3.35 \(\int \cos ^9(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=105 \[ -\frac {a^2 \sin ^7(c+d x)}{9 d}+\frac {7 a^2 \sin ^5(c+d x)}{15 d}-\frac {7 a^2 \sin ^3(c+d x)}{9 d}+\frac {7 a^2 \sin (c+d x)}{9 d}-\frac {2 i \cos ^9(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d} \]

[Out]

7/9*a^2*sin(d*x+c)/d-7/9*a^2*sin(d*x+c)^3/d+7/15*a^2*sin(d*x+c)^5/d-1/9*a^2*sin(d*x+c)^7/d-2/9*I*cos(d*x+c)^9*
(a^2+I*a^2*tan(d*x+c))/d

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3496, 2633} \[ -\frac {a^2 \sin ^7(c+d x)}{9 d}+\frac {7 a^2 \sin ^5(c+d x)}{15 d}-\frac {7 a^2 \sin ^3(c+d x)}{9 d}+\frac {7 a^2 \sin (c+d x)}{9 d}-\frac {2 i \cos ^9(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(7*a^2*Sin[c + d*x])/(9*d) - (7*a^2*Sin[c + d*x]^3)/(9*d) + (7*a^2*Sin[c + d*x]^5)/(15*d) - (a^2*Sin[c + d*x]^
7)/(9*d) - (((2*I)/9)*Cos[c + d*x]^9*(a^2 + I*a^2*Tan[c + d*x]))/d

Rule 2633

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 3496

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*} \int \cos ^9(c+d x) (a+i a \tan (c+d x))^2 \, dx &=-\frac {2 i \cos ^9(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac {1}{9} \left (7 a^2\right ) \int \cos ^7(c+d x) \, dx\\ &=-\frac {2 i \cos ^9(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}-\frac {\left (7 a^2\right ) \operatorname {Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,-\sin (c+d x)\right )}{9 d}\\ &=\frac {7 a^2 \sin (c+d x)}{9 d}-\frac {7 a^2 \sin ^3(c+d x)}{9 d}+\frac {7 a^2 \sin ^5(c+d x)}{15 d}-\frac {a^2 \sin ^7(c+d x)}{9 d}-\frac {2 i \cos ^9(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.36, size = 133, normalized size = 1.27 \[ \frac {a^2 (-525 \sin (c+d x)+567 \sin (3 (c+d x))+75 \sin (5 (c+d x))+7 \sin (7 (c+d x))-1050 i \cos (c+d x)+378 i \cos (3 (c+d x))+30 i \cos (5 (c+d x))+2 i \cos (7 (c+d x))) (\cos (2 (c+2 d x))+i \sin (2 (c+2 d x)))}{2880 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(a^2*((-1050*I)*Cos[c + d*x] + (378*I)*Cos[3*(c + d*x)] + (30*I)*Cos[5*(c + d*x)] + (2*I)*Cos[7*(c + d*x)] - 5
25*Sin[c + d*x] + 567*Sin[3*(c + d*x)] + 75*Sin[5*(c + d*x)] + 7*Sin[7*(c + d*x)])*(Cos[2*(c + 2*d*x)] + I*Sin
[2*(c + 2*d*x)]))/(2880*d*(Cos[d*x] + I*Sin[d*x])^2)

________________________________________________________________________________________

fricas [A]  time = 0.60, size = 118, normalized size = 1.12 \[ \frac {{\left (-5 i \, a^{2} e^{\left (14 i \, d x + 14 i \, c\right )} - 45 i \, a^{2} e^{\left (12 i \, d x + 12 i \, c\right )} - 189 i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} - 525 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 1575 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 945 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 105 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 9 i \, a^{2}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{5760 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/5760*(-5*I*a^2*e^(14*I*d*x + 14*I*c) - 45*I*a^2*e^(12*I*d*x + 12*I*c) - 189*I*a^2*e^(10*I*d*x + 10*I*c) - 52
5*I*a^2*e^(8*I*d*x + 8*I*c) - 1575*I*a^2*e^(6*I*d*x + 6*I*c) + 945*I*a^2*e^(4*I*d*x + 4*I*c) + 105*I*a^2*e^(2*
I*d*x + 2*I*c) + 9*I*a^2)*e^(-5*I*d*x - 5*I*c)/d

________________________________________________________________________________________

giac [B]  time = 2.12, size = 669, normalized size = 6.37 \[ -\frac {18585 \, a^{2} e^{\left (9 i \, d x + 3 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 37170 \, a^{2} e^{\left (7 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 18585 \, a^{2} e^{\left (5 i \, d x - i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 14625 \, a^{2} e^{\left (9 i \, d x + 3 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 29250 \, a^{2} e^{\left (7 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 14625 \, a^{2} e^{\left (5 i \, d x - i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 18585 \, a^{2} e^{\left (9 i \, d x + 3 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 37170 \, a^{2} e^{\left (7 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 18585 \, a^{2} e^{\left (5 i \, d x - i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 14625 \, a^{2} e^{\left (9 i \, d x + 3 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 29250 \, a^{2} e^{\left (7 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 14625 \, a^{2} e^{\left (5 i \, d x - i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 3960 \, a^{2} e^{\left (9 i \, d x + 3 i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 7920 \, a^{2} e^{\left (7 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 3960 \, a^{2} e^{\left (5 i \, d x - i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 3960 \, a^{2} e^{\left (9 i \, d x + 3 i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 7920 \, a^{2} e^{\left (7 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 3960 \, a^{2} e^{\left (5 i \, d x - i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 80 i \, a^{2} e^{\left (18 i \, d x + 12 i \, c\right )} + 880 i \, a^{2} e^{\left (16 i \, d x + 10 i \, c\right )} + 4544 i \, a^{2} e^{\left (14 i \, d x + 8 i \, c\right )} + 15168 i \, a^{2} e^{\left (12 i \, d x + 6 i \, c\right )} + 45024 i \, a^{2} e^{\left (10 i \, d x + 4 i \, c\right )} + 43680 i \, a^{2} e^{\left (8 i \, d x + 2 i \, c\right )} - 18624 i \, a^{2} e^{\left (4 i \, d x - 2 i \, c\right )} - 1968 i \, a^{2} e^{\left (2 i \, d x - 4 i \, c\right )} - 6720 i \, a^{2} e^{\left (6 i \, d x\right )} - 144 i \, a^{2} e^{\left (-6 i \, c\right )}}{92160 \, {\left (d e^{\left (9 i \, d x + 3 i \, c\right )} + 2 \, d e^{\left (7 i \, d x + i \, c\right )} + d e^{\left (5 i \, d x - i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/92160*(18585*a^2*e^(9*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) + 1) + 37170*a^2*e^(7*I*d*x + I*c)*log(I*e^(I*d*
x + I*c) + 1) + 18585*a^2*e^(5*I*d*x - I*c)*log(I*e^(I*d*x + I*c) + 1) + 14625*a^2*e^(9*I*d*x + 3*I*c)*log(I*e
^(I*d*x + I*c) - 1) + 29250*a^2*e^(7*I*d*x + I*c)*log(I*e^(I*d*x + I*c) - 1) + 14625*a^2*e^(5*I*d*x - I*c)*log
(I*e^(I*d*x + I*c) - 1) - 18585*a^2*e^(9*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 37170*a^2*e^(7*I*d*x + I
*c)*log(-I*e^(I*d*x + I*c) + 1) - 18585*a^2*e^(5*I*d*x - I*c)*log(-I*e^(I*d*x + I*c) + 1) - 14625*a^2*e^(9*I*d
*x + 3*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 29250*a^2*e^(7*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) - 1) - 14625*a^2*
e^(5*I*d*x - I*c)*log(-I*e^(I*d*x + I*c) - 1) - 3960*a^2*e^(9*I*d*x + 3*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 792
0*a^2*e^(7*I*d*x + I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 3960*a^2*e^(5*I*d*x - I*c)*log(I*e^(I*d*x) + e^(-I*c)) +
3960*a^2*e^(9*I*d*x + 3*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 7920*a^2*e^(7*I*d*x + I*c)*log(-I*e^(I*d*x) + e^(
-I*c)) + 3960*a^2*e^(5*I*d*x - I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 80*I*a^2*e^(18*I*d*x + 12*I*c) + 880*I*a^2*
e^(16*I*d*x + 10*I*c) + 4544*I*a^2*e^(14*I*d*x + 8*I*c) + 15168*I*a^2*e^(12*I*d*x + 6*I*c) + 45024*I*a^2*e^(10
*I*d*x + 4*I*c) + 43680*I*a^2*e^(8*I*d*x + 2*I*c) - 18624*I*a^2*e^(4*I*d*x - 2*I*c) - 1968*I*a^2*e^(2*I*d*x -
4*I*c) - 6720*I*a^2*e^(6*I*d*x) - 144*I*a^2*e^(-6*I*c))/(d*e^(9*I*d*x + 3*I*c) + 2*d*e^(7*I*d*x + I*c) + d*e^(
5*I*d*x - I*c))

________________________________________________________________________________________

maple [A]  time = 0.52, size = 131, normalized size = 1.25 \[ \frac {-a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{8}\left (d x +c \right )\right )}{9}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{63}\right )-\frac {2 i a^{2} \left (\cos ^{9}\left (d x +c \right )\right )}{9}+\frac {a^{2} \left (\frac {128}{35}+\cos ^{8}\left (d x +c \right )+\frac {8 \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {48 \left (\cos ^{4}\left (d x +c \right )\right )}{35}+\frac {64 \left (\cos ^{2}\left (d x +c \right )\right )}{35}\right ) \sin \left (d x +c \right )}{9}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^2,x)

[Out]

1/d*(-a^2*(-1/9*sin(d*x+c)*cos(d*x+c)^8+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))
-2/9*I*a^2*cos(d*x+c)^9+1/9*a^2*(128/35+cos(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)*s
in(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 119, normalized size = 1.13 \[ -\frac {70 i \, a^{2} \cos \left (d x + c\right )^{9} - {\left (35 \, \sin \left (d x + c\right )^{9} - 135 \, \sin \left (d x + c\right )^{7} + 189 \, \sin \left (d x + c\right )^{5} - 105 \, \sin \left (d x + c\right )^{3}\right )} a^{2} - {\left (35 \, \sin \left (d x + c\right )^{9} - 180 \, \sin \left (d x + c\right )^{7} + 378 \, \sin \left (d x + c\right )^{5} - 420 \, \sin \left (d x + c\right )^{3} + 315 \, \sin \left (d x + c\right )\right )} a^{2}}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/315*(70*I*a^2*cos(d*x + c)^9 - (35*sin(d*x + c)^9 - 135*sin(d*x + c)^7 + 189*sin(d*x + c)^5 - 105*sin(d*x +
c)^3)*a^2 - (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*sin(d*x +
c))*a^2)/d

________________________________________________________________________________________

mupad [B]  time = 5.08, size = 330, normalized size = 3.14 \[ \frac {2\,a^2\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2{}\mathrm {i}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {1024\,a^2\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )}{9\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^9}-\frac {8\,a^2\,\left (5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-12{}\mathrm {i}\right )}{3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2}-\frac {512\,a^2\,\left (8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9{}\mathrm {i}\right )}{9\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^8}+\frac {128\,a^2\,\left (19\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-24{}\mathrm {i}\right )}{3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7}-\frac {64\,a^2\,\left (19\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-35{}\mathrm {i}\right )}{5\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4}+\frac {56\,a^2\,\left (19\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-40{}\mathrm {i}\right )}{15\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3}-\frac {128\,a^2\,\left (59\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-84{}\mathrm {i}\right )}{9\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6}+\frac {32\,a^2\,\left (781\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1260{}\mathrm {i}\right )}{45\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

(2*a^2*(tan(c/2 + (d*x)/2) - 2i))/(d*(tan(c/2 + (d*x)/2)^2 + 1)) + (1024*a^2*(tan(c/2 + (d*x)/2) - 1i))/(9*d*(
tan(c/2 + (d*x)/2)^2 + 1)^9) - (8*a^2*(5*tan(c/2 + (d*x)/2) - 12i))/(3*d*(tan(c/2 + (d*x)/2)^2 + 1)^2) - (512*
a^2*(8*tan(c/2 + (d*x)/2) - 9i))/(9*d*(tan(c/2 + (d*x)/2)^2 + 1)^8) + (128*a^2*(19*tan(c/2 + (d*x)/2) - 24i))/
(3*d*(tan(c/2 + (d*x)/2)^2 + 1)^7) - (64*a^2*(19*tan(c/2 + (d*x)/2) - 35i))/(5*d*(tan(c/2 + (d*x)/2)^2 + 1)^4)
+ (56*a^2*(19*tan(c/2 + (d*x)/2) - 40i))/(15*d*(tan(c/2 + (d*x)/2)^2 + 1)^3) - (128*a^2*(59*tan(c/2 + (d*x)/2
) - 84i))/(9*d*(tan(c/2 + (d*x)/2)^2 + 1)^6) + (32*a^2*(781*tan(c/2 + (d*x)/2) - 1260i))/(45*d*(tan(c/2 + (d*x
)/2)^2 + 1)^5)

________________________________________________________________________________________

sympy [A]  time = 0.78, size = 316, normalized size = 3.01 \[ \begin {cases} \frac {\left (- 126663739519795200 i a^{2} d^{7} e^{18 i c} e^{9 i d x} - 1139973655678156800 i a^{2} d^{7} e^{16 i c} e^{7 i d x} - 4787889353848258560 i a^{2} d^{7} e^{14 i c} e^{5 i d x} - 13299692649578496000 i a^{2} d^{7} e^{12 i c} e^{3 i d x} - 39899077948735488000 i a^{2} d^{7} e^{10 i c} e^{i d x} + 23939446769241292800 i a^{2} d^{7} e^{8 i c} e^{- i d x} + 2659938529915699200 i a^{2} d^{7} e^{6 i c} e^{- 3 i d x} + 227994731135631360 i a^{2} d^{7} e^{4 i c} e^{- 5 i d x}\right ) e^{- 9 i c}}{145916627926804070400 d^{8}} & \text {for}\: 145916627926804070400 d^{8} e^{9 i c} \neq 0 \\\frac {x \left (a^{2} e^{14 i c} + 7 a^{2} e^{12 i c} + 21 a^{2} e^{10 i c} + 35 a^{2} e^{8 i c} + 35 a^{2} e^{6 i c} + 21 a^{2} e^{4 i c} + 7 a^{2} e^{2 i c} + a^{2}\right ) e^{- 5 i c}}{128} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**9*(a+I*a*tan(d*x+c))**2,x)

[Out]

Piecewise(((-126663739519795200*I*a**2*d**7*exp(18*I*c)*exp(9*I*d*x) - 1139973655678156800*I*a**2*d**7*exp(16*
I*c)*exp(7*I*d*x) - 4787889353848258560*I*a**2*d**7*exp(14*I*c)*exp(5*I*d*x) - 13299692649578496000*I*a**2*d**
7*exp(12*I*c)*exp(3*I*d*x) - 39899077948735488000*I*a**2*d**7*exp(10*I*c)*exp(I*d*x) + 23939446769241292800*I*
a**2*d**7*exp(8*I*c)*exp(-I*d*x) + 2659938529915699200*I*a**2*d**7*exp(6*I*c)*exp(-3*I*d*x) + 2279947311356313
60*I*a**2*d**7*exp(4*I*c)*exp(-5*I*d*x))*exp(-9*I*c)/(145916627926804070400*d**8), Ne(145916627926804070400*d*
*8*exp(9*I*c), 0)), (x*(a**2*exp(14*I*c) + 7*a**2*exp(12*I*c) + 21*a**2*exp(10*I*c) + 35*a**2*exp(8*I*c) + 35*
a**2*exp(6*I*c) + 21*a**2*exp(4*I*c) + 7*a**2*exp(2*I*c) + a**2)*exp(-5*I*c)/128, True))

________________________________________________________________________________________