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

Optimal result

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

[Out]

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)*co
s(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

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {4171, 4102, 4103, 4081, 3855} \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {5 a^4 (7 A+8 B+4 C) \sin (c+d x)}{8 d}-\frac {(35 A+32 B-12 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+\frac {1}{8} a^4 x (35 A+48 B+52 C)+\frac {a^4 (B+4 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {(7 A+8 B+4 C) \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{8 d}+\frac {a (A+B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^4}{4 d} \]

[In]

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]

[Out]

(a^4*(35*A + 48*B + 52*C)*x)/8 + (a^4*(B + 4*C)*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(7*A + 8*B + 4*C)*Sin[c + d*
x])/(8*d) + (a*(A + B)*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + (A*Cos[c + d*x]^3*(a + a*Se
c[c + d*x])^4*Sin[c + d*x])/(4*d) + ((7*A + 8*B + 4*C)*Cos[c + d*x]*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(
8*d) - ((35*A + 32*B - 12*C)*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(24*d)

Rule 3855

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

Rule 4081

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] + Dist[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 4102

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]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[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 4103

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] + Dist[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 4171

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] - Dist[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])

Rubi steps \begin{align*} \text {integral}& = \frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{4 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (4 a (A+B)-a (A-4 C) \sec (c+d x)) \, dx}{4 a} \\ & = \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 {\int \cos ^2(c+d x) (a+a \sec (c+d x))^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}{12 a} \\ & = \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 {\int \cos (c+d x) (a+a \sec (c+d x))^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}{24 a} \\ & = \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}+\frac {\int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^4 (7 A+8 B+4 C)+24 a^4 (B+4 C) \sec (c+d x)\right ) \, dx}{24 a} \\ & = \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}-\frac {\int \left (-3 a^5 (35 A+48 B+52 C)-24 a^5 (B+4 C) \sec (c+d x)\right ) \, dx}{24 a} \\ & = \frac {1}{8} a^4 (35 A+48 B+52 C) x+\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}+\left (a^4 (B+4 C)\right ) \int \sec (c+d x) \, 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} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 10.03 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.95 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^4 \cos ^2(c+d x) (1+\cos (c+d x))^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (12 (35 A+48 B+52 C) x-\frac {96 (B+4 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {96 (B+4 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {24 (28 A+27 B+16 C) \cos (d x) \sin (c)}{d}+\frac {24 (7 A+4 B+C) \cos (2 d x) \sin (2 c)}{d}+\frac {8 (4 A+B) \cos (3 d x) \sin (3 c)}{d}+\frac {3 A \cos (4 d x) \sin (4 c)}{d}+\frac {24 (28 A+27 B+16 C) \cos (c) \sin (d x)}{d}+\frac {24 (7 A+4 B+C) \cos (2 c) \sin (2 d x)}{d}+\frac {8 (4 A+B) \cos (3 c) \sin (3 d x)}{d}+\frac {3 A \cos (4 c) \sin (4 d x)}{d}+\frac {96 C \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {96 C \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{768 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \]

[In]

Integrate[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]

[Out]

(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[Cos[(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*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (96*C*S
in[(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)]))

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.77

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

[In]

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,method=_RETURNVERBOSE)

[Out]

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*s
in(5*d*x+5*c)+105/88*x*d*(A+48/35*B+52/35*C)*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)

none

Time = 0.28 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.82 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (35 \, A + 48 \, B + 52 \, C\right )} a^{4} d x \cos \left (d x + c\right ) + 12 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, A a^{4} \cos \left (d x + c\right )^{4} + 8 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (27 \, A + 16 \, B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 32 \, {\left (5 \, A + 5 \, B + 3 \, C\right )} a^{4} \cos \left (d x + c\right ) + 24 \, C a^{4}\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \]

[In]

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")

[Out]

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

[In]

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)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.34 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {128 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 144 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 96 \, {\left (d x + c\right )} A a^{4} + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} - 96 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 384 \, {\left (d x + c\right )} B a^{4} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 576 \, {\left (d x + c\right )} C a^{4} - 48 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 192 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 384 \, A a^{4} \sin \left (d x + c\right ) - 576 \, B a^{4} \sin \left (d x + c\right ) - 384 \, C a^{4} \sin \left (d x + c\right ) - 96 \, C a^{4} \tan \left (d x + c\right )}{96 \, d} \]

[In]

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")

[Out]

-1/96*(128*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 3*(12*d*x + 12*c + sin(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*sin(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)

none

Time = 0.38 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.53 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {\frac {48 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, {\left (35 \, A a^{4} + 48 \, B a^{4} + 52 \, C a^{4}\right )} {\left (d x + c\right )} - 24 \, {\left (B a^{4} + 4 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 24 \, {\left (B a^{4} + 4 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 84 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 385 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 424 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 276 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 511 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 520 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 300 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 279 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 216 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 108 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

[In]

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")

[Out]

-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
+ 511*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 + 27
9*A*a^4*tan(1/2*d*x + 1/2*c) + 216*B*a^4*tan(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 = 17.69 (sec) , antiderivative size = 1244, normalized size of antiderivative = 5.73 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

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)

[Out]

- (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 + 52*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 + 132
00*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 + 1
680*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 + 14840*A*B*C*a^12))*(35*A + 48*B + 52*C))/(4*d) - (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*(B
+ 4*C)*(140*A*a^4 + 224*B*a^4 + 336*C*a^4))*(B + 4*C)*1i + 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*(B + 4*C)*(140*A*a^4 + 224*B*a^4 + 33
6*C*a^4))*(B + 4*C)*1i)/(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*(B + 4*C)*(140*A*a^4 + 224*B*a^4 + 336
*C*a^4))*(B + 4*C) - 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*(B + 4*C)*(140*A*a^4 + 224*B*a^4 + 336*C*a^4))*(B + 4*C) + 14840*A*B*C*a^12
))*(B + 4*C)*2i)/d