\(\int \frac {\cos ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^2} \, dx\) [807]

Optimal result

Integrand size = 40, antiderivative size = 261 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {\left (a^2 B+6 b^2 B-4 a b C\right ) x}{2 a^4}-\frac {2 b^2 \left (4 a^2 b B-3 b^3 B-3 a^3 C+2 a b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}-\frac {\left (2 a^2 b B-3 b^3 B-a^3 C+2 a b^2 C\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2 B-3 b^2 B+2 a b C\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b (b B-a C) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 1.56 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.70 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {2 \left (a^2 B+6 b^2 B-4 a b C\right ) (c+d x)-\frac {8 b^2 \left (-4 a^2 b B+3 b^3 B+3 a^3 C-2 a b^2 C\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+4 a (-2 b B+a C) \sin (c+d x)-\frac {4 a b^3 (-b B+a C) \sin (c+d x)}{(a-b) (a+b) (b+a \cos (c+d x))}+a^2 B \sin (2 (c+d x))}{4 a^4 d} \] Input:

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

Output:

(2*(a^2*B + 6*b^2*B - 4*a*b*C)*(c + d*x) - (8*b^2*(-4*a^2*b*B + 3*b^3*B + 
3*a^3*C - 2*a*b^2*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]) 
/(a^2 - b^2)^(3/2) + 4*a*(-2*b*B + a*C)*Sin[c + d*x] - (4*a*b^3*(-(b*B) + 
a*C)*Sin[c + d*x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])) + a^2*B*Sin[2*(c 
 + d*x)])/(4*a^4*d)
 

Rubi [A] (verified)

Time = 1.86 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.05, 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, 4518, 25, 3042, 4592, 3042, 4592, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4518

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4592

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4592

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4407

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(b*(b*B - a*C)*Cos[c + d*x]*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + 
d*x])) + (((a^2*B - 3*b^2*B + 2*a*b*C)*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) 
- (-((((a^2 - b^2)*(a^2*B + 6*b^2*B - 4*a*b*C)*x)/a - (4*b^2*(4*a^2*b*B - 
3*b^3*B - 3*a^3*C + 2*a*b^2*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*B - 3*b^3*B - a^ 
3*C + 2*a*b^2*C)*Sin[c + d*x])/(a*d))/(2*a))/(a*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 4518

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

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 = 0.82 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (\left (-\frac {1}{2} B \,a^{2}-2 B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {1}{2} B \,a^{2}-2 B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (B \,a^{2}+6 B \,b^{2}-4 C a b \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}+\frac {2 b^{2} \left (-\frac {a b \left (B b -C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (4 B \,a^{2} b -3 B \,b^{3}-3 a^{3} C +2 C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}}{d}\)	\(270\)
default	\(\frac {\frac {\frac {2 \left (\left (-\frac {1}{2} B \,a^{2}-2 B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {1}{2} B \,a^{2}-2 B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (B \,a^{2}+6 B \,b^{2}-4 C a b \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}+\frac {2 b^{2} \left (-\frac {a b \left (B b -C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (4 B \,a^{2} b -3 B \,b^{3}-3 a^{3} C +2 C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}}{d}\)	\(270\)
risch	\(\frac {B x}{2 a^{2}}+\frac {3 x B \,b^{2}}{a^{4}}-\frac {2 x C b}{a^{3}}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} B b}{a^{3} d}-\frac {i B \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a^{2} d}+\frac {i B \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{2} d}-\frac {2 i b^{3} \left (-B b +C a \right ) \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{a^{4} \left (a^{2}-b^{2}\right ) d \left ({\mathrm e}^{2 i \left (d x +c \right )} a +2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {i {\mathrm e}^{i \left (d x +c \right )} B b}{a^{3} d}+\frac {4 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 ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}-\frac {3 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}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{4}}-\frac {3 b^{2} \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}}\, \left (a +b \right ) \left (a -b \right ) d a}+\frac {2 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 ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}-\frac {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 \,b^{3}}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {3 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}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{4}}+\frac {3 b^{2} \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}}\, \left (a +b \right ) \left (a -b \right ) d a}-\frac {2 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 ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}\)	\(918\)

Input:

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

Output:

1/d*(2/a^4*(((-1/2*B*a^2-2*B*a*b+C*a^2)*tan(1/2*d*x+1/2*c)^3+(1/2*B*a^2-2* 
B*a*b+C*a^2)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^2+1/2*(B*a^2+6*B 
*b^2-4*C*a*b)*arctan(tan(1/2*d*x+1/2*c)))+2*b^2/a^4*(-a*b*(B*b-C*a)/(a^2-b 
^2)*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) 
-(4*B*a^2*b-3*B*b^3-3*C*a^3+2*C*a*b^2)/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arc 
tanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 971, normalized size of antiderivative = 3.72 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, 
 algorithm="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 [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.30 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (3 \, C a^{3} b^{2} - 4 \, B a^{2} b^{3} - 2 \, C a b^{4} + 3 \, B b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {4 \, {\left (C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}} + \frac {{\left (B a^{2} - 4 \, C a b + 6 \, B b^{2}\right )} {\left (d x + c\right )}}{a^{4}} - \frac {2 \, {\left (B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}}}{2 \, d} \] Input:

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

Output:

1/2*(4*(3*C*a^3*b^2 - 4*B*a^2*b^3 - 2*C*a*b^4 + 3*B*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^6 - a^4*b^2)*sqrt(-a^2 + b^2)) + 
 4*(C*a*b^3*tan(1/2*d*x + 1/2*c) - B*b^4*tan(1/2*d*x + 1/2*c))/((a^5 - a^3 
*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)) + (B* 
a^2 - 4*C*a*b + 6*B*b^2)*(d*x + c)/a^4 - 2*(B*a*tan(1/2*d*x + 1/2*c)^3 - 2 
*C*a*tan(1/2*d*x + 1/2*c)^3 + 4*B*b*tan(1/2*d*x + 1/2*c)^3 - B*a*tan(1/2*d 
*x + 1/2*c) - 2*C*a*tan(1/2*d*x + 1/2*c) + 4*B*b*tan(1/2*d*x + 1/2*c))/((t 
an(1/2*d*x + 1/2*c)^2 + 1)^2*a^3))/d
 

Mupad [B] (verification not implemented)

Time = 21.36 (sec) , antiderivative size = 6730, normalized size of antiderivative = 25.79 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan(((((8*tan(c/2 + (d*x)/2)*(B^2*a^10 + 72*B^2*b^10 - 72*B^2*a*b^9 - 2* 
B^2*a^9*b - 120*B^2*a^2*b^8 + 120*B^2*a^3*b^7 + 17*B^2*a^4*b^6 - 26*B^2*a^ 
5*b^5 + 23*B^2*a^6*b^4 - 20*B^2*a^7*b^3 + 11*B^2*a^8*b^2 + 32*C^2*a^2*b^8 
- 32*C^2*a^3*b^7 - 64*C^2*a^4*b^6 + 64*C^2*a^5*b^5 + 20*C^2*a^6*b^4 - 32*C 
^2*a^7*b^3 + 16*C^2*a^8*b^2 - 96*B*C*a*b^9 - 8*B*C*a^9*b + 96*B*C*a^2*b^8 
+ 176*B*C*a^3*b^7 - 176*B*C*a^4*b^6 - 40*B*C*a^5*b^5 + 64*B*C*a^6*b^4 - 40 
*B*C*a^7*b^3 + 16*B*C*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (((8*( 
2*B*a^15 - 12*B*a^8*b^7 + 6*B*a^9*b^6 + 28*B*a^10*b^5 - 14*B*a^11*b^4 - 16 
*B*a^12*b^3 + 6*B*a^13*b^2 + 8*C*a^9*b^6 - 4*C*a^10*b^5 - 20*C*a^11*b^4 + 
12*C*a^12*b^3 + 12*C*a^13*b^2 - 8*C*a^14*b))/(a^11*b + a^12 - a^9*b^3 - a^ 
10*b^2) - (4*tan(c/2 + (d*x)/2)*(B*a^2*1i + B*b^2*6i - C*a*b*4i)*(8*a^13*b 
 - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/(a^4*( 
a^8*b + a^9 - a^6*b^3 - a^7*b^2)))*(B*a^2*1i + B*b^2*6i - C*a*b*4i))/(2*a^ 
4))*(B*a^2*1i + B*b^2*6i - C*a*b*4i)*1i)/(2*a^4) + (((8*tan(c/2 + (d*x)/2) 
*(B^2*a^10 + 72*B^2*b^10 - 72*B^2*a*b^9 - 2*B^2*a^9*b - 120*B^2*a^2*b^8 + 
120*B^2*a^3*b^7 + 17*B^2*a^4*b^6 - 26*B^2*a^5*b^5 + 23*B^2*a^6*b^4 - 20*B^ 
2*a^7*b^3 + 11*B^2*a^8*b^2 + 32*C^2*a^2*b^8 - 32*C^2*a^3*b^7 - 64*C^2*a^4* 
b^6 + 64*C^2*a^5*b^5 + 20*C^2*a^6*b^4 - 32*C^2*a^7*b^3 + 16*C^2*a^8*b^2 - 
96*B*C*a*b^9 - 8*B*C*a^9*b + 96*B*C*a^2*b^8 + 176*B*C*a^3*b^7 - 176*B*C*a^ 
4*b^6 - 40*B*C*a^5*b^5 + 64*B*C*a^6*b^4 - 40*B*C*a^7*b^3 + 16*B*C*a^8*b...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1134, normalized size of antiderivative = 4.34 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(12*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**4*b**2*c - 16*sqrt( - a**2 + b**2)*ata 
n((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((tan((c + d*x)/2)*a - tan((c 
+ d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)*a**2*b**4*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))*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))*a**3*b**3*c - 16*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 - 8*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**5*c + 12*sqrt( - a**2 + b**2)*at 
an((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*b**7 + 
2*cos(c + d*x)*sin(c + d*x)*a**7*c - 3*cos(c + d*x)*sin(c + d*x)*a**6*b**2 
 - 4*cos(c + d*x)*sin(c + d*x)*a**5*b**2*c + 6*cos(c + d*x)*sin(c + d*x)*a 
**4*b**4 + 2*cos(c + d*x)*sin(c + d*x)*a**3*b**4*c - 3*cos(c + d*x)*sin(c 
+ d*x)*a**2*b**6 + cos(c + d*x)*a**7*b*c + cos(c + d*x)*a**7*b*d*x - 4*cos 
(c + d*x)*a**6*b*c**2 - 4*cos(c + d*x)*a**6*b*c*d*x + 4*cos(c + d*x)*a**5* 
b**3*c + 4*cos(c + d*x)*a**5*b**3*d*x + 8*cos(c + d*x)*a**4*b**3*c**2 + 8* 
cos(c + d*x)*a**4*b**3*c*d*x - 11*cos(c + d*x)*a**3*b**5*c - 11*cos(c + d* 
x)*a**3*b**5*d*x - 4*cos(c + d*x)*a**2*b**5*c**2 - 4*cos(c + d*x)*a**2*...