\(\int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\) [318]

Optimal result

Integrand size = 31, antiderivative size = 178 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=-\frac {\left (a^2+2 b^2\right ) (A b-a B) x}{2 a^4}+\frac {2 b^3 (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (2 a^2 A+3 A b^2-3 a b B\right ) \sin (c+d x)}{3 a^3 d}-\frac {(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 a d} \] Output:

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

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {6 \left (a^2+2 b^2\right ) (-A b+a B) (c+d x)-\frac {24 b^3 (A b-a B) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+3 a \left (3 a^2 A+4 A b^2-4 a b B\right ) \sin (c+d x)+3 a^2 (-A b+a B) \sin (2 (c+d x))+a^3 A \sin (3 (c+d x))}{12 a^4 d} \] Input:

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

Output:

(6*(a^2 + 2*b^2)*(-(A*b) + a*B)*(c + d*x) - (24*b^3*(A*b - a*B)*ArcTanh[(( 
-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 3*a*(3*a^2*A 
 + 4*A*b^2 - 4*a*b*B)*Sin[c + d*x] + 3*a^2*(-(A*b) + a*B)*Sin[2*(c + d*x)] 
 + a^3*A*Sin[3*(c + d*x)])/(12*a^4*d)
 

Rubi [A] (verified)

Time = 1.36 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4522, 3042, 4592, 3042, 4592, 27, 3042, 4407, 3042, 4318, 3042, 3138, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4522

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4592

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4592

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4407

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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 4318

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[1/b   Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, 
f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4407

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
 (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a   Int[Csc[e + f* 
x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0]
 

rule 4522

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

rule 4592

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

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.36

method	result	size
derivativedivides	\(\frac {\frac {2 b^{3} \left (A b -B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \left (\frac {\left (-a^{3} A -\frac {1}{2} A \,a^{2} b -A a \,b^{2}+\frac {1}{2} B \,a^{3}+B \,a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {2}{3} a^{3} A -2 A a \,b^{2}+2 B \,a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{3} A -A a \,b^{2}+B \,a^{2} b +\frac {1}{2} A \,a^{2} b -\frac {1}{2} B \,a^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {\left (A \,a^{2} b +2 A \,b^{3}-B \,a^{3}-2 B a \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{a^{4}}}{d}\)	\(242\)
default	\(\frac {\frac {2 b^{3} \left (A b -B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \left (\frac {\left (-a^{3} A -\frac {1}{2} A \,a^{2} b -A a \,b^{2}+\frac {1}{2} B \,a^{3}+B \,a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {2}{3} a^{3} A -2 A a \,b^{2}+2 B \,a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{3} A -A a \,b^{2}+B \,a^{2} b +\frac {1}{2} A \,a^{2} b -\frac {1}{2} B \,a^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {\left (A \,a^{2} b +2 A \,b^{3}-B \,a^{3}-2 B a \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{a^{4}}}{d}\)	\(242\)
risch	\(-\frac {x A b}{2 a^{2}}-\frac {x A \,b^{3}}{a^{4}}+\frac {x B}{2 a}+\frac {x B \,b^{2}}{a^{3}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,b^{2}}{2 a^{3} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} B b}{2 a^{2} d}-\frac {3 i A \,{\mathrm e}^{i \left (d x +c \right )}}{8 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{2}}{2 a^{3} d}+\frac {3 i A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} B b}{2 a^{2} d}+\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}-\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}+\frac {A \sin \left (3 d x +3 c \right )}{12 a d}-\frac {\sin \left (2 d x +2 c \right ) A b}{4 a^{2} d}+\frac {\sin \left (2 d x +2 c \right ) B}{4 a d}\)	\(511\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 547, normalized size of antiderivative = 3.07 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\left [\frac {3 \, {\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} d x - 3 \, {\left (B a b^{3} - A b^{4}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (4 \, A a^{5} - 6 \, B a^{4} b + 2 \, A a^{3} b^{2} + 6 \, B a^{2} b^{3} - 6 \, A a b^{4} + 2 \, {\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}, \frac {3 \, {\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} d x - 6 \, {\left (B a b^{3} - A b^{4}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (4 \, A a^{5} - 6 \, B a^{4} b + 2 \, A a^{3} b^{2} + 6 \, B a^{2} b^{3} - 6 \, A a b^{4} + 2 \, {\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}\right ] \] Input:

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

Output:

[1/6*(3*(B*a^5 - A*a^4*b + B*a^3*b^2 - A*a^2*b^3 - 2*B*a*b^4 + 2*A*b^5)*d* 
x - 3*(B*a*b^3 - A*b^4)*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2 
*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) 
 + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) + (4*A*a^ 
5 - 6*B*a^4*b + 2*A*a^3*b^2 + 6*B*a^2*b^3 - 6*A*a*b^4 + 2*(A*a^5 - A*a^3*b 
^2)*cos(d*x + c)^2 + 3*(B*a^5 - A*a^4*b - B*a^3*b^2 + A*a^2*b^3)*cos(d*x + 
 c))*sin(d*x + c))/((a^6 - a^4*b^2)*d), 1/6*(3*(B*a^5 - A*a^4*b + B*a^3*b^ 
2 - A*a^2*b^3 - 2*B*a*b^4 + 2*A*b^5)*d*x - 6*(B*a*b^3 - A*b^4)*sqrt(-a^2 + 
 b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + 
 c))) + (4*A*a^5 - 6*B*a^4*b + 2*A*a^3*b^2 + 6*B*a^2*b^3 - 6*A*a*b^4 + 2*( 
A*a^5 - A*a^3*b^2)*cos(d*x + c)^2 + 3*(B*a^5 - A*a^4*b - B*a^3*b^2 + A*a^2 
*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),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*a^2-4*b^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. 360 vs. \(2 (161) = 322\).

Time = 0.21 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.02 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\frac {3 \, {\left (B a^{3} - A a^{2} b + 2 \, B a b^{2} - 2 \, A b^{3}\right )} {\left (d x + c\right )}}{a^{4}} - \frac {12 \, {\left (B a b^{3} - A b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{4}} + \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}}}{6 \, d} \] Input:

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

Output:

1/6*(3*(B*a^3 - A*a^2*b + 2*B*a*b^2 - 2*A*b^3)*(d*x + c)/a^4 - 12*(B*a*b^3 
 - A*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*t 
an(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/(sqrt(-a^ 
2 + b^2)*a^4) + 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^2*tan(1/2*d*x + 
1/2*c)^5 + 3*A*a*b*tan(1/2*d*x + 1/2*c)^5 - 6*B*a*b*tan(1/2*d*x + 1/2*c)^5 
 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*B* 
a*b*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan 
(1/2*d*x + 1/2*c) + 3*B*a^2*tan(1/2*d*x + 1/2*c) - 3*A*a*b*tan(1/2*d*x + 1 
/2*c) - 6*B*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*d*x + 1/2*c))/((tan 
(1/2*d*x + 1/2*c)^2 + 1)^3*a^3))/d
 

Mupad [B] (verification not implemented)

Time = 16.75 (sec) , antiderivative size = 4572, normalized size of antiderivative = 25.69 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

((tan(c/2 + (d*x)/2)*(2*A*a^2 + 2*A*b^2 + B*a^2 - A*a*b - 2*B*a*b))/a^3 + 
(tan(c/2 + (d*x)/2)^5*(2*A*a^2 + 2*A*b^2 - B*a^2 + A*a*b - 2*B*a*b))/a^3 + 
 (4*tan(c/2 + (d*x)/2)^3*(A*a^2 + 3*A*b^2 - 3*B*a*b))/(3*a^3))/(d*(3*tan(c 
/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) - (a 
tan((((a^2 + 2*b^2)*(A*b - B*a)*((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - B^2*a^ 
9 - 16*A^2*a*b^8 + 3*B^2*a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2* 
a^4*b^5 - 7*A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2*a^2*b^7 - 16 
*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 13*B^2*a^6*b^3 - 7*B^2*a^ 
7*b^2 - 16*A*B*a*b^8 + 2*A*B*a^8*b + 32*A*B*a^2*b^7 - 32*A*B*a^3*b^6 + 32* 
A*B*a^4*b^5 - 26*A*B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^7*b^2))/a^6 - ((a^ 
2 + 2*b^2)*(A*b - B*a)*((8*(2*B*a^13 - 4*A*a^8*b^5 + 6*A*a^9*b^4 - 2*A*a^1 
0*b^3 + 2*A*a^11*b^2 + 4*B*a^9*b^4 - 6*B*a^10*b^3 + 2*B*a^11*b^2 - 2*A*a^1 
2*b - 2*B*a^12*b))/a^9 - (tan(c/2 + (d*x)/2)*(a^2 + 2*b^2)*(A*b - B*a)*(8* 
a^10*b + 8*a^8*b^3 - 16*a^9*b^2)*4i)/a^10)*1i)/(2*a^4)))/(2*a^4) + ((a^2 + 
 2*b^2)*(A*b - B*a)*((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a 
*b^8 + 3*B^2*a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2*a^4*b^5 - 7* 
A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2*a^2*b^7 - 16*B^2*a^3*b^6 
 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 13*B^2*a^6*b^3 - 7*B^2*a^7*b^2 - 16*A 
*B*a*b^8 + 2*A*B*a^8*b + 32*A*B*a^2*b^7 - 32*A*B*a^3*b^6 + 32*A*B*a^4*b^5 
- 26*A*B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^7*b^2))/a^6 + ((a^2 + 2*b^2...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.13 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\sin \left (d x +c \right ) \left (-\sin \left (d x +c \right )^{2}+3\right )}{3 d} \] Input:

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

Output:

(sin(c + d*x)*( - sin(c + d*x)**2 + 3))/(3*d)