\(\int \frac {\cos ^4(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\) [800]

Optimal result

Integrand size = 40, antiderivative size = 178 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=-\frac {\left (a^2+2 b^2\right ) (b B-a C) x}{2 a^4}+\frac {2 b^3 (b B-a C) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (2 a^2 B+3 b^2 B-3 a b C\right ) \sin (c+d x)}{3 a^3 d}-\frac {(b B-a C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {B \cos ^2(c+d x) \sin (c+d x)}{3 a d} \] Output:

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

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {6 \left (a^2+2 b^2\right ) (-b B+a C) (c+d x)-\frac {24 b^3 (b B-a C) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+3 a \left (3 a^2 B+4 b^2 B-4 a b C\right ) \sin (c+d x)+3 a^2 (-b B+a C) \sin (2 (c+d x))+a^3 B \sin (3 (c+d x))}{12 a^4 d} \] Input:

Integrate[(Cos[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[ 
c + d*x]),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.51 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.10, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4560, 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.

\(\displaystyle \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4522

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4592

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4592

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4407

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

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

Input:

Int[(Cos[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d* 
x]),x]
 

Output:

(B*Cos[c + d*x]^2*Sin[c + d*x])/(3*a*d) - ((3*(b*B - a*C)*Cos[c + d*x]*Sin 
[c + d*x])/(2*a*d) - ((-3*(((a^2 + 2*b^2)*(b*B - a*C)*x)/a - (4*b^3*(b*B - 
 a*C)*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*B + 3*b^2*B - 3*a*b*C)*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 4560

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

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 = 0.78 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.36

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

Input:

int(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x,method=_ 
RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 547, normalized size of antiderivative = 3.07 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\left [\frac {3 \, {\left (C a^{5} - B a^{4} b + C a^{3} b^{2} - B a^{2} b^{3} - 2 \, C a b^{4} + 2 \, B b^{5}\right )} d x - 3 \, {\left (C a b^{3} - B b^{4}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (4 \, B a^{5} - 6 \, C a^{4} b + 2 \, B a^{3} b^{2} + 6 \, C a^{2} b^{3} - 6 \, B a b^{4} + 2 \, {\left (B a^{5} - B a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (C a^{5} - B a^{4} b - C a^{3} b^{2} + B a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}, \frac {3 \, {\left (C a^{5} - B a^{4} b + C a^{3} b^{2} - B a^{2} b^{3} - 2 \, C a b^{4} + 2 \, B b^{5}\right )} d x - 6 \, {\left (C a b^{3} - B b^{4}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (4 \, B a^{5} - 6 \, C a^{4} b + 2 \, B a^{3} b^{2} + 6 \, C a^{2} b^{3} - 6 \, B a b^{4} + 2 \, {\left (B a^{5} - B a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (C a^{5} - B a^{4} b - C a^{3} b^{2} + B a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - a^{4} b^{2}\right )} d}\right ] \] Input:

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, a 
lgorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \] Input:

integrate(cos(d*x+c)**4*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c)),x)
 

Output:

Integral((B + C*sec(c + d*x))*cos(c + d*x)**4*sec(c + d*x)/(a + b*sec(c + 
d*x)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, a 
lgorithm="maxima")
 

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 = 0.30 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.02 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\frac {3 \, {\left (C a^{3} - B a^{2} b + 2 \, C a b^{2} - 2 \, B b^{3}\right )} {\left (d x + c\right )}}{a^{4}} - \frac {12 \, {\left (C a b^{3} - B b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{4}} + \frac {2 \, {\left (6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}}}{6 \, d} \] Input:

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, a 
lgorithm="giac")
 

Output:

1/6*(3*(C*a^3 - B*a^2*b + 2*C*a*b^2 - 2*B*b^3)*(d*x + c)/a^4 - 12*(C*a*b^3 
 - B*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*B*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^2*tan(1/2*d*x + 
1/2*c)^5 + 3*B*a*b*tan(1/2*d*x + 1/2*c)^5 - 6*C*a*b*tan(1/2*d*x + 1/2*c)^5 
 + 6*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*C* 
a*b*tan(1/2*d*x + 1/2*c)^3 + 12*B*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^2*tan 
(1/2*d*x + 1/2*c) + 3*C*a^2*tan(1/2*d*x + 1/2*c) - 3*B*a*b*tan(1/2*d*x + 1 
/2*c) - 6*C*a*b*tan(1/2*d*x + 1/2*c) + 6*B*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 = 17.72 (sec) , antiderivative size = 4572, normalized size of antiderivative = 25.69 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \] Input:

int((cos(c + d*x)^4*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d* 
x)),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.22 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {-12 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) a \,b^{3} c +12 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) b^{5}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{5} c -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4} b^{2}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b^{2} c +3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{4}-2 \sin \left (d x +c \right )^{3} a^{5} b +2 \sin \left (d x +c \right )^{3} a^{3} b^{3}+6 \sin \left (d x +c \right ) a^{5} b -6 \sin \left (d x +c \right ) a^{4} b c +6 \sin \left (d x +c \right ) a^{2} b^{3} c -6 \sin \left (d x +c \right ) a \,b^{5}+3 a^{5} c^{2}+3 a^{5} c d x -3 a^{4} b^{2} c -3 a^{4} b^{2} d x +3 a^{3} b^{2} c^{2}+3 a^{3} b^{2} c d x -3 a^{2} b^{4} c -3 a^{2} b^{4} d x -6 a \,b^{4} c^{2}-6 a \,b^{4} c d x +6 b^{6} c +6 b^{6} d x}{6 a^{4} d \left (a^{2}-b^{2}\right )} \] Input:

int(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x)
 

Output:

( - 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*b**3*c + 12*sqrt( - a**2 + b**2)*atan((tan((c + d 
*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*b**5 + 3*cos(c + d*x) 
*sin(c + d*x)*a**5*c - 3*cos(c + d*x)*sin(c + d*x)*a**4*b**2 - 3*cos(c + d 
*x)*sin(c + d*x)*a**3*b**2*c + 3*cos(c + d*x)*sin(c + d*x)*a**2*b**4 - 2*s 
in(c + d*x)**3*a**5*b + 2*sin(c + d*x)**3*a**3*b**3 + 6*sin(c + d*x)*a**5* 
b - 6*sin(c + d*x)*a**4*b*c + 6*sin(c + d*x)*a**2*b**3*c - 6*sin(c + d*x)* 
a*b**5 + 3*a**5*c**2 + 3*a**5*c*d*x - 3*a**4*b**2*c - 3*a**4*b**2*d*x + 3* 
a**3*b**2*c**2 + 3*a**3*b**2*c*d*x - 3*a**2*b**4*c - 3*a**2*b**4*d*x - 6*a 
*b**4*c**2 - 6*a*b**4*c*d*x + 6*b**6*c + 6*b**6*d*x)/(6*a**4*d*(a**2 - b** 
2))