\(\int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\) [307]

Optimal result

Integrand size = 31, antiderivative size = 198 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {1}{2} a \left (4 a^2 A b+8 A b^3+a^3 B+12 a b^2 B\right ) x+\frac {b^3 (A b+4 a B) \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 \left (2 a^2 A+9 A b^2+9 a b B\right ) \sin (c+d x)}{3 d}+\frac {a (2 A b+a B) \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac {b^2 \left (8 a A b+3 a^2 B-6 b^2 B\right ) \tan (c+d x)}{6 d} \] Output:

1/2*a*(4*A*a^2*b+8*A*b^3+B*a^3+12*B*a*b^2)*x+b^3*(A*b+4*B*a)*arctanh(sin(d 
*x+c))/d+1/3*a^2*(2*A*a^2+9*A*b^2+9*B*a*b)*sin(d*x+c)/d+1/2*a*(2*A*b+B*a)* 
cos(d*x+c)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/3*a*A*cos(d*x+c)^2*(a+b*sec(d 
*x+c))^3*sin(d*x+c)/d-1/6*b^2*(8*A*a*b+3*B*a^2-6*B*b^2)*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 2.68 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.30 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {6 a \left (4 a^2 A b+8 A b^3+a^3 B+12 a b^2 B\right ) (c+d x)-12 b^3 (A b+4 a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 b^3 (A b+4 a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {12 b^4 B \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {12 b^4 B \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+3 a^2 \left (3 a^2 A+24 A b^2+16 a b B\right ) \sin (c+d x)+3 a^3 (4 A b+a B) \sin (2 (c+d x))+a^4 A \sin (3 (c+d x))}{12 d} \] Input:

Integrate[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]
 

Output:

(6*a*(4*a^2*A*b + 8*A*b^3 + a^3*B + 12*a*b^2*B)*(c + d*x) - 12*b^3*(A*b + 
4*a*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*b^3*(A*b + 4*a*B)*Log 
[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (12*b^4*B*Sin[(c + d*x)/2])/(Cos[( 
c + d*x)/2] - Sin[(c + d*x)/2]) + (12*b^4*B*Sin[(c + d*x)/2])/(Cos[(c + d* 
x)/2] + Sin[(c + d*x)/2]) + 3*a^2*(3*a^2*A + 24*A*b^2 + 16*a*b*B)*Sin[c + 
d*x] + 3*a^3*(4*A*b + a*B)*Sin[2*(c + d*x)] + a^4*A*Sin[3*(c + d*x)])/(12* 
d)
 

Rubi [A] (verified)

Time = 1.48 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4513, 25, 3042, 4582, 3042, 4564, 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.

\(\displaystyle \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4513

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4582

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \cos (c+d x) (a+b \sec (c+d x)) \left (-b \left (3 B a^2+8 A b a-6 b^2 B\right ) \sec ^2(c+d x)+\left (3 B a^3+8 A b a^2+18 b^2 B a+6 A b^3\right ) \sec (c+d x)+2 a \left (2 A a^2+9 b B a+9 A b^2\right )\right )dx+\frac {3 a (a B+2 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-b \left (3 B a^2+8 A b a-6 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (3 B a^3+8 A b a^2+18 b^2 B a+6 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a \left (2 A a^2+9 b B a+9 A b^2\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {3 a (a B+2 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 4564

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int \cos (c+d x) \left (6 (A b+4 a B) \sec ^2(c+d x) b^3+2 a^2 \left (2 A a^2+9 b B a+9 A b^2\right )+3 a \left (B a^3+4 A b a^2+12 b^2 B a+8 A b^3\right ) \sec (c+d x)\right )dx-\frac {b^2 \left (3 a^2 B+8 a A b-6 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {3 a (a B+2 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int \frac {6 (A b+4 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^3+2 a^2 \left (2 A a^2+9 b B a+9 A b^2\right )+3 a \left (B a^3+4 A b a^2+12 b^2 B a+8 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {b^2 \left (3 a^2 B+8 a A b-6 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {3 a (a B+2 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 4535

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int \cos (c+d x) \left (6 (A b+4 a B) \sec ^2(c+d x) b^3+2 a^2 \left (2 A a^2+9 b B a+9 A b^2\right )\right )dx+3 a \left (a^3 B+4 a^2 A b+12 a b^2 B+8 A b^3\right ) \int 1dx-\frac {b^2 \left (3 a^2 B+8 a A b-6 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {3 a (a B+2 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int \cos (c+d x) \left (6 (A b+4 a B) \sec ^2(c+d x) b^3+2 a^2 \left (2 A a^2+9 b B a+9 A b^2\right )\right )dx-\frac {b^2 \left (3 a^2 B+8 a A b-6 b^2 B\right ) \tan (c+d x)}{d}+3 a x \left (a^3 B+4 a^2 A b+12 a b^2 B+8 A b^3\right )\right )+\frac {3 a (a B+2 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int \frac {6 (A b+4 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^3+2 a^2 \left (2 A a^2+9 b B a+9 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {b^2 \left (3 a^2 B+8 a A b-6 b^2 B\right ) \tan (c+d x)}{d}+3 a x \left (a^3 B+4 a^2 A b+12 a b^2 B+8 A b^3\right )\right )+\frac {3 a (a B+2 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 4533

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (6 b^3 (4 a B+A b) \int \sec (c+d x)dx+\frac {2 a^2 \left (2 a^2 A+9 a b B+9 A b^2\right ) \sin (c+d x)}{d}-\frac {b^2 \left (3 a^2 B+8 a A b-6 b^2 B\right ) \tan (c+d x)}{d}+3 a x \left (a^3 B+4 a^2 A b+12 a b^2 B+8 A b^3\right )\right )+\frac {3 a (a B+2 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (6 b^3 (4 a B+A b) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 a^2 \left (2 a^2 A+9 a b B+9 A b^2\right ) \sin (c+d x)}{d}-\frac {b^2 \left (3 a^2 B+8 a A b-6 b^2 B\right ) \tan (c+d x)}{d}+3 a x \left (a^3 B+4 a^2 A b+12 a b^2 B+8 A b^3\right )\right )+\frac {3 a (a B+2 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {2 a^2 \left (2 a^2 A+9 a b B+9 A b^2\right ) \sin (c+d x)}{d}-\frac {b^2 \left (3 a^2 B+8 a A b-6 b^2 B\right ) \tan (c+d x)}{d}+3 a x \left (a^3 B+4 a^2 A b+12 a b^2 B+8 A b^3\right )+\frac {6 b^3 (4 a B+A b) \text {arctanh}(\sin (c+d x))}{d}\right )+\frac {3 a (a B+2 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d}\)

Input:

Int[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]
 

Output:

(a*A*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + ((3*a*(2* 
A*b + a*B)*Cos[c + d*x]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + (3*a* 
(4*a^2*A*b + 8*A*b^3 + a^3*B + 12*a*b^2*B)*x + (6*b^3*(A*b + 4*a*B)*ArcTan 
h[Sin[c + d*x]])/d + (2*a^2*(2*a^2*A + 9*A*b^2 + 9*a*b*B)*Sin[c + d*x])/d 
- (b^2*(8*a*A*b + 3*a^2*B - 6*b^2*B)*Tan[c + d*x])/d)/2)/3
 

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 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 4513

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

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 4564

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[(-b)*C*Csc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f*x])^ 
n/(f*(n + 2))), x] + Simp[1/(n + 2)   Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 
2) + (B*a*(n + 2) + b*(C*(n + 1) + A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*( 
n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, 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]
 

Maple [A] (verified)

Time = 1.19 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 A \,a^{3} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b B \sin \left (d x +c \right )+6 A \,a^{2} b^{2} \sin \left (d x +c \right )+6 B \,a^{2} b^{2} \left (d x +c \right )+4 A a \,b^{3} \left (d x +c \right )+4 a \,b^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,b^{4} \tan \left (d x +c \right )}{d}\)	\(189\)
default	\(\frac {\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 A \,a^{3} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b B \sin \left (d x +c \right )+6 A \,a^{2} b^{2} \sin \left (d x +c \right )+6 B \,a^{2} b^{2} \left (d x +c \right )+4 A a \,b^{3} \left (d x +c \right )+4 a \,b^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,b^{4} \tan \left (d x +c \right )}{d}\)	\(189\)
parallelrisch	\(\frac {-24 b^{3} \cos \left (d x +c \right ) \left (A b +4 B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+24 b^{3} \cos \left (d x +c \right ) \left (A b +4 B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (10 a^{4} A +72 A \,a^{2} b^{2}+48 a^{3} b B \right ) \sin \left (2 d x +2 c \right )+\left (12 A \,a^{3} b +3 B \,a^{4}\right ) \sin \left (3 d x +3 c \right )+a^{4} A \sin \left (4 d x +4 c \right )+48 x a \left (A \,a^{2} b +2 A \,b^{3}+\frac {1}{4} B \,a^{3}+3 B a \,b^{2}\right ) d \cos \left (d x +c \right )+12 \sin \left (d x +c \right ) \left (A \,a^{3} b +\frac {1}{4} B \,a^{4}+2 B \,b^{4}\right )}{24 d \cos \left (d x +c \right )}\)	\(212\)
risch	\(2 A \,a^{3} b x +4 A a \,b^{3} x +\frac {B \,a^{4} x}{2}+6 B \,a^{2} b^{2} x -\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B \,a^{4}}{8 d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} a^{3} b B}{d}+\frac {3 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} A \,a^{3} b}{2 d}+\frac {2 i B \,b^{4}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} A \,a^{3} b}{2 d}-\frac {3 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A \,a^{2} b^{2}}{d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B \,a^{4}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{2} b^{2}}{d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} a^{3} b B}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{4}}{d}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{3} B}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{4}}{d}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{3} B}{d}+\frac {a^{4} A \sin \left (3 d x +3 c \right )}{12 d}\)	\(365\)
norman	\(\frac {\left (2 A \,a^{3} b +4 A a \,b^{3}+\frac {1}{2} B \,a^{4}+6 B \,a^{2} b^{2}\right ) x +\left (-6 A \,a^{3} b -12 A a \,b^{3}-\frac {3}{2} B \,a^{4}-18 B \,a^{2} b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-6 A \,a^{3} b -12 A a \,b^{3}-\frac {3}{2} B \,a^{4}-18 B \,a^{2} b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-2 A \,a^{3} b -4 A a \,b^{3}-\frac {1}{2} B \,a^{4}-6 B \,a^{2} b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-2 A \,a^{3} b -4 A a \,b^{3}-\frac {1}{2} B \,a^{4}-6 B \,a^{2} b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (2 A \,a^{3} b +4 A a \,b^{3}+\frac {1}{2} B \,a^{4}+6 B \,a^{2} b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (6 A \,a^{3} b +12 A a \,b^{3}+\frac {3}{2} B \,a^{4}+18 B \,a^{2} b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (6 A \,a^{3} b +12 A a \,b^{3}+\frac {3}{2} B \,a^{4}+18 B \,a^{2} b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {\left (2 a^{4} A -4 A \,a^{3} b +12 A \,a^{2} b^{2}-B \,a^{4}+8 a^{3} b B -2 B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {\left (2 a^{4} A +4 A \,a^{3} b +12 A \,a^{2} b^{2}+B \,a^{4}+8 a^{3} b B +2 B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (26 a^{4} A -60 A \,a^{3} b -36 A \,a^{2} b^{2}-15 B \,a^{4}-24 a^{3} b B +18 B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}+\frac {\left (26 a^{4} A +60 A \,a^{3} b -36 A \,a^{2} b^{2}+15 B \,a^{4}-24 a^{3} b B -18 B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {8 a^{2} \left (A \,a^{2}-6 A \,b^{2}-4 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {4 a^{2} \left (5 A \,a^{2}-12 A a b +18 A \,b^{2}-3 B \,a^{2}+12 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {4 a^{2} \left (5 A \,a^{2}+12 A a b +18 A \,b^{2}+3 B \,a^{2}+12 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}+\frac {b^{3} \left (A b +4 B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b^{3} \left (A b +4 B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(781\)

Input:

int(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.99 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (B a^{4} + 4 \, A a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, A a^{4} \cos \left (d x + c\right )^{3} + 6 \, B b^{4} + 3 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (A a^{4} + 6 \, B a^{3} b + 9 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \] Input:

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="f 
ricas")
 

Output:

1/6*(3*(B*a^4 + 4*A*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3)*d*x*cos(d*x + c) + 3 
*(4*B*a*b^3 + A*b^4)*cos(d*x + c)*log(sin(d*x + c) + 1) - 3*(4*B*a*b^3 + A 
*b^4)*cos(d*x + c)*log(-sin(d*x + c) + 1) + (2*A*a^4*cos(d*x + c)^3 + 6*B* 
b^4 + 3*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)^2 + 4*(A*a^4 + 6*B*a^3*b + 9*A*a^ 
2*b^2)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**3*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.99 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} b - 72 \, {\left (d x + c\right )} B a^{2} b^{2} - 48 \, {\left (d x + c\right )} A a b^{3} - 24 \, B a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 48 \, B a^{3} b \sin \left (d x + c\right ) - 72 \, A a^{2} b^{2} \sin \left (d x + c\right ) - 12 \, B b^{4} \tan \left (d x + c\right )}{12 \, d} \] Input:

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="m 
axima")
 

Output:

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 3*(2*d*x + 2*c + sin(2* 
d*x + 2*c))*B*a^4 - 12*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3*b - 72*(d*x 
+ c)*B*a^2*b^2 - 48*(d*x + c)*A*a*b^3 - 24*B*a*b^3*(log(sin(d*x + c) + 1) 
- log(sin(d*x + c) - 1)) - 6*A*b^4*(log(sin(d*x + c) + 1) - log(sin(d*x + 
c) - 1)) - 48*B*a^3*b*sin(d*x + c) - 72*A*a^2*b^2*sin(d*x + c) - 12*B*b^4* 
tan(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.87 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=-\frac {\frac {12 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, {\left (B a^{4} + 4 \, A a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} {\left (d x + c\right )} - 6 \, {\left (4 \, B a b^{3} + A b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 6 \, {\left (4 \, B a b^{3} + A b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 48 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="g 
iac")
 

Output:

-1/6*(12*B*b^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(B*a^ 
4 + 4*A*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3)*(d*x + c) - 6*(4*B*a*b^3 + A*b^4 
)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 6*(4*B*a*b^3 + A*b^4)*log(abs(tan(1 
/2*d*x + 1/2*c) - 1)) - 2*(6*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^4*tan(1/ 
2*d*x + 1/2*c)^5 - 12*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 24*B*a^3*b*tan(1/2* 
d*x + 1/2*c)^5 + 36*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*A*a^4*tan(1/2*d*x 
 + 1/2*c)^3 + 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 72*A*a^2*b^2*tan(1/2*d*x 
 + 1/2*c)^3 + 6*A*a^4*tan(1/2*d*x + 1/2*c) + 3*B*a^4*tan(1/2*d*x + 1/2*c) 
+ 12*A*a^3*b*tan(1/2*d*x + 1/2*c) + 24*B*a^3*b*tan(1/2*d*x + 1/2*c) + 36*A 
*a^2*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
 

Mupad [B] (verification not implemented)

Time = 12.64 (sec) , antiderivative size = 2523, normalized size of antiderivative = 12.74 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

int(cos(c + d*x)^3*(A + B/cos(c + d*x))*(a + b/cos(c + d*x))^4,x)
 

Output:

- (tan(c/2 + (d*x)/2)^7*(2*A*a^4 - B*a^4 - 2*B*b^4 + 12*A*a^2*b^2 - 4*A*a^ 
3*b + 8*B*a^3*b) - tan(c/2 + (d*x)/2)*(2*A*a^4 + B*a^4 + 2*B*b^4 + 12*A*a^ 
2*b^2 + 4*A*a^3*b + 8*B*a^3*b) + tan(c/2 + (d*x)/2)^3*((2*A*a^4)/3 + B*a^4 
 - 6*B*b^4 - 12*A*a^2*b^2 + 4*A*a^3*b - 8*B*a^3*b) + tan(c/2 + (d*x)/2)^5* 
(B*a^4 - (2*A*a^4)/3 - 6*B*b^4 + 12*A*a^2*b^2 + 4*A*a^3*b + 8*B*a^3*b))/(d 
*(2*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2)^6 - tan(c/2 + (d*x)/2)^8 + 
 1)) - (atan(((A*b^4 + 4*B*a*b^3)*((A*b^4 + 4*B*a*b^3)*(32*A*b^4 + 16*B*a^ 
4 + 192*B*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*B*a*b^3) + tan(c/2 + (d 
*x)/2)*(32*A^2*b^8 + 8*B^2*a^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A 
^2*a^6*b^2 + 512*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 192*B^2*a^6*b^2 + 256*A* 
B*a*b^7 + 64*A*B*a^7*b + 1536*A*B*a^3*b^5 + 896*A*B*a^5*b^3))*1i - (A*b^4 
+ 4*B*a*b^3)*((A*b^4 + 4*B*a*b^3)*(32*A*b^4 + 16*B*a^4 + 192*B*a^2*b^2 + 1 
28*A*a*b^3 + 64*A*a^3*b + 128*B*a*b^3) - tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 
8*B^2*a^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 512*B^2* 
a^2*b^6 + 1152*B^2*a^4*b^4 + 192*B^2*a^6*b^2 + 256*A*B*a*b^7 + 64*A*B*a^7* 
b + 1536*A*B*a^3*b^5 + 896*A*B*a^5*b^3))*1i)/((A*b^4 + 4*B*a*b^3)*((A*b^4 
+ 4*B*a*b^3)*(32*A*b^4 + 16*B*a^4 + 192*B*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3 
*b + 128*B*a*b^3) + tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*B^2*a^8 + 512*A^2*a 
^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 512*B^2*a^2*b^6 + 1152*B^2*a^ 
4*b^4 + 192*B^2*a^6*b^2 + 256*A*B*a*b^7 + 64*A*B*a^7*b + 1536*A*B*a^3*b...
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.08 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {-30 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{4}+30 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{4}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{5}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{5}+60 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b^{2}+15 \cos \left (d x +c \right ) a^{4} b c +15 \cos \left (d x +c \right ) a^{4} b d x +60 \cos \left (d x +c \right ) a^{2} b^{3} c +60 \cos \left (d x +c \right ) a^{2} b^{3} d x -15 \sin \left (d x +c \right )^{3} a^{4} b +15 \sin \left (d x +c \right ) a^{4} b +6 \sin \left (d x +c \right ) b^{5}}{6 \cos \left (d x +c \right ) d} \] Input:

int(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)
 

Output:

( - 30*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**4 + 30*cos(c + d*x)*log 
(tan((c + d*x)/2) + 1)*a*b**4 - 2*cos(c + d*x)*sin(c + d*x)**3*a**5 + 6*co 
s(c + d*x)*sin(c + d*x)*a**5 + 60*cos(c + d*x)*sin(c + d*x)*a**3*b**2 + 15 
*cos(c + d*x)*a**4*b*c + 15*cos(c + d*x)*a**4*b*d*x + 60*cos(c + d*x)*a**2 
*b**3*c + 60*cos(c + d*x)*a**2*b**3*d*x - 15*sin(c + d*x)**3*a**4*b + 15*s 
in(c + d*x)*a**4*b + 6*sin(c + d*x)*b**5)/(6*cos(c + d*x)*d)