\(\int (a+b \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\) [553]

Optimal result

Integrand size = 33, antiderivative size = 219 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=2 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) x+\frac {a^2 \left (12 A b^2+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac {a b^3 (9 A-4 C) \cos (c+d x) \sin (c+d x)}{3 d}-\frac {b^2 (15 A-2 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

2*a*b*(2*A*b^2+(2*a^2+b^2)*C)*x+1/2*a^2*(12*A*b^2+a^2*(A+2*C))*arctanh(sin(d*x+c))/d-1/6*b^2*(a^2*(39*A-34*C)-
2*b^2*(3*A+2*C))*sin(d*x+c)/d-1/3*a*b^3*(9*A-4*C)*cos(d*x+c)*sin(d*x+c)/d-1/6*b^2*(15*A-2*C)*(a+b*cos(d*x+c))^
2*sin(d*x+c)/d+2*A*b*(a+b*cos(d*x+c))^3*tan(d*x+c)/d+1/2*A*(a+b*cos(d*x+c))^4*sec(d*x+c)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3127, 3126, 3128, 3112, 3102, 2814, 3855} \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {a^2 \left (a^2 (A+2 C)+12 A b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}+2 a b x \left (C \left (2 a^2+b^2\right )+2 A b^2\right )-\frac {a b^3 (9 A-4 C) \sin (c+d x) \cos (c+d x)}{3 d}-\frac {b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{6 d}+\frac {2 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d} \]

[In]

Int[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

2*a*b*(2*A*b^2 + (2*a^2 + b^2)*C)*x + (a^2*(12*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(2*d) - (b^2*(a^2
*(39*A - 34*C) - 2*b^2*(3*A + 2*C))*Sin[c + d*x])/(6*d) - (a*b^3*(9*A - 4*C)*Cos[c + d*x]*Sin[c + d*x])/(3*d)
- (b^2*(15*A - 2*C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(6*d) + (2*A*b*(a + b*Cos[c + d*x])^3*Tan[c + d*x])/d
+ (A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]*Tan[c + d*x])/(2*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 3102

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

Rule 3112

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[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a +
b*Sin[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*
c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, 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 3128

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e
+ f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+b \cos (c+d x))^3 \left (4 A b+a (A+2 C) \cos (c+d x)-b (3 A-2 C) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx \\ & = \frac {2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+b \cos (c+d x))^2 \left (12 A b^2+a^2 (A+2 C)-2 a b (A-2 C) \cos (c+d x)-b^2 (15 A-2 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = -\frac {b^2 (15 A-2 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{6} \int (a+b \cos (c+d x)) \left (3 a \left (12 A b^2+a^2 (A+2 C)\right )-b \left (3 a^2 (A-6 C)-2 b^2 (3 A+2 C)\right ) \cos (c+d x)-4 a b^2 (9 A-4 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = -\frac {a b^3 (9 A-4 C) \cos (c+d x) \sin (c+d x)}{3 d}-\frac {b^2 (15 A-2 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{12} \int \left (6 a^2 \left (12 A b^2+a^2 (A+2 C)\right )+24 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x)-2 b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = -\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac {a b^3 (9 A-4 C) \cos (c+d x) \sin (c+d x)}{3 d}-\frac {b^2 (15 A-2 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{12} \int \left (6 a^2 \left (12 A b^2+a^2 (A+2 C)\right )+24 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx \\ & = 2 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) x-\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac {a b^3 (9 A-4 C) \cos (c+d x) \sin (c+d x)}{3 d}-\frac {b^2 (15 A-2 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \left (a^2 \left (12 A b^2+a^2 (A+2 C)\right )\right ) \int \sec (c+d x) \, dx \\ & = 2 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) x+\frac {a^2 \left (12 A b^2+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac {a b^3 (9 A-4 C) \cos (c+d x) \sin (c+d x)}{3 d}-\frac {b^2 (15 A-2 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.84 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.47 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {24 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) (c+d x)-6 a^2 \left (12 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 a^2 \left (12 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {3 a^4 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {48 a^3 A b \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {3 a^4 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {48 a^3 A b \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+3 b^2 \left (4 A b^2+3 \left (8 a^2+b^2\right ) C\right ) \sin (c+d x)+12 a b^3 C \sin (2 (c+d x))+b^4 C \sin (3 (c+d x))}{12 d} \]

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

(24*a*b*(2*A*b^2 + (2*a^2 + b^2)*C)*(c + d*x) - 6*a^2*(12*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] + 6*a^2*(12*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (3*a^4*A)/(Cos[(c +
d*x)/2] - Sin[(c + d*x)/2])^2 + (48*a^3*A*b*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (3*a^4*A
)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (48*a^3*A*b*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]
) + 3*b^2*(4*A*b^2 + 3*(8*a^2 + b^2)*C)*Sin[c + d*x] + 12*a*b^3*C*Sin[2*(c + d*x)] + b^4*C*Sin[3*(c + d*x)])/(
12*d)

Maple [A] (verified)

Time = 9.10 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.89

method	result	size
parts	\(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (A \,b^{4}+6 C \,a^{2} b^{2}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (4 A a \,b^{3}+4 C \,a^{3} b \right ) \left (d x +c \right )}{d}+\frac {\left (6 A \,a^{2} b^{2}+C \,a^{4}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \,b^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {4 A \,a^{3} b \tan \left (d x +c \right )}{d}+\frac {4 C a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(196\)
derivativedivides	\(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 A \,a^{3} b \tan \left (d x +c \right )+4 C \,a^{3} b \left (d x +c \right )+6 A \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 C \sin \left (d x +c \right ) a^{2} b^{2}+4 A a \,b^{3} \left (d x +c \right )+4 C a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right ) b^{4}+\frac {C \,b^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(198\)
default	\(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 A \,a^{3} b \tan \left (d x +c \right )+4 C \,a^{3} b \left (d x +c \right )+6 A \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 C \sin \left (d x +c \right ) a^{2} b^{2}+4 A a \,b^{3} \left (d x +c \right )+4 C a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right ) b^{4}+\frac {C \,b^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(198\)
parallelrisch	\(\frac {-12 \left (1+\cos \left (2 d x +2 c \right )\right ) a^{2} \left (12 A \,b^{2}+a^{2} \left (A +2 C \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 \left (1+\cos \left (2 d x +2 c \right )\right ) a^{2} \left (12 A \,b^{2}+a^{2} \left (A +2 C \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+96 x b \left (a^{2} C +\left (A +\frac {C}{2}\right ) b^{2}\right ) d a \cos \left (2 d x +2 c \right )+12 \left (6 C \,a^{2} b^{2}+b^{4} \left (A +\frac {11 C}{12}\right )\right ) \sin \left (3 d x +3 c \right )+24 \left (4 A \,a^{3} b +C a \,b^{3}\right ) \sin \left (2 d x +2 c \right )+12 C \sin \left (4 d x +4 c \right ) a \,b^{3}+C \sin \left (5 d x +5 c \right ) b^{4}+12 \left (2 a^{4} A +6 C \,a^{2} b^{2}+\left (A +\frac {5 C}{6}\right ) b^{4}\right ) \sin \left (d x +c \right )+96 x b \left (a^{2} C +\left (A +\frac {C}{2}\right ) b^{2}\right ) d a}{24 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(277\)
risch	\(4 x A a \,b^{3}+4 C \,a^{3} b x +2 a \,b^{3} C x +\frac {i C \,b^{4} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,b^{4}}{2 d}-\frac {i C a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{2 d}-\frac {i C \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C \,a^{2} b^{2}}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{4}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C \,b^{4}}{8 d}-\frac {i A \,a^{3} \left ({\mathrm e}^{3 i \left (d x +c \right )} a -8 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a \,{\mathrm e}^{i \left (d x +c \right )}-8 b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {i C a \,b^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C \,b^{4}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{2} b^{2}}{d}+\frac {a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\)	\(423\)
norman	\(\frac {\left (4 A a \,b^{3}+4 C \,a^{3} b +2 C a \,b^{3}\right ) x +\left (-40 A a \,b^{3}-40 C \,a^{3} b -20 C a \,b^{3}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-16 A a \,b^{3}-16 C \,a^{3} b -8 C a \,b^{3}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-16 A a \,b^{3}-16 C \,a^{3} b -8 C a \,b^{3}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 A a \,b^{3}+4 C \,a^{3} b +2 C a \,b^{3}\right ) x \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (16 A a \,b^{3}+16 C \,a^{3} b +8 C a \,b^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (16 A a \,b^{3}+16 C \,a^{3} b +8 C a \,b^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (16 A a \,b^{3}+16 C \,a^{3} b +8 C a \,b^{3}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (16 A a \,b^{3}+16 C \,a^{3} b +8 C a \,b^{3}\right ) x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (a^{4} A -8 A \,a^{3} b +2 A \,b^{4}+12 C \,a^{2} b^{2}-4 C a \,b^{3}+2 C \,b^{4}\right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (a^{4} A +8 A \,a^{3} b +2 A \,b^{4}+12 C \,a^{2} b^{2}+4 C a \,b^{3}+2 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (21 a^{4} A -120 A \,a^{3} b +18 A \,b^{4}+108 C \,a^{2} b^{2}-12 C a \,b^{3}+10 C \,b^{4}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (21 a^{4} A +120 A \,a^{3} b +18 A \,b^{4}+108 C \,a^{2} b^{2}+12 C a \,b^{3}+10 C \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (63 a^{4} A -216 A \,a^{3} b +6 A \,b^{4}+36 C \,a^{2} b^{2}+36 C a \,b^{3}-2 C \,b^{4}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (63 a^{4} A +216 A \,a^{3} b +6 A \,b^{4}+36 C \,a^{2} b^{2}-36 C a \,b^{3}-2 C \,b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (105 a^{4} A -120 A \,a^{3} b -30 A \,b^{4}-180 C \,a^{2} b^{2}+36 C a \,b^{3}-14 C \,b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (105 a^{4} A +120 A \,a^{3} b -30 A \,b^{4}-180 C \,a^{2} b^{2}-36 C a \,b^{3}-14 C \,b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a^{2} \left (A \,a^{2}+12 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (A \,a^{2}+12 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(872\)

[In]

int((a+cos(d*x+c)*b)^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.96 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {24 \, {\left (2 \, C a^{3} b + {\left (2 \, A + C\right )} a b^{3}\right )} d x \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (A + 2 \, C\right )} a^{4} + 12 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (A + 2 \, C\right )} a^{4} + 12 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C b^{4} \cos \left (d x + c\right )^{4} + 12 \, C a b^{3} \cos \left (d x + c\right )^{3} + 24 \, A a^{3} b \cos \left (d x + c\right ) + 3 \, A a^{4} + 2 \, {\left (18 \, C a^{2} b^{2} + {\left (3 \, A + 2 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \]

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

1/12*(24*(2*C*a^3*b + (2*A + C)*a*b^3)*d*x*cos(d*x + c)^2 + 3*((A + 2*C)*a^4 + 12*A*a^2*b^2)*cos(d*x + c)^2*lo
g(sin(d*x + c) + 1) - 3*((A + 2*C)*a^4 + 12*A*a^2*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*C*b^4*cos(
d*x + c)^4 + 12*C*a*b^3*cos(d*x + c)^3 + 24*A*a^3*b*cos(d*x + c) + 3*A*a^4 + 2*(18*C*a^2*b^2 + (3*A + 2*C)*b^4
)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.01 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {48 \, {\left (d x + c\right )} C a^{3} b + 48 \, {\left (d x + c\right )} A a b^{3} + 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{3} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{4} - 3 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, A a^{2} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 72 \, C a^{2} b^{2} \sin \left (d x + c\right ) + 12 \, A b^{4} \sin \left (d x + c\right ) + 48 \, A a^{3} b \tan \left (d x + c\right )}{12 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

1/12*(48*(d*x + c)*C*a^3*b + 48*(d*x + c)*A*a*b^3 + 12*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a*b^3 - 4*(sin(d*x +
c)^3 - 3*sin(d*x + c))*C*b^4 - 3*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)) + 6*C*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*A*a^2*b^2*(log(sin(d*x + c) + 1
) - log(sin(d*x + c) - 1)) + 72*C*a^2*b^2*sin(d*x + c) + 12*A*b^4*sin(d*x + c) + 48*A*a^3*b*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.81 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {12 \, {\left (2 \, C a^{3} b + 2 \, A a b^{3} + C a b^{3}\right )} {\left (d x + c\right )} + 3 \, {\left (A a^{4} + 2 \, C a^{4} + 12 \, A a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (A a^{4} + 2 \, C a^{4} + 12 \, A a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {6 \, {\left (A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {4 \, {\left (18 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="giac")

[Out]

1/6*(12*(2*C*a^3*b + 2*A*a*b^3 + C*a*b^3)*(d*x + c) + 3*(A*a^4 + 2*C*a^4 + 12*A*a^2*b^2)*log(abs(tan(1/2*d*x +
1/2*c) + 1)) - 3*(A*a^4 + 2*C*a^4 + 12*A*a^2*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 6*(A*a^4*tan(1/2*d*x +
1/2*c)^3 - 8*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 + A*a^4*tan(1/2*d*x + 1/2*c) + 8*A*a^3*b*tan(1/2*d*x + 1/2*c))/(t
an(1/2*d*x + 1/2*c)^2 - 1)^2 + 4*(18*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*A
*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*C*b^4*tan(1/2*d*x + 1/2*c)^5 + 36*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*b^4*t
an(1/2*d*x + 1/2*c)^3 + 2*C*b^4*tan(1/2*d*x + 1/2*c)^3 + 18*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 6*C*a*b^3*tan(1/2
*d*x + 1/2*c) + 3*A*b^4*tan(1/2*d*x + 1/2*c) + 3*C*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

Mupad [B] (verification not implemented)

Time = 3.53 (sec) , antiderivative size = 2658, normalized size of antiderivative = 12.14 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Too large to display} \]

[In]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^4)/cos(c + d*x)^3,x)

[Out]

(tan(c/2 + (d*x)/2)^5*(6*A*a^4 - 4*A*b^4 + (4*C*b^4)/3 - 24*C*a^2*b^2) + tan(c/2 + (d*x)/2)^3*(4*A*a^4 - (8*C*
b^4)/3 + 16*A*a^3*b - 8*C*a*b^3) + tan(c/2 + (d*x)/2)^7*(4*A*a^4 - (8*C*b^4)/3 - 16*A*a^3*b + 8*C*a*b^3) + tan
(c/2 + (d*x)/2)*(A*a^4 + 2*A*b^4 + 2*C*b^4 + 12*C*a^2*b^2 + 8*A*a^3*b + 4*C*a*b^3) + tan(c/2 + (d*x)/2)^9*(A*a
^4 + 2*A*b^4 + 2*C*b^4 + 12*C*a^2*b^2 - 8*A*a^3*b - 4*C*a*b^3))/(d*(tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2
)^4 - 2*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (atan(((((A*a^4)/2 + C*a^4
+ 6*A*a^2*b^2)*(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64*C*a*b^3 + 128*C*a^3*b) + tan(c/2 + (d*
x)/2)*(8*A^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C
^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2))*((A*a^4)/2
+ C*a^4 + 6*A*a^2*b^2)*1i - (((A*a^4)/2 + C*a^4 + 6*A*a^2*b^2)*(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a*
b^3 + 64*C*a*b^3 + 128*C*a^3*b) - tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*
b^4 + 192*A^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1
024*A*C*a^4*b^4 + 384*A*C*a^6*b^2))*((A*a^4)/2 + C*a^4 + 6*A*a^2*b^2)*1i)/((((A*a^4)/2 + C*a^4 + 6*A*a^2*b^2)*
(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64*C*a*b^3 + 128*C*a^3*b) + tan(c/2 + (d*x)/2)*(8*A^2*a^8
+ 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512
*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2))*((A*a^4)/2 + C*a^4 + 6*A*a^
2*b^2) + (((A*a^4)/2 + C*a^4 + 6*A*a^2*b^2)*(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64*C*a*b^3 +
128*C*a^3*b) - tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b
^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 3
84*A*C*a^6*b^2))*((A*a^4)/2 + C*a^4 + 6*A*a^2*b^2) - 256*C^3*a^11*b + 6144*A^3*a^4*b^8 - 9216*A^3*a^5*b^7 + 51
2*A^3*a^6*b^6 - 1536*A^3*a^7*b^5 - 64*A^3*a^9*b^3 + 256*C^3*a^6*b^6 + 1024*C^3*a^8*b^4 - 128*C^3*a^9*b^3 + 102
4*C^3*a^10*b^2 - 256*A*C^2*a^11*b - 64*A^2*C*a^11*b + 1536*A*C^2*a^4*b^8 + 7296*A*C^2*a^6*b^6 - 1536*A*C^2*a^7
*b^5 + 8704*A*C^2*a^8*b^4 - 3456*A*C^2*a^9*b^3 + 512*A*C^2*a^10*b^2 + 6144*A^2*C*a^4*b^8 - 4608*A^2*C*a^5*b^7
+ 13824*A^2*C*a^6*b^6 - 13056*A^2*C*a^7*b^5 + 1024*A^2*C*a^8*b^4 - 1824*A^2*C*a^9*b^3))*(A*a^4*1i + C*a^4*2i +
A*a^2*b^2*12i))/d + (4*a*b*atan((2*a*b*(tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A
^2*a^4*b^4 + 192*A^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*
b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2) - a*b*(2*A*b^2 + 2*C*a^2 + C*b^2)*(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b
^2 + 128*A*a*b^3 + 64*C*a*b^3 + 128*C*a^3*b)*2i)*(2*A*b^2 + 2*C*a^2 + C*b^2) + 2*a*b*(tan(c/2 + (d*x)/2)*(8*A^
2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4
+ 512*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2) + a*b*(2*A*b^2 + 2*C*a^
2 + C*b^2)*(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64*C*a*b^3 + 128*C*a^3*b)*2i)*(2*A*b^2 + 2*C*a
^2 + C*b^2))/(256*C^3*a^11*b - 6144*A^3*a^4*b^8 + 9216*A^3*a^5*b^7 - 512*A^3*a^6*b^6 + 1536*A^3*a^7*b^5 + 64*A
^3*a^9*b^3 - 256*C^3*a^6*b^6 - 1024*C^3*a^8*b^4 + 128*C^3*a^9*b^3 - 1024*C^3*a^10*b^2 + a*b*(tan(c/2 + (d*x)/2
)*(8*A^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a
^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2) - a*b*(2*A*b^2 +
2*C*a^2 + C*b^2)*(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64*C*a*b^3 + 128*C*a^3*b)*2i)*(2*A*b^2
+ 2*C*a^2 + C*b^2)*2i - a*b*(tan(c/2 + (d*x)/2)*(8*A^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 +
192*A^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024*A
*C*a^4*b^4 + 384*A*C*a^6*b^2) + a*b*(2*A*b^2 + 2*C*a^2 + C*b^2)*(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a
*b^3 + 64*C*a*b^3 + 128*C*a^3*b)*2i)*(2*A*b^2 + 2*C*a^2 + C*b^2)*2i + 256*A*C^2*a^11*b + 64*A^2*C*a^11*b - 153
6*A*C^2*a^4*b^8 - 7296*A*C^2*a^6*b^6 + 1536*A*C^2*a^7*b^5 - 8704*A*C^2*a^8*b^4 + 3456*A*C^2*a^9*b^3 - 512*A*C^
2*a^10*b^2 - 6144*A^2*C*a^4*b^8 + 4608*A^2*C*a^5*b^7 - 13824*A^2*C*a^6*b^6 + 13056*A^2*C*a^7*b^5 - 1024*A^2*C*
a^8*b^4 + 1824*A^2*C*a^9*b^3))*(2*A*b^2 + 2*C*a^2 + C*b^2))/d