\(\int \frac {(A+C \cos ^2(c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [582]

Optimal result

Integrand size = 31, antiderivative size = 211 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {b \left (5 a^2 A b^2-2 A b^4-3 a^4 (2 A+C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac {A \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (2 A b^4-a^4 C-a^2 b^2 (5 A+2 C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

b*(5*A*a^2*b^2-2*A*b^4-3*a^4*(2*A+C))*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*(A*b^2+C*a^2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^2-1/2*(2*A*b^4- 
a^4*C-a^2*b^2*(5*A+2*C))*sin(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 4.78 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.94 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\cos (c+d x) (C \cos (c+d x)+A \sec (c+d x)) \left (-4 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 b \left (-5 a^2 A b^2+2 A b^4+3 a^4 (2 A+C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (b \sin (c)+(-a+b \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}}\right ) (i \cos (c)+\sin (c))}{\left (a^2-b^2\right )^2 \sqrt {\left (-a^2+b^2\right ) (\cos (c)-i \sin (c))^2}}-\frac {a \sec (c) \left (\left (2 a^2+b^2\right ) \left (-2 A b^4+a^4 C+a^2 b^2 (5 A+2 C)\right ) \sin (c)+b \left (-a \left (-7 A b^4+4 a^4 C+a^2 b^2 (16 A+5 C)\right ) \sin (d x)+b \left (a b \left (-A b^2+a^2 (4 A+3 C)\right ) \sin (2 c+d x)-\left (-2 A b^4+a^4 C+a^2 b^2 (5 A+2 C)\right ) \sin (c+2 d x)\right )\right )\right )}{b \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}\right )}{2 a^3 d (2 A+C+C \cos (2 (c+d x)))} \] Input:

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^3,x]
 

Output:

(Cos[c + d*x]*(C*Cos[c + d*x] + A*Sec[c + d*x])*(-4*A*Log[Cos[(c + d*x)/2] 
 - Sin[(c + d*x)/2]] + 4*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (4*b 
*(-5*a^2*A*b^2 + 2*A*b^4 + 3*a^4*(2*A + C))*ArcTan[((I*Cos[c] + Sin[c])*(b 
*Sin[c] + (-a + b*Cos[c])*Tan[(d*x)/2]))/Sqrt[-((a^2 - b^2)*(Cos[c] - I*Si 
n[c])^2)]]*(I*Cos[c] + Sin[c]))/((a^2 - b^2)^2*Sqrt[(-a^2 + b^2)*(Cos[c] - 
 I*Sin[c])^2]) - (a*Sec[c]*((2*a^2 + b^2)*(-2*A*b^4 + a^4*C + a^2*b^2*(5*A 
 + 2*C))*Sin[c] + b*(-(a*(-7*A*b^4 + 4*a^4*C + a^2*b^2*(16*A + 5*C))*Sin[d 
*x]) + b*(a*b*(-(A*b^2) + a^2*(4*A + 3*C))*Sin[2*c + d*x] - (-2*A*b^4 + a^ 
4*C + a^2*b^2*(5*A + 2*C))*Sin[c + 2*d*x]))))/(b*(a^2 - b^2)^2*(a + b*Cos[ 
c + d*x])^2)))/(2*a^3*d*(2*A + C + C*Cos[2*(c + d*x)]))
 

Rubi [A] (verified)

Time = 1.26 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.22, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3535, 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.

\(\displaystyle \int \frac {\sec (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4257

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

Input:

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^3,x]
 

Output:

((A*b^2 + a^2*C)*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) 
+ (((2*b*(5*a^2*A*b^2 - 2*A*b^4 - 3*a^4*(2*A + C))*ArcTan[(Sqrt[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)) - ((2*A*b^4 - a^4*C - 
a^2*b^2*(5*A + 2*C))*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 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 3535

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S 
in[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[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n 
+ 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d 
*(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, 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 = 0.85 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.48

method	result	size
derivativedivides	\(\frac {-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}-\frac {2 \left (\frac {-\frac {\left (6 A \,a^{2} b^{2}+a A \,b^{3}-2 A \,b^{4}+2 a^{4} C +a^{3} b C +2 C \,a^{2} b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a^{2}+2 a b +b^{2}\right ) \left (a -b \right )}-\frac {a \left (6 A \,a^{2} b^{2}-a A \,b^{3}-2 A \,b^{4}+2 a^{4} C -a^{3} b C +2 C \,a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a -b \right )^{2} \left (a +b \right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {b \left (6 A \,a^{4}-5 A \,a^{2} b^{2}+2 A \,b^{4}+3 a^{4} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}}{d}\)	\(313\)
default	\(\frac {-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}-\frac {2 \left (\frac {-\frac {\left (6 A \,a^{2} b^{2}+a A \,b^{3}-2 A \,b^{4}+2 a^{4} C +a^{3} b C +2 C \,a^{2} b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a^{2}+2 a b +b^{2}\right ) \left (a -b \right )}-\frac {a \left (6 A \,a^{2} b^{2}-a A \,b^{3}-2 A \,b^{4}+2 a^{4} C -a^{3} b C +2 C \,a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a -b \right )^{2} \left (a +b \right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {b \left (6 A \,a^{4}-5 A \,a^{2} b^{2}+2 A \,b^{4}+3 a^{4} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}}{d}\)	\(313\)
risch	\(\frac {i \left (4 A \,a^{3} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-A a \,b^{5} {\mathrm e}^{3 i \left (d x +c \right )}+3 C \,a^{3} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+10 A \,a^{4} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+A \,a^{2} b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-2 A \,b^{6} {\mathrm e}^{2 i \left (d x +c \right )}+2 C \,a^{6} {\mathrm e}^{2 i \left (d x +c \right )}+5 C \,a^{4} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 C \,a^{2} b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+16 A \,a^{3} b^{3} {\mathrm e}^{i \left (d x +c \right )}-7 A a \,b^{5} {\mathrm e}^{i \left (d x +c \right )}+4 C \,a^{5} b \,{\mathrm e}^{i \left (d x +c \right )}+5 C \,a^{3} b^{3} {\mathrm e}^{i \left (d x +c \right )}+5 a^{2} A \,b^{4}-2 A \,b^{6}+a^{4} b^{2} C +2 C \,a^{2} b^{4}\right )}{b \left (a^{2}-b^{2}\right )^{2} d \,a^{2} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2}}-\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {5 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}-\frac {b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{3}}-\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {5 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}+\frac {b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{3}}+\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}\)	\(1036\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^3,x,method=_RETURNVERBO 
SE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 619 vs. \(2 (197) = 394\).

Time = 4.58 (sec) , antiderivative size = 1308, normalized size of antiderivative = 6.20 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^3,x, algorithm="f 
ricas")
 

Output:

[-1/4*((3*(2*A + C)*a^6*b - 5*A*a^4*b^3 + 2*A*a^2*b^5 + (3*(2*A + C)*a^4*b 
^3 - 5*A*a^2*b^5 + 2*A*b^7)*cos(d*x + c)^2 + 2*(3*(2*A + C)*a^5*b^2 - 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(-sin(d*x + c) + 1) - 2*(2*C*a^8 + (6*A - C)*a^6*b^ 
2 - (9*A + C)*a^4*b^4 + 3*A*a^2*b^6 + (C*a^7*b + (5*A + C)*a^5*b^3 - (7*A 
+ 2*C)*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*((3*(2*A + C)*a^6*b - 5*A*a^4*b^3 + 2*A*a^2*b^5 + (3*(2*A + C)*a 
^4*b^3 - 5*A*a^2*b^5 + 2*A*b^7)*cos(d*x + c)^2 + 2*(3*(2*A + C)*a^5*b^2 - 
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(d*x + c))) - (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(...
 

Sympy [F]

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

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

Output:

Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)/(a + b*cos(c + d*x))**3, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^3,x, algorithm="m 
axima")
 

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. 504 vs. \(2 (197) = 394\).

Time = 0.20 (sec) , antiderivative size = 504, normalized size of antiderivative = 2.39 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^3,x, algorithm="g 
iac")
 

Output:

-((6*A*a^4*b + 3*C*a^4*b - 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 - (2*C*a^5*tan(1/2*d*x + 1/2*c)^3 - C*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 + C*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 - 2*C*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 + 2 
*C*a^5*tan(1/2*d*x + 1/2*c) + C*a^4*b*tan(1/2*d*x + 1/2*c) + 6*A*a^3*b^2*t 
an(1/2*d*x + 1/2*c) + C*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) + 2*C*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*ta 
n(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 = 5.75 (sec) , antiderivative size = 6574, normalized size of antiderivative = 31.16 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

(A*atan(((A*((A*((8*(4*A*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 
 + 6*C*a^9*b^6 - 6*C*a^10*b^5 - 12*C*a^11*b^4 + 12*C*a^12*b^3 + 6*C*a^13*b 
^2 - 12*A*a^14*b - 6*C*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 + 4 
8*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 - (8*tan(c/2 + (d*x)/2)*(4*A^2*a^10 + 8*A^2*b^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 + 9*C^2*a^8*b^2 + 12*A*C 
*a^4*b^6 - 30*A*C*a^6*b^4 + 36*A*C*a^8*b^2))/(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))*1i)/a^3 - (A*((A*( 
(8*(4*A*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 + 6*C*a^9*b^6 - 
6*C*a^10*b^5 - 12*C*a^11*b^4 + 12*C*a^12*b^3 + 6*C*a^13*b^2 - 12*A*a^14*b 
- 6*C*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^1 
0 + 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^...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 1747, normalized size of antiderivative = 8.28 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq 
rt(a**2 - b**2))*cos(c + d*x)*a**5*b**2 - 12*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)*a**4*b 
**2*c + 20*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)*a**3*b**4 - 8*sqrt(a**2 - b**2)*atan((ta 
n((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a*b 
**6 + 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/ 
sqrt(a**2 - b**2))*sin(c + d*x)**2*a**4*b**3 + 6*sqrt(a**2 - b**2)*atan((t 
an((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2 
*a**3*b**3*c - 10*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d* 
x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**2*b**5 + 4*sqrt(a**2 - b**2 
)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c 
+ d*x)**2*b**7 - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + 
d*x)/2)*b)/sqrt(a**2 - b**2))*a**6*b - 6*sqrt(a**2 - b**2)*atan((tan((c + 
d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**5*b*c - 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**4*b**3 - 6*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**3*c + 6*sqrt(a**2 - b**2)*atan((tan((c + d 
*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**2*b**5 - 4*sqrt(a**2 
- b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2...