\(\int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [444]

Optimal result
Mathematica [A] (warning: unable to verify)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 41, antiderivative size = 217 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} a^4 (35 A+48 B+52 C) x+\frac {a^4 (B+4 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (7 A+8 B+4 C) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac {(7 A+8 B+4 C) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{8 d}-\frac {(35 A+32 B-12 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d} \] Output:

1/8*a^4*(35*A+48*B+52*C)*x+a^4*(B+4*C)*arctanh(sin(d*x+c))/d+5/8*a^4*(7*A+ 
8*B+4*C)*sin(d*x+c)/d+1/3*a*(A+B)*cos(d*x+c)^2*(a+a*sec(d*x+c))^3*sin(d*x+ 
c)/d+1/4*A*cos(d*x+c)^3*(a+a*sec(d*x+c))^4*sin(d*x+c)/d+1/8*(7*A+8*B+4*C)* 
cos(d*x+c)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/d-1/24*(35*A+32*B-12*C)*(a^4+ 
a^4*sec(d*x+c))*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 9.29 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.95 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^4 \cos ^2(c+d x) (1+\cos (c+d x))^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (12 (35 A+48 B+52 C) x-\frac {96 (B+4 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {96 (B+4 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {24 (28 A+27 B+16 C) \cos (d x) \sin (c)}{d}+\frac {24 (7 A+4 B+C) \cos (2 d x) \sin (2 c)}{d}+\frac {8 (4 A+B) \cos (3 d x) \sin (3 c)}{d}+\frac {3 A \cos (4 d x) \sin (4 c)}{d}+\frac {24 (28 A+27 B+16 C) \cos (c) \sin (d x)}{d}+\frac {24 (7 A+4 B+C) \cos (2 c) \sin (2 d x)}{d}+\frac {8 (4 A+B) \cos (3 c) \sin (3 d x)}{d}+\frac {3 A \cos (4 c) \sin (4 d x)}{d}+\frac {96 C \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {96 C \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{768 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \] Input:

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

Output:

(a^4*Cos[c + d*x]^2*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(A + B*Sec[c + 
 d*x] + C*Sec[c + d*x]^2)*(12*(35*A + 48*B + 52*C)*x - (96*(B + 4*C)*Log[C 
os[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (96*(B + 4*C)*Log[Cos[(c + d*x)/2 
] + Sin[(c + d*x)/2]])/d + (24*(28*A + 27*B + 16*C)*Cos[d*x]*Sin[c])/d + ( 
24*(7*A + 4*B + C)*Cos[2*d*x]*Sin[2*c])/d + (8*(4*A + B)*Cos[3*d*x]*Sin[3* 
c])/d + (3*A*Cos[4*d*x]*Sin[4*c])/d + (24*(28*A + 27*B + 16*C)*Cos[c]*Sin[ 
d*x])/d + (24*(7*A + 4*B + C)*Cos[2*c]*Sin[2*d*x])/d + (8*(4*A + B)*Cos[3* 
c]*Sin[3*d*x])/d + (3*A*Cos[4*c]*Sin[4*d*x])/d + (96*C*Sin[(d*x)/2])/(d*(C 
os[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (96*C*Sin[(d* 
x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/( 
768*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {3042, 4574, 3042, 4505, 3042, 4505, 3042, 4506, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4505

\(\displaystyle \frac {\frac {1}{3} \int \cos ^2(c+d x) (\sec (c+d x) a+a)^3 \left (3 a^2 (7 A+8 B+4 C)-a^2 (7 A+4 B-12 C) \sec (c+d x)\right )dx+\frac {4 a^2 (A+B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (3 a^2 (7 A+8 B+4 C)-a^2 (7 A+4 B-12 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {4 a^2 (A+B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d}\)

\(\Big \downarrow \) 4505

\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int \cos (c+d x) (\sec (c+d x) a+a)^2 \left (2 a^3 (35 A+44 B+36 C)-a^3 (35 A+32 B-12 C) \sec (c+d x)\right )dx+\frac {3 a^3 (7 A+8 B+4 C) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\right )+\frac {4 a^2 (A+B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a^3 (35 A+44 B+36 C)-a^3 (35 A+32 B-12 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {3 a^3 (7 A+8 B+4 C) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\right )+\frac {4 a^2 (A+B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d}\)

\(\Big \downarrow \) 4506

\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (\int 3 \cos (c+d x) (\sec (c+d x) a+a) \left (5 (7 A+8 B+4 C) a^4+8 (B+4 C) \sec (c+d x) a^4\right )dx-\frac {(35 A+32 B-12 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}\right )+\frac {3 a^3 (7 A+8 B+4 C) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\right )+\frac {4 a^2 (A+B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (3 \int \cos (c+d x) (\sec (c+d x) a+a) \left (5 (7 A+8 B+4 C) a^4+8 (B+4 C) \sec (c+d x) a^4\right )dx-\frac {(35 A+32 B-12 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}\right )+\frac {3 a^3 (7 A+8 B+4 C) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\right )+\frac {4 a^2 (A+B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (3 \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (5 (7 A+8 B+4 C) a^4+8 (B+4 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^4\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(35 A+32 B-12 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}\right )+\frac {3 a^3 (7 A+8 B+4 C) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\right )+\frac {4 a^2 (A+B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d}\)

\(\Big \downarrow \) 4484

\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (3 \left (\frac {5 a^5 (7 A+8 B+4 C) \sin (c+d x)}{d}-\int \left (-\left ((35 A+48 B+52 C) a^5\right )-8 (B+4 C) \sec (c+d x) a^5\right )dx\right )-\frac {(35 A+32 B-12 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}\right )+\frac {3 a^3 (7 A+8 B+4 C) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\right )+\frac {4 a^2 (A+B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {4 a^2 (A+B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}+\frac {1}{3} \left (\frac {1}{2} \left (3 \left (\frac {5 a^5 (7 A+8 B+4 C) \sin (c+d x)}{d}+a^5 x (35 A+48 B+52 C)+\frac {8 a^5 (B+4 C) \text {arctanh}(\sin (c+d x))}{d}\right )-\frac {(35 A+32 B-12 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}\right )+\frac {3 a^3 (7 A+8 B+4 C) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\right )}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d}\)

Input:

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

Output:

(A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/(4*d) + ((4*a^2*(A 
+ B)*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + ((3*a^3*( 
7*A + 8*B + 4*C)*Cos[c + d*x]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + 
 (-(((35*A + 32*B - 12*C)*(a^5 + a^5*Sec[c + d*x])*Sin[c + d*x])/d) + 3*(a 
^5*(35*A + 48*B + 52*C)*x + (8*a^5*(B + 4*C)*ArcTanh[Sin[c + d*x]])/d + (5 
*a^5*(7*A + 8*B + 4*C)*Sin[c + d*x])/d))/2)/3)/(4*a)
 

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 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 4505
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot 
[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim 
p[b/(a*d*n)   Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim 
p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] 
/; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 
] && GtQ[m, 1/2] && LtQ[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 = 1.48 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.77

method result size
parallelrisch \(\frac {11 a^{4} \left (-\frac {3 \cos \left (d x +c \right ) \left (B +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{11}+\frac {3 \cos \left (d x +c \right ) \left (B +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{11}+\left (A +\frac {41 B}{44}+\frac {6 C}{11}\right ) \sin \left (2 d x +2 c \right )+\frac {3 \left (\frac {57 A}{32}+B +\frac {C}{4}\right ) \sin \left (3 d x +3 c \right )}{22}+\frac {\left (A +\frac {B}{4}\right ) \sin \left (4 d x +4 c \right )}{22}+\frac {3 A \sin \left (5 d x +5 c \right )}{704}+\frac {105 x \left (A +\frac {48 B}{35}+\frac {52 C}{35}\right ) d \cos \left (d x +c \right )}{88}+\frac {21 \sin \left (d x +c \right ) \left (A +\frac {4 B}{7}+\frac {9 C}{7}\right )}{88}\right )}{3 d \cos \left (d x +c \right )}\) \(167\)
derivativedivides \(\frac {a^{4} A \left (d x +c \right )+B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} C \tan \left (d x +c \right )+4 a^{4} A \sin \left (d x +c \right )+4 B \,a^{4} \left (d x +c \right )+4 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 B \,a^{4} \sin \left (d x +c \right )+6 a^{4} C \left (d x +c \right )+\frac {4 a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} C \sin \left (d x +c \right )+a^{4} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) \(289\)
default \(\frac {a^{4} A \left (d x +c \right )+B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} C \tan \left (d x +c \right )+4 a^{4} A \sin \left (d x +c \right )+4 B \,a^{4} \left (d x +c \right )+4 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 B \,a^{4} \sin \left (d x +c \right )+6 a^{4} C \left (d x +c \right )+\frac {4 a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} C \sin \left (d x +c \right )+a^{4} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) \(289\)
risch \(\frac {35 a^{4} A x}{8}+6 a^{4} x B +\frac {13 a^{4} x C}{2}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B \,a^{4}}{2 d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a^{4} C}{8 d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B \,a^{4}}{2 d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a^{4} C}{8 d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} a^{4} C}{d}+\frac {2 i a^{4} C}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {7 i a^{4} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {27 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{4}}{8 d}-\frac {7 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {7 i a^{4} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} a^{4} C}{d}+\frac {27 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{4}}{8 d}+\frac {7 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}+\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}-\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {a^{4} A \sin \left (4 d x +4 c \right )}{32 d}+\frac {a^{4} A \sin \left (3 d x +3 c \right )}{3 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{4}}{12 d}\) \(415\)

Input:

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,meth 
od=_RETURNVERBOSE)
 

Output:

11/3*a^4*(-3/11*cos(d*x+c)*(B+4*C)*ln(tan(1/2*d*x+1/2*c)-1)+3/11*cos(d*x+c 
)*(B+4*C)*ln(tan(1/2*d*x+1/2*c)+1)+(A+41/44*B+6/11*C)*sin(2*d*x+2*c)+3/22* 
(57/32*A+B+1/4*C)*sin(3*d*x+3*c)+1/22*(A+1/4*B)*sin(4*d*x+4*c)+3/704*A*sin 
(5*d*x+5*c)+105/88*x*(A+48/35*B+52/35*C)*d*cos(d*x+c)+21/88*sin(d*x+c)*(A+ 
4/7*B+9/7*C))/d/cos(d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.82 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (35 \, A + 48 \, B + 52 \, C\right )} a^{4} d x \cos \left (d x + c\right ) + 12 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, A a^{4} \cos \left (d x + c\right )^{4} + 8 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (27 \, A + 16 \, B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 32 \, {\left (5 \, A + 5 \, B + 3 \, C\right )} a^{4} \cos \left (d x + c\right ) + 24 \, C a^{4}\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \] Input:

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

Output:

1/24*(3*(35*A + 48*B + 52*C)*a^4*d*x*cos(d*x + c) + 12*(B + 4*C)*a^4*cos(d 
*x + c)*log(sin(d*x + c) + 1) - 12*(B + 4*C)*a^4*cos(d*x + c)*log(-sin(d*x 
 + c) + 1) + (6*A*a^4*cos(d*x + c)^4 + 8*(4*A + B)*a^4*cos(d*x + c)^3 + 3* 
(27*A + 16*B + 4*C)*a^4*cos(d*x + c)^2 + 32*(5*A + 5*B + 3*C)*a^4*cos(d*x 
+ c) + 24*C*a^4)*sin(d*x + c))/(d*cos(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.34 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {128 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 144 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 96 \, {\left (d x + c\right )} A a^{4} + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} - 96 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 384 \, {\left (d x + c\right )} B a^{4} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 576 \, {\left (d x + c\right )} C a^{4} - 48 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 192 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 384 \, A a^{4} \sin \left (d x + c\right ) - 576 \, B a^{4} \sin \left (d x + c\right ) - 384 \, C a^{4} \sin \left (d x + c\right ) - 96 \, C a^{4} \tan \left (d x + c\right )}{96 \, d} \] Input:

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

Output:

-1/96*(128*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 3*(12*d*x + 12*c + si 
n(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 - 144*(2*d*x + 2*c + sin(2*d*x 
+ 2*c))*A*a^4 - 96*(d*x + c)*A*a^4 + 32*(sin(d*x + c)^3 - 3*sin(d*x + c))* 
B*a^4 - 96*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 - 384*(d*x + c)*B*a^4 - 
24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 - 576*(d*x + c)*C*a^4 - 48*B*a^4 
*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 192*C*a^4*(log(sin(d*x 
+ c) + 1) - log(sin(d*x + c) - 1)) - 384*A*a^4*sin(d*x + c) - 576*B*a^4*si 
n(d*x + c) - 384*C*a^4*sin(d*x + c) - 96*C*a^4*tan(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.53 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {\frac {48 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, {\left (35 \, A a^{4} + 48 \, B a^{4} + 52 \, C a^{4}\right )} {\left (d x + c\right )} - 24 \, {\left (B a^{4} + 4 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 24 \, {\left (B a^{4} + 4 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 84 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 385 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 424 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 276 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 511 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 520 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 300 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 279 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 216 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 108 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \] Input:

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

Output:

-1/24*(48*C*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(35* 
A*a^4 + 48*B*a^4 + 52*C*a^4)*(d*x + c) - 24*(B*a^4 + 4*C*a^4)*log(abs(tan( 
1/2*d*x + 1/2*c) + 1)) + 24*(B*a^4 + 4*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) 
 - 1)) - 2*(105*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 120*B*a^4*tan(1/2*d*x + 1/2 
*c)^7 + 84*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 385*A*a^4*tan(1/2*d*x + 1/2*c)^5 
 + 424*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 276*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 5 
11*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 520*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 300*C 
*a^4*tan(1/2*d*x + 1/2*c)^3 + 279*A*a^4*tan(1/2*d*x + 1/2*c) + 216*B*a^4*t 
an(1/2*d*x + 1/2*c) + 108*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c 
)^2 + 1)^4)/d
 

Mupad [B] (verification not implemented)

Time = 13.46 (sec) , antiderivative size = 1244, normalized size of antiderivative = 5.73 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

- (tan(c/2 + (d*x)/2)^5*((21*A*a^4)/2 + 8*B*a^4 - 10*C*a^4) + tan(c/2 + (d 
*x)/2)^9*((35*A*a^4)/4 + 10*B*a^4 + 5*C*a^4) - tan(c/2 + (d*x)/2)^3*((58*A 
*a^4)/3 + (76*B*a^4)/3 + 24*C*a^4) + tan(c/2 + (d*x)/2)^7*((70*A*a^4)/3 + 
(76*B*a^4)/3 + 8*C*a^4) - tan(c/2 + (d*x)/2)*((93*A*a^4)/4 + 18*B*a^4 + 11 
*C*a^4))/(d*(3*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + 
 (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 - tan(c/2 + (d*x)/2)^10 + 1)) - (a^4* 
atan(((a^4*(tan(c/2 + (d*x)/2)*((1225*A^2*a^8)/2 + 1184*B^2*a^8 + 1864*C^2 
*a^8 + 1680*A*B*a^8 + 1820*A*C*a^8 + 2752*B*C*a^8) - (a^4*(35*A + 48*B + 5 
2*C)*(140*A*a^4 + 224*B*a^4 + 336*C*a^4)*1i)/8)*(35*A + 48*B + 52*C))/8 + 
(a^4*(tan(c/2 + (d*x)/2)*((1225*A^2*a^8)/2 + 1184*B^2*a^8 + 1864*C^2*a^8 + 
 1680*A*B*a^8 + 1820*A*C*a^8 + 2752*B*C*a^8) + (a^4*(35*A + 48*B + 52*C)*( 
140*A*a^4 + 224*B*a^4 + 336*C*a^4)*1i)/8)*(35*A + 48*B + 52*C))/8)/(1920*B 
^3*a^12 + 4160*C^3*a^12 + 3080*A*B^2*a^12 + 1225*A^2*B*a^12 + 10080*A*C^2* 
a^12 + 4900*A^2*C*a^12 + 13200*B*C^2*a^12 + 10720*B^2*C*a^12 - (a^4*(tan(c 
/2 + (d*x)/2)*((1225*A^2*a^8)/2 + 1184*B^2*a^8 + 1864*C^2*a^8 + 1680*A*B*a 
^8 + 1820*A*C*a^8 + 2752*B*C*a^8) - (a^4*(35*A + 48*B + 52*C)*(140*A*a^4 + 
 224*B*a^4 + 336*C*a^4)*1i)/8)*(35*A + 48*B + 52*C)*1i)/8 + (a^4*(tan(c/2 
+ (d*x)/2)*((1225*A^2*a^8)/2 + 1184*B^2*a^8 + 1864*C^2*a^8 + 1680*A*B*a^8 
+ 1820*A*C*a^8 + 2752*B*C*a^8) + (a^4*(35*A + 48*B + 52*C)*(140*A*a^4 + 22 
4*B*a^4 + 336*C*a^4)*1i)/8)*(35*A + 48*B + 52*C)*1i)/8 + 14840*A*B*C*a^...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.45 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{4} \left (-24 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -96 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +24 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +96 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a -8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +192 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +168 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +96 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c +105 \cos \left (d x +c \right ) a c +105 \cos \left (d x +c \right ) a d x +144 \cos \left (d x +c \right ) b c +144 \cos \left (d x +c \right ) b d x +156 \cos \left (d x +c \right ) c^{2}+156 \cos \left (d x +c \right ) c d x +6 \sin \left (d x +c \right )^{5} a -93 \sin \left (d x +c \right )^{3} a -48 \sin \left (d x +c \right )^{3} b -12 \sin \left (d x +c \right )^{3} c +87 \sin \left (d x +c \right ) a +48 \sin \left (d x +c \right ) b +36 \sin \left (d x +c \right ) c \right )}{24 \cos \left (d x +c \right ) d} \] Input:

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

Output:

(a**4*( - 24*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b - 96*cos(c + d*x)*lo 
g(tan((c + d*x)/2) - 1)*c + 24*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b + 
96*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*c - 32*cos(c + d*x)*sin(c + d*x) 
**3*a - 8*cos(c + d*x)*sin(c + d*x)**3*b + 192*cos(c + d*x)*sin(c + d*x)*a 
 + 168*cos(c + d*x)*sin(c + d*x)*b + 96*cos(c + d*x)*sin(c + d*x)*c + 105* 
cos(c + d*x)*a*c + 105*cos(c + d*x)*a*d*x + 144*cos(c + d*x)*b*c + 144*cos 
(c + d*x)*b*d*x + 156*cos(c + d*x)*c**2 + 156*cos(c + d*x)*c*d*x + 6*sin(c 
 + d*x)**5*a - 93*sin(c + d*x)**3*a - 48*sin(c + d*x)**3*b - 12*sin(c + d* 
x)**3*c + 87*sin(c + d*x)*a + 48*sin(c + d*x)*b + 36*sin(c + d*x)*c))/(24* 
cos(c + d*x)*d)