3.2.64 \(\int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [164]

Optimal. Leaf size=156 \[ \frac {10 \sin (c+d x)}{33 a^4 d}-\frac {20 \sin ^3(c+d x)}{99 a^4 d}+\frac {2 \sin ^5(c+d x)}{33 a^4 d}+\frac {i \cos ^3(c+d x)}{11 d (a+i a \tan (c+d x))^4}+\frac {7 i \cos ^3(c+d x)}{99 a d (a+i a \tan (c+d x))^3}+\frac {4 i \cos ^5(c+d x)}{33 d \left (a^4+i a^4 \tan (c+d x)\right )} \]

[Out]

10/33*sin(d*x+c)/a^4/d-20/99*sin(d*x+c)^3/a^4/d+2/33*sin(d*x+c)^5/a^4/d+1/11*I*cos(d*x+c)^3/d/(a+I*a*tan(d*x+c
))^4+7/99*I*cos(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^3+4/33*I*cos(d*x+c)^5/d/(a^4+I*a^4*tan(d*x+c))

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Rubi [A]
time = 0.12, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3583, 3581, 2713} \begin {gather*} \frac {2 \sin ^5(c+d x)}{33 a^4 d}-\frac {20 \sin ^3(c+d x)}{99 a^4 d}+\frac {10 \sin (c+d x)}{33 a^4 d}+\frac {4 i \cos ^5(c+d x)}{33 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {7 i \cos ^3(c+d x)}{99 a d (a+i a \tan (c+d x))^3}+\frac {i \cos ^3(c+d x)}{11 d (a+i a \tan (c+d x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(10*Sin[c + d*x])/(33*a^4*d) - (20*Sin[c + d*x]^3)/(99*a^4*d) + (2*Sin[c + d*x]^5)/(33*a^4*d) + ((I/11)*Cos[c
+ d*x]^3)/(d*(a + I*a*Tan[c + d*x])^4) + (((7*I)/99)*Cos[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*x])^3) + (((4*I)/
33)*Cos[c + d*x]^5)/(d*(a^4 + I*a^4*Tan[c + d*x]))

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 3581

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

Rule 3583

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(b*f*(m + 2*n))), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx &=\frac {i \cos ^3(c+d x)}{11 d (a+i a \tan (c+d x))^4}+\frac {7 \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx}{11 a}\\ &=\frac {i \cos ^3(c+d x)}{11 d (a+i a \tan (c+d x))^4}+\frac {7 i \cos ^3(c+d x)}{99 a d (a+i a \tan (c+d x))^3}+\frac {14 \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{33 a^2}\\ &=\frac {i \cos ^3(c+d x)}{11 d (a+i a \tan (c+d x))^4}+\frac {7 i \cos ^3(c+d x)}{99 a d (a+i a \tan (c+d x))^3}+\frac {4 i \cos ^5(c+d x)}{33 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {10 \int \cos ^5(c+d x) \, dx}{33 a^4}\\ &=\frac {i \cos ^3(c+d x)}{11 d (a+i a \tan (c+d x))^4}+\frac {7 i \cos ^3(c+d x)}{99 a d (a+i a \tan (c+d x))^3}+\frac {4 i \cos ^5(c+d x)}{33 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {10 \text {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{33 a^4 d}\\ &=\frac {10 \sin (c+d x)}{33 a^4 d}-\frac {20 \sin ^3(c+d x)}{99 a^4 d}+\frac {2 \sin ^5(c+d x)}{33 a^4 d}+\frac {i \cos ^3(c+d x)}{11 d (a+i a \tan (c+d x))^4}+\frac {7 i \cos ^3(c+d x)}{99 a d (a+i a \tan (c+d x))^3}+\frac {4 i \cos ^5(c+d x)}{33 d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.55, size = 117, normalized size = 0.75 \begin {gather*} -\frac {i \sec ^4(c+d x) (-924 \cos (c+d x)-1188 \cos (3 (c+d x))+308 \cos (5 (c+d x))+12 \cos (7 (c+d x))-231 i \sin (c+d x)-891 i \sin (3 (c+d x))+385 i \sin (5 (c+d x))+21 i \sin (7 (c+d x)))}{6336 a^4 d (-i+\tan (c+d x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((-1/6336*I)*Sec[c + d*x]^4*(-924*Cos[c + d*x] - 1188*Cos[3*(c + d*x)] + 308*Cos[5*(c + d*x)] + 12*Cos[7*(c +
d*x)] - (231*I)*Sin[c + d*x] - (891*I)*Sin[3*(c + d*x)] + (385*I)*Sin[5*(c + d*x)] + (21*I)*Sin[7*(c + d*x)]))
/(a^4*d*(-I + Tan[c + d*x])^4)

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Maple [A]
time = 0.30, size = 240, normalized size = 1.54

method	result	size
risch	\(\frac {7 i {\mathrm e}^{-5 i \left (d x +c \right )}}{128 a^{4} d}+\frac {3 i {\mathrm e}^{-7 i \left (d x +c \right )}}{128 a^{4} d}+\frac {7 i {\mathrm e}^{-9 i \left (d x +c \right )}}{1152 a^{4} d}+\frac {i {\mathrm e}^{-11 i \left (d x +c \right )}}{1408 a^{4} d}+\frac {7 i \cos \left (d x +c \right )}{64 a^{4} d}+\frac {7 \sin \left (d x +c \right )}{32 a^{4} d}+\frac {17 i \cos \left (3 d x +3 c \right )}{192 a^{4} d}+\frac {3 \sin \left (3 d x +3 c \right )}{32 a^{4} d}\)	\(138\)
derivativedivides	\(\frac {-\frac {i}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}-\frac {1}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{3}}+\frac {2}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16 i}+\frac {8 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{10}}-\frac {67 i}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {44 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {385 i}{6 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {201 i}{32 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {16}{11 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{11}}+\frac {208}{9 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {61}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {105}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {267}{16 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {15}{8 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{a^{4} d}\)	\(240\)
default	\(\frac {-\frac {i}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}-\frac {1}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{3}}+\frac {2}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16 i}+\frac {8 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{10}}-\frac {67 i}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {44 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {385 i}{6 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {201 i}{32 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {16}{11 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{11}}+\frac {208}{9 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {61}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {105}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {267}{16 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {15}{8 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{a^{4} d}\)	\(240\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

2/d/a^4*(-1/64*I/(tan(1/2*d*x+1/2*c)+I)^2-1/96/(tan(1/2*d*x+1/2*c)+I)^3+1/16/(tan(1/2*d*x+1/2*c)+I)+4*I/(-I+ta
n(1/2*d*x+1/2*c))^10-67/4*I/(-I+tan(1/2*d*x+1/2*c))^4-22*I/(-I+tan(1/2*d*x+1/2*c))^8+385/12*I/(-I+tan(1/2*d*x+
1/2*c))^6+201/64*I/(-I+tan(1/2*d*x+1/2*c))^2-8/11/(-I+tan(1/2*d*x+1/2*c))^11+104/9/(-I+tan(1/2*d*x+1/2*c))^9-6
1/2/(-I+tan(1/2*d*x+1/2*c))^7+105/4/(-I+tan(1/2*d*x+1/2*c))^5-267/32/(-I+tan(1/2*d*x+1/2*c))^3+15/16/(-I+tan(1
/2*d*x+1/2*c)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [A]
time = 0.37, size = 96, normalized size = 0.62 \begin {gather*} \frac {{\left (-33 i \, e^{\left (14 i \, d x + 14 i \, c\right )} - 693 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 2079 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 1155 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 693 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 297 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 77 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 9 i\right )} e^{\left (-11 i \, d x - 11 i \, c\right )}}{12672 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/12672*(-33*I*e^(14*I*d*x + 14*I*c) - 693*I*e^(12*I*d*x + 12*I*c) + 2079*I*e^(10*I*d*x + 10*I*c) + 1155*I*e^(
8*I*d*x + 8*I*c) + 693*I*e^(6*I*d*x + 6*I*c) + 297*I*e^(4*I*d*x + 4*I*c) + 77*I*e^(2*I*d*x + 2*I*c) + 9*I)*e^(
-11*I*d*x - 11*I*c)/(a^4*d)

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Sympy [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (136) = 272\).
time = 0.46, size = 299, normalized size = 1.92 \begin {gather*} \begin {cases} \frac {\left (- 167196136166129664 i a^{28} d^{7} e^{39 i c} e^{3 i d x} - 3511118859488722944 i a^{28} d^{7} e^{37 i c} e^{i d x} + 10533356578466168832 i a^{28} d^{7} e^{35 i c} e^{- i d x} + 5851864765814538240 i a^{28} d^{7} e^{33 i c} e^{- 3 i d x} + 3511118859488722944 i a^{28} d^{7} e^{31 i c} e^{- 5 i d x} + 1504765225495166976 i a^{28} d^{7} e^{29 i c} e^{- 7 i d x} + 390124317720969216 i a^{28} d^{7} e^{27 i c} e^{- 9 i d x} + 45598946227126272 i a^{28} d^{7} e^{25 i c} e^{- 11 i d x}\right ) e^{- 36 i c}}{64203316287793790976 a^{32} d^{8}} & \text {for}\: a^{32} d^{8} e^{36 i c} \neq 0 \\\frac {x \left (e^{14 i c} + 7 e^{12 i c} + 21 e^{10 i c} + 35 e^{8 i c} + 35 e^{6 i c} + 21 e^{4 i c} + 7 e^{2 i c} + 1\right ) e^{- 11 i c}}{128 a^{4}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a+I*a*tan(d*x+c))**4,x)

[Out]

Piecewise(((-167196136166129664*I*a**28*d**7*exp(39*I*c)*exp(3*I*d*x) - 3511118859488722944*I*a**28*d**7*exp(3
7*I*c)*exp(I*d*x) + 10533356578466168832*I*a**28*d**7*exp(35*I*c)*exp(-I*d*x) + 5851864765814538240*I*a**28*d*
*7*exp(33*I*c)*exp(-3*I*d*x) + 3511118859488722944*I*a**28*d**7*exp(31*I*c)*exp(-5*I*d*x) + 150476522549516697
6*I*a**28*d**7*exp(29*I*c)*exp(-7*I*d*x) + 390124317720969216*I*a**28*d**7*exp(27*I*c)*exp(-9*I*d*x) + 4559894
6227126272*I*a**28*d**7*exp(25*I*c)*exp(-11*I*d*x))*exp(-36*I*c)/(64203316287793790976*a**32*d**8), Ne(a**32*d
**8*exp(36*I*c), 0)), (x*(exp(14*I*c) + 7*exp(12*I*c) + 21*exp(10*I*c) + 35*exp(8*I*c) + 35*exp(6*I*c) + 21*ex
p(4*I*c) + 7*exp(2*I*c) + 1)*exp(-11*I*c)/(128*a**4), True))

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Giac [A]
time = 0.86, size = 197, normalized size = 1.26 \begin {gather*} \frac {\frac {33 \, {\left (12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 21 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 11\right )}}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )}^{3}} + \frac {5940 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 39501 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 141075 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 313236 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 479556 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 516054 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 397914 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 214500 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 79024 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 17765 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2155}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{11}}}{3168 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/3168*(33*(12*tan(1/2*d*x + 1/2*c)^2 + 21*I*tan(1/2*d*x + 1/2*c) - 11)/(a^4*(tan(1/2*d*x + 1/2*c) + I)^3) + (
5940*tan(1/2*d*x + 1/2*c)^10 - 39501*I*tan(1/2*d*x + 1/2*c)^9 - 141075*tan(1/2*d*x + 1/2*c)^8 + 313236*I*tan(1
/2*d*x + 1/2*c)^7 + 479556*tan(1/2*d*x + 1/2*c)^6 - 516054*I*tan(1/2*d*x + 1/2*c)^5 - 397914*tan(1/2*d*x + 1/2
*c)^4 + 214500*I*tan(1/2*d*x + 1/2*c)^3 + 79024*tan(1/2*d*x + 1/2*c)^2 - 17765*I*tan(1/2*d*x + 1/2*c) - 2155)/
(a^4*(tan(1/2*d*x + 1/2*c) - I)^11))/d

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Mupad [B]
time = 5.61, size = 216, normalized size = 1.38 \begin {gather*} -\frac {\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {269\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}-\frac {1307\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{64}+\frac {1307\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{64}-\frac {1099\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{32}+\frac {203\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{32}-\frac {21\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )}{64}+\frac {21\,\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )}{64}+\frac {\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,231{}\mathrm {i}}{16}-\frac {\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,231{}\mathrm {i}}{16}+\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,33{}\mathrm {i}-\cos \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\,5{}\mathrm {i}+\frac {\cos \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,3{}\mathrm {i}}{16}-\frac {\cos \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,3{}\mathrm {i}}{16}\right )\,2{}\mathrm {i}}{99\,a^4\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^{11}\,{\left (\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

-(cos(c/2 + (d*x)/2)*((cos((3*c)/2 + (3*d*x)/2)*231i)/16 - (cos((5*c)/2 + (5*d*x)/2)*231i)/16 + cos((7*c)/2 +
(7*d*x)/2)*33i - cos((9*c)/2 + (9*d*x)/2)*5i + (cos((11*c)/2 + (11*d*x)/2)*3i)/16 - (cos((13*c)/2 + (13*d*x)/2
)*3i)/16 + (269*sin(c/2 + (d*x)/2))/16 - (1307*sin((3*c)/2 + (3*d*x)/2))/64 + (1307*sin((5*c)/2 + (5*d*x)/2))/
64 - (1099*sin((7*c)/2 + (7*d*x)/2))/32 + (203*sin((9*c)/2 + (9*d*x)/2))/32 - (21*sin((11*c)/2 + (11*d*x)/2))/
64 + (21*sin((13*c)/2 + (13*d*x)/2))/64)*2i)/(99*a^4*d*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2)*1i)^11*(cos(c/
2 + (d*x)/2)*1i + sin(c/2 + (d*x)/2))^3)

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