 
*(2*B*a^6*b - 6*A*a^5*b^2 + B*a^4*b^3 + 5*A*a^3*b^4 - 2*A*a*b^6)*cos(d*x + 
 c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(...
 

Sympy [F]

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

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

Output:

Integral((A + B*cos(c + d*x))*sec(c + d*x)/(a + b*cos(c + d*x))**3, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))^3,x, algorithm="max 
ima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (199) = 398\).

Time = 0.25 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.25 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {{\left (2 \, B a^{5} - 6 \, A a^{4} b + B a^{3} b^{2} + 5 \, A a^{2} b^{3} - 2 \, A b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {4 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}}}{d} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))^3,x, algorithm="gia 
c")
 

Output:

((2*B*a^5 - 6*A*a^4*b + B*a^3*b^2 + 5*A*a^2*b^3 - 2*A*b^5)*(pi*floor(1/2*( 
d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*c) - b*tan 
(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^7 - 2*a^5*b^2 + a^3*b^4)*sqrt(a^2 
 - b^2)) + A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - A*log(abs(tan(1/2*d* 
x + 1/2*c) - 1))/a^3 - (4*B*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 6*A*a^3*b^2*tan 
(1/2*d*x + 1/2*c)^3 - 3*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 5*A*a^2*b^3*tan 
(1/2*d*x + 1/2*c)^3 - B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 3*A*a*b^4*tan(1/2 
*d*x + 1/2*c)^3 - 2*A*b^5*tan(1/2*d*x + 1/2*c)^3 + 4*B*a^4*b*tan(1/2*d*x + 
 1/2*c) - 6*A*a^3*b^2*tan(1/2*d*x + 1/2*c) + 3*B*a^3*b^2*tan(1/2*d*x + 1/2 
*c) - 5*A*a^2*b^3*tan(1/2*d*x + 1/2*c) - B*a^2*b^3*tan(1/2*d*x + 1/2*c) + 
3*A*a*b^4*tan(1/2*d*x + 1/2*c) + 2*A*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - 2*a 
^4*b^2 + a^2*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a 
 + b)^2))/d
 

Mupad [B] (verification not implemented)

Time = 31.37 (sec) , antiderivative size = 6913, normalized size of antiderivative = 32.30 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

((tan(c/2 + (d*x)/2)^3*(2*A*b^4 - 6*A*a^2*b^2 + B*a^2*b^2 - A*a*b^3 + 4*B* 
a^3*b))/((a^2*b - a^3)*(a + b)^2) - (tan(c/2 + (d*x)/2)*(2*A*b^4 - 6*A*a^2 
*b^2 - B*a^2*b^2 + A*a*b^3 + 4*B*a^3*b))/((a + b)*(a^4 - 2*a^3*b + a^2*b^2 
)))/(d*(2*a*b + tan(c/2 + (d*x)/2)^2*(2*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^ 
4*(a^2 - 2*a*b + b^2) + a^2 + b^2)) - (A*atan(((A*((8*tan(c/2 + (d*x)/2)*( 
4*A^2*a^10 + 8*A^2*b^10 + 4*B^2*a^10 - 8*A^2*a*b^9 - 8*A^2*a^9*b - 32*A^2* 
a^2*b^8 + 32*A^2*a^3*b^7 + 57*A^2*a^4*b^6 - 48*A^2*a^5*b^5 - 52*A^2*a^6*b^ 
4 + 32*A^2*a^7*b^3 + 24*A^2*a^8*b^2 + B^2*a^6*b^4 + 4*B^2*a^8*b^2 - 24*A*B 
*a^9*b - 4*A*B*a^3*b^7 + 2*A*B*a^5*b^5 + 8*A*B*a^7*b^3))/(a^10*b + a^11 - 
a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2) + (A*(( 
8*(4*A*a^15 + 4*B*a^15 - 4*A*a^6*b^9 + 2*A*a^7*b^8 + 18*A*a^8*b^7 - 4*A*a^ 
9*b^6 - 36*A*a^10*b^5 + 6*A*a^11*b^4 + 34*A*a^12*b^3 - 8*A*a^13*b^2 - 2*B* 
a^8*b^7 + 2*B*a^9*b^6 + 6*B*a^12*b^3 - 6*B*a^13*b^2 - 12*A*a^14*b - 4*B*a^ 
14*b))/(a^12*b + a^13 - a^6*b^7 - a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10 
*b^3 - 3*a^11*b^2) + (8*A*tan(c/2 + (d*x)/2)*(8*a^15*b - 8*a^6*b^10 + 8*a^ 
7*b^9 + 32*a^8*b^8 - 32*a^9*b^7 - 48*a^10*b^6 + 48*a^11*b^5 + 32*a^12*b^4 
- 32*a^13*b^3 - 8*a^14*b^2))/(a^3*(a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a 
^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2))))/a^3)*1i)/a^3 + (A*((8*tan(c 
/2 + (d*x)/2)*(4*A^2*a^10 + 8*A^2*b^10 + 4*B^2*a^10 - 8*A^2*a*b^9 - 8*A^2* 
a^9*b - 32*A^2*a^2*b^8 + 32*A^2*a^3*b^7 + 57*A^2*a^4*b^6 - 48*A^2*a^5*b...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 568, normalized size of antiderivative = 2.65 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t(a**2 - b**2))*cos(c + d*x)*a**2*b**2 + 2*sqrt(a**2 - b**2)*atan((tan((c 
+ d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*b**4 - 4 
*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a** 
2 - b**2))*a**3*b + 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c 
+ d*x)/2)*b)/sqrt(a**2 - b**2))*a*b**3 - cos(c + d*x)*log(tan((c + d*x)/2) 
 - 1)*a**4*b + 2*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b**3 - cos(c 
+ d*x)*log(tan((c + d*x)/2) - 1)*b**5 + cos(c + d*x)*log(tan((c + d*x)/2) 
+ 1)*a**4*b - 2*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b**3 + cos(c + 
 d*x)*log(tan((c + d*x)/2) + 1)*b**5 - log(tan((c + d*x)/2) - 1)*a**5 + 2* 
log(tan((c + d*x)/2) - 1)*a**3*b**2 - log(tan((c + d*x)/2) - 1)*a*b**4 + l 
og(tan((c + d*x)/2) + 1)*a**5 - 2*log(tan((c + d*x)/2) + 1)*a**3*b**2 + lo 
g(tan((c + d*x)/2) + 1)*a*b**4 + sin(c + d*x)*a**3*b**2 - sin(c + d*x)*a*b 
**4)/(a**2*d*(cos(c + d*x)*a**4*b - 2*cos(c + d*x)*a**2*b**3 + cos(c + d*x 
)*b**5 + a**5 - 2*a**3*b**2 + a*b**4))