\(\int (a+a \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\) [36]

Optimal result

Integrand size = 33, antiderivative size = 207 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=a^4 C x+\frac {a^4 (7 A+12 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^4 (7 A+10 C) \tan (c+d x)}{2 d}+\frac {(7 A+8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d} \]

[Out]

a^4*C*x+1/2*a^4*(7*A+12*C)*arctanh(sin(d*x+c))/d+1/2*a^4*(7*A+10*C)*tan(d*x+c)/d+1/6*(7*A+8*C)*(a^4+a^4*cos(d*
x+c))*sec(d*x+c)*tan(d*x+c)/d+1/15*(7*A+5*C)*(a^2+a^2*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/d+1/5*a*A*(a+a*cos
(d*x+c))^3*sec(d*x+c)^3*tan(d*x+c)/d+1/5*A*(a+a*cos(d*x+c))^4*sec(d*x+c)^4*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3123, 3054, 3047, 3100, 2814, 3855} \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a^4 (7 A+12 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^4 (7 A+10 C) \tan (c+d x)}{2 d}+\frac {(7 A+8 C) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{6 d}+a^4 C x+\frac {(7 A+5 C) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{15 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^4}{5 d}+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]

[In]

Int[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

a^4*C*x + (a^4*(7*A + 12*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a^4*(7*A + 10*C)*Tan[c + d*x])/(2*d) + ((7*A + 8*C
)*(a^4 + a^4*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(6*d) + ((7*A + 5*C)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c +
d*x]^2*Tan[c + d*x])/(15*d) + (a*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/(5*d) + (A*(a + a*Cos[c
+ d*x])^4*Sec[c + d*x]^4*Tan[c + d*x])/(5*d)

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3047

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 3054

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-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] - Dist[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 3100

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] + Dist[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 3123

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Si
n[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x]
)^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n
+ 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2,
0])

Rule 3855

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

Rubi steps \begin{align*} \text {integral}& = \frac {A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x))^4 (4 a A+5 a C \cos (c+d x)) \sec ^5(c+d x) \, dx}{5 a} \\ & = \frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x))^3 \left (4 a^2 (7 A+5 C)+20 a^2 C \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a} \\ & = \frac {(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x))^2 \left (20 a^3 (7 A+8 C)+60 a^3 C \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a} \\ & = \frac {(7 A+8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x)) \left (60 a^4 (7 A+10 C)+120 a^4 C \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{120 a} \\ & = \frac {(7 A+8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int \left (60 a^5 (7 A+10 C)+\left (120 a^5 C+60 a^5 (7 A+10 C)\right ) \cos (c+d x)+120 a^5 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx}{120 a} \\ & = \frac {a^4 (7 A+10 C) \tan (c+d x)}{2 d}+\frac {(7 A+8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int \left (60 a^5 (7 A+12 C)+120 a^5 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{120 a} \\ & = a^4 C x+\frac {a^4 (7 A+10 C) \tan (c+d x)}{2 d}+\frac {(7 A+8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{2} \left (a^4 (7 A+12 C)\right ) \int \sec (c+d x) \, dx \\ & = a^4 C x+\frac {a^4 (7 A+12 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^4 (7 A+10 C) \tan (c+d x)}{2 d}+\frac {(7 A+8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.09 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.94 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=a^4 C x+\frac {7 a^4 A \text {arctanh}(\sin (c+d x))}{2 d}+\frac {6 a^4 C \text {arctanh}(\sin (c+d x))}{d}+\frac {8 a^4 A \tan (c+d x)}{d}+\frac {7 a^4 C \tan (c+d x)}{d}+\frac {7 a^4 A \sec (c+d x) \tan (c+d x)}{2 d}+\frac {2 a^4 C \sec (c+d x) \tan (c+d x)}{d}+\frac {a^4 A \sec ^3(c+d x) \tan (c+d x)}{d}+\frac {8 a^4 A \tan ^3(c+d x)}{3 d}+\frac {a^4 C \tan ^3(c+d x)}{3 d}+\frac {a^4 A \tan ^5(c+d x)}{5 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

a^4*C*x + (7*a^4*A*ArcTanh[Sin[c + d*x]])/(2*d) + (6*a^4*C*ArcTanh[Sin[c + d*x]])/d + (8*a^4*A*Tan[c + d*x])/d
+ (7*a^4*C*Tan[c + d*x])/d + (7*a^4*A*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (2*a^4*C*Sec[c + d*x]*Tan[c + d*x])/
d + (a^4*A*Sec[c + d*x]^3*Tan[c + d*x])/d + (8*a^4*A*Tan[c + d*x]^3)/(3*d) + (a^4*C*Tan[c + d*x]^3)/(3*d) + (a
^4*A*Tan[c + d*x]^5)/(5*d)

Maple [A] (verified)

Time = 10.96 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.13

method	result	size
parts	\(-\frac {a^{4} A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (a^{4} A +6 C \,a^{4}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (4 a^{4} A +4 C \,a^{4}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (6 a^{4} A +C \,a^{4}\right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C \,a^{4} \left (d x +c \right )}{d}+\frac {4 C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{4}}{d}+\frac {4 a^{4} A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(233\)
parallelrisch	\(\frac {70 \left (-\frac {3 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {12 C}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {3 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {12 C}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {3 d x C \cos \left (3 d x +3 c \right )}{14}+\frac {3 d x C \cos \left (5 d x +5 c \right )}{70}+\left (\frac {33 A}{35}+\frac {12 C}{35}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {32 C}{35}+\frac {11 A}{10}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {3 A}{10}+\frac {6 C}{35}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {83 A}{350}+\frac {2 C}{7}\right ) \sin \left (5 d x +5 c \right )+\frac {3 d x C \cos \left (d x +c \right )}{7}+\sin \left (d x +c \right ) \left (\frac {22 C}{35}+A \right )\right ) a^{4}}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(256\)
derivativedivides	\(\frac {a^{4} A \tan \left (d x +c \right )+C \,a^{4} \left (d x +c \right )+4 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 C \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-6 a^{4} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+6 C \,a^{4} \tan \left (d x +c \right )+4 a^{4} A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+4 C \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a^{4} A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-C \,a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(265\)
default	\(\frac {a^{4} A \tan \left (d x +c \right )+C \,a^{4} \left (d x +c \right )+4 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 C \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-6 a^{4} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+6 C \,a^{4} \tan \left (d x +c \right )+4 a^{4} A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+4 C \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a^{4} A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-C \,a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(265\)
risch	\(a^{4} C x -\frac {i a^{4} \left (105 A \,{\mathrm e}^{9 i \left (d x +c \right )}+60 C \,{\mathrm e}^{9 i \left (d x +c \right )}-30 A \,{\mathrm e}^{8 i \left (d x +c \right )}-180 C \,{\mathrm e}^{8 i \left (d x +c \right )}+330 A \,{\mathrm e}^{7 i \left (d x +c \right )}+120 C \,{\mathrm e}^{7 i \left (d x +c \right )}-480 A \,{\mathrm e}^{6 i \left (d x +c \right )}-780 C \,{\mathrm e}^{6 i \left (d x +c \right )}-1180 A \,{\mathrm e}^{4 i \left (d x +c \right )}-1220 C \,{\mathrm e}^{4 i \left (d x +c \right )}-330 A \,{\mathrm e}^{3 i \left (d x +c \right )}-120 C \,{\mathrm e}^{3 i \left (d x +c \right )}-800 A \,{\mathrm e}^{2 i \left (d x +c \right )}-820 C \,{\mathrm e}^{2 i \left (d x +c \right )}-105 A \,{\mathrm e}^{i \left (d x +c \right )}-60 C \,{\mathrm e}^{i \left (d x +c \right )}-166 A -200 C \right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {7 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {7 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\)	\(317\)

[In]

int((a+cos(d*x+c)*a)^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x,method=_RETURNVERBOSE)

[Out]

-a^4*A/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+(A*a^4+6*C*a^4)/d*tan(d*x+c)+(4*A*a^4+4*C*a^4)/
d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-(6*A*a^4+C*a^4)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+
c)+C*a^4/d*(d*x+c)+4/d*C*ln(sec(d*x+c)+tan(d*x+c))*a^4+4*a^4*A/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+
c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.86 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {60 \, C a^{4} d x \cos \left (d x + c\right )^{5} + 15 \, {\left (7 \, A + 12 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (7 \, A + 12 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (83 \, A + 100 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 15 \, {\left (7 \, A + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (34 \, A + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 30 \, A a^{4} \cos \left (d x + c\right ) + 6 \, A a^{4}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{5}} \]

[In]

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/60*(60*C*a^4*d*x*cos(d*x + c)^5 + 15*(7*A + 12*C)*a^4*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(7*A + 12*C)
*a^4*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(2*(83*A + 100*C)*a^4*cos(d*x + c)^4 + 15*(7*A + 4*C)*a^4*cos(d
*x + c)^3 + 2*(34*A + 5*C)*a^4*cos(d*x + c)^2 + 30*A*a^4*cos(d*x + c) + 6*A*a^4)*sin(d*x + c))/(d*cos(d*x + c)
^5)

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.52 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {4 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{4} + 120 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 20 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} + 60 \, {\left (d x + c\right )} C a^{4} - 15 \, A a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 60 \, A a^{4} \tan \left (d x + c\right ) + 360 \, C a^{4} \tan \left (d x + c\right )}{60 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/60*(4*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^4 + 120*(tan(d*x + c)^3 + 3*tan(d*x + c))
*A*a^4 + 20*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^4 + 60*(d*x + c)*C*a^4 - 15*A*a^4*(2*(3*sin(d*x + c)^3 - 5*s
in(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 60
*A*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 60*C*a^4*(2*sin
(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 120*C*a^4*(log(sin(d*x + c)
+ 1) - log(sin(d*x + c) - 1)) + 60*A*a^4*tan(d*x + c) + 360*C*a^4*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.24 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {30 \, {\left (d x + c\right )} C a^{4} + 15 \, {\left (7 \, A a^{4} + 12 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (7 \, A a^{4} + 12 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 150 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 490 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 680 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 896 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1180 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 790 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 920 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 375 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 270 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{30 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/30*(30*(d*x + c)*C*a^4 + 15*(7*A*a^4 + 12*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(7*A*a^4 + 12*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)^9 + 150*C*a^4*tan(1/2*d*x + 1/2*c)^9
- 490*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 680*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 896*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 118
0*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 790*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 920*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 375*A*a
^4*tan(1/2*d*x + 1/2*c) + 270*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

Mupad [B] (verification not implemented)

Time = 1.22 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.34 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {7\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,C\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {83\,A\,a^4\,\sin \left (c+d\,x\right )}{15\,d\,\cos \left (c+d\,x\right )}+\frac {7\,A\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {34\,A\,a^4\,\sin \left (c+d\,x\right )}{15\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^4}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{5\,d\,{\cos \left (c+d\,x\right )}^5}+\frac {20\,C\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,C\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3} \]

[In]

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^4)/cos(c + d*x)^6,x)

[Out]

(7*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (12*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (83*A*a^4*sin(c + d*x))/(15*d*cos(c + d*x))
+ (7*A*a^4*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (34*A*a^4*sin(c + d*x))/(15*d*cos(c + d*x)^3) + (A*a^4*sin(c
+ d*x))/(d*cos(c + d*x)^4) + (A*a^4*sin(c + d*x))/(5*d*cos(c + d*x)^5) + (20*C*a^4*sin(c + d*x))/(3*d*cos(c +
d*x)) + (2*C*a^4*sin(c + d*x))/(d*cos(c + d*x)^2) + (C*a^4*sin(c + d*x))/(3*d*cos(c + d*x)^3)