\(\int (a+b \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\) [556]

   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 = 33, antiderivative size = 250 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=b^4 C x+\frac {a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right ) \tan (c+d x)}{15 d}+\frac {a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac {\left (3 A b^2+a^2 (4 A+5 C)\right ) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {A b (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d} \]

[Out]

b^4*C*x+1/2*a*b*(4*b^2*(A+2*C)+a^2*(3*A+4*C))*arctanh(sin(d*x+c))/d+1/15*(6*A*b^4+2*a^4*(4*A+5*C)+a^2*b^2*(56*
A+85*C))*tan(d*x+c)/d+1/30*a*b*(6*A*b^2+a^2*(29*A+40*C))*sec(d*x+c)*tan(d*x+c)/d+1/15*(3*A*b^2+a^2*(4*A+5*C))*
(a+b*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/d+1/5*A*b*(a+b*cos(d*x+c))^3*sec(d*x+c)^3*tan(d*x+c)/d+1/5*A*(a+b*c
os(d*x+c))^4*sec(d*x+c)^4*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3127, 3126, 3110, 3100, 2814, 3855} \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a b \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \tan (c+d x) \sec (c+d x)}{30 d}+\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{15 d}+\frac {\left (2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)+6 A b^4\right ) \tan (c+d x)}{15 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^4}{5 d}+\frac {A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^3}{5 d}+b^4 C x \]

[In]

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

[Out]

b^4*C*x + (a*b*(4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/(2*d) + ((6*A*b^4 + 2*a^4*(4*A + 5*C
) + a^2*b^2*(56*A + 85*C))*Tan[c + d*x])/(15*d) + (a*b*(6*A*b^2 + a^2*(29*A + 40*C))*Sec[c + d*x]*Tan[c + d*x]
)/(30*d) + ((3*A*b^2 + a^2*(4*A + 5*C))*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(15*d) + (A*b*(a +
 b*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/(5*d) + (A*(a + b*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 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 3127

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/(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*c*m + a*d*(n + 1)) - (A*d*(a*d*(n
 + 2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(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, 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 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))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (a+b \cos (c+d x))^3 \left (4 A b+a (4 A+5 C) \cos (c+d x)+5 b C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx \\ & = \frac {A b (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int (a+b \cos (c+d x))^2 \left (4 \left (3 A b^2+a^2 (4 A+5 C)\right )+4 a b (7 A+10 C) \cos (c+d x)+20 b^2 C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx \\ & = \frac {\left (3 A b^2+a^2 (4 A+5 C)\right ) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {A b (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{60} \int (a+b \cos (c+d x)) \left (4 b \left (6 A b^2+a^2 (29 A+40 C)\right )+4 a \left (9 b^2 (3 A+5 C)+2 a^2 (4 A+5 C)\right ) \cos (c+d x)+60 b^3 C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac {\left (3 A b^2+a^2 (4 A+5 C)\right ) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {A b (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{120} \int \left (-8 \left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right )-60 a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \cos (c+d x)-120 b^4 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx \\ & = \frac {\left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right ) \tan (c+d x)}{15 d}+\frac {a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac {\left (3 A b^2+a^2 (4 A+5 C)\right ) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {A b (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{120} \int \left (-60 a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right )-120 b^4 C \cos (c+d x)\right ) \sec (c+d x) \, dx \\ & = b^4 C x+\frac {\left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right ) \tan (c+d x)}{15 d}+\frac {a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac {\left (3 A b^2+a^2 (4 A+5 C)\right ) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {A b (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{2} \left (a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right )\right ) \int \sec (c+d x) \, dx \\ & = b^4 C x+\frac {a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right ) \tan (c+d x)}{15 d}+\frac {a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac {\left (3 A b^2+a^2 (4 A+5 C)\right ) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {A b (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.68 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {30 b^4 C d x+15 a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \text {arctanh}(\sin (c+d x))+15 \left (2 \left (A b^4+a^4 (A+C)+6 a^2 b^2 (A+C)\right )+a b \left (4 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x)+2 a^3 A b \sec ^3(c+d x)\right ) \tan (c+d x)+10 a^2 \left (6 A b^2+a^2 (2 A+C)\right ) \tan ^3(c+d x)+6 a^4 A \tan ^5(c+d x)}{30 d} \]

[In]

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

[Out]

(30*b^4*C*d*x + 15*a*b*(4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]] + 15*(2*(A*b^4 + a^4*(A + C)
+ 6*a^2*b^2*(A + C)) + a*b*(4*A*b^2 + a^2*(3*A + 4*C))*Sec[c + d*x] + 2*a^3*A*b*Sec[c + d*x]^3)*Tan[c + d*x] +
 10*a^2*(6*A*b^2 + a^2*(2*A + C))*Tan[c + d*x]^3 + 6*a^4*A*Tan[c + d*x]^5)/(30*d)

Maple [A] (verified)

Time = 11.34 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.97

method result size
parts \(-\frac {a^{4} A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A \,b^{4}+6 C \,a^{2} b^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (4 A a \,b^{3}+4 C \,a^{3} b \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (6 A \,a^{2} b^{2}+C \,a^{4}\right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C \,b^{4} \left (d x +c \right )}{d}+\frac {4 A \,a^{3} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {4 C a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) \(243\)
derivativedivides \(\frac {-a^{4} A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-C \,a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+4 A \,a^{3} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+4 C \,a^{3} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-6 A \,a^{2} b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+6 C \tan \left (d x +c \right ) a^{2} b^{2}+4 A a \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 C a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \tan \left (d x +c \right ) b^{4}+C \,b^{4} \left (d x +c \right )}{d}\) \(275\)
default \(\frac {-a^{4} A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-C \,a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+4 A \,a^{3} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+4 C \,a^{3} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-6 A \,a^{2} b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+6 C \tan \left (d x +c \right ) a^{2} b^{2}+4 A a \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 C a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \tan \left (d x +c \right ) b^{4}+C \,b^{4} \left (d x +c \right )}{d}\) \(275\)
parallelrisch \(\frac {-45 \left (\left (A +\frac {4 C}{3}\right ) a^{2}+\frac {4 b^{2} \left (A +2 C \right )}{3}\right ) b \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+45 \left (\left (A +\frac {4 C}{3}\right ) a^{2}+\frac {4 b^{2} \left (A +2 C \right )}{3}\right ) b \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+150 C \,b^{4} d x \cos \left (3 d x +3 c \right )+30 C \,b^{4} d x \cos \left (5 d x +5 c \right )+\left (\left (80 A +100 C \right ) a^{4}+600 b^{2} \left (A +\frac {9 C}{10}\right ) a^{2}+90 A \,b^{4}\right ) \sin \left (3 d x +3 c \right )+\left (\left (16 A +20 C \right ) a^{4}+120 \left (A +\frac {3 C}{2}\right ) b^{2} a^{2}+30 A \,b^{4}\right ) \sin \left (5 d x +5 c \right )+420 b \left (a^{2} \left (A +\frac {4 C}{7}\right )+\frac {4 A \,b^{2}}{7}\right ) a \sin \left (2 d x +2 c \right )+90 \left (\left (A +\frac {4 C}{3}\right ) a^{2}+\frac {4 A \,b^{2}}{3}\right ) b a \sin \left (4 d x +4 c \right )+300 C \,b^{4} d x \cos \left (d x +c \right )+160 \sin \left (d x +c \right ) \left (a^{4} \left (A +\frac {C}{2}\right )+3 \left (A +\frac {3 C}{4}\right ) a^{2} b^{2}+\frac {3 A \,b^{4}}{8}\right )}{30 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) \(393\)
risch \(b^{4} C x -\frac {i \left (-120 A \,a^{2} b^{2}-16 a^{4} A -210 A \,a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}+120 A a \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-360 A \,a^{2} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-120 A a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-600 A \,a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-120 C \,a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}-180 C \,a^{2} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+120 C \,a^{3} b \,{\mathrm e}^{7 i \left (d x +c \right )}+45 A \,a^{3} b \,{\mathrm e}^{9 i \left (d x +c \right )}+60 C \,a^{3} b \,{\mathrm e}^{9 i \left (d x +c \right )}-180 C \,a^{2} b^{2}-80 A \,a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-30 A \,b^{4} {\mathrm e}^{8 i \left (d x +c \right )}-160 A \,a^{4} {\mathrm e}^{4 i \left (d x +c \right )}-180 A \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-100 C \,a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-120 A \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-120 A \,b^{4} {\mathrm e}^{6 i \left (d x +c \right )}-140 C \,a^{4} {\mathrm e}^{4 i \left (d x +c \right )}-60 C \,a^{4} {\mathrm e}^{6 i \left (d x +c \right )}-20 C \,a^{4}-30 A \,b^{4}-1080 C \,a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-720 C \,a^{2} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+60 A a \,b^{3} {\mathrm e}^{9 i \left (d x +c \right )}+210 A \,a^{3} b \,{\mathrm e}^{7 i \left (d x +c \right )}-60 C \,a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-45 A \,a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-60 A a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}-720 C \,a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-840 A \,a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 A \,a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {2 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {2 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}+\frac {4 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {3 A \,a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {2 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {2 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}-\frac {4 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) \(701\)

[In]

int((a+cos(d*x+c)*b)^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*b^4+6*C*a^2*b^2)/d*tan(d*x+c)+(4*A*a*b^3+4*C
*a^3*b)/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^2*b^2+C*a^4)/d*(-2/3-1/3*sec(d*x+c)
^2)*tan(d*x+c)+C*b^4/d*(d*x+c)+4*A*a^3*b/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)+t
an(d*x+c)))+4*C*a*b^3/d*ln(sec(d*x+c)+tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.00 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {60 \, C b^{4} d x \cos \left (d x + c\right )^{5} + 15 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{3} b + 4 \, {\left (A + 2 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{3} b + 4 \, {\left (A + 2 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (30 \, A a^{3} b \cos \left (d x + c\right ) + 6 \, A a^{4} + 2 \, {\left (2 \, {\left (4 \, A + 5 \, C\right )} a^{4} + 30 \, {\left (2 \, A + 3 \, C\right )} a^{2} b^{2} + 15 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{3} b + 4 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{4} + 30 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{5}} \]

[In]

integrate((a+b*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*b^4*d*x*cos(d*x + c)^5 + 15*((3*A + 4*C)*a^3*b + 4*(A + 2*C)*a*b^3)*cos(d*x + c)^5*log(sin(d*x + c)
 + 1) - 15*((3*A + 4*C)*a^3*b + 4*(A + 2*C)*a*b^3)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(30*A*a^3*b*cos(d
*x + c) + 6*A*a^4 + 2*(2*(4*A + 5*C)*a^4 + 30*(2*A + 3*C)*a^2*b^2 + 15*A*b^4)*cos(d*x + c)^4 + 15*((3*A + 4*C)
*a^3*b + 4*A*a*b^3)*cos(d*x + c)^3 + 2*((4*A + 5*C)*a^4 + 30*A*a^2*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d
*x + c)^5)

Sympy [F(-1)]

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

[In]

integrate((a+b*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 = 0.21 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.30 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {4 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{4} + 20 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} + 120 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} b^{2} + 60 \, {\left (d x + c\right )} C b^{4} - 15 \, A a^{3} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{3} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, C a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 360 \, C a^{2} b^{2} \tan \left (d x + c\right ) + 60 \, A b^{4} \tan \left (d x + c\right )}{60 \, d} \]

[In]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 778 vs. \(2 (238) = 476\).

Time = 0.39 (sec) , antiderivative size = 778, normalized size of antiderivative = 3.11 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Too large to display} \]

[In]

integrate((a+b*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*b^4 + 15*(3*A*a^3*b + 4*C*a^3*b + 4*A*a*b^3 + 8*C*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1
)) - 15*(3*A*a^3*b + 4*C*a^3*b + 4*A*a*b^3 + 8*C*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(30*A*a^4*tan(1
/2*d*x + 1/2*c)^9 + 30*C*a^4*tan(1/2*d*x + 1/2*c)^9 - 75*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 60*C*a^3*b*tan(1/2*d
*x + 1/2*c)^9 + 180*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 180*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*A*a*b^3*tan(1
/2*d*x + 1/2*c)^9 + 30*A*b^4*tan(1/2*d*x + 1/2*c)^9 - 40*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 80*C*a^4*tan(1/2*d*x +
 1/2*c)^7 + 30*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 120*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 480*A*a^2*b^2*tan(1/2*d*x
 + 1/2*c)^7 - 720*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 120*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 120*A*b^4*tan(1/2*d*
x + 1/2*c)^7 + 116*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 100*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 600*A*a^2*b^2*tan(1/2*d*x
 + 1/2*c)^5 + 1080*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 180*A*b^4*tan(1/2*d*x + 1/2*c)^5 - 40*A*a^4*tan(1/2*d*x
+ 1/2*c)^3 - 80*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 30*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 120*C*a^3*b*tan(1/2*d*x + 1
/2*c)^3 - 480*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 720*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 120*A*a*b^3*tan(1/2*d*
x + 1/2*c)^3 - 120*A*b^4*tan(1/2*d*x + 1/2*c)^3 + 30*A*a^4*tan(1/2*d*x + 1/2*c) + 30*C*a^4*tan(1/2*d*x + 1/2*c
) + 75*A*a^3*b*tan(1/2*d*x + 1/2*c) + 60*C*a^3*b*tan(1/2*d*x + 1/2*c) + 180*A*a^2*b^2*tan(1/2*d*x + 1/2*c) + 1
80*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 60*A*a*b^3*tan(1/2*d*x + 1/2*c) + 30*A*b^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 = 4.42 (sec) , antiderivative size = 1738, normalized size of antiderivative = 6.95 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

((A*a^4*sin(3*c + 3*d*x))/6 + (A*a^4*sin(5*c + 5*d*x))/30 + (3*A*b^4*sin(3*c + 3*d*x))/16 + (A*b^4*sin(5*c + 5
*d*x))/16 + (5*C*a^4*sin(3*c + 3*d*x))/24 + (C*a^4*sin(5*c + 5*d*x))/24 + (A*a^4*sin(c + d*x))/3 + (A*b^4*sin(
c + d*x))/8 + (C*a^4*sin(c + d*x))/6 + (5*C*b^4*cos(c + d*x)*atan((9*A^2*a^6*sin(c/2 + (d*x)/2) + 16*C^2*a^6*s
in(c/2 + (d*x)/2) + 4*C^2*b^6*sin(c/2 + (d*x)/2) + 16*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 24*A^2*a^4*b^2*sin(c/2
+ (d*x)/2) + 64*C^2*a^2*b^4*sin(c/2 + (d*x)/2) + 64*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*a^6*sin(c/2 + (d*x
)/2) + 64*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 80*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(9*A^2*a^6 +
 16*C^2*a^6 + 4*C^2*b^6 + 16*A^2*a^2*b^4 + 24*A^2*a^4*b^2 + 64*C^2*a^2*b^4 + 64*C^2*a^4*b^2 + 24*A*C*a^6 + 64*
A*C*a^2*b^4 + 80*A*C*a^4*b^2))))/4 + (A*a*b^3*sin(2*c + 2*d*x))/2 + (7*A*a^3*b*sin(2*c + 2*d*x))/8 + (A*a*b^3*
sin(4*c + 4*d*x))/4 + (3*A*a^3*b*sin(4*c + 4*d*x))/16 + A*a^2*b^2*sin(c + d*x) + (C*a^3*b*sin(2*c + 2*d*x))/2
+ (C*a^3*b*sin(4*c + 4*d*x))/4 + (3*C*a^2*b^2*sin(c + d*x))/4 + (5*C*b^4*atan((9*A^2*a^6*sin(c/2 + (d*x)/2) +
16*C^2*a^6*sin(c/2 + (d*x)/2) + 4*C^2*b^6*sin(c/2 + (d*x)/2) + 16*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 24*A^2*a^4*
b^2*sin(c/2 + (d*x)/2) + 64*C^2*a^2*b^4*sin(c/2 + (d*x)/2) + 64*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*a^6*si
n(c/2 + (d*x)/2) + 64*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 80*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*
(9*A^2*a^6 + 16*C^2*a^6 + 4*C^2*b^6 + 16*A^2*a^2*b^4 + 24*A^2*a^4*b^2 + 64*C^2*a^2*b^4 + 64*C^2*a^4*b^2 + 24*A
*C*a^6 + 64*A*C*a^2*b^4 + 80*A*C*a^4*b^2)))*cos(3*c + 3*d*x))/8 + (C*b^4*atan((9*A^2*a^6*sin(c/2 + (d*x)/2) +
16*C^2*a^6*sin(c/2 + (d*x)/2) + 4*C^2*b^6*sin(c/2 + (d*x)/2) + 16*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 24*A^2*a^4*
b^2*sin(c/2 + (d*x)/2) + 64*C^2*a^2*b^4*sin(c/2 + (d*x)/2) + 64*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*a^6*si
n(c/2 + (d*x)/2) + 64*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 80*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*
(9*A^2*a^6 + 16*C^2*a^6 + 4*C^2*b^6 + 16*A^2*a^2*b^4 + 24*A^2*a^4*b^2 + 64*C^2*a^2*b^4 + 64*C^2*a^4*b^2 + 24*A
*C*a^6 + 64*A*C*a^2*b^4 + 80*A*C*a^4*b^2)))*cos(5*c + 5*d*x))/8 + (5*A*a^2*b^2*sin(3*c + 3*d*x))/4 + (A*a^2*b^
2*sin(5*c + 5*d*x))/4 + (9*C*a^2*b^2*sin(3*c + 3*d*x))/8 + (3*C*a^2*b^2*sin(5*c + 5*d*x))/8 + (5*A*a*b^3*atanh
(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4 + (15*A*a^3*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (
d*x)/2))*cos(3*c + 3*d*x))/16 + (A*a*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/4 + (3
*A*a^3*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/16 + (5*C*a*b^3*atanh(sin(c/2 + (d*x)/
2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2 + (5*C*a^3*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c +
 3*d*x))/4 + (C*a*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/2 + (C*a^3*b*atanh(sin(c/
2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/4 + (5*A*a*b^3*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/
2 + (d*x)/2)))/2 + (15*A*a^3*b*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 + 5*C*a*b^3*cos(c
+ d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (5*C*a^3*b*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2
 + (d*x)/2)))/2)/(d*((5*cos(c + d*x))/8 + (5*cos(3*c + 3*d*x))/16 + cos(5*c + 5*d*x)/16))