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

Optimal result

Integrand size = 31, antiderivative size = 198 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {7 a^4 (4 A+5 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^4 (83 A+100 B) \tan (c+d x)}{15 d}+\frac {a^4 (244 A+275 B) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(26 A+25 B) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(8 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d} \]

[Out]

7/8*a^4*(4*A+5*B)*arctanh(sin(d*x+c))/d+1/15*a^4*(83*A+100*B)*tan(d*x+c)/d+1/120*a^4*(244*A+275*B)*sec(d*x+c)*
tan(d*x+c)/d+1/30*(26*A+25*B)*(a^4+a^4*cos(d*x+c))*sec(d*x+c)^2*tan(d*x+c)/d+1/20*(8*A+5*B)*(a^2+a^2*cos(d*x+c
))^2*sec(d*x+c)^3*tan(d*x+c)/d+1/5*a*A*(a+a*cos(d*x+c))^3*sec(d*x+c)^4*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {3054, 3047, 3100, 2827, 3852, 8, 3855} \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {7 a^4 (4 A+5 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^4 (83 A+100 B) \tan (c+d x)}{15 d}+\frac {a^4 (244 A+275 B) \tan (c+d x) \sec (c+d x)}{120 d}+\frac {(26 A+25 B) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{30 d}+\frac {(8 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{20 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]

[In]

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

[Out]

(7*a^4*(4*A + 5*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^4*(83*A + 100*B)*Tan[c + d*x])/(15*d) + (a^4*(244*A + 275
*B)*Sec[c + d*x]*Tan[c + d*x])/(120*d) + ((26*A + 25*B)*(a^4 + a^4*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/
(30*d) + ((8*A + 5*B)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (a*A*(a + a*Cos[c + d*x
])^3*Sec[c + d*x]^4*Tan[c + d*x])/(5*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[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+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (a+a \cos (c+d x))^3 (a (8 A+5 B)+a (A+5 B) \cos (c+d x)) \sec ^5(c+d x) \, dx \\ & = \frac {(8 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int (a+a \cos (c+d x))^2 \left (2 a^2 (26 A+25 B)+a^2 (12 A+25 B) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx \\ & = \frac {(26 A+25 B) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(8 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{60} \int (a+a \cos (c+d x)) \left (a^3 (244 A+275 B)+a^3 (88 A+125 B) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {(26 A+25 B) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(8 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{60} \int \left (a^4 (244 A+275 B)+\left (a^4 (88 A+125 B)+a^4 (244 A+275 B)\right ) \cos (c+d x)+a^4 (88 A+125 B) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {a^4 (244 A+275 B) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(26 A+25 B) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(8 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{120} \int \left (8 a^4 (83 A+100 B)+105 a^4 (4 A+5 B) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx \\ & = \frac {a^4 (244 A+275 B) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(26 A+25 B) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(8 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{8} \left (7 a^4 (4 A+5 B)\right ) \int \sec (c+d x) \, dx+\frac {1}{15} \left (a^4 (83 A+100 B)\right ) \int \sec ^2(c+d x) \, dx \\ & = \frac {7 a^4 (4 A+5 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^4 (244 A+275 B) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(26 A+25 B) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(8 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {\left (a^4 (83 A+100 B)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d} \\ & = \frac {7 a^4 (4 A+5 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^4 (83 A+100 B) \tan (c+d x)}{15 d}+\frac {a^4 (244 A+275 B) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(26 A+25 B) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(8 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.34 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.10 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {7 a^4 A \text {arctanh}(\sin (c+d x))}{2 d}+\frac {35 a^4 B \text {arctanh}(\sin (c+d x))}{8 d}+\frac {8 a^4 A \tan (c+d x)}{d}+\frac {8 a^4 B \tan (c+d x)}{d}+\frac {7 a^4 A \sec (c+d x) \tan (c+d x)}{2 d}+\frac {27 a^4 B \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^4 A \sec ^3(c+d x) \tan (c+d x)}{d}+\frac {a^4 B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {8 a^4 A \tan ^3(c+d x)}{3 d}+\frac {4 a^4 B \tan ^3(c+d x)}{3 d}+\frac {a^4 A \tan ^5(c+d x)}{5 d} \]

[In]

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

[Out]

(7*a^4*A*ArcTanh[Sin[c + d*x]])/(2*d) + (35*a^4*B*ArcTanh[Sin[c + d*x]])/(8*d) + (8*a^4*A*Tan[c + d*x])/d + (8
*a^4*B*Tan[c + d*x])/d + (7*a^4*A*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (27*a^4*B*Sec[c + d*x]*Tan[c + d*x])/(8*d
) + (a^4*A*Sec[c + d*x]^3*Tan[c + d*x])/d + (a^4*B*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (8*a^4*A*Tan[c + d*x]^
3)/(3*d) + (4*a^4*B*Tan[c + d*x]^3)/(3*d) + (a^4*A*Tan[c + d*x]^5)/(5*d)

Maple [A] (verified)

Time = 5.59 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.10

method	result	size
parallelrisch	\(\frac {70 \left (-\frac {3 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {5 B}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {3 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {5 B}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\left (\frac {33 A}{35}+\frac {93 B}{140}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {11 A}{10}+\frac {38 B}{35}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {3 A}{10}+\frac {81 B}{280}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {83 A}{350}+\frac {2 B}{7}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +\frac {4 B}{5}\right )\right ) a^{4}}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(217\)
parts	\(-\frac {a^{4} A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (a^{4} A +4 B \,a^{4}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (4 a^{4} A +B \,a^{4}\right ) \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {\left (4 a^{4} A +6 B \,a^{4}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (6 a^{4} A +4 B \,a^{4}\right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{4}}{d}\)	\(227\)
derivativedivides	\(\frac {a^{4} A \tan \left (d x +c \right )+B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 B \,a^{4} \tan \left (d x +c \right )-6 a^{4} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+6 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-4 B \,a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-a^{4} A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+B \,a^{4} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(303\)
default	\(\frac {a^{4} A \tan \left (d x +c \right )+B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 B \,a^{4} \tan \left (d x +c \right )-6 a^{4} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+6 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-4 B \,a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-a^{4} A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+B \,a^{4} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(303\)
risch	\(-\frac {i a^{4} \left (420 A \,{\mathrm e}^{9 i \left (d x +c \right )}+405 B \,{\mathrm e}^{9 i \left (d x +c \right )}-120 A \,{\mathrm e}^{8 i \left (d x +c \right )}-480 B \,{\mathrm e}^{8 i \left (d x +c \right )}+1320 A \,{\mathrm e}^{7 i \left (d x +c \right )}+930 B \,{\mathrm e}^{7 i \left (d x +c \right )}-1920 A \,{\mathrm e}^{6 i \left (d x +c \right )}-2880 B \,{\mathrm e}^{6 i \left (d x +c \right )}-4720 A \,{\mathrm e}^{4 i \left (d x +c \right )}-5120 B \,{\mathrm e}^{4 i \left (d x +c \right )}-1320 A \,{\mathrm e}^{3 i \left (d x +c \right )}-930 B \,{\mathrm e}^{3 i \left (d x +c \right )}-3200 A \,{\mathrm e}^{2 i \left (d x +c \right )}-3520 B \,{\mathrm e}^{2 i \left (d x +c \right )}-420 A \,{\mathrm e}^{i \left (d x +c \right )}-405 B \,{\mathrm e}^{i \left (d x +c \right )}-664 A -800 B \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {7 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}-\frac {7 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}\)	\(311\)

[In]

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

[Out]

70/3*(-3/2*(1/10*cos(5*d*x+5*c)+1/2*cos(3*d*x+3*c)+cos(d*x+c))*(A+5/4*B)*ln(tan(1/2*d*x+1/2*c)-1)+3/2*(1/10*co
s(5*d*x+5*c)+1/2*cos(3*d*x+3*c)+cos(d*x+c))*(A+5/4*B)*ln(tan(1/2*d*x+1/2*c)+1)+(33/35*A+93/140*B)*sin(2*d*x+2*
c)+(11/10*A+38/35*B)*sin(3*d*x+3*c)+(3/10*A+81/280*B)*sin(4*d*x+4*c)+(83/350*A+2/7*B)*sin(5*d*x+5*c)+sin(d*x+c
)*(A+4/5*B))*a^4/d/(cos(5*d*x+5*c)+5*cos(3*d*x+3*c)+10*cos(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.83 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {105 \, {\left (4 \, A + 5 \, B\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (4 \, A + 5 \, B\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (83 \, A + 100 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} + 15 \, {\left (28 \, A + 27 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} + 16 \, {\left (17 \, A + 10 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 30 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 24 \, A a^{4}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

[In]

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

[Out]

1/240*(105*(4*A + 5*B)*a^4*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 105*(4*A + 5*B)*a^4*cos(d*x + c)^5*log(-sin(
d*x + c) + 1) + 2*(8*(83*A + 100*B)*a^4*cos(d*x + c)^4 + 15*(28*A + 27*B)*a^4*cos(d*x + c)^3 + 16*(17*A + 10*B
)*a^4*cos(d*x + c)^2 + 30*(4*A + B)*a^4*cos(d*x + c) + 24*A*a^4)*sin(d*x + c))/(d*cos(d*x + c)^5)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (186) = 372\).

Time = 0.22 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.90 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{4} + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 320 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} - 60 \, A a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 240 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 360 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{4} \tan \left (d x + c\right ) + 960 \, B a^{4} \tan \left (d x + c\right )}{240 \, d} \]

[In]

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

[Out]

1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^4 + 480*(tan(d*x + c)^3 + 3*tan(d*x + c
))*A*a^4 + 320*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 - 60*A*a^4*(2*(3*sin(d*x + c)^3 - 5*sin(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)) - 15*B*a^4*(2*(3*sin(d
*x + c)^3 - 5*sin(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)) - 240*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)) -
360*B*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*B*a^4*(
log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 240*A*a^4*tan(d*x + c) + 960*B*a^4*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.24 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {105 \, {\left (4 \, A a^{4} + 5 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, {\left (4 \, A a^{4} + 5 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (420 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 525 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1960 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 2450 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3584 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4480 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3160 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3950 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1500 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1395 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

[In]

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

[Out]

1/120*(105*(4*A*a^4 + 5*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(4*A*a^4 + 5*B*a^4)*log(abs(tan(1/2*d*
x + 1/2*c) - 1)) - 2*(420*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 525*B*a^4*tan(1/2*d*x + 1/2*c)^9 - 1960*A*a^4*tan(1/2
*d*x + 1/2*c)^7 - 2450*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 3584*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 4480*B*a^4*tan(1/2*d
*x + 1/2*c)^5 - 3160*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 3950*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 1500*A*a^4*tan(1/2*d*x
+ 1/2*c) + 1395*B*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 = 2.94 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.13 \[ \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {7\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+5\,B\right )}{4\,d}-\frac {\left (7\,A\,a^4+\frac {35\,B\,a^4}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {98\,A\,a^4}{3}-\frac {245\,B\,a^4}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {896\,A\,a^4}{15}+\frac {224\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {158\,A\,a^4}{3}-\frac {395\,B\,a^4}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (25\,A\,a^4+\frac {93\,B\,a^4}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

(7*a^4*atanh(tan(c/2 + (d*x)/2))*(4*A + 5*B))/(4*d) - (tan(c/2 + (d*x)/2)*(25*A*a^4 + (93*B*a^4)/4) + tan(c/2
+ (d*x)/2)^9*(7*A*a^4 + (35*B*a^4)/4) - tan(c/2 + (d*x)/2)^7*((98*A*a^4)/3 + (245*B*a^4)/6) - tan(c/2 + (d*x)/
2)^3*((158*A*a^4)/3 + (395*B*a^4)/6) + tan(c/2 + (d*x)/2)^5*((896*A*a^4)/15 + (224*B*a^4)/3))/(d*(5*tan(c/2 +
(d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^1
0 - 1))