\(\int \frac {\cos ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\) [319]

Optimal result

Integrand size = 31, antiderivative size = 240 \[ \int \frac {\cos ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\left (3 a^4 A+4 a^2 A b^2+8 A b^4-4 a^3 b B-8 a b^3 B\right ) x}{8 a^5}-\frac {2 b^4 (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 \sqrt {a-b} \sqrt {a+b} d}-\frac {\left (2 a^2+3 b^2\right ) (A b-a B) \sin (c+d x)}{3 a^4 d}+\frac {\left (3 a^2 A+4 A b^2-4 a b B\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac {(A b-a B) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {A \cos ^3(c+d x) \sin (c+d x)}{4 a d} \] Output:

1/8*(3*A*a^4+4*A*a^2*b^2+8*A*b^4-4*B*a^3*b-8*B*a*b^3)*x/a^5-2*b^4*(A*b-B*a 
)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^5/(a-b)^(1/2)/(a+b 
)^(1/2)/d-1/3*(2*a^2+3*b^2)*(A*b-B*a)*sin(d*x+c)/a^4/d+1/8*(3*A*a^2+4*A*b^ 
2-4*B*a*b)*cos(d*x+c)*sin(d*x+c)/a^3/d-1/3*(A*b-B*a)*cos(d*x+c)^2*sin(d*x+ 
c)/a^2/d+1/4*A*cos(d*x+c)^3*sin(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {12 \left (3 a^4 A+4 a^2 A b^2+8 A b^4-4 a^3 b B-8 a b^3 B\right ) (c+d x)+\frac {192 b^4 (A b-a B) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+24 a \left (3 a^2+4 b^2\right ) (-A b+a B) \sin (c+d x)+24 a^2 \left (a^2 A+A b^2-a b B\right ) \sin (2 (c+d x))+8 a^3 (-A b+a B) \sin (3 (c+d x))+3 a^4 A \sin (4 (c+d x))}{96 a^5 d} \] Input:

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

Output:

(12*(3*a^4*A + 4*a^2*A*b^2 + 8*A*b^4 - 4*a^3*b*B - 8*a*b^3*B)*(c + d*x) + 
(192*b^4*(A*b - a*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]) 
/Sqrt[a^2 - b^2] + 24*a*(3*a^2 + 4*b^2)*(-(A*b) + a*B)*Sin[c + d*x] + 24*a 
^2*(a^2*A + A*b^2 - a*b*B)*Sin[2*(c + d*x)] + 8*a^3*(-(A*b) + a*B)*Sin[3*( 
c + d*x)] + 3*a^4*A*Sin[4*(c + d*x)])/(96*a^5*d)
 

Rubi [A] (verified)

Time = 1.97 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.10, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.516, Rules used = {3042, 4522, 3042, 4592, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4522

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4592

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\int \frac {\cos ^2(c+d x) \left (-8 b (A b-a B) \sec ^2(c+d x)+a (A b+8 a B) \sec (c+d x)+3 \left (3 A a^2+4 b (A b-a B)\right )\right )}{a+b \sec (c+d x)}dx}{3 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\int \frac {-8 b (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (A b+8 a B) \csc \left (c+d x+\frac {\pi }{2}\right )+3 \left (3 A a^2+4 b (A b-a B)\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{4 a}\)

\(\Big \downarrow \) 4592

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (3 a^2 A-4 a b B+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {\cos (c+d x) \left (-3 b \left (3 A a^2-4 b B a+4 A b^2\right ) \sec ^2(c+d x)-a \left (9 A a^2+4 b B a-4 A b^2\right ) \sec (c+d x)+8 \left (2 a^2+3 b^2\right ) (A b-a B)\right )}{a+b \sec (c+d x)}dx}{2 a}}{3 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (3 a^2 A-4 a b B+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\int \frac {-3 b \left (3 A a^2-4 b B a+4 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-a \left (9 A a^2+4 b B a-4 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+8 \left (2 a^2+3 b^2\right ) (A b-a B)}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}}{3 a}}{4 a}\)

\(\Big \downarrow \) 4592

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (3 a^2 A-4 a b B+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 \left (2 a^2+3 b^2\right ) (A b-a B) \sin (c+d x)}{a d}-\frac {\int \frac {3 \left (3 A a^4-4 b B a^3+4 A b^2 a^2-8 b^3 B a+b \left (3 A a^2-4 b B a+4 A b^2\right ) \sec (c+d x) a+8 A b^4\right )}{a+b \sec (c+d x)}dx}{a}}{2 a}}{3 a}}{4 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (3 a^2 A-4 a b B+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 \left (2 a^2+3 b^2\right ) (A b-a B) \sin (c+d x)}{a d}-\frac {3 \int \frac {3 A a^4-4 b B a^3+4 A b^2 a^2-8 b^3 B a+b \left (3 A a^2-4 b B a+4 A b^2\right ) \sec (c+d x) a+8 A b^4}{a+b \sec (c+d x)}dx}{a}}{2 a}}{3 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (3 a^2 A-4 a b B+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 \left (2 a^2+3 b^2\right ) (A b-a B) \sin (c+d x)}{a d}-\frac {3 \int \frac {3 A a^4-4 b B a^3+4 A b^2 a^2-8 b^3 B a+b \left (3 A a^2-4 b B a+4 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+8 A b^4}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{2 a}}{3 a}}{4 a}\)

\(\Big \downarrow \) 4407

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (3 a^2 A-4 a b B+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 \left (2 a^2+3 b^2\right ) (A b-a B) \sin (c+d x)}{a d}-\frac {3 \left (\frac {x \left (3 a^4 A-4 a^3 b B+4 a^2 A b^2-8 a b^3 B+8 A b^4\right )}{a}-\frac {8 b^4 (A b-a B) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (3 a^2 A-4 a b B+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 \left (2 a^2+3 b^2\right ) (A b-a B) \sin (c+d x)}{a d}-\frac {3 \left (\frac {x \left (3 a^4 A-4 a^3 b B+4 a^2 A b^2-8 a b^3 B+8 A b^4\right )}{a}-\frac {8 b^4 (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\)

\(\Big \downarrow \) 4318

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (3 a^2 A-4 a b B+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 \left (2 a^2+3 b^2\right ) (A b-a B) \sin (c+d x)}{a d}-\frac {3 \left (\frac {x \left (3 a^4 A-4 a^3 b B+4 a^2 A b^2-8 a b^3 B+8 A b^4\right )}{a}-\frac {8 b^3 (A b-a B) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (3 a^2 A-4 a b B+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 \left (2 a^2+3 b^2\right ) (A b-a B) \sin (c+d x)}{a d}-\frac {3 \left (\frac {x \left (3 a^4 A-4 a^3 b B+4 a^2 A b^2-8 a b^3 B+8 A b^4\right )}{a}-\frac {8 b^3 (A b-a B) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (3 a^2 A-4 a b B+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 \left (2 a^2+3 b^2\right ) (A b-a B) \sin (c+d x)}{a d}-\frac {3 \left (\frac {x \left (3 a^4 A-4 a^3 b B+4 a^2 A b^2-8 a b^3 B+8 A b^4\right )}{a}-\frac {16 b^3 (A b-a B) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}}{2 a}}{3 a}}{4 a}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {A \sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 a d}-\frac {\frac {3 \left (3 a^2 A-4 a b B+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\frac {8 \left (2 a^2+3 b^2\right ) (A b-a B) \sin (c+d x)}{a d}-\frac {3 \left (\frac {x \left (3 a^4 A-4 a^3 b B+4 a^2 A b^2-8 a b^3 B+8 A b^4\right )}{a}-\frac {16 b^4 (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}}{4 a}\)

Input:

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

Output:

(A*Cos[c + d*x]^3*Sin[c + d*x])/(4*a*d) - ((4*(A*b - a*B)*Cos[c + d*x]^2*S 
in[c + d*x])/(3*a*d) - ((3*(3*a^2*A + 4*A*b^2 - 4*a*b*B)*Cos[c + d*x]*Sin[ 
c + d*x])/(2*a*d) - ((-3*(((3*a^4*A + 4*a^2*A*b^2 + 8*A*b^4 - 4*a^3*b*B - 
8*a*b^3*B)*x)/a - (16*b^4*(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 + (8*(2*a^2 + 3*b^2)*(A 
*b - a*B)*Sin[c + d*x])/(a*d))/(2*a))/(3*a))/(4*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 = 0.86 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.59

method	result	size
derivativedivides	\(\frac {-\frac {2 b^{4} \left (A b -B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{5} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\frac {2 \left (\left (-\frac {5}{8} a^{4} A -A \,a^{3} b -\frac {1}{2} A \,a^{2} b^{2}-A a \,b^{3}+B \,a^{4}+\frac {1}{2} a^{3} b B +B \,a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (\frac {3}{8} a^{4} A -\frac {5}{3} A \,a^{3} b -3 A a \,b^{3}+\frac {5}{3} B \,a^{4}+3 B \,a^{2} b^{2}-\frac {1}{2} A \,a^{2} b^{2}+\frac {1}{2} a^{3} b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {3}{8} a^{4} A +\frac {1}{2} A \,a^{2} b^{2}-\frac {1}{2} a^{3} b B -\frac {5}{3} A \,a^{3} b -3 A a \,b^{3}+\frac {5}{3} B \,a^{4}+3 B \,a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {5}{8} a^{4} A +\frac {1}{2} A \,a^{2} b^{2}-\frac {1}{2} a^{3} b B -A \,a^{3} b -A a \,b^{3}+B \,a^{4}+B \,a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {\left (3 a^{4} A +4 A \,a^{2} b^{2}+8 A \,b^{4}-4 a^{3} b B -8 a \,b^{3} B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{5}}}{d}\)	\(381\)
default	\(\frac {-\frac {2 b^{4} \left (A b -B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{5} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\frac {2 \left (\left (-\frac {5}{8} a^{4} A -A \,a^{3} b -\frac {1}{2} A \,a^{2} b^{2}-A a \,b^{3}+B \,a^{4}+\frac {1}{2} a^{3} b B +B \,a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (\frac {3}{8} a^{4} A -\frac {5}{3} A \,a^{3} b -3 A a \,b^{3}+\frac {5}{3} B \,a^{4}+3 B \,a^{2} b^{2}-\frac {1}{2} A \,a^{2} b^{2}+\frac {1}{2} a^{3} b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {3}{8} a^{4} A +\frac {1}{2} A \,a^{2} b^{2}-\frac {1}{2} a^{3} b B -\frac {5}{3} A \,a^{3} b -3 A a \,b^{3}+\frac {5}{3} B \,a^{4}+3 B \,a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {5}{8} a^{4} A +\frac {1}{2} A \,a^{2} b^{2}-\frac {1}{2} a^{3} b B -A \,a^{3} b -A a \,b^{3}+B \,a^{4}+B \,a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {\left (3 a^{4} A +4 A \,a^{2} b^{2}+8 A \,b^{4}-4 a^{3} b B -8 a \,b^{3} B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{5}}}{d}\)	\(381\)
risch	\(\frac {3 A x}{8 a}+\frac {x A \,b^{2}}{2 a^{3}}+\frac {x A \,b^{4}}{a^{5}}-\frac {b B x}{2 a^{2}}-\frac {x \,b^{3} B}{a^{4}}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B}{8 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,b^{2}}{2 a^{3} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{3}}{2 a^{4} d}+\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A b}{8 a^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,b^{3}}{2 a^{4} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,b^{2}}{2 a^{3} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B}{8 a d}-\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A b}{8 a^{2} d}+\frac {b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,a^{5}}-\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}-\frac {b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,a^{5}}+\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{\sqrt {a^{2}-b^{2}}\, a}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,a^{4}}+\frac {A \sin \left (4 d x +4 c \right )}{32 a d}-\frac {\sin \left (3 d x +3 c \right ) A b}{12 a^{2} d}+\frac {\sin \left (3 d x +3 c \right ) B}{12 a d}+\frac {\sin \left (2 d x +2 c \right ) A}{4 a d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{2}}{4 a^{3} d}-\frac {\sin \left (2 d x +2 c \right ) B b}{4 a^{2} d}\)	\(627\)

Input:

int(cos(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE 
)
 

Output:

1/d*(-2*b^4*(A*b-B*a)/a^5/((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^5*(((-5/8*a^4*A-A*a^3*b-1/2*A*a^2*b^2-A*a*b^ 
3+B*a^4+1/2*a^3*b*B+B*a^2*b^2)*tan(1/2*d*x+1/2*c)^7+(3/8*a^4*A-5/3*A*a^3*b 
-3*A*a*b^3+5/3*B*a^4+3*B*a^2*b^2-1/2*A*a^2*b^2+1/2*a^3*b*B)*tan(1/2*d*x+1/ 
2*c)^5+(-3/8*a^4*A+1/2*A*a^2*b^2-1/2*a^3*b*B-5/3*A*a^3*b-3*A*a*b^3+5/3*B*a 
^4+3*B*a^2*b^2)*tan(1/2*d*x+1/2*c)^3+(5/8*a^4*A+1/2*A*a^2*b^2-1/2*a^3*b*B- 
A*a^3*b-A*a*b^3+B*a^4+B*a^2*b^2)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c) 
^2)^4+1/8*(3*A*a^4+4*A*a^2*b^2+8*A*b^4-4*B*a^3*b-8*B*a*b^3)*arctan(tan(1/2 
*d*x+1/2*c))))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 685, normalized size of antiderivative = 2.85 \[ \int \frac {\cos ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\left [\frac {3 \, {\left (3 \, A a^{6} - 4 \, B a^{5} b + A a^{4} b^{2} - 4 \, B a^{3} b^{3} + 4 \, A a^{2} b^{4} + 8 \, B a b^{5} - 8 \, A b^{6}\right )} d x - 12 \, {\left (B a b^{4} - A b^{5}\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 (16 \, B a^{6} - 16 \, A a^{5} b + 8 \, B a^{4} b^{2} - 8 \, A a^{3} b^{3} - 24 \, B a^{2} b^{4} + 24 \, A a b^{5} + 6 \, {\left (A a^{6} - A a^{4} b^{2}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (B a^{6} - A a^{5} b - B a^{4} b^{2} + A a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A a^{6} - 4 \, B a^{5} b + A a^{4} b^{2} + 4 \, B a^{3} b^{3} - 4 \, A a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{7} - a^{5} b^{2}\right )} d}, \frac {3 \, {\left (3 \, A a^{6} - 4 \, B a^{5} b + A a^{4} b^{2} - 4 \, B a^{3} b^{3} + 4 \, A a^{2} b^{4} + 8 \, B a b^{5} - 8 \, A b^{6}\right )} d x + 24 \, {\left (B a b^{4} - A b^{5}\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 (16 \, B a^{6} - 16 \, A a^{5} b + 8 \, B a^{4} b^{2} - 8 \, A a^{3} b^{3} - 24 \, B a^{2} b^{4} + 24 \, A a b^{5} + 6 \, {\left (A a^{6} - A a^{4} b^{2}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (B a^{6} - A a^{5} b - B a^{4} b^{2} + A a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A a^{6} - 4 \, B a^{5} b + A a^{4} b^{2} + 4 \, B a^{3} b^{3} - 4 \, A a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{7} - a^{5} b^{2}\right )} d}\right ] \] Input:

integrate(cos(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="fri 
cas")
 

Output:

[1/24*(3*(3*A*a^6 - 4*B*a^5*b + A*a^4*b^2 - 4*B*a^3*b^3 + 4*A*a^2*b^4 + 8* 
B*a*b^5 - 8*A*b^6)*d*x - 12*(B*a*b^4 - A*b^5)*sqrt(a^2 - b^2)*log((2*a*b*c 
os(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)) + (16*B*a^6 - 16*A*a^5*b + 8*B*a^4*b^2 - 8*A*a^3*b^3 - 24*B*a 
^2*b^4 + 24*A*a*b^5 + 6*(A*a^6 - A*a^4*b^2)*cos(d*x + c)^3 + 8*(B*a^6 - A* 
a^5*b - B*a^4*b^2 + A*a^3*b^3)*cos(d*x + c)^2 + 3*(3*A*a^6 - 4*B*a^5*b + A 
*a^4*b^2 + 4*B*a^3*b^3 - 4*A*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/((a^7 - 
a^5*b^2)*d), 1/24*(3*(3*A*a^6 - 4*B*a^5*b + A*a^4*b^2 - 4*B*a^3*b^3 + 4*A* 
a^2*b^4 + 8*B*a*b^5 - 8*A*b^6)*d*x + 24*(B*a*b^4 - A*b^5)*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))) 
 + (16*B*a^6 - 16*A*a^5*b + 8*B*a^4*b^2 - 8*A*a^3*b^3 - 24*B*a^2*b^4 + 24* 
A*a*b^5 + 6*(A*a^6 - A*a^4*b^2)*cos(d*x + c)^3 + 8*(B*a^6 - A*a^5*b - B*a^ 
4*b^2 + A*a^3*b^3)*cos(d*x + c)^2 + 3*(3*A*a^6 - 4*B*a^5*b + A*a^4*b^2 + 4 
*B*a^3*b^3 - 4*A*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/((a^7 - a^5*b^2)*d)]
 

Sympy [F]

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

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

Maxima [F(-2)]

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

integrate(cos(d*x+c)^4*(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. 642 vs. \(2 (221) = 442\).

Time = 0.24 (sec) , antiderivative size = 642, normalized size of antiderivative = 2.68 \[ \int \frac {\cos ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="gia 
c")
 

Output:

1/24*(3*(3*A*a^4 - 4*B*a^3*b + 4*A*a^2*b^2 - 8*B*a*b^3 + 8*A*b^4)*(d*x + c 
)/a^5 + 48*(B*a*b^4 - 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)))/(sqrt(-a^2 + b^2)*a^5) - 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 24 
*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 24*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 12*B*a 
^2*b*tan(1/2*d*x + 1/2*c)^7 + 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 24*B*a*b 
^2*tan(1/2*d*x + 1/2*c)^7 + 24*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 9*A*a^3*tan( 
1/2*d*x + 1/2*c)^5 - 40*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*A*a^2*b*tan(1/2* 
d*x + 1/2*c)^5 - 12*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 12*A*a*b^2*tan(1/2*d* 
x + 1/2*c)^5 - 72*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*A*b^3*tan(1/2*d*x + 
1/2*c)^5 + 9*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^3*tan(1/2*d*x + 1/2*c)^ 
3 + 40*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 12*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 
- 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 72*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 
72*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^3*tan(1/2*d*x + 1/2*c) - 24*B*a^3 
*tan(1/2*d*x + 1/2*c) + 24*A*a^2*b*tan(1/2*d*x + 1/2*c) + 12*B*a^2*b*tan(1 
/2*d*x + 1/2*c) - 12*A*a*b^2*tan(1/2*d*x + 1/2*c) - 24*B*a*b^2*tan(1/2*d*x 
 + 1/2*c) + 24*A*b^3*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4 
*a^4))/d
 

Mupad [B] (verification not implemented)

Time = 18.50 (sec) , antiderivative size = 5903, normalized size of antiderivative = 24.60 \[ \int \frac {\cos ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

((tan(c/2 + (d*x)/2)*(5*A*a^3 - 8*A*b^3 + 8*B*a^3 + 4*A*a*b^2 - 8*A*a^2*b 
+ 8*B*a*b^2 - 4*B*a^2*b))/(4*a^4) - (tan(c/2 + (d*x)/2)^7*(5*A*a^3 + 8*A*b 
^3 - 8*B*a^3 + 4*A*a*b^2 + 8*A*a^2*b - 8*B*a*b^2 - 4*B*a^2*b))/(4*a^4) - ( 
tan(c/2 + (d*x)/2)^3*(9*A*a^3 + 72*A*b^3 - 40*B*a^3 - 12*A*a*b^2 + 40*A*a^ 
2*b - 72*B*a*b^2 + 12*B*a^2*b))/(12*a^4) + (tan(c/2 + (d*x)/2)^5*(9*A*a^3 
- 72*A*b^3 + 40*B*a^3 - 12*A*a*b^2 - 40*A*a^2*b + 72*B*a*b^2 + 12*B*a^2*b) 
)/(12*a^4))/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/ 
2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (atan(((((((12*A*a^16 + 32*A 
*a^10*b^6 - 48*A*a^11*b^5 + 16*A*a^12*b^4 - 4*A*a^13*b^3 + 4*A*a^14*b^2 - 
32*B*a^11*b^5 + 48*B*a^12*b^4 - 16*B*a^13*b^3 + 16*B*a^14*b^2 - 12*A*a^15* 
b - 16*B*a^15*b)/a^12 - (tan(c/2 + (d*x)/2)*(128*a^12*b + 128*a^10*b^3 - 2 
56*a^11*b^2)*(A*a^4*3i + A*b^4*8i + A*a^2*b^2*4i - B*a*b^3*8i - B*a^3*b*4i 
))/(16*a^13))*(A*a^4*3i + A*b^4*8i + A*a^2*b^2*4i - B*a*b^3*8i - B*a^3*b*4 
i))/(8*a^5) + (tan(c/2 + (d*x)/2)*(9*A^2*a^11 - 128*A^2*b^11 + 256*A^2*a*b 
^10 - 27*A^2*a^10*b - 256*A^2*a^2*b^9 + 256*A^2*a^3*b^8 - 256*A^2*a^4*b^7 
+ 256*A^2*a^5*b^6 - 216*A^2*a^6*b^5 + 136*A^2*a^7*b^4 - 81*A^2*a^8*b^3 + 5 
1*A^2*a^9*b^2 - 128*B^2*a^2*b^9 + 256*B^2*a^3*b^8 - 256*B^2*a^4*b^7 + 256* 
B^2*a^5*b^6 - 208*B^2*a^6*b^5 + 112*B^2*a^7*b^4 - 48*B^2*a^8*b^3 + 16*B^2* 
a^9*b^2 + 256*A*B*a*b^10 - 24*A*B*a^10*b - 512*A*B*a^2*b^9 + 512*A*B*a^3*b 
^8 - 512*A*B*a^4*b^7 + 464*A*B*a^5*b^6 - 368*A*B*a^6*b^5 + 264*A*B*a^7*...
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.17 \[ \int \frac {\cos ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+5 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x}{8 d} \] Input:

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

Output:

( - 2*cos(c + d*x)*sin(c + d*x)**3 + 5*cos(c + d*x)*sin(c + d*x) + 3*d*x)/ 
(8*d)