\(\int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\) [78]

Optimal result

Integrand size = 31, antiderivative size = 173 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {1}{8} a^4 (35 A+48 B) x+\frac {a^4 B \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (7 A+8 B) \sin (c+d x)}{8 d}+\frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(7 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 d}+\frac {(35 A+32 B) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {4102, 4081, 3855} \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {5 a^4 (7 A+8 B) \sin (c+d x)}{8 d}+\frac {(35 A+32 B) \sin (c+d x) \cos (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+\frac {1}{8} a^4 x (35 A+48 B)+\frac {a^4 B \text {arctanh}(\sin (c+d x))}{d}+\frac {(7 A+4 B) \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{12 d}+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{4} \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (a (7 A+4 B)+4 a B \sec (c+d x)) \, dx \\ & = \frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(7 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 d}+\frac {1}{12} \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (a^2 (35 A+32 B)+12 a^2 B \sec (c+d x)\right ) \, dx \\ & = \frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(7 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 d}+\frac {(35 A+32 B) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}+\frac {1}{24} \int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^3 (7 A+8 B)+24 a^3 B \sec (c+d x)\right ) \, dx \\ & = \frac {5 a^4 (7 A+8 B) \sin (c+d x)}{8 d}+\frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(7 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 d}+\frac {(35 A+32 B) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}-\frac {1}{24} \int \left (-3 a^4 (35 A+48 B)-24 a^4 B \sec (c+d x)\right ) \, dx \\ & = \frac {1}{8} a^4 (35 A+48 B) x+\frac {5 a^4 (7 A+8 B) \sin (c+d x)}{8 d}+\frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(7 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 d}+\frac {(35 A+32 B) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}+\left (a^4 B\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{8} a^4 (35 A+48 B) x+\frac {a^4 B \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (7 A+8 B) \sin (c+d x)}{8 d}+\frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(7 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 d}+\frac {(35 A+32 B) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {a^4 \left (420 A d x+576 B d x-96 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+96 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+24 (28 A+27 B) \sin (c+d x)+24 (7 A+4 B) \sin (2 (c+d x))+32 A \sin (3 (c+d x))+8 B \sin (3 (c+d x))+3 A \sin (4 (c+d x))\right )}{96 d} \]

[In]

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

[Out]

(a^4*(420*A*d*x + 576*B*d*x - 96*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 96*B*Log[Cos[(c + d*x)/2] + Sin[
(c + d*x)/2]] + 24*(28*A + 27*B)*Sin[c + d*x] + 24*(7*A + 4*B)*Sin[2*(c + d*x)] + 32*A*Sin[3*(c + d*x)] + 8*B*
Sin[3*(c + d*x)] + 3*A*Sin[4*(c + d*x)]))/(96*d)

Maple [A] (verified)

Time = 2.60 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.63

method	result	size
parallelrisch	\(\frac {\left (-32 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+32 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 \left (7 A +4 B \right ) \sin \left (2 d x +2 c \right )+\frac {8 \left (4 A +B \right ) \sin \left (3 d x +3 c \right )}{3}+A \sin \left (4 d x +4 c \right )+8 \left (28 A +27 B \right ) \sin \left (d x +c \right )+140 d \left (A +\frac {48 B}{35}\right ) x \right ) a^{4}}{32 d}\)	\(109\)
derivativedivides	\(\frac {a^{4} A \left (d x +c \right )+B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \sin \left (d x +c \right )+4 B \,a^{4} \left (d x +c \right )+6 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 B \,a^{4} \sin \left (d x +c \right )+\frac {4 a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(208\)
default	\(\frac {a^{4} A \left (d x +c \right )+B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \sin \left (d x +c \right )+4 B \,a^{4} \left (d x +c \right )+6 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 B \,a^{4} \sin \left (d x +c \right )+\frac {4 a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(208\)
risch	\(\frac {35 a^{4} A x}{8}+6 a^{4} x B -\frac {7 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {27 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{4}}{8 d}+\frac {7 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {27 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{4}}{8 d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {a^{4} A \sin \left (4 d x +4 c \right )}{32 d}+\frac {a^{4} A \sin \left (3 d x +3 c \right )}{3 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{4}}{12 d}+\frac {7 \sin \left (2 d x +2 c \right ) a^{4} A}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{4}}{d}\)	\(224\)
norman	\(\frac {\left (\frac {35}{8} a^{4} A +6 B \,a^{4}\right ) x +\left (-\frac {35}{2} a^{4} A -24 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {35}{2} a^{4} A -24 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {35}{8} a^{4} A +6 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (\frac {105}{4} a^{4} A +36 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {5 a^{4} \left (A -24 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {25 a^{4} \left (5 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}-\frac {7 a^{4} \left (5 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 d}-\frac {19 a^{4} \left (7 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {5 a^{4} \left (7 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4 d}+\frac {3 a^{4} \left (31 A +24 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {5 a^{4} \left (103 A +136 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 d}-\frac {a^{4} \left (605 A +344 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {B \,a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {B \,a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(398\)

[In]

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

[Out]

1/32*(-32*B*ln(tan(1/2*d*x+1/2*c)-1)+32*B*ln(tan(1/2*d*x+1/2*c)+1)+8*(7*A+4*B)*sin(2*d*x+2*c)+8/3*(4*A+B)*sin(
3*d*x+3*c)+A*sin(4*d*x+4*c)+8*(28*A+27*B)*sin(d*x+c)+140*d*(A+48/35*B)*x)*a^4/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.68 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (35 \, A + 48 \, B\right )} a^{4} d x + 12 \, B a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, B a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, A a^{4} \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{2} + 3 \, {\left (27 \, A + 16 \, B\right )} a^{4} \cos \left (d x + c\right ) + 160 \, {\left (A + B\right )} a^{4}\right )} \sin \left (d x + c\right )}{24 \, d} \]

[In]

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

[Out]

1/24*(3*(35*A + 48*B)*a^4*d*x + 12*B*a^4*log(sin(d*x + c) + 1) - 12*B*a^4*log(-sin(d*x + c) + 1) + (6*A*a^4*co
s(d*x + c)^3 + 8*(4*A + B)*a^4*cos(d*x + c)^2 + 3*(27*A + 16*B)*a^4*cos(d*x + c) + 160*(A + B)*a^4)*sin(d*x +
c))/d

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.18 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=-\frac {128 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 144 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 96 \, {\left (d x + c\right )} A a^{4} + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} - 96 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 384 \, {\left (d x + c\right )} B a^{4} - 48 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 384 \, A a^{4} \sin \left (d x + c\right ) - 576 \, B a^{4} \sin \left (d x + c\right )}{96 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),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 - 48*B*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))/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.24 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {24 \, B a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 24 \, B a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (35 \, A a^{4} + 48 \, B a^{4}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (105 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 385 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 424 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 511 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 520 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 279 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 216 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

[In]

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

[Out]

1/24*(24*B*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 24*B*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(35*A*a^4
+ 48*B*a^4)*(d*x + c) + 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 + 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 + 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 + 279*A*a^4*tan(1/2*d*x + 1/2*c) + 216*B*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 = 13.91 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.09 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {105\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+144\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+24\,B\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+21\,A\,a^4\,\sin \left (2\,c+2\,d\,x\right )+4\,A\,a^4\,\sin \left (3\,c+3\,d\,x\right )+\frac {3\,A\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{8}+12\,B\,a^4\,\sin \left (2\,c+2\,d\,x\right )+B\,a^4\,\sin \left (3\,c+3\,d\,x\right )+84\,A\,a^4\,\sin \left (c+d\,x\right )+81\,B\,a^4\,\sin \left (c+d\,x\right )}{12\,d} \]

[In]

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

[Out]

(105*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 144*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
+ 24*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 21*A*a^4*sin(2*c + 2*d*x) + 4*A*a^4*sin(3*c + 3*d*x
) + (3*A*a^4*sin(4*c + 4*d*x))/8 + 12*B*a^4*sin(2*c + 2*d*x) + B*a^4*sin(3*c + 3*d*x) + 84*A*a^4*sin(c + d*x)
+ 81*B*a^4*sin(c + d*x))/(12*d)