\(\int \cos ^2(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [422]

Optimal result

Integrand size = 41, antiderivative size = 128 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a^2 (3 A+4 B+2 C) x+\frac {a^2 (B+2 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (3 A+2 B-2 C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(329\) vs. \(2(128)=256\).

Time = 4.40 (sec) , antiderivative size = 329, normalized size of antiderivative = 2.57 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \cos ^2(c+d x) (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 (3 A+4 B+2 C) x-\frac {4 (B+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 (B+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 (2 A+B) \cos (d x) \sin (c)}{d}+\frac {A \cos (2 d x) \sin (2 c)}{d}+\frac {4 (2 A+B) \cos (c) \sin (d x)}{d}+\frac {A \cos (2 c) \sin (2 d x)}{d}+\frac {4 C \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 C \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{8 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \] Input:

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

Output:

(a^2*Cos[c + d*x]^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(A + B*Sec[c + 
 d*x] + C*Sec[c + d*x]^2)*(2*(3*A + 4*B + 2*C)*x - (4*(B + 2*C)*Log[Cos[(c 
 + d*x)/2] - Sin[(c + d*x)/2]])/d + (4*(B + 2*C)*Log[Cos[(c + d*x)/2] + Si 
n[(c + d*x)/2]])/d + (4*(2*A + B)*Cos[d*x]*Sin[c])/d + (A*Cos[2*d*x]*Sin[2 
*c])/d + (4*(2*A + B)*Cos[c]*Sin[d*x])/d + (A*Cos[2*c]*Sin[2*d*x])/d + (4* 
C*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x) 
/2])) + (4*C*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Si 
n[(c + d*x)/2]))))/(8*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3042, 4574, 3042, 4506, 3042, 4484, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

\(\displaystyle \frac {\int \cos (c+d x) (\sec (c+d x) a+a)^2 (2 a (A+B)-a (A-2 C) \sec (c+d x))dx}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4484

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(A*Cos[c + d*x]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + (a^3*(3*A + 4 
*B + 2*C)*x + (2*a^3*(B + 2*C)*ArcTanh[Sin[c + d*x]])/d + (a^3*(3*A + 2*B 
- 2*C)*Sin[c + d*x])/d - ((A - 2*C)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x]) 
/d)/(2*a)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4484

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

rule 4506

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

rule 4574

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/(b*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[ 
e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] 
, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] 
&&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
 

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.97

method	result	size
parallelrisch	\(\frac {a^{2} \left (-\cos \left (d x +c \right ) \left (B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\cos \left (d x +c \right ) \left (B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (A +\frac {B}{2}\right ) \sin \left (2 d x +2 c \right )+\frac {A \sin \left (3 d x +3 c \right )}{8}+\frac {3 \left (A +\frac {4 B}{3}+\frac {2 C}{3}\right ) x d \cos \left (d x +c \right )}{2}+\frac {\sin \left (d x +c \right ) \left (A +8 C \right )}{8}\right )}{d \cos \left (d x +c \right )}\)	\(124\)
derivativedivides	\(\frac {a^{2} A \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{2} \tan \left (d x +c \right )+2 a^{2} A \sin \left (d x +c \right )+2 B \,a^{2} \left (d x +c \right )+2 C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \sin \left (d x +c \right )+C \,a^{2} \left (d x +c \right )}{d}\)	\(137\)
default	\(\frac {a^{2} A \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{2} \tan \left (d x +c \right )+2 a^{2} A \sin \left (d x +c \right )+2 B \,a^{2} \left (d x +c \right )+2 C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \sin \left (d x +c \right )+C \,a^{2} \left (d x +c \right )}{d}\)	\(137\)
risch	\(\frac {3 a^{2} A x}{2}+2 a^{2} B x +a^{2} x C -\frac {i a^{2} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i a^{2} A \,{\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {i B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{2} A \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2}}{2 d}+\frac {i a^{2} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i C \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\)	\(246\)
norman	\(\frac {\left (-\frac {3}{2} a^{2} A -2 B \,a^{2}-C \,a^{2}\right ) x +\left (-\frac {3}{2} a^{2} A -2 B \,a^{2}-C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {3}{2} a^{2} A +2 B \,a^{2}+C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3}{2} a^{2} A +2 B \,a^{2}+C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-3 a^{2} A -4 B \,a^{2}-2 C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (3 a^{2} A +4 B \,a^{2}+2 C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {a^{2} \left (3 A +2 B -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {4 a^{2} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {4 a^{2} \left (3 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {2 a^{2} \left (3 A -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a^{2} \left (5 A +2 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}+\frac {a^{2} \left (B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{2} \left (B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(389\)

Input:

int(cos(d*x+c)^2*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,meth 
od=_RETURNVERBOSE)
 

Output:

a^2*(-cos(d*x+c)*(B+2*C)*ln(tan(1/2*d*x+1/2*c)-1)+cos(d*x+c)*(B+2*C)*ln(ta 
n(1/2*d*x+1/2*c)+1)+(A+1/2*B)*sin(2*d*x+2*c)+1/8*A*sin(3*d*x+3*c)+3/2*(A+4 
/3*B+2/3*C)*x*d*cos(d*x+c)+1/8*sin(d*x+c)*(A+8*C))/d/cos(d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.02 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (3 \, A + 4 \, B + 2 \, C\right )} a^{2} d x \cos \left (d x + c\right ) + {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (A a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, C a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \] Input:

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), 
x, algorithm="fricas")
 

Output:

1/2*((3*A + 4*B + 2*C)*a^2*d*x*cos(d*x + c) + (B + 2*C)*a^2*cos(d*x + c)*l 
og(sin(d*x + c) + 1) - (B + 2*C)*a^2*cos(d*x + c)*log(-sin(d*x + c) + 1) + 
 (A*a^2*cos(d*x + c)^2 + 2*(2*A + B)*a^2*cos(d*x + c) + 2*C*a^2)*sin(d*x + 
 c))/(d*cos(d*x + c))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.18 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 4 \, {\left (d x + c\right )} A a^{2} + 8 \, {\left (d x + c\right )} B a^{2} + 4 \, {\left (d x + c\right )} C a^{2} + 2 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a^{2} \sin \left (d x + c\right ) + 4 \, B a^{2} \sin \left (d x + c\right ) + 4 \, C a^{2} \tan \left (d x + c\right )}{4 \, d} \] Input:

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), 
x, algorithm="maxima")
 

Output:

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 4*(d*x + c)*A*a^2 + 8*(d*x + 
 c)*B*a^2 + 4*(d*x + c)*C*a^2 + 2*B*a^2*(log(sin(d*x + c) + 1) - log(sin(d 
*x + c) - 1)) + 4*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 
8*A*a^2*sin(d*x + c) + 4*B*a^2*sin(d*x + c) + 4*C*a^2*tan(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.55 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {\frac {4 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (3 \, A a^{2} + 4 \, B a^{2} + 2 \, C a^{2}\right )} {\left (d x + c\right )} - 2 \, {\left (B a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (B a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \] Input:

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), 
x, algorithm="giac")
 

Output:

-1/2*(4*C*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (3*A*a^2 
 + 4*B*a^2 + 2*C*a^2)*(d*x + c) - 2*(B*a^2 + 2*C*a^2)*log(abs(tan(1/2*d*x 
+ 1/2*c) + 1)) + 2*(B*a^2 + 2*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 
2*(3*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 5*A*a 
^2*tan(1/2*d*x + 1/2*c) + 2*B*a^2*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 = 12.65 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.81 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+4\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}+2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}}{d}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )+\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{8}+C\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \] Input:

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

Output:

(3*A*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 4*B*a^2*atan(sin(c/ 
2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - B*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos( 
c/2 + (d*x)/2))*2i + 2*C*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - 
 C*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*4i)/d + (A*a^2*sin 
(2*c + 2*d*x) + (A*a^2*sin(3*c + 3*d*x))/8 + (B*a^2*sin(2*c + 2*d*x))/2 + 
(A*a^2*sin(c + d*x))/8 + C*a^2*sin(c + d*x))/(d*cos(c + d*x))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.74 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{2} \left (-2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -4 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +4 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c +4 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +3 \cos \left (d x +c \right ) a c +3 \cos \left (d x +c \right ) a d x +4 \cos \left (d x +c \right ) b c +4 \cos \left (d x +c \right ) b d x +2 \cos \left (d x +c \right ) c^{2}+2 \cos \left (d x +c \right ) c d x -\sin \left (d x +c \right )^{3} a +\sin \left (d x +c \right ) a +2 \sin \left (d x +c \right ) c \right )}{2 \cos \left (d x +c \right ) d} \] Input:

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

Output:

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