\(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^7(c+d x) \, dx\) [963]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 41, antiderivative size = 336 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {\left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {\left (8 a^3 B+30 a b^2 B+5 b^3 (2 A+3 C)+6 a^2 b (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {\left (A b^3+4 a^3 B+12 a b^2 B+3 a^2 b (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a \left (6 A b^2+42 a b B+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d} \]

[Out]

1/16*(18*B*a^2*b+8*B*b^3+6*a*b^2*(3*A+4*C)+a^3*(5*A+6*C))*arctanh(sin(d*x+c))/d+1/15*(8*B*a^3+30*B*a*b^2+5*b^3
*(2*A+3*C)+6*a^2*b*(4*A+5*C))*tan(d*x+c)/d+1/16*(18*B*a^2*b+8*B*b^3+6*a*b^2*(3*A+4*C)+a^3*(5*A+6*C))*sec(d*x+c
)*tan(d*x+c)/d+1/15*(A*b^3+4*B*a^3+12*B*a*b^2+3*a^2*b*(4*A+5*C))*sec(d*x+c)^2*tan(d*x+c)/d+1/120*a*(6*A*b^2+42
*B*a*b+5*a^2*(5*A+6*C))*sec(d*x+c)^3*tan(d*x+c)/d+1/10*(A*b+2*B*a)*(a+b*cos(d*x+c))^2*sec(d*x+c)^4*tan(d*x+c)/
d+1/6*A*(a+b*cos(d*x+c))^3*sec(d*x+c)^5*tan(d*x+c)/d

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {3126, 3110, 3100, 2827, 3853, 3855, 3852, 8} \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {a \tan (c+d x) \sec ^3(c+d x) \left (5 a^2 (5 A+6 C)+42 a b B+6 A b^2\right )}{120 d}+\frac {\left (a^3 (5 A+6 C)+18 a^2 b B+6 a b^2 (3 A+4 C)+8 b^3 B\right ) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {\tan (c+d x) \left (8 a^3 B+6 a^2 b (4 A+5 C)+30 a b^2 B+5 b^3 (2 A+3 C)\right )}{15 d}+\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^3 B+3 a^2 b (4 A+5 C)+12 a b^2 B+A b^3\right )}{15 d}+\frac {\tan (c+d x) \sec (c+d x) \left (a^3 (5 A+6 C)+18 a^2 b B+6 a b^2 (3 A+4 C)+8 b^3 B\right )}{16 d}+\frac {(2 a B+A b) \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{10 d}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^3}{6 d} \]

[In]

Int[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

((18*a^2*b*B + 8*b^3*B + 6*a*b^2*(3*A + 4*C) + a^3*(5*A + 6*C))*ArcTanh[Sin[c + d*x]])/(16*d) + ((8*a^3*B + 30
*a*b^2*B + 5*b^3*(2*A + 3*C) + 6*a^2*b*(4*A + 5*C))*Tan[c + d*x])/(15*d) + ((18*a^2*b*B + 8*b^3*B + 6*a*b^2*(3
*A + 4*C) + a^3*(5*A + 6*C))*Sec[c + d*x]*Tan[c + d*x])/(16*d) + ((A*b^3 + 4*a^3*B + 12*a*b^2*B + 3*a^2*b*(4*A
 + 5*C))*Sec[c + d*x]^2*Tan[c + d*x])/(15*d) + (a*(6*A*b^2 + 42*a*b*B + 5*a^2*(5*A + 6*C))*Sec[c + d*x]^3*Tan[
c + d*x])/(120*d) + ((A*b + 2*a*B)*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(10*d) + (A*(a + b*Cos[
c + d*x])^3*Sec[c + d*x]^5*Tan[c + d*x])/(6*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 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 3126

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^m*((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*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) +
(c*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*
c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n +
1)))*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] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -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+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{6} \int (a+b \cos (c+d x))^2 \left (3 (A b+2 a B)+(5 a A+6 b B+6 a C) \cos (c+d x)+2 b (A+3 C) \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx \\ & = \frac {(A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{30} \int (a+b \cos (c+d x)) \left (6 A b^2+42 a b B+5 a^2 (5 A+6 C)+\left (24 a^2 B+30 b^2 B+a b (47 A+60 C)\right ) \cos (c+d x)+2 b (8 A b+6 a B+15 b C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx \\ & = \frac {a \left (6 A b^2+42 a b B+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{120} \int \left (-24 \left (A b^3+4 a^3 B+12 a b^2 B+3 a^2 b (4 A+5 C)\right )-15 \left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \cos (c+d x)-8 b^2 (8 A b+6 a B+15 b C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx \\ & = \frac {\left (A b^3+4 a^3 B+12 a b^2 B+3 a^2 b (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a \left (6 A b^2+42 a b B+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{360} \int \left (-45 \left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right )-24 \left (8 a^3 B+30 a b^2 B+5 b^3 (2 A+3 C)+6 a^2 b (4 A+5 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {\left (A b^3+4 a^3 B+12 a b^2 B+3 a^2 b (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a \left (6 A b^2+42 a b B+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{15} \left (-8 a^3 B-30 a b^2 B-5 b^3 (2 A+3 C)-6 a^2 b (4 A+5 C)\right ) \int \sec ^2(c+d x) \, dx-\frac {1}{8} \left (-18 a^2 b B-8 b^3 B-6 a b^2 (3 A+4 C)-a^3 (5 A+6 C)\right ) \int \sec ^3(c+d x) \, dx \\ & = \frac {\left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {\left (A b^3+4 a^3 B+12 a b^2 B+3 a^2 b (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a \left (6 A b^2+42 a b B+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{16} \left (-18 a^2 b B-8 b^3 B-6 a b^2 (3 A+4 C)-a^3 (5 A+6 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (8 a^3 B+30 a b^2 B+5 b^3 (2 A+3 C)+6 a^2 b (4 A+5 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d} \\ & = \frac {\left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {\left (8 a^3 B+30 a b^2 B+5 b^3 (2 A+3 C)+6 a^2 b (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {\left (A b^3+4 a^3 B+12 a b^2 B+3 a^2 b (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a \left (6 A b^2+42 a b B+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {(A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.95 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.75 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {15 \left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 \left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \sec (c+d x)+10 a \left (18 A b^2+18 a b B+a^2 (5 A+6 C)\right ) \sec ^3(c+d x)+40 a^3 A \sec ^5(c+d x)+16 \left (15 \left (a^3 B+3 a b^2 B+3 a^2 b (A+C)+b^3 (A+C)\right )+5 \left (A b^3+2 a^3 B+3 a b^2 B+3 a^2 b (2 A+C)\right ) \tan ^2(c+d x)+3 a^2 (3 A b+a B) \tan ^4(c+d x)\right )\right )}{240 d} \]

[In]

Integrate[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

(15*(18*a^2*b*B + 8*b^3*B + 6*a*b^2*(3*A + 4*C) + a^3*(5*A + 6*C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(1
8*a^2*b*B + 8*b^3*B + 6*a*b^2*(3*A + 4*C) + a^3*(5*A + 6*C))*Sec[c + d*x] + 10*a*(18*A*b^2 + 18*a*b*B + a^2*(5
*A + 6*C))*Sec[c + d*x]^3 + 40*a^3*A*Sec[c + d*x]^5 + 16*(15*(a^3*B + 3*a*b^2*B + 3*a^2*b*(A + C) + b^3*(A + C
)) + 5*(A*b^3 + 2*a^3*B + 3*a*b^2*B + 3*a^2*b*(2*A + C))*Tan[c + d*x]^2 + 3*a^2*(3*A*b + a*B)*Tan[c + d*x]^4))
)/(240*d)

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 283, normalized size of antiderivative = 0.84

method result size
parts \(\frac {A \,a^{3} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}-\frac {\left (3 A \,a^{2} b +B \,a^{3}\right ) \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (B \,b^{3}+3 C a \,b^{2}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (A \,b^{3}+3 B a \,b^{2}+3 a^{2} b C \right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 a A \,b^{2}+3 B \,a^{2} b +a^{3} C \right ) \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {C \,b^{3} \tan \left (d x +c \right )}{d}\) \(283\)
derivativedivides \(\frac {A \,a^{3} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )-B \,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 )+a^{3} C \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 (-\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 )+3 B \,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 a^{2} b C \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 (-\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 \,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 C 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 {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \,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 )+C \tan \left (d x +c \right ) b^{3}}{d}\) \(444\)
default \(\frac {A \,a^{3} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )-B \,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 )+a^{3} C \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 (-\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 )+3 B \,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 a^{2} b C \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 (-\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 \,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 C 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 {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \,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 )+C \tan \left (d x +c \right ) b^{3}}{d}\) \(444\)
parallelrisch \(\frac {-450 \left (\left (A +\frac {6 C}{5}\right ) a^{3}+\frac {18 B \,a^{2} b}{5}+\frac {18 \left (A +\frac {4 C}{3}\right ) b^{2} a}{5}+\frac {8 B \,b^{3}}{5}\right ) \left (\frac {5}{3}+\frac {\cos \left (6 d x +6 c \right )}{6}+\cos \left (4 d x +4 c \right )+\frac {5 \cos \left (2 d x +2 c \right )}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+450 \left (\left (A +\frac {6 C}{5}\right ) a^{3}+\frac {18 B \,a^{2} b}{5}+\frac {18 \left (A +\frac {4 C}{3}\right ) b^{2} a}{5}+\frac {8 B \,b^{3}}{5}\right ) \left (\frac {5}{3}+\frac {\cos \left (6 d x +6 c \right )}{6}+\cos \left (4 d x +4 c \right )+\frac {5 \cos \left (2 d x +2 c \right )}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (1920 B \,a^{3}+5760 a^{2} \left (A +\frac {3 C}{4}\right ) b +4320 B a \,b^{2}+1440 \left (A +\frac {5 C}{6}\right ) b^{3}\right ) \sin \left (2 d x +2 c \right )+\left (\left (850 A +1020 C \right ) a^{3}+3060 B \,a^{2} b +3060 \left (A +\frac {12 C}{17}\right ) b^{2} a +720 B \,b^{3}\right ) \sin \left (3 d x +3 c \right )+\left (768 B \,a^{3}+2304 b \left (A +\frac {5 C}{4}\right ) a^{2}+2880 B a \,b^{2}+960 b^{3} \left (A +C \right )\right ) \sin \left (4 d x +4 c \right )+\left (\left (150 A +180 C \right ) a^{3}+540 B \,a^{2} b +540 \left (A +\frac {4 C}{3}\right ) b^{2} a +240 B \,b^{3}\right ) \sin \left (5 d x +5 c \right )+\left (128 B \,a^{3}+384 b \left (A +\frac {5 C}{4}\right ) a^{2}+480 B a \,b^{2}+160 \left (A +\frac {3 C}{2}\right ) b^{3}\right ) \sin \left (6 d x +6 c \right )+1980 \sin \left (d x +c \right ) \left (\left (A +\frac {14 C}{33}\right ) a^{3}+\frac {14 B \,a^{2} b}{11}+\frac {14 \left (A +\frac {4 C}{7}\right ) b^{2} a}{11}+\frac {8 B \,b^{3}}{33}\right )}{240 d \left (10+\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )\right )}\) \(471\)
risch \(\text {Expression too large to display}\) \(1245\)

[In]

int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x,method=_RETURNVERBOSE)

[Out]

A*a^3/d*(-(-1/6*sec(d*x+c)^5-5/24*sec(d*x+c)^3-5/16*sec(d*x+c))*tan(d*x+c)+5/16*ln(sec(d*x+c)+tan(d*x+c)))-(3*
A*a^2*b+B*a^3)/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+(B*b^3+3*C*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)))-(A*b^3+3*B*a*b^2+3*C*a^2*b)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(3*A*a
*b^2+3*B*a^2*b+C*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)))+C*b^3/d
*tan(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.02 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {15 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 18 \, B a^{2} b + 6 \, {\left (3 \, A + 4 \, C\right )} a b^{2} + 8 \, B b^{3}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 18 \, B a^{2} b + 6 \, {\left (3 \, A + 4 \, C\right )} a b^{2} + 8 \, B b^{3}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (8 \, B a^{3} + 6 \, {\left (4 \, A + 5 \, C\right )} a^{2} b + 30 \, B a b^{2} + 5 \, {\left (2 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} + 15 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 18 \, B a^{2} b + 6 \, {\left (3 \, A + 4 \, C\right )} a b^{2} + 8 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} + 40 \, A a^{3} + 16 \, {\left (4 \, B a^{3} + 3 \, {\left (4 \, A + 5 \, C\right )} a^{2} b + 15 \, B a b^{2} + 5 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 18 \, B a^{2} b + 18 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 48 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="fricas")

[Out]

1/480*(15*((5*A + 6*C)*a^3 + 18*B*a^2*b + 6*(3*A + 4*C)*a*b^2 + 8*B*b^3)*cos(d*x + c)^6*log(sin(d*x + c) + 1)
- 15*((5*A + 6*C)*a^3 + 18*B*a^2*b + 6*(3*A + 4*C)*a*b^2 + 8*B*b^3)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*
(16*(8*B*a^3 + 6*(4*A + 5*C)*a^2*b + 30*B*a*b^2 + 5*(2*A + 3*C)*b^3)*cos(d*x + c)^5 + 15*((5*A + 6*C)*a^3 + 18
*B*a^2*b + 6*(3*A + 4*C)*a*b^2 + 8*B*b^3)*cos(d*x + c)^4 + 40*A*a^3 + 16*(4*B*a^3 + 3*(4*A + 5*C)*a^2*b + 15*B
*a*b^2 + 5*A*b^3)*cos(d*x + c)^3 + 10*((5*A + 6*C)*a^3 + 18*B*a^2*b + 18*A*a*b^2)*cos(d*x + c)^2 + 48*(B*a^3 +
 3*A*a^2*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^6)

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**7,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 565, normalized size of antiderivative = 1.68 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {32 \, {\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 )} B a^{3} + 96 \, {\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^{2} b + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} b + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{3} - 5 \, A a^{3} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, C 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 )} - 90 \, B 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 )} - 90 \, A a b^{2} {\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 )} - 360 \, C 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 )} - 120 \, B 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 )} + 480 \, C b^{3} \tan \left (d x + c\right )}{480 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="maxima")

[Out]

1/480*(32*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a^3 + 96*(3*tan(d*x + c)^5 + 10*tan(d*x +
 c)^3 + 15*tan(d*x + c))*A*a^2*b + 480*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2*b + 480*(tan(d*x + c)^3 + 3*tan
(d*x + c))*B*a*b^2 + 160*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b^3 - 5*A*a^3*(2*(15*sin(d*x + c)^5 - 40*sin(d*x
+ c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1
) + 15*log(sin(d*x + c) - 1)) - 30*C*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)) - 90*B*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(sin(d*x + c) - 1)) - 90*A*a*b^2
*(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)) - 360*C*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)) - 120*B*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))
 + 480*C*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. 1370 vs. \(2 (322) = 644\).

Time = 0.41 (sec) , antiderivative size = 1370, normalized size of antiderivative = 4.08 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\text {Too large to display} \]

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="giac")

[Out]

1/240*(15*(5*A*a^3 + 6*C*a^3 + 18*B*a^2*b + 18*A*a*b^2 + 24*C*a*b^2 + 8*B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) +
1)) - 15*(5*A*a^3 + 6*C*a^3 + 18*B*a^2*b + 18*A*a*b^2 + 24*C*a*b^2 + 8*B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1
)) + 2*(165*A*a^3*tan(1/2*d*x + 1/2*c)^11 - 240*B*a^3*tan(1/2*d*x + 1/2*c)^11 + 150*C*a^3*tan(1/2*d*x + 1/2*c)
^11 - 720*A*a^2*b*tan(1/2*d*x + 1/2*c)^11 + 450*B*a^2*b*tan(1/2*d*x + 1/2*c)^11 - 720*C*a^2*b*tan(1/2*d*x + 1/
2*c)^11 + 450*A*a*b^2*tan(1/2*d*x + 1/2*c)^11 - 720*B*a*b^2*tan(1/2*d*x + 1/2*c)^11 + 360*C*a*b^2*tan(1/2*d*x
+ 1/2*c)^11 - 240*A*b^3*tan(1/2*d*x + 1/2*c)^11 + 120*B*b^3*tan(1/2*d*x + 1/2*c)^11 - 240*C*b^3*tan(1/2*d*x +
1/2*c)^11 + 25*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 560*B*a^3*tan(1/2*d*x + 1/2*c)^9 - 210*C*a^3*tan(1/2*d*x + 1/2*c
)^9 + 1680*A*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 630*B*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 2640*C*a^2*b*tan(1/2*d*x + 1/
2*c)^9 - 630*A*a*b^2*tan(1/2*d*x + 1/2*c)^9 + 2640*B*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 1080*C*a*b^2*tan(1/2*d*x +
 1/2*c)^9 + 880*A*b^3*tan(1/2*d*x + 1/2*c)^9 - 360*B*b^3*tan(1/2*d*x + 1/2*c)^9 + 1200*C*b^3*tan(1/2*d*x + 1/2
*c)^9 + 450*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 1248*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*C*a^3*tan(1/2*d*x + 1/2*c)^7
 - 3744*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 180*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 4320*C*a^2*b*tan(1/2*d*x + 1/2*c
)^7 + 180*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 4320*B*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 720*C*a*b^2*tan(1/2*d*x + 1/2
*c)^7 - 1440*A*b^3*tan(1/2*d*x + 1/2*c)^7 + 240*B*b^3*tan(1/2*d*x + 1/2*c)^7 - 2400*C*b^3*tan(1/2*d*x + 1/2*c)
^7 + 450*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 1248*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 60*C*a^3*tan(1/2*d*x + 1/2*c)^5 +
3744*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 180*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 4320*C*a^2*b*tan(1/2*d*x + 1/2*c)^5
 + 180*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 4320*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 720*C*a*b^2*tan(1/2*d*x + 1/2*c)
^5 + 1440*A*b^3*tan(1/2*d*x + 1/2*c)^5 + 240*B*b^3*tan(1/2*d*x + 1/2*c)^5 + 2400*C*b^3*tan(1/2*d*x + 1/2*c)^5
+ 25*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 560*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 210*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 1680
*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 630*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 2640*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 6
30*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 2640*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 1080*C*a*b^2*tan(1/2*d*x + 1/2*c)^3
- 880*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 360*B*b^3*tan(1/2*d*x + 1/2*c)^3 - 1200*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 16
5*A*a^3*tan(1/2*d*x + 1/2*c) + 240*B*a^3*tan(1/2*d*x + 1/2*c) + 150*C*a^3*tan(1/2*d*x + 1/2*c) + 720*A*a^2*b*t
an(1/2*d*x + 1/2*c) + 450*B*a^2*b*tan(1/2*d*x + 1/2*c) + 720*C*a^2*b*tan(1/2*d*x + 1/2*c) + 450*A*a*b^2*tan(1/
2*d*x + 1/2*c) + 720*B*a*b^2*tan(1/2*d*x + 1/2*c) + 360*C*a*b^2*tan(1/2*d*x + 1/2*c) + 240*A*b^3*tan(1/2*d*x +
 1/2*c) + 120*B*b^3*tan(1/2*d*x + 1/2*c) + 240*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d

Mupad [B] (verification not implemented)

Time = 4.13 (sec) , antiderivative size = 766, normalized size of antiderivative = 2.28 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,A\,a^3}{16}+\frac {B\,b^3}{2}+\frac {3\,C\,a^3}{8}+\frac {9\,A\,a\,b^2}{8}+\frac {9\,B\,a^2\,b}{8}+\frac {3\,C\,a\,b^2}{2}\right )}{\frac {5\,A\,a^3}{4}+2\,B\,b^3+\frac {3\,C\,a^3}{2}+\frac {9\,A\,a\,b^2}{2}+\frac {9\,B\,a^2\,b}{2}+6\,C\,a\,b^2}\right )\,\left (\frac {5\,A\,a^3}{8}+B\,b^3+\frac {3\,C\,a^3}{4}+\frac {9\,A\,a\,b^2}{4}+\frac {9\,B\,a^2\,b}{4}+3\,C\,a\,b^2\right )}{d}+\frac {\left (\frac {11\,A\,a^3}{8}-2\,A\,b^3-2\,B\,a^3+B\,b^3+\frac {5\,C\,a^3}{4}-2\,C\,b^3+\frac {15\,A\,a\,b^2}{4}-6\,A\,a^2\,b-6\,B\,a\,b^2+\frac {15\,B\,a^2\,b}{4}+3\,C\,a\,b^2-6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {5\,A\,a^3}{24}+\frac {22\,A\,b^3}{3}+\frac {14\,B\,a^3}{3}-3\,B\,b^3-\frac {7\,C\,a^3}{4}+10\,C\,b^3-\frac {21\,A\,a\,b^2}{4}+14\,A\,a^2\,b+22\,B\,a\,b^2-\frac {21\,B\,a^2\,b}{4}-9\,C\,a\,b^2+22\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {15\,A\,a^3}{4}-12\,A\,b^3-\frac {52\,B\,a^3}{5}+2\,B\,b^3+\frac {C\,a^3}{2}-20\,C\,b^3+\frac {3\,A\,a\,b^2}{2}-\frac {156\,A\,a^2\,b}{5}-36\,B\,a\,b^2+\frac {3\,B\,a^2\,b}{2}+6\,C\,a\,b^2-36\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {15\,A\,a^3}{4}+12\,A\,b^3+\frac {52\,B\,a^3}{5}+2\,B\,b^3+\frac {C\,a^3}{2}+20\,C\,b^3+\frac {3\,A\,a\,b^2}{2}+\frac {156\,A\,a^2\,b}{5}+36\,B\,a\,b^2+\frac {3\,B\,a^2\,b}{2}+6\,C\,a\,b^2+36\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {5\,A\,a^3}{24}-\frac {22\,A\,b^3}{3}-\frac {14\,B\,a^3}{3}-3\,B\,b^3-\frac {7\,C\,a^3}{4}-10\,C\,b^3-\frac {21\,A\,a\,b^2}{4}-14\,A\,a^2\,b-22\,B\,a\,b^2-\frac {21\,B\,a^2\,b}{4}-9\,C\,a\,b^2-22\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {11\,A\,a^3}{8}+2\,A\,b^3+2\,B\,a^3+B\,b^3+\frac {5\,C\,a^3}{4}+2\,C\,b^3+\frac {15\,A\,a\,b^2}{4}+6\,A\,a^2\,b+6\,B\,a\,b^2+\frac {15\,B\,a^2\,b}{4}+3\,C\,a\,b^2+6\,C\,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 )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

int(((a + b*cos(c + d*x))^3*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^7,x)

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((5*A*a^3)/16 + (B*b^3)/2 + (3*C*a^3)/8 + (9*A*a*b^2)/8 + (9*B*a^2*b)/8 + (3*C*a*
b^2)/2))/((5*A*a^3)/4 + 2*B*b^3 + (3*C*a^3)/2 + (9*A*a*b^2)/2 + (9*B*a^2*b)/2 + 6*C*a*b^2))*((5*A*a^3)/8 + B*b
^3 + (3*C*a^3)/4 + (9*A*a*b^2)/4 + (9*B*a^2*b)/4 + 3*C*a*b^2))/d + (tan(c/2 + (d*x)/2)*((11*A*a^3)/8 + 2*A*b^3
 + 2*B*a^3 + B*b^3 + (5*C*a^3)/4 + 2*C*b^3 + (15*A*a*b^2)/4 + 6*A*a^2*b + 6*B*a*b^2 + (15*B*a^2*b)/4 + 3*C*a*b
^2 + 6*C*a^2*b) + tan(c/2 + (d*x)/2)^11*((11*A*a^3)/8 - 2*A*b^3 - 2*B*a^3 + B*b^3 + (5*C*a^3)/4 - 2*C*b^3 + (1
5*A*a*b^2)/4 - 6*A*a^2*b - 6*B*a*b^2 + (15*B*a^2*b)/4 + 3*C*a*b^2 - 6*C*a^2*b) - tan(c/2 + (d*x)/2)^3*((22*A*b
^3)/3 - (5*A*a^3)/24 + (14*B*a^3)/3 + 3*B*b^3 + (7*C*a^3)/4 + 10*C*b^3 + (21*A*a*b^2)/4 + 14*A*a^2*b + 22*B*a*
b^2 + (21*B*a^2*b)/4 + 9*C*a*b^2 + 22*C*a^2*b) + tan(c/2 + (d*x)/2)^9*((5*A*a^3)/24 + (22*A*b^3)/3 + (14*B*a^3
)/3 - 3*B*b^3 - (7*C*a^3)/4 + 10*C*b^3 - (21*A*a*b^2)/4 + 14*A*a^2*b + 22*B*a*b^2 - (21*B*a^2*b)/4 - 9*C*a*b^2
 + 22*C*a^2*b) + tan(c/2 + (d*x)/2)^5*((15*A*a^3)/4 + 12*A*b^3 + (52*B*a^3)/5 + 2*B*b^3 + (C*a^3)/2 + 20*C*b^3
 + (3*A*a*b^2)/2 + (156*A*a^2*b)/5 + 36*B*a*b^2 + (3*B*a^2*b)/2 + 6*C*a*b^2 + 36*C*a^2*b) + tan(c/2 + (d*x)/2)
^7*((15*A*a^3)/4 - 12*A*b^3 - (52*B*a^3)/5 + 2*B*b^3 + (C*a^3)/2 - 20*C*b^3 + (3*A*a*b^2)/2 - (156*A*a^2*b)/5
- 36*B*a*b^2 + (3*B*a^2*b)/2 + 6*C*a*b^2 - 36*C*a^2*b))/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 -
 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1))