\(\int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx\) [17]

Optimal result

Integrand size = 31, antiderivative size = 113 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {a^2 (2 A+3 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^2 (5 A+6 B) \tan (c+d x)}{3 d}+\frac {a^2 (4 A+3 B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 d} \] Output:

1/2*a^2*(2*A+3*B)*arctanh(sin(d*x+c))/d+1/3*a^2*(5*A+6*B)*tan(d*x+c)/d+1/6 
*a^2*(4*A+3*B)*sec(d*x+c)*tan(d*x+c)/d+1/3*A*(a^2+a^2*cos(d*x+c))*sec(d*x+ 
c)^2*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.64 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {a^2 \left (6 B \coth ^{-1}(\sin (c+d x))+3 (2 A+B) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (12 (A+B)+3 (2 A+B) \sec (c+d x)+2 A \tan ^2(c+d x)\right )\right )}{6 d} \] Input:

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

Output:

(a^2*(6*B*ArcCoth[Sin[c + d*x]] + 3*(2*A + B)*ArcTanh[Sin[c + d*x]] + Tan[ 
c + d*x]*(12*(A + B) + 3*(2*A + B)*Sec[c + d*x] + 2*A*Tan[c + d*x]^2)))/(6 
*d)
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 3454, 3042, 3447, 3042, 3500, 3042, 3227, 3042, 4254, 24, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \left (2 (5 A+6 B) a^2+3 (2 A+3 B) \cos (c+d x) a^2\right ) \sec ^2(c+d x)dx+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {2 (5 A+6 B) a^2+3 (2 A+3 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (2 a^2 (5 A+6 B) \int \sec ^2(c+d x)dx+3 a^2 (2 A+3 B) \int \sec (c+d x)dx\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 a^2 (2 A+3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 a^2 (5 A+6 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 a^2 (2 A+3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {2 a^2 (5 A+6 B) \int 1d(-\tan (c+d x))}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 a^2 (2 A+3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 a^2 (5 A+6 B) \tan (c+d x)}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {3 a^2 (2 A+3 B) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a^2 (5 A+6 B) \tan (c+d x)}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\)

Input:

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

Output:

(A*(a^2 + a^2*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((a^2*(4* 
A + 3*B)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + ((3*a^2*(2*A + 3*B)*ArcTanh[Si 
n[c + d*x]])/d + (2*a^2*(5*A + 6*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 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 3454

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-b^2)*(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)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + 
 a*d))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp 
[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B 
*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] 
&& GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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 4257

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

Maple [A] (verified)

Time = 10.77 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.06

method	result	size
parts	\(-\frac {a^{2} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (a^{2} A +2 a^{2} B \right ) \tan \left (d x +c \right )}{d}+\frac {\left (2 a^{2} A +a^{2} B \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{2}}{d}\)	\(120\)
derivativedivides	\(\frac {a^{2} A \tan \left (d x +c \right )+a^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a^{2} B \tan \left (d x +c \right )-a^{2} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{2} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(145\)
default	\(\frac {a^{2} A \tan \left (d x +c \right )+a^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a^{2} B \tan \left (d x +c \right )-a^{2} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{2} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(145\)
parallelrisch	\(\frac {2 a^{2} \left (-\frac {3 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) \left (A +\frac {3 B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {3 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) \left (A +\frac {3 B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\left (A +\frac {B}{2}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {5 A}{6}+B \right ) \sin \left (3 d x +3 c \right )+\frac {3 \left (A +\frac {2 B}{3}\right ) \sin \left (d x +c \right )}{2}\right )}{d \left (3 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )\right )}\)	\(147\)
risch	\(-\frac {i a^{2} \left (6 A \,{\mathrm e}^{5 i \left (d x +c \right )}+3 B \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A \,{\mathrm e}^{4 i \left (d x +c \right )}-12 B \,{\mathrm e}^{4 i \left (d x +c \right )}-24 A \,{\mathrm e}^{2 i \left (d x +c \right )}-24 B \,{\mathrm e}^{2 i \left (d x +c \right )}-6 A \,{\mathrm e}^{i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-10 A -12 B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a^{2} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}-\frac {a^{2} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}\)	\(214\)
norman	\(\frac {-\frac {2 a^{2} \left (2 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a^{2} \left (2 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {a^{2} \left (2 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {2 a^{2} \left (2 A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {a^{2} \left (6 A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a^{2} \left (38 A +21 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {a^{2} \left (2 A +3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (2 A +3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(242\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.11 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (2 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (5 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, A a^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:

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

Output:

1/12*(3*(2*A + 3*B)*a^2*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*A + 3* 
B)*a^2*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(5*A + 6*B)*a^2*cos(d* 
x + c)^2 + 3*(2*A + B)*a^2*cos(d*x + c) + 2*A*a^2)*sin(d*x + c))/(d*cos(d* 
x + c)^3)
 

Sympy [F]

\[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=a^{2} \left (\int A \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 A \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 B \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \] Input:

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

Output:

a**2*(Integral(A*sec(c + d*x)**4, x) + Integral(2*A*cos(c + d*x)*sec(c + d 
*x)**4, x) + Integral(A*cos(c + d*x)**2*sec(c + d*x)**4, x) + Integral(B*c 
os(c + d*x)*sec(c + d*x)**4, x) + Integral(2*B*cos(c + d*x)**2*sec(c + d*x 
)**4, x) + Integral(B*cos(c + d*x)**3*sec(c + d*x)**4, x))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.54 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} - 6 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{2} \tan \left (d x + c\right ) + 24 \, B a^{2} \tan \left (d x + c\right )}{12 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.58 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (2 \, A a^{2} + 3 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a^{2} + 3 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:

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

Output:

1/6*(3*(2*A*a^2 + 3*B*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a^2 
 + 3*B*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*a^2*tan(1/2*d*x + 
1/2*c)^5 + 9*B*a^2*tan(1/2*d*x + 1/2*c)^5 - 16*A*a^2*tan(1/2*d*x + 1/2*c)^ 
3 - 24*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 18*A*a^2*tan(1/2*d*x + 1/2*c) + 15*B 
*a^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 = 43.13 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.28 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {2\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+\frac {3\,B}{2}\right )}{d}-\frac {\left (2\,A\,a^2+3\,B\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {16\,A\,a^2}{3}-8\,B\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,A\,a^2+5\,B\,a^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:

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

Output:

(2*a^2*atanh(tan(c/2 + (d*x)/2))*(A + (3*B)/2))/d - (tan(c/2 + (d*x)/2)*(6 
*A*a^2 + 5*B*a^2) + tan(c/2 + (d*x)/2)^5*(2*A*a^2 + 3*B*a^2) - tan(c/2 + ( 
d*x)/2)^3*((16*A*a^2)/3 + 8*B*a^2))/(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 = 0.18 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.65 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {a^{2} \left (-6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +10 \sin \left (d x +c \right )^{3} a +12 \sin \left (d x +c \right )^{3} b -12 \sin \left (d x +c \right ) a -12 \sin \left (d x +c \right ) b \right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

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

Output:

(a**2*( - 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - 9*c 
os(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 6*cos(c + d*x)*l 
og(tan((c + d*x)/2) - 1)*a + 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b + 
6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 9*cos(c + d*x 
)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 6*cos(c + d*x)*log(tan((c 
+ d*x)/2) + 1)*a - 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b - 6*cos(c + 
d*x)*sin(c + d*x)*a - 3*cos(c + d*x)*sin(c + d*x)*b + 10*sin(c + d*x)**3*a 
 + 12*sin(c + d*x)**3*b - 12*sin(c + d*x)*a - 12*sin(c + d*x)*b))/(6*cos(c 
 + d*x)*d*(sin(c + d*x)**2 - 1))