\(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx\) [239]

Optimal result

Integrand size = 31, antiderivative size = 236 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {\left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (8 a^3 A+30 a A b^2+30 a^2 b B+15 b^3 B\right ) \tan (c+d x)}{15 d}+\frac {\left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \left (4 a^2 A+12 A b^2+15 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (7 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d} \]

[Out]

1/8*(9*A*a^2*b+4*A*b^3+3*B*a^3+12*B*a*b^2)*arctanh(sin(d*x+c))/d+1/15*(8*A*a^3+30*A*a*b^2+30*B*a^2*b+15*B*b^3)
*tan(d*x+c)/d+1/8*(9*A*a^2*b+4*A*b^3+3*B*a^3+12*B*a*b^2)*sec(d*x+c)*tan(d*x+c)/d+1/15*a*(4*A*a^2+12*A*b^2+15*B
*a*b)*sec(d*x+c)^2*tan(d*x+c)/d+1/20*a^2*(7*A*b+5*B*a)*sec(d*x+c)^3*tan(d*x+c)/d+1/5*a*A*(a+b*cos(d*x+c))^2*se
c(d*x+c)^4*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3068, 3110, 3100, 2827, 3853, 3855, 3852, 8} \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {a \left (4 a^2 A+15 a b B+12 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac {a^2 (5 a B+7 A b) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {\left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (8 a^3 A+30 a^2 b B+30 a A b^2+15 b^3 B\right ) \tan (c+d x)}{15 d}+\frac {\left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

[In]

Int[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^6,x]

[Out]

((9*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*ArcTanh[Sin[c + d*x]])/(8*d) + ((8*a^3*A + 30*a*A*b^2 + 30*a^2*b
*B + 15*b^3*B)*Tan[c + d*x])/(15*d) + ((9*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*Sec[c + d*x]*Tan[c + d*x])
/(8*d) + (a*(4*a^2*A + 12*A*b^2 + 15*a*b*B)*Sec[c + d*x]^2*Tan[c + d*x])/(15*d) + (a^2*(7*A*b + 5*a*B)*Sec[c +
d*x]^3*Tan[c + d*x])/(20*d) + (a*A*(a + b*Cos[c + d*x])^2*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 3068

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*c - a*d))*(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)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Si
n[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c -
(A*b + a*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*
d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 1] && LtQ[n, -1]

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 3110

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

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

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+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (a+b \cos (c+d x)) \left (a (7 A b+5 a B)+\left (4 a^2 A+5 A b^2+10 a b B\right ) \cos (c+d x)+b (2 a A+5 b B) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx \\ & = \frac {a^2 (7 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{20} \int \left (-4 a \left (4 a^2 A+12 A b^2+15 a b B\right )-5 \left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \cos (c+d x)-4 b^2 (2 a A+5 b B) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx \\ & = \frac {a \left (4 a^2 A+12 A b^2+15 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (7 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{60} \int \left (-15 \left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right )-4 \left (8 a^3 A+30 a A b^2+30 a^2 b B+15 b^3 B\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {a \left (4 a^2 A+12 A b^2+15 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (7 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{4} \left (-9 a^2 A b-4 A b^3-3 a^3 B-12 a b^2 B\right ) \int \sec ^3(c+d x) \, dx-\frac {1}{15} \left (-8 a^3 A-30 a A b^2-30 a^2 b B-15 b^3 B\right ) \int \sec ^2(c+d x) \, dx \\ & = \frac {\left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \left (4 a^2 A+12 A b^2+15 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (7 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{8} \left (-9 a^2 A b-4 A b^3-3 a^3 B-12 a b^2 B\right ) \int \sec (c+d x) \, dx-\frac {\left (8 a^3 A+30 a A b^2+30 a^2 b B+15 b^3 B\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d} \\ & = \frac {\left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (8 a^3 A+30 a A b^2+30 a^2 b B+15 b^3 B\right ) \tan (c+d x)}{15 d}+\frac {\left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \left (4 a^2 A+12 A b^2+15 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (7 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.10 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.77 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {15 \left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 \left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \sec (c+d x)+30 a^2 (3 A b+a B) \sec ^3(c+d x)+8 \left (15 \left (a^3 A+3 a A b^2+3 a^2 b B+b^3 B\right )+5 a \left (2 a^2 A+3 A b^2+3 a b B\right ) \tan ^2(c+d x)+3 a^3 A \tan ^4(c+d x)\right )\right )}{120 d} \]

[In]

Integrate[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^6,x]

[Out]

(15*(9*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(9*a^2*A*b + 4*A*b^3
+ 3*a^3*B + 12*a*b^2*B)*Sec[c + d*x] + 30*a^2*(3*A*b + a*B)*Sec[c + d*x]^3 + 8*(15*(a^3*A + 3*a*A*b^2 + 3*a^2
*b*B + b^3*B) + 5*a*(2*a^2*A + 3*A*b^2 + 3*a*b*B)*Tan[c + d*x]^2 + 3*a^3*A*Tan[c + d*x]^4)))/(120*d)

Maple [A] (verified)

Time = 6.20 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.85

method	result	size
parts	\(-\frac {A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A \,b^{3}+3 B a \,b^{2}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (3 A a \,b^{2}+3 B \,a^{2} b \right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 A \,a^{2} b +B \,a^{3}\right ) \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {B \,b^{3} \tan \left (d x +c \right )}{d}\)	\(200\)
derivativedivides	\(\frac {-A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 A \,a^{2} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 B \,a^{2} b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-3 A a \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 B a \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \tan \left (d x +c \right ) b^{3}}{d}\)	\(275\)
default	\(\frac {-A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 A \,a^{2} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 B \,a^{2} b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-3 A a \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 B a \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \tan \left (d x +c \right ) b^{3}}{d}\)	\(275\)
parallelrisch	\(\frac {-135 \left (A \,a^{2} b +\frac {4}{9} A \,b^{3}+\frac {1}{3} B \,a^{3}+\frac {4}{3} B a \,b^{2}\right ) \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+135 \left (A \,a^{2} b +\frac {4}{9} A \,b^{3}+\frac {1}{3} B \,a^{3}+\frac {4}{3} B a \,b^{2}\right ) \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (1260 A \,a^{2} b +240 A \,b^{3}+420 B \,a^{3}+720 B a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+\left (320 A \,a^{3}+1200 A a \,b^{2}+1200 B \,a^{2} b +360 B \,b^{3}\right ) \sin \left (3 d x +3 c \right )+\left (270 A \,a^{2} b +120 A \,b^{3}+90 B \,a^{3}+360 B a \,b^{2}\right ) \sin \left (4 d x +4 c \right )+\left (64 A \,a^{3}+240 A a \,b^{2}+240 B \,a^{2} b +120 B \,b^{3}\right ) \sin \left (5 d x +5 c \right )+640 \left (A \,a^{3}+\frac {3}{2} A a \,b^{2}+\frac {3}{2} B \,a^{2} b +\frac {3}{8} B \,b^{3}\right ) \sin \left (d x +c \right )}{120 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(358\)
risch	\(-\frac {i \left (-120 B \,b^{3}-64 A \,a^{3}-240 A a \,b^{2}-240 B \,a^{2} b -135 A \,a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-180 B a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-720 A a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-720 B \,a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-1680 A a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+630 A \,a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+360 B a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-630 A \,a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-360 B a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+135 A \,a^{2} b \,{\mathrm e}^{9 i \left (d x +c \right )}+180 B a \,b^{2} {\mathrm e}^{9 i \left (d x +c \right )}-1680 B \,a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-1200 A a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-1200 B \,a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-60 A \,b^{3} {\mathrm e}^{i \left (d x +c \right )}+60 A \,b^{3} {\mathrm e}^{9 i \left (d x +c \right )}-480 B \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-320 A \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+120 A \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+45 B \,a^{3} {\mathrm e}^{9 i \left (d x +c \right )}+210 B \,a^{3} {\mathrm e}^{7 i \left (d x +c \right )}-210 B \,a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-120 A \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-45 B \,a^{3} {\mathrm e}^{i \left (d x +c \right )}-480 B \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-720 B \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-640 A \,a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-120 B \,b^{3} {\mathrm e}^{8 i \left (d x +c \right )}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {9 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{3}}{2 d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a \,b^{2}}{2 d}+\frac {9 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{3}}{2 d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a \,b^{2}}{2 d}\)	\(662\)

[In]

int((a+cos(d*x+c)*b)^3*(A+B*cos(d*x+c))*sec(d*x+c)^6,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.06 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, B a^{3} + 9 \, A a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, B a^{3} + 9 \, A a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (8 \, A a^{3} + 30 \, B a^{2} b + 30 \, A a b^{2} + 15 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} + 24 \, A a^{3} + 15 \, {\left (3 \, B a^{3} + 9 \, A a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A a^{3} + 15 \, B a^{2} b + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/240*(15*(3*B*a^3 + 9*A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3*B*a^3 + 9*
A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(8*(8*A*a^3 + 30*B*a^2*b + 30*A*a*b^
2 + 15*B*b^3)*cos(d*x + c)^4 + 24*A*a^3 + 15*(3*B*a^3 + 9*A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*cos(d*x + c)^3 + 8*(
4*A*a^3 + 15*B*a^2*b + 15*A*a*b^2)*cos(d*x + c)^2 + 30*(B*a^3 + 3*A*a^2*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(
d*x + c)^5)

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c))**3*(A+B*cos(d*x+c))*sec(d*x+c)**6,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.44 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{2} - 15 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 45 \, A a^{2} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, B a b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, B b^{3} \tan \left (d x + c\right )}{240 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^3*(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^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c
))*B*a^2*b + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b^2 - 15*B*a^3*(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)) - 45*A*a^2*b*(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(si
n(d*x + c) - 1)) - 180*B*a*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c)
- 1)) - 60*A*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 240*
B*b^3*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 722 vs. \(2 (224) = 448\).

Time = 0.36 (sec) , antiderivative size = 722, normalized size of antiderivative = 3.06 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\text {Too large to display} \]

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/120*(15*(3*B*a^3 + 9*A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*B*a^3 + 9*A*
a^2*b + 12*B*a*b^2 + 4*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a^3*tan(1/2*d*x + 1/2*c)^9 - 75*B*
a^3*tan(1/2*d*x + 1/2*c)^9 - 225*A*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*B*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*A*a
*b^2*tan(1/2*d*x + 1/2*c)^9 - 180*B*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*A*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*B*b^3
*tan(1/2*d*x + 1/2*c)^9 - 160*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 30*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 90*A*a^2*b*tan(
1/2*d*x + 1/2*c)^7 - 960*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 360*B*a*b^2*tan
(1/2*d*x + 1/2*c)^7 + 120*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 480*B*b^3*tan(1/2*d*x + 1/2*c)^7 + 464*A*a^3*tan(1/2*
d*x + 1/2*c)^5 + 1200*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 1200*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 720*B*b^3*tan(1/2
*d*x + 1/2*c)^5 - 160*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 30*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 90*A*a^2*b*tan(1/2*d*x
+ 1/2*c)^3 - 960*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 360*B*a*b^2*tan(1/2*d*x
+ 1/2*c)^3 - 120*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 480*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*A*a^3*tan(1/2*d*x + 1/
2*c) + 75*B*a^3*tan(1/2*d*x + 1/2*c) + 225*A*a^2*b*tan(1/2*d*x + 1/2*c) + 360*B*a^2*b*tan(1/2*d*x + 1/2*c) + 3
60*A*a*b^2*tan(1/2*d*x + 1/2*c) + 180*B*a*b^2*tan(1/2*d*x + 1/2*c) + 60*A*b^3*tan(1/2*d*x + 1/2*c) + 120*B*b^3
*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 = 4.44 (sec) , antiderivative size = 470, normalized size of antiderivative = 1.99 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,B\,a^3}{8}+\frac {9\,A\,a^2\,b}{8}+\frac {3\,B\,a\,b^2}{2}+\frac {A\,b^3}{2}\right )}{\frac {3\,B\,a^3}{2}+\frac {9\,A\,a^2\,b}{2}+6\,B\,a\,b^2+2\,A\,b^3}\right )\,\left (\frac {3\,B\,a^3}{4}+\frac {9\,A\,a^2\,b}{4}+3\,B\,a\,b^2+A\,b^3\right )}{d}-\frac {\left (2\,A\,a^3-A\,b^3-\frac {5\,B\,a^3}{4}+2\,B\,b^3+6\,A\,a\,b^2-\frac {15\,A\,a^2\,b}{4}-3\,B\,a\,b^2+6\,B\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,A\,b^3-\frac {8\,A\,a^3}{3}+\frac {B\,a^3}{2}-8\,B\,b^3-16\,A\,a\,b^2+\frac {3\,A\,a^2\,b}{2}+6\,B\,a\,b^2-16\,B\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^3}{15}+20\,B\,a^2\,b+20\,A\,a\,b^2+12\,B\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^3}{3}-2\,A\,b^3-\frac {B\,a^3}{2}-8\,B\,b^3-16\,A\,a\,b^2-\frac {3\,A\,a^2\,b}{2}-6\,B\,a\,b^2-16\,B\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^3+A\,b^3+\frac {5\,B\,a^3}{4}+2\,B\,b^3+6\,A\,a\,b^2+\frac {15\,A\,a^2\,b}{4}+3\,B\,a\,b^2+6\,B\,a^2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^3)/cos(c + d*x)^6,x)

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((A*b^3)/2 + (3*B*a^3)/8 + (9*A*a^2*b)/8 + (3*B*a*b^2)/2))/(2*A*b^3 + (3*B*a^3)/2
+ (9*A*a^2*b)/2 + 6*B*a*b^2))*(A*b^3 + (3*B*a^3)/4 + (9*A*a^2*b)/4 + 3*B*a*b^2))/d - (tan(c/2 + (d*x)/2)*(2*A
*a^3 + A*b^3 + (5*B*a^3)/4 + 2*B*b^3 + 6*A*a*b^2 + (15*A*a^2*b)/4 + 3*B*a*b^2 + 6*B*a^2*b) + tan(c/2 + (d*x)/2
)^5*((116*A*a^3)/15 + 12*B*b^3 + 20*A*a*b^2 + 20*B*a^2*b) + tan(c/2 + (d*x)/2)^9*(2*A*a^3 - A*b^3 - (5*B*a^3)/
4 + 2*B*b^3 + 6*A*a*b^2 - (15*A*a^2*b)/4 - 3*B*a*b^2 + 6*B*a^2*b) - tan(c/2 + (d*x)/2)^3*((8*A*a^3)/3 + 2*A*b^
3 + (B*a^3)/2 + 8*B*b^3 + 16*A*a*b^2 + (3*A*a^2*b)/2 + 6*B*a*b^2 + 16*B*a^2*b) - tan(c/2 + (d*x)/2)^7*((8*A*a^
3)/3 - 2*A*b^3 - (B*a^3)/2 + 8*B*b^3 + 16*A*a*b^2 - (3*A*a^2*b)/2 - 6*B*a*b^2 + 16*B*a^2*b))/(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)^10
- 1))