3.5.83 \(\int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n \, dx\) [483]

Optimal. Leaf size=269 \[ \frac {i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac {4 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (8-6 n+n^2\right )}-\frac {12 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (2-n) (4-n) n}+\frac {24 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{3+n}}{a^3 d (4-n) n \left (4-n^2\right )}-\frac {24 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{4+n}}{a^4 d n \left (64-20 n^2+n^4\right )} \]

[Out]

I*(e*sec(d*x+c))^(-4-n)*(a+I*a*tan(d*x+c))^n/d/(4-n)+4*I*(e*sec(d*x+c))^(-4-n)*(a+I*a*tan(d*x+c))^(1+n)/a/d/(n
^2-6*n+8)-12*I*(e*sec(d*x+c))^(-4-n)*(a+I*a*tan(d*x+c))^(2+n)/a^2/d/(2-n)/(4-n)/n+24*I*(e*sec(d*x+c))^(-4-n)*(
a+I*a*tan(d*x+c))^(3+n)/a^3/d/(4-n)/n/(-n^2+4)-24*I*(e*sec(d*x+c))^(-4-n)*(a+I*a*tan(d*x+c))^(4+n)/a^4/d/n/(n^
4-20*n^2+64)

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Rubi [A]
time = 0.29, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3585, 3569} \begin {gather*} -\frac {24 i (a+i a \tan (c+d x))^{n+4} (e \sec (c+d x))^{-n-4}}{a^4 d n \left (n^4-20 n^2+64\right )}+\frac {24 i (a+i a \tan (c+d x))^{n+3} (e \sec (c+d x))^{-n-4}}{a^3 d (4-n) n \left (4-n^2\right )}-\frac {12 i (a+i a \tan (c+d x))^{n+2} (e \sec (c+d x))^{-n-4}}{a^2 d (2-n) (4-n) n}+\frac {4 i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-n-4}}{a d \left (n^2-6 n+8\right )}+\frac {i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-4}}{d (4-n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^n,x]

[Out]

(I*(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^n)/(d*(4 - n)) + ((4*I)*(e*Sec[c + d*x])^(-4 - n)*(a + I*a
*Tan[c + d*x])^(1 + n))/(a*d*(8 - 6*n + n^2)) - ((12*I)*(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^(2 +
n))/(a^2*d*(2 - n)*(4 - n)*n) + ((24*I)*(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^(3 + n))/(a^3*d*(4 -
n)*n*(4 - n^2)) - ((24*I)*(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^(4 + n))/(a^4*d*n*(64 - 20*n^2 + n^
4))

Rule 3569

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

Rule 3585

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, n}, x] && EqQ[a^2 + b^2, 0] && IL
tQ[Simplify[m + n], 0] && NeQ[m + 2*n, 0]

Rubi steps

\begin {align*} \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n \, dx &=\frac {i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac {4 \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n} \, dx}{a (4-n)}\\ &=\frac {i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac {4 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (8-6 n+n^2\right )}+\frac {12 \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{2+n} \, dx}{a^2 (2-n) (4-n)}\\ &=\frac {i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac {4 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (8-6 n+n^2\right )}-\frac {12 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (2-n) (4-n) n}-\frac {24 \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{3+n} \, dx}{a^3 (2-n) (4-n) n}\\ &=\frac {i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac {4 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (8-6 n+n^2\right )}-\frac {12 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (2-n) (4-n) n}+\frac {24 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{3+n}}{a^3 d (2-n) (4-n) n (2+n)}+\frac {24 \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{4+n} \, dx}{a^4 (2-n) (4-n) n (2+n)}\\ &=\frac {i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac {4 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (8-6 n+n^2\right )}-\frac {12 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (2-n) (4-n) n}+\frac {24 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{3+n}}{a^3 d (2-n) (4-n) n (2+n)}-\frac {24 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{4+n}}{a^4 d n \left (64-20 n^2+n^4\right )}\\ \end {align*}

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Mathematica [A]
time = 0.73, size = 165, normalized size = 0.61 \begin {gather*} -\frac {i (e \sec (c+d x))^{-n} \left (192-60 n^2+3 n^4+4 n^2 \left (-16+n^2\right ) \cos (2 (c+d x))+n^2 \left (-4+n^2\right ) \cos (4 (c+d x))+128 i n \sin (2 (c+d x))-8 i n^3 \sin (2 (c+d x))+16 i n \sin (4 (c+d x))-4 i n^3 \sin (4 (c+d x))\right ) (a+i a \tan (c+d x))^n}{8 d e^4 (-4+n) (-2+n) n (2+n) (4+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^n,x]

[Out]

((-1/8*I)*(192 - 60*n^2 + 3*n^4 + 4*n^2*(-16 + n^2)*Cos[2*(c + d*x)] + n^2*(-4 + n^2)*Cos[4*(c + d*x)] + (128*
I)*n*Sin[2*(c + d*x)] - (8*I)*n^3*Sin[2*(c + d*x)] + (16*I)*n*Sin[4*(c + d*x)] - (4*I)*n^3*Sin[4*(c + d*x)])*(
a + I*a*Tan[c + d*x])^n)/(d*e^4*(-4 + n)*(-2 + n)*n*(2 + n)*(4 + n)*(e*Sec[c + d*x])^n)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.43, size = 5823, normalized size = 21.65

method	result	size
risch	\(\text {Expression too large to display}\)	\(5823\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(-4-n)*(a+I*a*tan(d*x+c))^n,x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [A]
time = 0.60, size = 430, normalized size = 1.60 \begin {gather*} \frac {{\left ({\left (-i \, a^{n} n^{4} + 4 i \, a^{n} n^{3} + 4 i \, a^{n} n^{2} - 16 i \, a^{n} n\right )} \cos \left ({\left (d x + c\right )} {\left (n + 4\right )}\right ) - 4 \, {\left (i \, a^{n} n^{4} - 2 i \, a^{n} n^{3} - 16 i \, a^{n} n^{2} + 32 i \, a^{n} n\right )} \cos \left ({\left (d x + c\right )} {\left (n + 2\right )}\right ) - 4 \, {\left (i \, a^{n} n^{4} + 2 i \, a^{n} n^{3} - 16 i \, a^{n} n^{2} - 32 i \, a^{n} n\right )} \cos \left ({\left (d x + c\right )} {\left (n - 2\right )}\right ) + {\left (-i \, a^{n} n^{4} - 4 i \, a^{n} n^{3} + 4 i \, a^{n} n^{2} + 16 i \, a^{n} n\right )} \cos \left ({\left (d x + c\right )} {\left (n - 4\right )}\right ) - 6 \, {\left (i \, a^{n} n^{4} - 20 i \, a^{n} n^{2} + 64 i \, a^{n}\right )} \cos \left ({\left (d x + c\right )} n\right ) + {\left (a^{n} n^{4} - 4 \, a^{n} n^{3} - 4 \, a^{n} n^{2} + 16 \, a^{n} n\right )} \sin \left ({\left (d x + c\right )} {\left (n + 4\right )}\right ) + 4 \, {\left (a^{n} n^{4} - 2 \, a^{n} n^{3} - 16 \, a^{n} n^{2} + 32 \, a^{n} n\right )} \sin \left ({\left (d x + c\right )} {\left (n + 2\right )}\right ) + 4 \, {\left (a^{n} n^{4} + 2 \, a^{n} n^{3} - 16 \, a^{n} n^{2} - 32 \, a^{n} n\right )} \sin \left ({\left (d x + c\right )} {\left (n - 2\right )}\right ) + {\left (a^{n} n^{4} + 4 \, a^{n} n^{3} - 4 \, a^{n} n^{2} - 16 \, a^{n} n\right )} \sin \left ({\left (d x + c\right )} {\left (n - 4\right )}\right ) + 6 \, {\left (a^{n} n^{4} - 20 \, a^{n} n^{2} + 64 \, a^{n}\right )} \sin \left ({\left (d x + c\right )} n\right )\right )} e^{\left (-n\right )}}{16 \, {\left (n^{5} e^{4} - 20 \, n^{3} e^{4} + 64 \, n e^{4}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(-4-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

1/16*((-I*a^n*n^4 + 4*I*a^n*n^3 + 4*I*a^n*n^2 - 16*I*a^n*n)*cos((d*x + c)*(n + 4)) - 4*(I*a^n*n^4 - 2*I*a^n*n^
3 - 16*I*a^n*n^2 + 32*I*a^n*n)*cos((d*x + c)*(n + 2)) - 4*(I*a^n*n^4 + 2*I*a^n*n^3 - 16*I*a^n*n^2 - 32*I*a^n*n
)*cos((d*x + c)*(n - 2)) + (-I*a^n*n^4 - 4*I*a^n*n^3 + 4*I*a^n*n^2 + 16*I*a^n*n)*cos((d*x + c)*(n - 4)) - 6*(I
*a^n*n^4 - 20*I*a^n*n^2 + 64*I*a^n)*cos((d*x + c)*n) + (a^n*n^4 - 4*a^n*n^3 - 4*a^n*n^2 + 16*a^n*n)*sin((d*x +
c)*(n + 4)) + 4*(a^n*n^4 - 2*a^n*n^3 - 16*a^n*n^2 + 32*a^n*n)*sin((d*x + c)*(n + 2)) + 4*(a^n*n^4 + 2*a^n*n^3
- 16*a^n*n^2 - 32*a^n*n)*sin((d*x + c)*(n - 2)) + (a^n*n^4 + 4*a^n*n^3 - 4*a^n*n^2 - 16*a^n*n)*sin((d*x + c)*
(n - 4)) + 6*(a^n*n^4 - 20*a^n*n^2 + 64*a^n)*sin((d*x + c)*n))*e^(-n)/((n^5*e^4 - 20*n^3*e^4 + 64*n*e^4)*d)

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Fricas [A]
time = 0.37, size = 334, normalized size = 1.24 \begin {gather*} \frac {{\left (-i \, n^{4} - 4 i \, n^{3} + 4 i \, n^{2} + {\left (-i \, n^{4} + 4 i \, n^{3} + 4 i \, n^{2} - 16 i \, n\right )} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, {\left (i \, n^{4} - 2 i \, n^{3} - 16 i \, n^{2} + 32 i \, n\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 6 \, {\left (i \, n^{4} - 20 i \, n^{2} + 64 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, {\left (i \, n^{4} + 2 i \, n^{3} - 16 i \, n^{2} - 32 i \, n\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i \, n\right )} \left (\frac {2 \, e^{\left (i \, d x + i \, c + 1\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{-n - 4} e^{\left (i \, d n x + i \, c n + n \log \left (a e^{\left (-1\right )}\right ) + n \log \left (\frac {2 \, e^{\left (i \, d x + i \, c + 1\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )\right )}}{d n^{5} - 20 \, d n^{3} + 64 \, d n + {\left (d n^{5} - 20 \, d n^{3} + 64 \, d n\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, {\left (d n^{5} - 20 \, d n^{3} + 64 \, d n\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (d n^{5} - 20 \, d n^{3} + 64 \, d n\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, {\left (d n^{5} - 20 \, d n^{3} + 64 \, d n\right )} e^{\left (2 i \, d x + 2 i \, c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(-4-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

(-I*n^4 - 4*I*n^3 + 4*I*n^2 + (-I*n^4 + 4*I*n^3 + 4*I*n^2 - 16*I*n)*e^(8*I*d*x + 8*I*c) - 4*(I*n^4 - 2*I*n^3 -
16*I*n^2 + 32*I*n)*e^(6*I*d*x + 6*I*c) - 6*(I*n^4 - 20*I*n^2 + 64*I)*e^(4*I*d*x + 4*I*c) - 4*(I*n^4 + 2*I*n^3
- 16*I*n^2 - 32*I*n)*e^(2*I*d*x + 2*I*c) + 16*I*n)*(2*e^(I*d*x + I*c + 1)/(e^(2*I*d*x + 2*I*c) + 1))^(-n - 4)
*e^(I*d*n*x + I*c*n + n*log(a*e^(-1)) + n*log(2*e^(I*d*x + I*c + 1)/(e^(2*I*d*x + 2*I*c) + 1)))/(d*n^5 - 20*d*
n^3 + 64*d*n + (d*n^5 - 20*d*n^3 + 64*d*n)*e^(8*I*d*x + 8*I*c) + 4*(d*n^5 - 20*d*n^3 + 64*d*n)*e^(6*I*d*x + 6*
I*c) + 6*(d*n^5 - 20*d*n^3 + 64*d*n)*e^(4*I*d*x + 4*I*c) + 4*(d*n^5 - 20*d*n^3 + 64*d*n)*e^(2*I*d*x + 2*I*c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e \sec {\left (c + d x \right )}\right )^{- n - 4} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(-4-n)*(a+I*a*tan(d*x+c))**n,x)

[Out]

Integral((e*sec(c + d*x))**(-n - 4)*(I*a*(tan(c + d*x) - I))**n, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(-4-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(-n - 4)*(I*a*tan(d*x + c) + a)^n, x)

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Mupad [B]
time = 9.90, size = 511, normalized size = 1.90 \begin {gather*} \frac {\left (2\,{\sin \left (2\,c+2\,d\,x\right )}^2+\sin \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {{\left (a-\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^n\,\left (-n^3-4\,n^2+4\,n+16\right )}{d\,\left (n^4\,1{}\mathrm {i}-n^2\,20{}\mathrm {i}+64{}\mathrm {i}\right )}+\frac {4\,{\left (a-\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^n\,\left (-2\,{\sin \left (c+d\,x\right )}^2+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+1\right )\,\left (-n^3-2\,n^2+16\,n+32\right )}{d\,\left (n^4\,1{}\mathrm {i}-n^2\,20{}\mathrm {i}+64{}\mathrm {i}\right )}+\frac {{\left (a-\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^n\,\left (-2\,{\sin \left (4\,c+4\,d\,x\right )}^2+\sin \left (8\,c+8\,d\,x\right )\,1{}\mathrm {i}+1\right )\,\left (-n^3+4\,n^2+4\,n-16\right )}{d\,\left (n^4\,1{}\mathrm {i}-n^2\,20{}\mathrm {i}+64{}\mathrm {i}\right )}+\frac {4\,{\left (a-\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^n\,\left (-2\,{\sin \left (3\,c+3\,d\,x\right )}^2+\sin \left (6\,c+6\,d\,x\right )\,1{}\mathrm {i}+1\right )\,\left (-n^3+2\,n^2+16\,n-32\right )}{d\,\left (n^4\,1{}\mathrm {i}-n^2\,20{}\mathrm {i}+64{}\mathrm {i}\right )}-\frac {{\left (a-\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^n\,\left (-2\,{\sin \left (2\,c+2\,d\,x\right )}^2+\sin \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}+1\right )\,\left (6\,n^4-120\,n^2+384\right )}{d\,n\,\left (n^4\,1{}\mathrm {i}-n^2\,20{}\mathrm {i}+64{}\mathrm {i}\right )}\right )}{16\,{\left (-\frac {e}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^{n+4}\,{\left ({\sin \left (c+d\,x\right )}^2-1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((sin(4*c + 4*d*x)*1i + 2*sin(2*c + 2*d*x)^2 - 1)*(((a - (a*sin(c + d*x)*1i)/(2*sin(c/2 + (d*x)/2)^2 - 1))^n*(
4*n - 4*n^2 - n^3 + 16))/(d*(n^4*1i - n^2*20i + 64i)) + (4*(a - (a*sin(c + d*x)*1i)/(2*sin(c/2 + (d*x)/2)^2 -
1))^n*(sin(2*c + 2*d*x)*1i - 2*sin(c + d*x)^2 + 1)*(16*n - 2*n^2 - n^3 + 32))/(d*(n^4*1i - n^2*20i + 64i)) + (
(a - (a*sin(c + d*x)*1i)/(2*sin(c/2 + (d*x)/2)^2 - 1))^n*(sin(8*c + 8*d*x)*1i - 2*sin(4*c + 4*d*x)^2 + 1)*(4*n
+ 4*n^2 - n^3 - 16))/(d*(n^4*1i - n^2*20i + 64i)) + (4*(a - (a*sin(c + d*x)*1i)/(2*sin(c/2 + (d*x)/2)^2 - 1))
^n*(sin(6*c + 6*d*x)*1i - 2*sin(3*c + 3*d*x)^2 + 1)*(16*n + 2*n^2 - n^3 - 32))/(d*(n^4*1i - n^2*20i + 64i)) -
((a - (a*sin(c + d*x)*1i)/(2*sin(c/2 + (d*x)/2)^2 - 1))^n*(sin(4*c + 4*d*x)*1i - 2*sin(2*c + 2*d*x)^2 + 1)*(6*
n^4 - 120*n^2 + 384))/(d*n*(n^4*1i - n^2*20i + 64i))))/(16*(-e/(2*sin(c/2 + (d*x)/2)^2 - 1))^(n + 4)*(sin(c +
d*x)^2 - 1)^2)

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