\(\int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx\) [8]

Optimal result

Integrand size = 29, antiderivative size = 86 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {a (A+B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (2 A+3 B) \tan (c+d x)}{3 d}+\frac {a (A+B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.65 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {a \left (3 (A+B) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (6 (A+B)+3 (A+B) \sec (c+d x)+2 A \tan ^2(c+d x)\right )\right )}{6 d} \] Input:

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

Output:

(a*(3*(A + B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*(A + B) + 3*(A + B)* 
Sec[c + d*x] + 2*A*Tan[c + d*x]^2)))/(6*d)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 3447, 3042, 3500, 3042, 3227, 3042, 4254, 24, 4255, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

\(\displaystyle \frac {1}{3} \int (3 a (A+B)+a (2 A+3 B) \cos (c+d x)) \sec ^3(c+d x)dx+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

(a*A*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((a*(2*A + 3*B)*Tan[c + d*x])/d 
+ 3*a*(A + B)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/( 
2*d)))/3
 

Defintions of rubi rules used

rule 24

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

rule 3042

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

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3447

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

rule 3500

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

rule 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

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 = 10.18 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94

method	result	size
parts	\(-\frac {A a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A a +B a \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {B a \tan \left (d x +c \right )}{d}\)	\(81\)
derivativedivides	\(\frac {A a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B a \tan \left (d x +c \right )-A a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(105\)
default	\(\frac {A a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B a \tan \left (d x +c \right )-A a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(105\)
parallelrisch	\(\frac {a \left (-\frac {3 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {3 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\left (A +B \right ) \sin \left (2 d x +2 c \right )+\left (\frac {2 A}{3}+B \right ) \sin \left (3 d x +3 c \right )+2 \sin \left (d x +c \right ) \left (A +\frac {B}{2}\right )\right )}{d \left (3 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )\right )}\)	\(138\)
norman	\(\frac {-\frac {4 a \left (A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {2 a \left (A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {2 a \left (7 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {3 \left (A +B \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (A +B \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {\left (A +B \right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (A +B \right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(182\)
risch	\(-\frac {i a \left (3 A \,{\mathrm e}^{5 i \left (d x +c \right )}+3 B \,{\mathrm e}^{5 i \left (d x +c \right )}-6 B \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 B \,{\mathrm e}^{2 i \left (d x +c \right )}-3 A \,{\mathrm e}^{i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-4 A -6 B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {A a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}-\frac {A a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}\)	\(193\)

Input:

int((a+a*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^4,x,method=_RETURNVERBOSE 
)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.22 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 2 \, A a\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="fri 
cas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.48 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a - 3 \, A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a \tan \left (d x + c\right )}{12 \, d} \] Input:

integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="max 
ima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.79 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:

integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="gia 
c")
 

Output:

1/6*(3*(A*a + B*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(A*a + B*a)*log( 
abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*A*a*tan(1/2*d*x + 1/2*c)^5 + 3*B*a*t 
an(1/2*d*x + 1/2*c)^5 - 4*A*a*tan(1/2*d*x + 1/2*c)^3 - 12*B*a*tan(1/2*d*x 
+ 1/2*c)^3 + 9*A*a*tan(1/2*d*x + 1/2*c) + 9*B*a*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 = 42.36 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.47 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+B\right )}{d}-\frac {\left (A\,a+B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,A\,a}{3}-4\,B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A\,a+3\,B\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:

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

Output:

(a*atanh(tan(c/2 + (d*x)/2))*(A + B))/d - (tan(c/2 + (d*x)/2)*(3*A*a + 3*B 
*a) + tan(c/2 + (d*x)/2)^5*(A*a + B*a) - tan(c/2 + (d*x)/2)^3*((4*A*a)/3 + 
 4*B*a))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + ( 
d*x)/2)^6 - 1))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 297, normalized size of antiderivative = 3.45 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {a \left (-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +4 \sin \left (d x +c \right )^{3} a +6 \sin \left (d x +c \right )^{3} b -6 \sin \left (d x +c \right ) a -6 \sin \left (d x +c \right ) b \right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

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

Output:

(a*( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - 3*cos( 
c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 3*cos(c + d*x)*log( 
tan((c + d*x)/2) - 1)*a + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b + 3*c 
os(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 3*cos(c + d*x)*l 
og(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 3*cos(c + d*x)*log(tan((c + d 
*x)/2) + 1)*a - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b - 3*cos(c + d*x 
)*sin(c + d*x)*a - 3*cos(c + d*x)*sin(c + d*x)*b + 4*sin(c + d*x)**3*a + 6 
*sin(c + d*x)**3*b - 6*sin(c + d*x)*a - 6*sin(c + d*x)*b))/(6*cos(c + d*x) 
*d*(sin(c + d*x)**2 - 1))