\(\int \cos ^5(c+d x) (a+b \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\) [670]

Optimal result

Integrand size = 33, antiderivative size = 250 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) x+\frac {b^4 C \text {arctanh}(\sin (c+d x))}{d}+\frac {\left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right ) \sin (c+d x)}{15 d}+\frac {a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \cos (c+d x) \sin (c+d x)}{30 d}+\frac {\left (3 A b^2+a^2 (4 A+5 C)\right ) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac {A b \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d} \] Output:

1/2*a*b*(4*b^2*(A+2*C)+a^2*(3*A+4*C))*x+b^4*C*arctanh(sin(d*x+c))/d+1/15*( 
6*A*b^4+2*a^4*(4*A+5*C)+a^2*b^2*(56*A+85*C))*sin(d*x+c)/d+1/30*a*b*(6*A*b^ 
2+a^2*(29*A+40*C))*cos(d*x+c)*sin(d*x+c)/d+1/15*(3*A*b^2+a^2*(4*A+5*C))*co 
s(d*x+c)^2*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/5*A*b*cos(d*x+c)^3*(a+b*sec(d 
*x+c))^3*sin(d*x+c)/d+1/5*A*cos(d*x+c)^4*(a+b*sec(d*x+c))^4*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 2.19 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.89 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {120 a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) (c+d x)-240 b^4 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+240 b^4 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+30 \left (8 A b^4+12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)\right ) \sin (c+d x)+240 a b \left (A b^2+a^2 (A+C)\right ) \sin (2 (c+d x))+5 a^2 \left (24 A b^2+a^2 (5 A+4 C)\right ) \sin (3 (c+d x))+30 a^3 A b \sin (4 (c+d x))+3 a^4 A \sin (5 (c+d x))}{240 d} \] Input:

Integrate[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]
 

Output:

(120*a*b*(4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*(c + d*x) - 240*b^4*C*Log[Cos 
[(c + d*x)/2] - Sin[(c + d*x)/2]] + 240*b^4*C*Log[Cos[(c + d*x)/2] + Sin[( 
c + d*x)/2]] + 30*(8*A*b^4 + 12*a^2*b^2*(3*A + 4*C) + a^4*(5*A + 6*C))*Sin 
[c + d*x] + 240*a*b*(A*b^2 + a^2*(A + C))*Sin[2*(c + d*x)] + 5*a^2*(24*A*b 
^2 + a^2*(5*A + 4*C))*Sin[3*(c + d*x)] + 30*a^3*A*b*Sin[4*(c + d*x)] + 3*a 
^4*A*Sin[5*(c + d*x)])/(240*d)
 

Rubi [A] (verified)

Time = 1.93 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.04, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 4583, 3042, 4582, 27, 3042, 4582, 3042, 4562, 25, 3042, 4535, 24, 3042, 4533, 3042, 4257}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4583

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4582

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int 4 \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left ((4 A+5 C) a^2+b (7 A+10 C) \sec (c+d x) a+3 A b^2+5 b^2 C \sec ^2(c+d x)\right )dx+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left ((4 A+5 C) a^2+b (7 A+10 C) \sec (c+d x) a+3 A b^2+5 b^2 C \sec ^2(c+d x)\right )dx+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left ((4 A+5 C) a^2+b (7 A+10 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a+3 A b^2+5 b^2 C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 4582

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (15 C \sec ^2(c+d x) b^3+\left ((29 A+40 C) a^2+6 A b^2\right ) b+a \left (2 (4 A+5 C) a^2+9 b^2 (3 A+5 C)\right ) \sec (c+d x)\right )dx+\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (15 C \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^3+\left ((29 A+40 C) a^2+6 A b^2\right ) b+a \left (2 (4 A+5 C) a^2+9 b^2 (3 A+5 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 4562

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}-\frac {1}{2} \int -\cos (c+d x) \left (30 C \sec ^2(c+d x) b^4+15 a \left ((3 A+4 C) a^2+4 b^2 (A+2 C)\right ) \sec (c+d x) b+2 \left (2 (4 A+5 C) a^4+b^2 (56 A+85 C) a^2+6 A b^4\right )\right )dx\right )+\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \int \cos (c+d x) \left (30 C \sec ^2(c+d x) b^4+15 a \left ((3 A+4 C) a^2+4 b^2 (A+2 C)\right ) \sec (c+d x) b+2 \left (2 (4 A+5 C) a^4+b^2 (56 A+85 C) a^2+6 A b^4\right )\right )dx+\frac {a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {30 C \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^4+15 a \left ((3 A+4 C) a^2+4 b^2 (A+2 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) b+2 \left (2 (4 A+5 C) a^4+b^2 (56 A+85 C) a^2+6 A b^4\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 4535

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a b \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \int 1dx+\int \cos (c+d x) \left (30 C \sec ^2(c+d x) b^4+2 \left (2 (4 A+5 C) a^4+b^2 (56 A+85 C) a^2+6 A b^4\right )\right )dx\right )+\frac {a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \left (\int \cos (c+d x) \left (30 C \sec ^2(c+d x) b^4+2 \left (2 (4 A+5 C) a^4+b^2 (56 A+85 C) a^2+6 A b^4\right )\right )dx+15 a b x \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )\right )+\frac {a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \left (\int \frac {30 C \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^4+2 \left (2 (4 A+5 C) a^4+b^2 (56 A+85 C) a^2+6 A b^4\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+15 a b x \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )\right )+\frac {a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 4533

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \left (30 b^4 C \int \sec (c+d x)dx+15 a b x \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+\frac {2 \left (2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)+6 A b^4\right ) \sin (c+d x)}{d}\right )+\frac {a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \left (30 b^4 C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+15 a b x \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+\frac {2 \left (2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)+6 A b^4\right ) \sin (c+d x)}{d}\right )+\frac {a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{5} \left (\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {1}{3} \left (\frac {a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} \left (15 a b x \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+\frac {2 \left (2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)+6 A b^4\right ) \sin (c+d x)}{d}+\frac {30 b^4 C \text {arctanh}(\sin (c+d x))}{d}\right )\right )+\frac {A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}\)

Input:

Int[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]
 

Output:

(A*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(5*d) + (((3*A*b^2 
+ a^2*(4*A + 5*C))*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(3* 
d) + (A*b*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/d + ((a*b*(6 
*A*b^2 + a^2*(29*A + 40*C))*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (15*a*b*(4* 
b^2*(A + 2*C) + a^2*(3*A + 4*C))*x + (30*b^4*C*ArcTanh[Sin[c + d*x]])/d + 
(2*(6*A*b^4 + 2*a^4*(4*A + 5*C) + a^2*b^2*(56*A + 85*C))*Sin[c + d*x])/d)/ 
2)/3)/5
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 25

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

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4533

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + 
Simp[(C*m + A*(m + 1))/(b^2*m)   Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr 
eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
 

rule 4535

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

rule 4562

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 
_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Si 
mp[1/(d*n)   Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B* 
b) + A*a*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[ 
{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]
 

rule 4582

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*((d*Csc[e 
 + f*x])^n/(f*n)), x] - Simp[1/(d*n)   Int[(a + b*Csc[e + f*x])^(m - 1)*(d* 
Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs 
c[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a 
, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
 

rule 4583

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

Maple [A] (verified)

Time = 1.59 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.78

method	result	size
parallelrisch	\(\frac {-240 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{4}+240 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{4}+240 \left (a^{2} \left (A +C \right )+A \,b^{2}\right ) a b \sin \left (2 d x +2 c \right )+\left (\left (25 A +20 C \right ) a^{4}+120 A \,a^{2} b^{2}\right ) \sin \left (3 d x +3 c \right )+30 A \,a^{3} b \sin \left (4 d x +4 c \right )+3 a^{4} A \sin \left (5 d x +5 c \right )+\left (\left (150 A +180 C \right ) a^{4}+1080 \left (A +\frac {4 C}{3}\right ) b^{2} a^{2}+240 A \,b^{4}\right ) \sin \left (d x +c \right )+360 \left (\left (A +\frac {4 C}{3}\right ) a^{2}+\frac {4 b^{2} \left (A +2 C \right )}{3}\right ) a b x d}{240 d}\)	\(195\)
derivativedivides	\(\frac {\frac {a^{4} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 A \,a^{3} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+4 a^{3} b C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \,a^{2} b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+6 C \,a^{2} b^{2} \sin \left (d x +c \right )+4 a A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 C a \,b^{3} \left (d x +c \right )+A \,b^{4} \sin \left (d x +c \right )+C \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(240\)
default	\(\frac {\frac {a^{4} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 A \,a^{3} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+4 a^{3} b C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \,a^{2} b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+6 C \,a^{2} b^{2} \sin \left (d x +c \right )+4 a A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 C a \,b^{3} \left (d x +c \right )+A \,b^{4} \sin \left (d x +c \right )+C \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(240\)
risch	\(\frac {3 a^{3} A b x}{2}+2 A a \,b^{3} x +2 C \,a^{3} b x +4 C a \,b^{3} x +\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{4}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{4} C}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{2} b^{2}}{d}+\frac {5 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{16 d}+\frac {9 i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{2} b^{2}}{4 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{4} C}{8 d}-\frac {5 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{16 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C \,a^{2} b^{2}}{d}-\frac {9 i {\mathrm e}^{i \left (d x +c \right )} A \,a^{2} b^{2}}{4 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,b^{4}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,b^{4}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,b^{4}}{d}+\frac {a^{4} A \sin \left (5 d x +5 c \right )}{80 d}+\frac {A \,a^{3} b \sin \left (4 d x +4 c \right )}{8 d}+\frac {5 a^{4} A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) A \,a^{2} b^{2}}{2 d}+\frac {\sin \left (3 d x +3 c \right ) a^{4} C}{12 d}+\frac {\sin \left (2 d x +2 c \right ) A \,a^{3} b}{d}+\frac {\sin \left (2 d x +2 c \right ) a A \,b^{3}}{d}+\frac {\sin \left (2 d x +2 c \right ) a^{3} b C}{d}\)	\(427\)

Input:

int(cos(d*x+c)^5*(a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x,method=_RETURNVER 
BOSE)
 

Output:

1/240*(-240*C*ln(tan(1/2*d*x+1/2*c)-1)*b^4+240*C*ln(tan(1/2*d*x+1/2*c)+1)* 
b^4+240*(a^2*(A+C)+A*b^2)*a*b*sin(2*d*x+2*c)+((25*A+20*C)*a^4+120*A*a^2*b^ 
2)*sin(3*d*x+3*c)+30*A*a^3*b*sin(4*d*x+4*c)+3*a^4*A*sin(5*d*x+5*c)+((150*A 
+180*C)*a^4+1080*(A+4/3*C)*b^2*a^2+240*A*b^4)*sin(d*x+c)+360*((A+4/3*C)*a^ 
2+4/3*b^2*(A+2*C))*a*b*x*d)/d
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.78 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, C b^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, C b^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{3} b + 4 \, {\left (A + 2 \, C\right )} a b^{3}\right )} d x + {\left (6 \, A a^{4} \cos \left (d x + c\right )^{4} + 30 \, A a^{3} b \cos \left (d x + c\right )^{3} + 4 \, {\left (4 \, A + 5 \, C\right )} a^{4} + 60 \, {\left (2 \, A + 3 \, C\right )} a^{2} b^{2} + 30 \, A b^{4} + 2 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{4} + 30 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{3} b + 4 \, A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{30 \, d} \] Input:

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm= 
"fricas")
 

Output:

1/30*(15*C*b^4*log(sin(d*x + c) + 1) - 15*C*b^4*log(-sin(d*x + c) + 1) + 1 
5*((3*A + 4*C)*a^3*b + 4*(A + 2*C)*a*b^3)*d*x + (6*A*a^4*cos(d*x + c)^4 + 
30*A*a^3*b*cos(d*x + c)^3 + 4*(4*A + 5*C)*a^4 + 60*(2*A + 3*C)*a^2*b^2 + 3 
0*A*b^4 + 2*((4*A + 5*C)*a^4 + 30*A*a^2*b^2)*cos(d*x + c)^2 + 15*((3*A + 4 
*C)*a^3*b + 4*A*a*b^3)*cos(d*x + c))*sin(d*x + c))/d
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**5*(a+b*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.96 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{4} - 40 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} b + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} b - 240 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{3} + 480 \, {\left (d x + c\right )} C a b^{3} + 60 \, C b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 720 \, C a^{2} b^{2} \sin \left (d x + c\right ) + 120 \, A b^{4} \sin \left (d x + c\right )}{120 \, d} \] Input:

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm= 
"maxima")
 

Output:

1/120*(8*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^4 - 
40*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^4 + 15*(12*d*x + 12*c + sin(4*d*x 
 + 4*c) + 8*sin(2*d*x + 2*c))*A*a^3*b + 120*(2*d*x + 2*c + sin(2*d*x + 2*c 
))*C*a^3*b - 240*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2*b^2 + 120*(2*d*x 
+ 2*c + sin(2*d*x + 2*c))*A*a*b^3 + 480*(d*x + c)*C*a*b^3 + 60*C*b^4*(log( 
sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 720*C*a^2*b^2*sin(d*x + c) + 
120*A*b^4*sin(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 753 vs. \(2 (238) = 476\).

Time = 0.33 (sec) , antiderivative size = 753, normalized size of antiderivative = 3.01 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm= 
"giac")
 

Output:

1/30*(30*C*b^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 30*C*b^4*log(abs(tan(1 
/2*d*x + 1/2*c) - 1)) + 15*(3*A*a^3*b + 4*C*a^3*b + 4*A*a*b^3 + 8*C*a*b^3) 
*(d*x + c) + 2*(30*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 30*C*a^4*tan(1/2*d*x + 1 
/2*c)^9 - 75*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 60*C*a^3*b*tan(1/2*d*x + 1/2 
*c)^9 + 180*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 180*C*a^2*b^2*tan(1/2*d*x + 
 1/2*c)^9 - 60*A*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 30*A*b^4*tan(1/2*d*x + 1/2 
*c)^9 + 40*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 80*C*a^4*tan(1/2*d*x + 1/2*c)^7 
- 30*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 120*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 
 480*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 720*C*a^2*b^2*tan(1/2*d*x + 1/2*c) 
^7 - 120*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 120*A*b^4*tan(1/2*d*x + 1/2*c)^7 
 + 116*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 100*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 6 
00*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 1080*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^ 
5 + 180*A*b^4*tan(1/2*d*x + 1/2*c)^5 + 40*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 8 
0*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 30*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 120*C 
*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 480*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 720 
*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 1 
20*A*b^4*tan(1/2*d*x + 1/2*c)^3 + 30*A*a^4*tan(1/2*d*x + 1/2*c) + 30*C*a^4 
*tan(1/2*d*x + 1/2*c) + 75*A*a^3*b*tan(1/2*d*x + 1/2*c) + 60*C*a^3*b*tan(1 
/2*d*x + 1/2*c) + 180*A*a^2*b^2*tan(1/2*d*x + 1/2*c) + 180*C*a^2*b^2*tan(1 
/2*d*x + 1/2*c) + 60*A*a*b^3*tan(1/2*d*x + 1/2*c) + 30*A*b^4*tan(1/2*d*...
 

Mupad [B] (verification not implemented)

Time = 14.66 (sec) , antiderivative size = 2241, normalized size of antiderivative = 8.96 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

int(cos(c + d*x)^5*(A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^4,x)
 

Output:

(tan(c/2 + (d*x)/2)*(2*A*a^4 + 2*A*b^4 + 2*C*a^4 + 12*A*a^2*b^2 + 12*C*a^2 
*b^2 + 4*A*a*b^3 + 5*A*a^3*b + 4*C*a^3*b) + tan(c/2 + (d*x)/2)^5*((116*A*a 
^4)/15 + 12*A*b^4 + (20*C*a^4)/3 + 40*A*a^2*b^2 + 72*C*a^2*b^2) + tan(c/2 
+ (d*x)/2)^9*(2*A*a^4 + 2*A*b^4 + 2*C*a^4 + 12*A*a^2*b^2 + 12*C*a^2*b^2 - 
4*A*a*b^3 - 5*A*a^3*b - 4*C*a^3*b) + tan(c/2 + (d*x)/2)^3*((8*A*a^4)/3 + 8 
*A*b^4 + (16*C*a^4)/3 + 32*A*a^2*b^2 + 48*C*a^2*b^2 + 8*A*a*b^3 + 2*A*a^3* 
b + 8*C*a^3*b) + tan(c/2 + (d*x)/2)^7*((8*A*a^4)/3 + 8*A*b^4 + (16*C*a^4)/ 
3 + 32*A*a^2*b^2 + 48*C*a^2*b^2 - 8*A*a*b^3 - 2*A*a^3*b - 8*C*a^3*b))/(d*( 
5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 
 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (C*b^4*atan((C*b 
^4*(tan(c/2 + (d*x)/2)*(32*C^2*b^8 + 128*A^2*a^2*b^6 + 192*A^2*a^4*b^4 + 7 
2*A^2*a^6*b^2 + 512*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 128*C^2*a^6*b^2 + 512* 
A*C*a^2*b^6 + 640*A*C*a^4*b^4 + 192*A*C*a^6*b^2) + C*b^4*(32*C*b^4 + 64*A* 
a*b^3 + 48*A*a^3*b + 128*C*a*b^3 + 64*C*a^3*b))*1i + C*b^4*(tan(c/2 + (d*x 
)/2)*(32*C^2*b^8 + 128*A^2*a^2*b^6 + 192*A^2*a^4*b^4 + 72*A^2*a^6*b^2 + 51 
2*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 128*C^2*a^6*b^2 + 512*A*C*a^2*b^6 + 640* 
A*C*a^4*b^4 + 192*A*C*a^6*b^2) - C*b^4*(32*C*b^4 + 64*A*a*b^3 + 48*A*a^3*b 
 + 128*C*a*b^3 + 64*C*a^3*b))*1i)/(1024*C^3*a^2*b^10 - 256*C^3*a*b^11 - 12 
8*C^3*a^3*b^9 + 1024*C^3*a^4*b^8 + 256*C^3*a^6*b^6 + C*b^4*(tan(c/2 + (d*x 
)/2)*(32*C^2*b^8 + 128*A^2*a^2*b^6 + 192*A^2*a^4*b^4 + 72*A^2*a^6*b^2 +...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.24 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{4} b +75 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4} b +60 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b c +60 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{3}-30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{4} c +30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{4} c +6 \sin \left (d x +c \right )^{5} a^{5}-20 \sin \left (d x +c \right )^{3} a^{5}-10 \sin \left (d x +c \right )^{3} a^{4} c -60 \sin \left (d x +c \right )^{3} a^{3} b^{2}+30 \sin \left (d x +c \right ) a^{5}+30 \sin \left (d x +c \right ) a^{4} c +180 \sin \left (d x +c \right ) a^{3} b^{2}+180 \sin \left (d x +c \right ) a^{2} b^{2} c +30 \sin \left (d x +c \right ) a \,b^{4}+45 a^{4} b c +45 a^{4} b d x +60 a^{3} b \,c^{2}+60 a^{3} b c d x +60 a^{2} b^{3} c +60 a^{2} b^{3} d x +120 a \,b^{3} c^{2}+120 a \,b^{3} c d x}{30 d} \] Input:

int(cos(d*x+c)^5*(a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x)
 

Output:

( - 30*cos(c + d*x)*sin(c + d*x)**3*a**4*b + 75*cos(c + d*x)*sin(c + d*x)* 
a**4*b + 60*cos(c + d*x)*sin(c + d*x)*a**3*b*c + 60*cos(c + d*x)*sin(c + d 
*x)*a**2*b**3 - 30*log(tan((c + d*x)/2) - 1)*b**4*c + 30*log(tan((c + d*x) 
/2) + 1)*b**4*c + 6*sin(c + d*x)**5*a**5 - 20*sin(c + d*x)**3*a**5 - 10*si 
n(c + d*x)**3*a**4*c - 60*sin(c + d*x)**3*a**3*b**2 + 30*sin(c + d*x)*a**5 
 + 30*sin(c + d*x)*a**4*c + 180*sin(c + d*x)*a**3*b**2 + 180*sin(c + d*x)* 
a**2*b**2*c + 30*sin(c + d*x)*a*b**4 + 45*a**4*b*c + 45*a**4*b*d*x + 60*a* 
*3*b*c**2 + 60*a**3*b*c*d*x + 60*a**2*b**3*c + 60*a**2*b**3*d*x + 120*a*b* 
*3*c**2 + 120*a*b**3*c*d*x)/(30*d)