\(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx\) [244]

Optimal result

Integrand size = 31, antiderivative size = 195 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {1}{2} b \left (12 a^2 A b+A b^3+8 a^3 B+4 a b^2 B\right ) x+\frac {a^3 (4 A b+a B) \text {arctanh}(\sin (c+d x))}{d}-\frac {b \left (6 a^3 A-12 a A b^2-17 a^2 b B-2 b^3 B\right ) \sin (c+d x)}{3 d}-\frac {b^2 \left (6 a^2 A-3 A b^2-8 a b B\right ) \cos (c+d x) \sin (c+d x)}{6 d}-\frac {b (3 a A-b B) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {a A (a+b \cos (c+d x))^3 \tan (c+d x)}{d} \] Output:

1/2*b*(12*A*a^2*b+A*b^3+8*B*a^3+4*B*a*b^2)*x+a^3*(4*A*b+B*a)*arctanh(sin(d 
*x+c))/d-1/3*b*(6*A*a^3-12*A*a*b^2-17*B*a^2*b-2*B*b^3)*sin(d*x+c)/d-1/6*b^ 
2*(6*A*a^2-3*A*b^2-8*B*a*b)*cos(d*x+c)*sin(d*x+c)/d-1/3*b*(3*A*a-B*b)*(a+b 
*cos(d*x+c))^2*sin(d*x+c)/d+a*A*(a+b*cos(d*x+c))^3*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 3.78 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.32 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {6 b \left (12 a^2 A b+A b^3+8 a^3 B+4 a b^2 B\right ) (c+d x)-12 a^3 (4 A b+a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a^3 (4 A b+a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {12 a^4 A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {12 a^4 A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+3 b^2 \left (16 a A b+24 a^2 B+3 b^2 B\right ) \sin (c+d x)+3 b^3 (A b+4 a B) \sin (2 (c+d x))+b^4 B \sin (3 (c+d x))}{12 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.40 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 3468, 3042, 3528, 3042, 3512, 3042, 3502, 27, 3042, 3214, 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.

\(\displaystyle \int \sec ^2(c+d x) (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3468

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3512

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \left (6 (4 A b+a B) a^3-2 b \left (6 A a^3-17 b B a^2-12 A b^2 a-2 b^3 B\right ) \cos ^2(c+d x)+3 b \left (8 B a^3+12 A b a^2+4 b^2 B a+A b^3\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {b^2 \left (6 a^2 A-8 a b B-3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 a A-b B) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {6 (4 A b+a B) a^3-2 b \left (6 A a^3-17 b B a^2-12 A b^2 a-2 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+3 b \left (8 B a^3+12 A b a^2+4 b^2 B a+A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {b^2 \left (6 a^2 A-8 a b B-3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 a A-b B) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int 3 \left (2 (4 A b+a B) a^3+b \left (8 B a^3+12 A b a^2+4 b^2 B a+A b^3\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 b \left (6 a^3 A-17 a^2 b B-12 a A b^2-2 b^3 B\right ) \sin (c+d x)}{d}\right )-\frac {b^2 \left (6 a^2 A-8 a b B-3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 a A-b B) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \int \left (2 (4 A b+a B) a^3+b \left (8 B a^3+12 A b a^2+4 b^2 B a+A b^3\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 b \left (6 a^3 A-17 a^2 b B-12 a A b^2-2 b^3 B\right ) \sin (c+d x)}{d}\right )-\frac {b^2 \left (6 a^2 A-8 a b B-3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 a A-b B) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \int \frac {2 (4 A b+a B) a^3+b \left (8 B a^3+12 A b a^2+4 b^2 B a+A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 b \left (6 a^3 A-17 a^2 b B-12 a A b^2-2 b^3 B\right ) \sin (c+d x)}{d}\right )-\frac {b^2 \left (6 a^2 A-8 a b B-3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 a A-b B) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (2 a^3 (a B+4 A b) \int \sec (c+d x)dx+b x \left (8 a^3 B+12 a^2 A b+4 a b^2 B+A b^3\right )\right )-\frac {2 b \left (6 a^3 A-17 a^2 b B-12 a A b^2-2 b^3 B\right ) \sin (c+d x)}{d}\right )-\frac {b^2 \left (6 a^2 A-8 a b B-3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 a A-b B) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (2 a^3 (a B+4 A b) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+b x \left (8 a^3 B+12 a^2 A b+4 a b^2 B+A b^3\right )\right )-\frac {2 b \left (6 a^3 A-17 a^2 b B-12 a A b^2-2 b^3 B\right ) \sin (c+d x)}{d}\right )-\frac {b^2 \left (6 a^2 A-8 a b B-3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 a A-b B) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (\frac {2 a^3 (a B+4 A b) \text {arctanh}(\sin (c+d x))}{d}+b x \left (8 a^3 B+12 a^2 A b+4 a b^2 B+A b^3\right )\right )-\frac {2 b \left (6 a^3 A-17 a^2 b B-12 a A b^2-2 b^3 B\right ) \sin (c+d x)}{d}\right )-\frac {b^2 \left (6 a^2 A-8 a b B-3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 a A-b B) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\)

Input:

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

Output:

-1/3*(b*(3*a*A - b*B)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/d + (-1/2*(b^2* 
(6*a^2*A - 3*A*b^2 - 8*a*b*B)*Cos[c + d*x]*Sin[c + d*x])/d + (3*(b*(12*a^2 
*A*b + A*b^3 + 8*a^3*B + 4*a*b^2*B)*x + (2*a^3*(4*A*b + a*B)*ArcTanh[Sin[c 
 + d*x]])/d) - (2*b*(6*a^3*A - 12*a*A*b^2 - 17*a^2*b*B - 2*b^3*B)*Sin[c + 
d*x])/d)/2)/3 + (a*A*(a + b*Cos[c + d*x])^3*Tan[c + d*x])/d
 

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 3042

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

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3468

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c 
 + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 
1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n 
+ 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a 
*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) 
 - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - 
a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] 
/; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2 
, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3512

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si 
n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[(a + b*Si 
n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + 
A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 
0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 3528

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

rule 4257

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

Maple [A] (verified)

Time = 16.24 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.84

method	result	size
parts	\(\frac {a^{4} A \tan \left (d x +c \right )}{d}+\frac {\left (A \,b^{4}+4 B a \,b^{3}\right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (4 A a \,b^{3}+6 B \,a^{2} b^{2}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (6 A \,a^{2} b^{2}+4 B \,a^{3} b \right ) \left (d x +c \right )}{d}+\frac {\left (4 A \,a^{3} b +B \,a^{4}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B \,b^{4} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3 d}\)	\(164\)
derivativedivides	\(\frac {a^{4} A \tan \left (d x +c \right )+B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 A \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B \,a^{3} b \left (d x +c \right )+6 A \,a^{2} b^{2} \left (d x +c \right )+6 B \sin \left (d x +c \right ) a^{2} b^{2}+4 A \sin \left (d x +c \right ) a \,b^{3}+4 B a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,b^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {B \,b^{4} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(189\)
default	\(\frac {a^{4} A \tan \left (d x +c \right )+B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 A \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B \,a^{3} b \left (d x +c \right )+6 A \,a^{2} b^{2} \left (d x +c \right )+6 B \sin \left (d x +c \right ) a^{2} b^{2}+4 A \sin \left (d x +c \right ) a \,b^{3}+4 B a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,b^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {B \,b^{4} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(189\)
parallelrisch	\(\frac {-96 \cos \left (d x +c \right ) \left (A b +\frac {B a}{4}\right ) a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+96 \cos \left (d x +c \right ) \left (A b +\frac {B a}{4}\right ) a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (48 A a \,b^{3}+72 B \,a^{2} b^{2}+10 B \,b^{4}\right ) \sin \left (2 d x +2 c \right )+\left (3 A \,b^{4}+12 B a \,b^{3}\right ) \sin \left (3 d x +3 c \right )+B \sin \left (4 d x +4 c \right ) b^{4}+144 \left (A \,a^{2} b +\frac {1}{12} A \,b^{3}+\frac {2}{3} a^{3} B +\frac {1}{3} B a \,b^{2}\right ) x d b \cos \left (d x +c \right )+24 \left (a^{4} A +\frac {1}{8} A \,b^{4}+\frac {1}{2} B a \,b^{3}\right ) \sin \left (d x +c \right )}{24 d \cos \left (d x +c \right )}\)	\(212\)
risch	\(6 A \,a^{2} b^{2} x +\frac {A \,b^{4} x}{2}+4 B \,a^{3} b x +2 B a \,b^{3} x +\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B a \,b^{3}}{2 d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} A a \,b^{3}}{d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2} b^{2}}{d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B \,b^{4}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{2} b^{2}}{d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B a \,b^{3}}{2 d}+\frac {2 i a^{4} A}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} A \,b^{4}}{8 d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} A \,b^{4}}{8 d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} A a \,b^{3}}{d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B \,b^{4}}{8 d}+\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {B \,b^{4} \sin \left (3 d x +3 c \right )}{12 d}\)	\(365\)
norman	\(\frac {\left (-6 A \,a^{2} b^{2}-\frac {1}{2} A \,b^{4}-4 B \,a^{3} b -2 B a \,b^{3}\right ) x +\left (-30 A \,a^{2} b^{2}-\frac {5}{2} A \,b^{4}-20 B \,a^{3} b -10 B a \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (6 A \,a^{2} b^{2}+\frac {1}{2} A \,b^{4}+4 B \,a^{3} b +2 B a \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (30 A \,a^{2} b^{2}+\frac {5}{2} A \,b^{4}+20 B \,a^{3} b +10 B a \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-24 A \,a^{2} b^{2}-2 A \,b^{4}-16 B \,a^{3} b -8 B a \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (24 A \,a^{2} b^{2}+2 A \,b^{4}+16 B \,a^{3} b +8 B a \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-\frac {\left (2 a^{4} A -8 A a \,b^{3}+A \,b^{4}-12 B \,a^{2} b^{2}+4 B a \,b^{3}-2 B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {\left (2 a^{4} A +8 A a \,b^{3}+A \,b^{4}+12 B \,a^{2} b^{2}+4 B a \,b^{3}+2 B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (30 a^{4} A -72 A a \,b^{3}+3 A \,b^{4}-108 B \,a^{2} b^{2}+12 B a \,b^{3}-10 B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {2 \left (30 a^{4} A -24 A a \,b^{3}-3 A \,b^{4}-36 B \,a^{2} b^{2}-12 B a \,b^{3}-2 B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {2 \left (30 a^{4} A +24 A a \,b^{3}-3 A \,b^{4}+36 B \,a^{2} b^{2}-12 B a \,b^{3}+2 B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {\left (30 a^{4} A +72 A a \,b^{3}+3 A \,b^{4}+108 B \,a^{2} b^{2}+12 B a \,b^{3}+10 B \,b^{4}\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 )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {a^{3} \left (4 A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{3} \left (4 A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(680\)

Input:

int((a+cos(d*x+c)*b)^4*(A+B*cos(d*x+c))*sec(d*x+c)^2,x,method=_RETURNVERBO 
SE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.01 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {3 \, {\left (8 \, B a^{3} b + 12 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, B b^{4} \cos \left (d x + c\right )^{3} + 6 \, A a^{4} + 3 \, {\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (9 \, B a^{2} b^{2} + 6 \, A a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \] Input:

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="f 
ricas")
 

Output:

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

Sympy [F]

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

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

Output:

Integral((A + B*cos(c + d*x))*(a + b*cos(c + d*x))**4*sec(c + d*x)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.01 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {48 \, {\left (d x + c\right )} B a^{3} b + 72 \, {\left (d x + c\right )} A a^{2} b^{2} + 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{3} + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{4} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{4} + 6 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a^{3} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 72 \, B a^{2} b^{2} \sin \left (d x + c\right ) + 48 \, A a b^{3} \sin \left (d x + c\right ) + 12 \, A a^{4} \tan \left (d x + c\right )}{12 \, d} \] Input:

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="m 
axima")
 

Output:

1/12*(48*(d*x + c)*B*a^3*b + 72*(d*x + c)*A*a^2*b^2 + 12*(2*d*x + 2*c + si 
n(2*d*x + 2*c))*B*a*b^3 + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*b^4 - 4*(si 
n(d*x + c)^3 - 3*sin(d*x + c))*B*b^4 + 6*B*a^4*(log(sin(d*x + c) + 1) - lo 
g(sin(d*x + c) - 1)) + 24*A*a^3*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c 
) - 1)) + 72*B*a^2*b^2*sin(d*x + c) + 48*A*a*b^3*sin(d*x + c) + 12*A*a^4*t 
an(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.90 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=-\frac {\frac {12 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, {\left (8 \, B a^{3} b + 12 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} {\left (d x + c\right )} - 6 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 6 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (36 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 48 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="g 
iac")
 

Output:

-1/6*(12*A*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(8*B* 
a^3*b + 12*A*a^2*b^2 + 4*B*a*b^3 + A*b^4)*(d*x + c) - 6*(B*a^4 + 4*A*a^3*b 
)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 6*(B*a^4 + 4*A*a^3*b)*log(abs(tan(1 
/2*d*x + 1/2*c) - 1)) - 2*(36*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 24*A*a*b^ 
3*tan(1/2*d*x + 1/2*c)^5 - 12*B*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 3*A*b^4*tan 
(1/2*d*x + 1/2*c)^5 + 6*B*b^4*tan(1/2*d*x + 1/2*c)^5 + 72*B*a^2*b^2*tan(1/ 
2*d*x + 1/2*c)^3 + 48*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 4*B*b^4*tan(1/2*d*x 
 + 1/2*c)^3 + 36*B*a^2*b^2*tan(1/2*d*x + 1/2*c) + 24*A*a*b^3*tan(1/2*d*x + 
 1/2*c) + 12*B*a*b^3*tan(1/2*d*x + 1/2*c) + 3*A*b^4*tan(1/2*d*x + 1/2*c) + 
 6*B*b^4*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 = 44.09 (sec) , antiderivative size = 2522, normalized size of antiderivative = 12.93 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.92 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {15 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) a \,b^{4}-30 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{4} b +30 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{4} b -2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{5}+60 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{3}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{5}+60 \cos \left (d x +c \right ) a^{3} b^{2} d x +15 \cos \left (d x +c \right ) a \,b^{4} d x +6 \sin \left (d x +c \right ) a^{5}}{6 \cos \left (d x +c \right ) d} \] Input:

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

Output:

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