\(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx\) [235]

Optimal result

Integrand size = 31, antiderivative size = 131 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {1}{2} b \left (6 a A b+6 a^2 B+b^2 B\right ) x+\frac {a^2 (3 A b+a B) \text {arctanh}(\sin (c+d x))}{d}-\frac {b \left (2 a^2 A-A b^2-3 a b B\right ) \sin (c+d x)}{d}-\frac {b^2 (2 a A-b B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a A (a+b \cos (c+d x))^2 \tan (c+d x)}{d} \] Output:

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

Mathematica [A] (verified)

Time = 2.63 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.66 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {2 b \left (6 a A b+6 a^2 B+b^2 B\right ) (c+d x)-4 a^2 (3 A b+a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a^2 (3 A b+a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 a^3 A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 a^3 A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+4 b^2 (A b+3 a B) \sin (c+d x)+b^3 B \sin (2 (c+d x))}{4 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3468, 3042, 3512, 3042, 3502, 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.

\(\displaystyle \int \sec ^2(c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3468

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3512

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

-1/2*(b^2*(2*a*A - b*B)*Cos[c + d*x]*Sin[c + d*x])/d + (b*(6*a*A*b + 6*a^2 
*B + b^2*B)*x + (2*a^2*(3*A*b + a*B)*ArcTanh[Sin[c + d*x]])/d - (2*b*(2*a^ 
2*A - A*b^2 - 3*a*b*B)*Sin[c + d*x])/d)/2 + (a*A*(a + b*Cos[c + d*x])^2*Ta 
n[c + d*x])/d
 

Defintions of rubi rules used

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 4257

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

Maple [A] (verified)

Time = 5.93 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.95

method	result	size
parts	\(\frac {a^{3} A \tan \left (d x +c \right )}{d}+\frac {\left (A \,b^{3}+3 B a \,b^{2}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (3 A a \,b^{2}+3 B \,a^{2} b \right ) \left (d x +c \right )}{d}+\frac {\left (3 A \,a^{2} b +a^{3} B \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(124\)
derivativedivides	\(\frac {A \tan \left (d x +c \right ) a^{3}+a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{2} b \left (d x +c \right )+3 A a \,b^{2} \left (d x +c \right )+3 B \sin \left (d x +c \right ) a \,b^{2}+A \sin \left (d x +c \right ) b^{3}+b^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(132\)
default	\(\frac {A \tan \left (d x +c \right ) a^{3}+a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{2} b \left (d x +c \right )+3 A a \,b^{2} \left (d x +c \right )+3 B \sin \left (d x +c \right ) a \,b^{2}+A \sin \left (d x +c \right ) b^{3}+b^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(132\)
parallelrisch	\(\frac {8 \left (-3 A \,a^{2} b -a^{3} B \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 \left (3 A \,a^{2} b +a^{3} B \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (A \,b^{3}+3 B a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+B \sin \left (3 d x +3 c \right ) b^{3}+24 x d \left (A a b +a^{2} B +\frac {1}{6} B \,b^{2}\right ) b \cos \left (d x +c \right )+\sin \left (d x +c \right ) \left (8 a^{3} A +b^{3} B \right )}{8 \cos \left (d x +c \right ) d}\)	\(166\)
risch	\(3 x A a \,b^{2}+3 x B \,a^{2} b +\frac {b^{3} B x}{2}-\frac {i b^{3} B \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,b^{3}}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B a \,b^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{3}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B a \,b^{2}}{2 d}+\frac {i b^{3} B \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i a^{3} A}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}\)	\(253\)
norman	\(\frac {\left (-3 A a \,b^{2}-3 B \,a^{2} b -\frac {1}{2} b^{3} B \right ) x +\left (-9 A a \,b^{2}-9 B \,a^{2} b -\frac {3}{2} b^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (3 A a \,b^{2}+3 B \,a^{2} b +\frac {1}{2} b^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (9 A a \,b^{2}+9 B \,a^{2} b +\frac {3}{2} b^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-6 A a \,b^{2}-6 B \,a^{2} b -b^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (6 A a \,b^{2}+6 B \,a^{2} b +b^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\frac {2 \left (6 a^{3} A -b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {4 \left (2 a^{3} A -A \,b^{3}-3 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {4 \left (2 a^{3} A +A \,b^{3}+3 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {\left (2 a^{3} A -2 A \,b^{3}-6 B a \,b^{2}+b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {\left (2 a^{3} A +2 A \,b^{3}+6 B a \,b^{2}+b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {a^{2} \left (3 A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{2} \left (3 A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(461\)

Input:

int((a+cos(d*x+c)*b)^3*(A+B*cos(d*x+c))*sec(d*x+c)^2,x,method=_RETURNVERBO 
SE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.16 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {{\left (6 \, B a^{2} b + 6 \, A a b^{2} + B b^{3}\right )} d x \cos \left (d x + c\right ) + {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (B b^{3} \cos \left (d x + c\right )^{2} + 2 \, A a^{3} + 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \] Input:

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="f 
ricas")
 

Output:

1/2*((6*B*a^2*b + 6*A*a*b^2 + B*b^3)*d*x*cos(d*x + c) + (B*a^3 + 3*A*a^2*b 
)*cos(d*x + c)*log(sin(d*x + c) + 1) - (B*a^3 + 3*A*a^2*b)*cos(d*x + c)*lo 
g(-sin(d*x + c) + 1) + (B*b^3*cos(d*x + c)^2 + 2*A*a^3 + 2*(3*B*a*b^2 + A* 
b^3)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))
 

Sympy [F]

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

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.10 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {12 \, {\left (d x + c\right )} B a^{2} b + 12 \, {\left (d x + c\right )} A a b^{2} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{3} + 2 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a b^{2} \sin \left (d x + c\right ) + 4 \, A b^{3} \sin \left (d x + c\right ) + 4 \, A a^{3} \tan \left (d x + c\right )}{4 \, d} \] Input:

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="m 
axima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.79 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=-\frac {\frac {4 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (6 \, B a^{2} b + 6 \, A a b^{2} + B b^{3}\right )} {\left (d x + c\right )} - 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \] Input:

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="g 
iac")
 

Output:

-1/2*(4*A*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (6*B*a^2 
*b + 6*A*a*b^2 + B*b^3)*(d*x + c) - 2*(B*a^3 + 3*A*a^2*b)*log(abs(tan(1/2* 
d*x + 1/2*c) + 1)) + 2*(B*a^3 + 3*A*a^2*b)*log(abs(tan(1/2*d*x + 1/2*c) - 
1)) - 2*(6*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*A*b^3*tan(1/2*d*x + 1/2*c)^3 
 - B*b^3*tan(1/2*d*x + 1/2*c)^3 + 6*B*a*b^2*tan(1/2*d*x + 1/2*c) + 2*A*b^3 
*tan(1/2*d*x + 1/2*c) + B*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^ 
2 + 1)^2)/d
 

Mupad [B] (verification not implemented)

Time = 42.03 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.80 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {B\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,A\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,B\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}-A\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}}{d}+\frac {\frac {A\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {B\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+A\,a^3\,\sin \left (c+d\,x\right )+\frac {B\,b^3\,\sin \left (c+d\,x\right )}{8}+\frac {3\,B\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\cos \left (c+d\,x\right )} \] Input:

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

Output:

(B*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - B*a^3*atan((sin(c/2 + 
 (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i + 6*A*a*b^2*atan(sin(c/2 + (d*x)/2)/c 
os(c/2 + (d*x)/2)) - A*a^2*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/ 
2))*6i + 6*B*a^2*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + ((A*b^ 
3*sin(2*c + 2*d*x))/2 + (B*b^3*sin(3*c + 3*d*x))/8 + A*a^3*sin(c + d*x) + 
(B*b^3*sin(c + d*x))/8 + (3*B*a*b^2*sin(2*c + 2*d*x))/2)/(d*cos(c + d*x))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {\cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) b^{4}-8 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3} b +8 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3} b +8 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{3}+12 \cos \left (d x +c \right ) a^{2} b^{2} d x +\cos \left (d x +c \right ) b^{4} d x +2 \sin \left (d x +c \right ) a^{4}}{2 \cos \left (d x +c \right ) d} \] Input:

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

Output:

(cos(c + d*x)**2*sin(c + d*x)*b**4 - 8*cos(c + d*x)*log(tan((c + d*x)/2) - 
 1)*a**3*b + 8*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**3*b + 8*cos(c + d 
*x)*sin(c + d*x)*a*b**3 + 12*cos(c + d*x)*a**2*b**2*d*x + cos(c + d*x)*b** 
4*d*x + 2*sin(c + d*x)*a**4)/(2*cos(c + d*x)*d)