3.724 \(\int \frac{\cos ^8(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=159 \[ -\frac{\cos ^9(c+d x)}{9 a^2 d}+\frac{3 \cos ^7(c+d x)}{7 a^2 d}-\frac{2 \cos ^5(c+d x)}{5 a^2 d}+\frac{\sin ^3(c+d x) \cos ^5(c+d x)}{4 a^2 d}+\frac{\sin (c+d x) \cos ^5(c+d x)}{8 a^2 d}-\frac{\sin (c+d x) \cos ^3(c+d x)}{32 a^2 d}-\frac{3 \sin (c+d x) \cos (c+d x)}{64 a^2 d}-\frac{3 x}{64 a^2} \]

[Out]

(-3*x)/(64*a^2) - (2*Cos[c + d*x]^5)/(5*a^2*d) + (3*Cos[c + d*x]^7)/(7*a^2*d) - Cos[c + d*x]^9/(9*a^2*d) - (3*
Cos[c + d*x]*Sin[c + d*x])/(64*a^2*d) - (Cos[c + d*x]^3*Sin[c + d*x])/(32*a^2*d) + (Cos[c + d*x]^5*Sin[c + d*x
])/(8*a^2*d) + (Cos[c + d*x]^5*Sin[c + d*x]^3)/(4*a^2*d)

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Rubi [A]  time = 0.363248, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2875, 2873, 2565, 14, 2568, 2635, 8, 270} \[ -\frac{\cos ^9(c+d x)}{9 a^2 d}+\frac{3 \cos ^7(c+d x)}{7 a^2 d}-\frac{2 \cos ^5(c+d x)}{5 a^2 d}+\frac{\sin ^3(c+d x) \cos ^5(c+d x)}{4 a^2 d}+\frac{\sin (c+d x) \cos ^5(c+d x)}{8 a^2 d}-\frac{\sin (c+d x) \cos ^3(c+d x)}{32 a^2 d}-\frac{3 \sin (c+d x) \cos (c+d x)}{64 a^2 d}-\frac{3 x}{64 a^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^8*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-3*x)/(64*a^2) - (2*Cos[c + d*x]^5)/(5*a^2*d) + (3*Cos[c + d*x]^7)/(7*a^2*d) - Cos[c + d*x]^9/(9*a^2*d) - (3*
Cos[c + d*x]*Sin[c + d*x])/(64*a^2*d) - (Cos[c + d*x]^3*Sin[c + d*x])/(32*a^2*d) + (Cos[c + d*x]^5*Sin[c + d*x
])/(8*a^2*d) + (Cos[c + d*x]^5*Sin[c + d*x]^3)/(4*a^2*d)

Rule 2875

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[((g*Cos[e + f*x])^(2*m + p)*(d*Sin[e + f*x])^n)/(a - b*Sin[e +
f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^8(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\int \cos ^4(c+d x) \sin ^3(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac{\int \left (a^2 \cos ^4(c+d x) \sin ^3(c+d x)-2 a^2 \cos ^4(c+d x) \sin ^4(c+d x)+a^2 \cos ^4(c+d x) \sin ^5(c+d x)\right ) \, dx}{a^4}\\ &=\frac{\int \cos ^4(c+d x) \sin ^3(c+d x) \, dx}{a^2}+\frac{\int \cos ^4(c+d x) \sin ^5(c+d x) \, dx}{a^2}-\frac{2 \int \cos ^4(c+d x) \sin ^4(c+d x) \, dx}{a^2}\\ &=\frac{\cos ^5(c+d x) \sin ^3(c+d x)}{4 a^2 d}-\frac{3 \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx}{4 a^2}-\frac{\operatorname{Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}-\frac{\operatorname{Subst}\left (\int x^4 \left (1-x^2\right )^2 \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=\frac{\cos ^5(c+d x) \sin (c+d x)}{8 a^2 d}+\frac{\cos ^5(c+d x) \sin ^3(c+d x)}{4 a^2 d}-\frac{\int \cos ^4(c+d x) \, dx}{8 a^2}-\frac{\operatorname{Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}-\frac{\operatorname{Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac{2 \cos ^5(c+d x)}{5 a^2 d}+\frac{3 \cos ^7(c+d x)}{7 a^2 d}-\frac{\cos ^9(c+d x)}{9 a^2 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{32 a^2 d}+\frac{\cos ^5(c+d x) \sin (c+d x)}{8 a^2 d}+\frac{\cos ^5(c+d x) \sin ^3(c+d x)}{4 a^2 d}-\frac{3 \int \cos ^2(c+d x) \, dx}{32 a^2}\\ &=-\frac{2 \cos ^5(c+d x)}{5 a^2 d}+\frac{3 \cos ^7(c+d x)}{7 a^2 d}-\frac{\cos ^9(c+d x)}{9 a^2 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{64 a^2 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{32 a^2 d}+\frac{\cos ^5(c+d x) \sin (c+d x)}{8 a^2 d}+\frac{\cos ^5(c+d x) \sin ^3(c+d x)}{4 a^2 d}-\frac{3 \int 1 \, dx}{64 a^2}\\ &=-\frac{3 x}{64 a^2}-\frac{2 \cos ^5(c+d x)}{5 a^2 d}+\frac{3 \cos ^7(c+d x)}{7 a^2 d}-\frac{\cos ^9(c+d x)}{9 a^2 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{64 a^2 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{32 a^2 d}+\frac{\cos ^5(c+d x) \sin (c+d x)}{8 a^2 d}+\frac{\cos ^5(c+d x) \sin ^3(c+d x)}{4 a^2 d}\\ \end{align*}

Mathematica [B]  time = 6.46633, size = 430, normalized size = 2.7 \[ -\frac{15120 d x \sin \left (\frac{c}{2}\right )-11340 \sin \left (\frac{c}{2}+d x\right )+11340 \sin \left (\frac{3 c}{2}+d x\right )-3360 \sin \left (\frac{5 c}{2}+3 d x\right )+3360 \sin \left (\frac{7 c}{2}+3 d x\right )-2520 \sin \left (\frac{7 c}{2}+4 d x\right )-2520 \sin \left (\frac{9 c}{2}+4 d x\right )+1008 \sin \left (\frac{9 c}{2}+5 d x\right )-1008 \sin \left (\frac{11 c}{2}+5 d x\right )+450 \sin \left (\frac{13 c}{2}+7 d x\right )-450 \sin \left (\frac{15 c}{2}+7 d x\right )+315 \sin \left (\frac{15 c}{2}+8 d x\right )+315 \sin \left (\frac{17 c}{2}+8 d x\right )-70 \sin \left (\frac{17 c}{2}+9 d x\right )+70 \sin \left (\frac{19 c}{2}+9 d x\right )+420 \cos \left (\frac{c}{2}\right ) (330 c+36 d x+7)+11340 \cos \left (\frac{c}{2}+d x\right )+11340 \cos \left (\frac{3 c}{2}+d x\right )+3360 \cos \left (\frac{5 c}{2}+3 d x\right )+3360 \cos \left (\frac{7 c}{2}+3 d x\right )-2520 \cos \left (\frac{7 c}{2}+4 d x\right )+2520 \cos \left (\frac{9 c}{2}+4 d x\right )-1008 \cos \left (\frac{9 c}{2}+5 d x\right )-1008 \cos \left (\frac{11 c}{2}+5 d x\right )-450 \cos \left (\frac{13 c}{2}+7 d x\right )-450 \cos \left (\frac{15 c}{2}+7 d x\right )+315 \cos \left (\frac{15 c}{2}+8 d x\right )-315 \cos \left (\frac{17 c}{2}+8 d x\right )+70 \cos \left (\frac{17 c}{2}+9 d x\right )+70 \cos \left (\frac{19 c}{2}+9 d x\right )+138600 c \sin \left (\frac{c}{2}\right )-78960 \sin \left (\frac{c}{2}\right )}{322560 a^2 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^8*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

-(420*(7 + 330*c + 36*d*x)*Cos[c/2] + 11340*Cos[c/2 + d*x] + 11340*Cos[(3*c)/2 + d*x] + 3360*Cos[(5*c)/2 + 3*d
*x] + 3360*Cos[(7*c)/2 + 3*d*x] - 2520*Cos[(7*c)/2 + 4*d*x] + 2520*Cos[(9*c)/2 + 4*d*x] - 1008*Cos[(9*c)/2 + 5
*d*x] - 1008*Cos[(11*c)/2 + 5*d*x] - 450*Cos[(13*c)/2 + 7*d*x] - 450*Cos[(15*c)/2 + 7*d*x] + 315*Cos[(15*c)/2
+ 8*d*x] - 315*Cos[(17*c)/2 + 8*d*x] + 70*Cos[(17*c)/2 + 9*d*x] + 70*Cos[(19*c)/2 + 9*d*x] - 78960*Sin[c/2] +
138600*c*Sin[c/2] + 15120*d*x*Sin[c/2] - 11340*Sin[c/2 + d*x] + 11340*Sin[(3*c)/2 + d*x] - 3360*Sin[(5*c)/2 +
3*d*x] + 3360*Sin[(7*c)/2 + 3*d*x] - 2520*Sin[(7*c)/2 + 4*d*x] - 2520*Sin[(9*c)/2 + 4*d*x] + 1008*Sin[(9*c)/2
+ 5*d*x] - 1008*Sin[(11*c)/2 + 5*d*x] + 450*Sin[(13*c)/2 + 7*d*x] - 450*Sin[(15*c)/2 + 7*d*x] + 315*Sin[(15*c)
/2 + 8*d*x] + 315*Sin[(17*c)/2 + 8*d*x] - 70*Sin[(17*c)/2 + 9*d*x] + 70*Sin[(19*c)/2 + 9*d*x])/(322560*a^2*d*(
Cos[c/2] + Sin[c/2]))

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Maple [B]  time = 0.121, size = 551, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x)

[Out]

-52/315/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9+3/32/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)-52/35/d/a^2/
(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^2+13/16/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^3-68
/35/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^4-155/16/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+
1/2*c)^5+4/5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^6+169/16/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan
(1/2*d*x+1/2*c)^7-164/5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^8+12/d/a^2/(1+tan(1/2*d*x+1/2*c)^2
)^9*tan(1/2*d*x+1/2*c)^10-169/16/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^11-44/3/d/a^2/(1+tan(1/2*
d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^12+155/16/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^13-4/d/a^2/(1
+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^14-13/16/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^15-3/
32/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^17-3/32/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.61691, size = 732, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/10080*((945*sin(d*x + c)/(cos(d*x + c) + 1) - 14976*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8190*sin(d*x + c)^
3/(cos(d*x + c) + 1)^3 - 19584*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 97650*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
+ 8064*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 106470*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 330624*sin(d*x + c)
^8/(cos(d*x + c) + 1)^8 + 120960*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 106470*sin(d*x + c)^11/(cos(d*x + c)
+ 1)^11 - 147840*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 97650*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 40320*s
in(d*x + c)^14/(cos(d*x + c) + 1)^14 - 8190*sin(d*x + c)^15/(cos(d*x + c) + 1)^15 - 945*sin(d*x + c)^17/(cos(d
*x + c) + 1)^17 - 1664)/(a^2 + 9*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 36*a^2*sin(d*x + c)^4/(cos(d*x + c)
+ 1)^4 + 84*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 126*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 126*a^2*s
in(d*x + c)^10/(cos(d*x + c) + 1)^10 + 84*a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 36*a^2*sin(d*x + c)^14/(
cos(d*x + c) + 1)^14 + 9*a^2*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 + a^2*sin(d*x + c)^18/(cos(d*x + c) + 1)^18
) - 945*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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Fricas [A]  time = 1.21003, size = 255, normalized size = 1.6 \begin{align*} -\frac{2240 \, \cos \left (d x + c\right )^{9} - 8640 \, \cos \left (d x + c\right )^{7} + 8064 \, \cos \left (d x + c\right )^{5} + 945 \, d x + 315 \,{\left (16 \, \cos \left (d x + c\right )^{7} - 24 \, \cos \left (d x + c\right )^{5} + 2 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{20160 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/20160*(2240*cos(d*x + c)^9 - 8640*cos(d*x + c)^7 + 8064*cos(d*x + c)^5 + 945*d*x + 315*(16*cos(d*x + c)^7 -
24*cos(d*x + c)^5 + 2*cos(d*x + c)^3 + 3*cos(d*x + c))*sin(d*x + c))/(a^2*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**8*sin(d*x+c)**3/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.2719, size = 312, normalized size = 1.96 \begin{align*} -\frac{\frac{945 \,{\left (d x + c\right )}}{a^{2}} + \frac{2 \,{\left (945 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{17} + 8190 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{15} + 40320 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{14} - 97650 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} + 147840 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{12} + 106470 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 120960 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} + 330624 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 106470 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 8064 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 97650 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 19584 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 8190 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 14976 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 945 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1664\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{9} a^{2}}}{20160 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/20160*(945*(d*x + c)/a^2 + 2*(945*tan(1/2*d*x + 1/2*c)^17 + 8190*tan(1/2*d*x + 1/2*c)^15 + 40320*tan(1/2*d*
x + 1/2*c)^14 - 97650*tan(1/2*d*x + 1/2*c)^13 + 147840*tan(1/2*d*x + 1/2*c)^12 + 106470*tan(1/2*d*x + 1/2*c)^1
1 - 120960*tan(1/2*d*x + 1/2*c)^10 + 330624*tan(1/2*d*x + 1/2*c)^8 - 106470*tan(1/2*d*x + 1/2*c)^7 - 8064*tan(
1/2*d*x + 1/2*c)^6 + 97650*tan(1/2*d*x + 1/2*c)^5 + 19584*tan(1/2*d*x + 1/2*c)^4 - 8190*tan(1/2*d*x + 1/2*c)^3
+ 14976*tan(1/2*d*x + 1/2*c)^2 - 945*tan(1/2*d*x + 1/2*c) + 1664)/((tan(1/2*d*x + 1/2*c)^2 + 1)^9*a^2))/d