\(\int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx\) [58]

Optimal result

Integrand size = 24, antiderivative size = 120 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {5 a^4 \sin (c+d x)}{21 d}-\frac {10 a^4 \sin ^3(c+d x)}{63 d}+\frac {a^4 \sin ^5(c+d x)}{21 d}-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^3}{9 d}-\frac {2 i \cos ^7(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{21 d} \]

[Out]

5/21*a^4*sin(d*x+c)/d-10/63*a^4*sin(d*x+c)^3/d+1/21*a^4*sin(d*x+c)^5/d-2/9*I*a*cos(d*x+c)^9*(a+I*a*tan(d*x+c))
^3/d-2/21*I*cos(d*x+c)^7*(a^4+I*a^4*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3577, 2713} \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4 \sin ^5(c+d x)}{21 d}-\frac {10 a^4 \sin ^3(c+d x)}{63 d}+\frac {5 a^4 \sin (c+d x)}{21 d}-\frac {2 i \cos ^7(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{21 d}-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^3}{9 d} \]

[In]

Int[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(5*a^4*Sin[c + d*x])/(21*d) - (10*a^4*Sin[c + d*x]^3)/(63*d) + (a^4*Sin[c + d*x]^5)/(21*d) - (((2*I)/9)*a*Cos[
c + d*x]^9*(a + I*a*Tan[c + d*x])^3)/d - (((2*I)/21)*Cos[c + d*x]^7*(a^4 + I*a^4*Tan[c + d*x]))/d

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3577

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(d
*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] - Dist[b^2*((m + 2*n - 2)/(d^2*m)), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^3}{9 d}+\frac {1}{3} a^2 \int \cos ^7(c+d x) (a+i a \tan (c+d x))^2 \, dx \\ & = -\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^3}{9 d}-\frac {2 i \cos ^7(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{21 d}+\frac {1}{21} \left (5 a^4\right ) \int \cos ^5(c+d x) \, dx \\ & = -\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^3}{9 d}-\frac {2 i \cos ^7(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{21 d}-\frac {\left (5 a^4\right ) \text {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{21 d} \\ & = \frac {5 a^4 \sin (c+d x)}{21 d}-\frac {10 a^4 \sin ^3(c+d x)}{63 d}+\frac {a^4 \sin ^5(c+d x)}{21 d}-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^3}{9 d}-\frac {2 i \cos ^7(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{21 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.80 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4 (-i \cos (4 (c+d x))+\sin (4 (c+d x))) \left (168 \cos (c+d x) \sqrt {\cos ^2(c+d x)}+4 \left (16+45 \sqrt {\cos ^2(c+d x)}\right ) \cos (3 (c+d x))+64 \cos (5 (c+d x))-28 \sqrt {\cos ^2(c+d x)} \cos (5 (c+d x))-42 i \sqrt {\cos ^2(c+d x)} \sin (c+d x)-64 i \sin (3 (c+d x))-135 i \sqrt {\cos ^2(c+d x)} \sin (3 (c+d x))-64 i \sin (5 (c+d x))+35 i \sqrt {\cos ^2(c+d x)} \sin (5 (c+d x))\right )}{1008 d \sqrt {\cos ^2(c+d x)}} \]

[In]

Integrate[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(a^4*((-I)*Cos[4*(c + d*x)] + Sin[4*(c + d*x)])*(168*Cos[c + d*x]*Sqrt[Cos[c + d*x]^2] + 4*(16 + 45*Sqrt[Cos[c
+ d*x]^2])*Cos[3*(c + d*x)] + 64*Cos[5*(c + d*x)] - 28*Sqrt[Cos[c + d*x]^2]*Cos[5*(c + d*x)] - (42*I)*Sqrt[Co
s[c + d*x]^2]*Sin[c + d*x] - (64*I)*Sin[3*(c + d*x)] - (135*I)*Sqrt[Cos[c + d*x]^2]*Sin[3*(c + d*x)] - (64*I)*
Sin[5*(c + d*x)] + (35*I)*Sqrt[Cos[c + d*x]^2]*Sin[5*(c + d*x)]))/(1008*d*Sqrt[Cos[c + d*x]^2])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (106 ) = 212\).

Time = 0.82 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.94

\[\frac {a^{4} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{9}-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{21}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{105}\right )-4 i a^{4} \left (-\frac {\left (\cos ^{7}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{9}-\frac {2 \left (\cos ^{7}\left (d x +c \right )\right )}{63}\right )-6 a^{4} \left (-\frac {\left (\cos ^{8}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{9}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{63}\right )-\frac {4 i a^{4} \left (\cos ^{9}\left (d x +c \right )\right )}{9}+\frac {a^{4} \left (\frac {128}{35}+\cos ^{8}\left (d x +c \right )+\frac {8 \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {48 \left (\cos ^{4}\left (d x +c \right )\right )}{35}+\frac {64 \left (\cos ^{2}\left (d x +c \right )\right )}{35}\right ) \sin \left (d x +c \right )}{9}}{d}\]

[In]

int(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^4,x)

[Out]

1/d*(a^4*(-1/9*sin(d*x+c)^3*cos(d*x+c)^6-1/21*sin(d*x+c)*cos(d*x+c)^6+1/105*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2
)*sin(d*x+c))-4*I*a^4*(-1/9*cos(d*x+c)^7*sin(d*x+c)^2-2/63*cos(d*x+c)^7)-6*a^4*(-1/9*cos(d*x+c)^8*sin(d*x+c)+1
/63*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))-4/9*I*a^4*cos(d*x+c)^9+1/9*a^4*(128/35+c
os(d*x+c)^8+8/7*cos(d*x+c)^6+48/35*cos(d*x+c)^4+64/35*cos(d*x+c)^2)*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.75 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {{\left (-7 i \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} - 45 i \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 126 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 210 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 315 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 63 i \, a^{4}\right )} e^{\left (-i \, d x - i \, c\right )}}{2016 \, d} \]

[In]

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/2016*(-7*I*a^4*e^(10*I*d*x + 10*I*c) - 45*I*a^4*e^(8*I*d*x + 8*I*c) - 126*I*a^4*e^(6*I*d*x + 6*I*c) - 210*I*
a^4*e^(4*I*d*x + 4*I*c) - 315*I*a^4*e^(2*I*d*x + 2*I*c) + 63*I*a^4)*e^(-I*d*x - I*c)/d

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (107) = 214\).

Time = 0.39 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.90 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\begin {cases} \frac {\left (- 176160768 i a^{4} d^{5} e^{10 i c} e^{9 i d x} - 1132462080 i a^{4} d^{5} e^{8 i c} e^{7 i d x} - 3170893824 i a^{4} d^{5} e^{6 i c} e^{5 i d x} - 5284823040 i a^{4} d^{5} e^{4 i c} e^{3 i d x} - 7927234560 i a^{4} d^{5} e^{2 i c} e^{i d x} + 1585446912 i a^{4} d^{5} e^{- i d x}\right ) e^{- i c}}{50734301184 d^{6}} & \text {for}\: d^{6} e^{i c} \neq 0 \\\frac {x \left (a^{4} e^{10 i c} + 5 a^{4} e^{8 i c} + 10 a^{4} e^{6 i c} + 10 a^{4} e^{4 i c} + 5 a^{4} e^{2 i c} + a^{4}\right ) e^{- i c}}{32} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**9*(a+I*a*tan(d*x+c))**4,x)

[Out]

Piecewise(((-176160768*I*a**4*d**5*exp(10*I*c)*exp(9*I*d*x) - 1132462080*I*a**4*d**5*exp(8*I*c)*exp(7*I*d*x) -
3170893824*I*a**4*d**5*exp(6*I*c)*exp(5*I*d*x) - 5284823040*I*a**4*d**5*exp(4*I*c)*exp(3*I*d*x) - 7927234560*
I*a**4*d**5*exp(2*I*c)*exp(I*d*x) + 1585446912*I*a**4*d**5*exp(-I*d*x))*exp(-I*c)/(50734301184*d**6), Ne(d**6*
exp(I*c), 0)), (x*(a**4*exp(10*I*c) + 5*a**4*exp(8*I*c) + 10*a**4*exp(6*I*c) + 10*a**4*exp(4*I*c) + 5*a**4*exp
(2*I*c) + a**4)*exp(-I*c)/32, True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.51 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {140 i \, a^{4} \cos \left (d x + c\right )^{9} + 20 i \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} a^{4} - {\left (35 \, \sin \left (d x + c\right )^{9} - 90 \, \sin \left (d x + c\right )^{7} + 63 \, \sin \left (d x + c\right )^{5}\right )} a^{4} - 6 \, {\left (35 \, \sin \left (d x + c\right )^{9} - 135 \, \sin \left (d x + c\right )^{7} + 189 \, \sin \left (d x + c\right )^{5} - 105 \, \sin \left (d x + c\right )^{3}\right )} a^{4} - {\left (35 \, \sin \left (d x + c\right )^{9} - 180 \, \sin \left (d x + c\right )^{7} + 378 \, \sin \left (d x + c\right )^{5} - 420 \, \sin \left (d x + c\right )^{3} + 315 \, \sin \left (d x + c\right )\right )} a^{4}}{315 \, d} \]

[In]

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/315*(140*I*a^4*cos(d*x + c)^9 + 20*I*(7*cos(d*x + c)^9 - 9*cos(d*x + c)^7)*a^4 - (35*sin(d*x + c)^9 - 90*si
n(d*x + c)^7 + 63*sin(d*x + c)^5)*a^4 - 6*(35*sin(d*x + c)^9 - 135*sin(d*x + c)^7 + 189*sin(d*x + c)^5 - 105*s
in(d*x + c)^3)*a^4 - (35*sin(d*x + c)^9 - 180*sin(d*x + c)^7 + 378*sin(d*x + c)^5 - 420*sin(d*x + c)^3 + 315*s
in(d*x + c))*a^4)/d

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1409 vs. \(2 (102) = 204\).

Time = 0.81 (sec) , antiderivative size = 1409, normalized size of antiderivative = 11.74 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/516096*(435267*a^4*e^(13*I*d*x + 7*I*c)*log(I*e^(I*d*x + I*c) + 1) + 2611602*a^4*e^(11*I*d*x + 5*I*c)*log(I*
e^(I*d*x + I*c) + 1) + 6529005*a^4*e^(9*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) + 1) + 8705340*a^4*e^(7*I*d*x + I
*c)*log(I*e^(I*d*x + I*c) + 1) + 6529005*a^4*e^(5*I*d*x - I*c)*log(I*e^(I*d*x + I*c) + 1) + 2611602*a^4*e^(3*I
*d*x - 3*I*c)*log(I*e^(I*d*x + I*c) + 1) + 435267*a^4*e^(I*d*x - 5*I*c)*log(I*e^(I*d*x + I*c) + 1) + 427896*a^
4*e^(13*I*d*x + 7*I*c)*log(I*e^(I*d*x + I*c) - 1) + 2567376*a^4*e^(11*I*d*x + 5*I*c)*log(I*e^(I*d*x + I*c) - 1
) + 6418440*a^4*e^(9*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) - 1) + 8557920*a^4*e^(7*I*d*x + I*c)*log(I*e^(I*d*x
+ I*c) - 1) + 6418440*a^4*e^(5*I*d*x - I*c)*log(I*e^(I*d*x + I*c) - 1) + 2567376*a^4*e^(3*I*d*x - 3*I*c)*log(I
*e^(I*d*x + I*c) - 1) + 427896*a^4*e^(I*d*x - 5*I*c)*log(I*e^(I*d*x + I*c) - 1) - 435267*a^4*e^(13*I*d*x + 7*I
*c)*log(-I*e^(I*d*x + I*c) + 1) - 2611602*a^4*e^(11*I*d*x + 5*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 6529005*a^4*e
^(9*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 8705340*a^4*e^(7*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) + 1) - 6
529005*a^4*e^(5*I*d*x - I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2611602*a^4*e^(3*I*d*x - 3*I*c)*log(-I*e^(I*d*x + I
*c) + 1) - 435267*a^4*e^(I*d*x - 5*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 427896*a^4*e^(13*I*d*x + 7*I*c)*log(-I*e
^(I*d*x + I*c) - 1) - 2567376*a^4*e^(11*I*d*x + 5*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 6418440*a^4*e^(9*I*d*x +
3*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 8557920*a^4*e^(7*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) - 1) - 6418440*a^4*e
^(5*I*d*x - I*c)*log(-I*e^(I*d*x + I*c) - 1) - 2567376*a^4*e^(3*I*d*x - 3*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 4
27896*a^4*e^(I*d*x - 5*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 7371*a^4*e^(13*I*d*x + 7*I*c)*log(I*e^(I*d*x) + e^(-
I*c)) - 44226*a^4*e^(11*I*d*x + 5*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 110565*a^4*e^(9*I*d*x + 3*I*c)*log(I*e^(I
*d*x) + e^(-I*c)) - 147420*a^4*e^(7*I*d*x + I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 110565*a^4*e^(5*I*d*x - I*c)*lo
g(I*e^(I*d*x) + e^(-I*c)) - 44226*a^4*e^(3*I*d*x - 3*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 7371*a^4*e^(I*d*x - 5*
I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 7371*a^4*e^(13*I*d*x + 7*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 44226*a^4*e^(1
1*I*d*x + 5*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 110565*a^4*e^(9*I*d*x + 3*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) +
147420*a^4*e^(7*I*d*x + I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 110565*a^4*e^(5*I*d*x - I*c)*log(-I*e^(I*d*x) + e^
(-I*c)) + 44226*a^4*e^(3*I*d*x - 3*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 7371*a^4*e^(I*d*x - 5*I*c)*log(-I*e^(I*
d*x) + e^(-I*c)) - 1792*I*a^4*e^(22*I*d*x + 16*I*c) - 22272*I*a^4*e^(20*I*d*x + 14*I*c) - 128256*I*a^4*e^(18*I
*d*x + 12*I*c) - 455936*I*a^4*e^(16*I*d*x + 10*I*c) - 1144320*I*a^4*e^(14*I*d*x + 8*I*c) - 2102784*I*a^4*e^(12
*I*d*x + 6*I*c) - 2742784*I*a^4*e^(10*I*d*x + 4*I*c) - 2382336*I*a^4*e^(8*I*d*x + 2*I*c) - 295680*I*a^4*e^(4*I
*d*x - 2*I*c) + 16128*I*a^4*e^(2*I*d*x - 4*I*c) - 1241856*I*a^4*e^(6*I*d*x) + 16128*I*a^4*e^(-6*I*c))/(d*e^(13
*I*d*x + 7*I*c) + 6*d*e^(11*I*d*x + 5*I*c) + 15*d*e^(9*I*d*x + 3*I*c) + 20*d*e^(7*I*d*x + I*c) + 15*d*e^(5*I*d
*x - I*c) + 6*d*e^(3*I*d*x - 3*I*c) + d*e^(I*d*x - 5*I*c))

Mupad [B] (verification not implemented)

Time = 6.09 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.21 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {2\,a^4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {89\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}-\frac {55\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{4}+\frac {55\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{4}-\frac {355\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{16}+\frac {35\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{16}-\frac {\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,21{}\mathrm {i}}{2}+\frac {\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,21{}\mathrm {i}}{2}-\frac {\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,87{}\mathrm {i}}{4}+\frac {\cos \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\,7{}\mathrm {i}}{4}\right )}{63\,d\,\left (\cos \left (4\,c+4\,d\,x\right )-\sin \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}\right )} \]

[In]

int(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^4,x)

[Out]

(2*a^4*cos(c/2 + (d*x)/2)*((cos((5*c)/2 + (5*d*x)/2)*21i)/2 - (cos((3*c)/2 + (3*d*x)/2)*21i)/2 - (cos((7*c)/2
+ (7*d*x)/2)*87i)/4 + (cos((9*c)/2 + (9*d*x)/2)*7i)/4 + (89*sin(c/2 + (d*x)/2))/8 - (55*sin((3*c)/2 + (3*d*x)/
2))/4 + (55*sin((5*c)/2 + (5*d*x)/2))/4 - (355*sin((7*c)/2 + (7*d*x)/2))/16 + (35*sin((9*c)/2 + (9*d*x)/2))/16
))/(63*d*(cos(4*c + 4*d*x) - sin(4*c + 4*d*x)*1i))