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

Optimal. Leaf size=133 \[ \frac{\cos ^7(c+d x)}{7 a^3 d}-\frac{\cos ^5(c+d x)}{a^3 d}+\frac{4 \cos ^3(c+d x)}{3 a^3 d}-\frac{\sin ^3(c+d x) \cos ^3(c+d x)}{2 a^3 d}-\frac{5 \sin (c+d x) \cos ^3(c+d x)}{8 a^3 d}+\frac{5 \sin (c+d x) \cos (c+d x)}{16 a^3 d}+\frac{5 x}{16 a^3} \]

[Out]

(5*x)/(16*a^3) + (4*Cos[c + d*x]^3)/(3*a^3*d) - Cos[c + d*x]^5/(a^3*d) + Cos[c + d*x]^7/(7*a^3*d) + (5*Cos[c +
d*x]*Sin[c + d*x])/(16*a^3*d) - (5*Cos[c + d*x]^3*Sin[c + d*x])/(8*a^3*d) - (Cos[c + d*x]^3*Sin[c + d*x]^3)/(
2*a^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.401399, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2875, 2873, 2568, 2635, 8, 2565, 14, 270} \[ \frac{\cos ^7(c+d x)}{7 a^3 d}-\frac{\cos ^5(c+d x)}{a^3 d}+\frac{4 \cos ^3(c+d x)}{3 a^3 d}-\frac{\sin ^3(c+d x) \cos ^3(c+d x)}{2 a^3 d}-\frac{5 \sin (c+d x) \cos ^3(c+d x)}{8 a^3 d}+\frac{5 \sin (c+d x) \cos (c+d x)}{16 a^3 d}+\frac{5 x}{16 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x)/(16*a^3) + (4*Cos[c + d*x]^3)/(3*a^3*d) - Cos[c + d*x]^5/(a^3*d) + Cos[c + d*x]^7/(7*a^3*d) + (5*Cos[c +
d*x]*Sin[c + d*x])/(16*a^3*d) - (5*Cos[c + d*x]^3*Sin[c + d*x])/(8*a^3*d) - (Cos[c + d*x]^3*Sin[c + d*x]^3)/(
2*a^3*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 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 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 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 ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \cos ^2(c+d x) \sin ^2(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac{\int \left (a^3 \cos ^2(c+d x) \sin ^2(c+d x)-3 a^3 \cos ^2(c+d x) \sin ^3(c+d x)+3 a^3 \cos ^2(c+d x) \sin ^4(c+d x)-a^3 \cos ^2(c+d x) \sin ^5(c+d x)\right ) \, dx}{a^6}\\ &=\frac{\int \cos ^2(c+d x) \sin ^2(c+d x) \, dx}{a^3}-\frac{\int \cos ^2(c+d x) \sin ^5(c+d x) \, dx}{a^3}-\frac{3 \int \cos ^2(c+d x) \sin ^3(c+d x) \, dx}{a^3}+\frac{3 \int \cos ^2(c+d x) \sin ^4(c+d x) \, dx}{a^3}\\ &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a^3 d}-\frac{\cos ^3(c+d x) \sin ^3(c+d x)}{2 a^3 d}+\frac{\int \cos ^2(c+d x) \, dx}{4 a^3}+\frac{3 \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx}{2 a^3}+\frac{\operatorname{Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\cos (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^3 d}\\ &=\frac{\cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{5 \cos ^3(c+d x) \sin (c+d x)}{8 a^3 d}-\frac{\cos ^3(c+d x) \sin ^3(c+d x)}{2 a^3 d}+\frac{\int 1 \, dx}{8 a^3}+\frac{3 \int \cos ^2(c+d x) \, dx}{8 a^3}+\frac{\operatorname{Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\cos (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{a^3 d}\\ &=\frac{x}{8 a^3}+\frac{4 \cos ^3(c+d x)}{3 a^3 d}-\frac{\cos ^5(c+d x)}{a^3 d}+\frac{\cos ^7(c+d x)}{7 a^3 d}+\frac{5 \cos (c+d x) \sin (c+d x)}{16 a^3 d}-\frac{5 \cos ^3(c+d x) \sin (c+d x)}{8 a^3 d}-\frac{\cos ^3(c+d x) \sin ^3(c+d x)}{2 a^3 d}+\frac{3 \int 1 \, dx}{16 a^3}\\ &=\frac{5 x}{16 a^3}+\frac{4 \cos ^3(c+d x)}{3 a^3 d}-\frac{\cos ^5(c+d x)}{a^3 d}+\frac{\cos ^7(c+d x)}{7 a^3 d}+\frac{5 \cos (c+d x) \sin (c+d x)}{16 a^3 d}-\frac{5 \cos ^3(c+d x) \sin (c+d x)}{8 a^3 d}-\frac{\cos ^3(c+d x) \sin ^3(c+d x)}{2 a^3 d}\\ \end{align*}

Mathematica [B]  time = 9.81987, size = 429, normalized size = 3.23 \[ \frac{840 d x \sin \left (\frac{c}{2}\right )-609 \sin \left (\frac{c}{2}+d x\right )+609 \sin \left (\frac{3 c}{2}+d x\right )-63 \sin \left (\frac{3 c}{2}+2 d x\right )-63 \sin \left (\frac{5 c}{2}+2 d x\right )-91 \sin \left (\frac{5 c}{2}+3 d x\right )+91 \sin \left (\frac{7 c}{2}+3 d x\right )-105 \sin \left (\frac{7 c}{2}+4 d x\right )-105 \sin \left (\frac{9 c}{2}+4 d x\right )+63 \sin \left (\frac{9 c}{2}+5 d x\right )-63 \sin \left (\frac{11 c}{2}+5 d x\right )+21 \sin \left (\frac{11 c}{2}+6 d x\right )+21 \sin \left (\frac{13 c}{2}+6 d x\right )-3 \sin \left (\frac{13 c}{2}+7 d x\right )+3 \sin \left (\frac{15 c}{2}+7 d x\right )-168 \cos \left (\frac{c}{2}\right ) (99 c-5 d x)+609 \cos \left (\frac{c}{2}+d x\right )+609 \cos \left (\frac{3 c}{2}+d x\right )-63 \cos \left (\frac{3 c}{2}+2 d x\right )+63 \cos \left (\frac{5 c}{2}+2 d x\right )+91 \cos \left (\frac{5 c}{2}+3 d x\right )+91 \cos \left (\frac{7 c}{2}+3 d x\right )-105 \cos \left (\frac{7 c}{2}+4 d x\right )+105 \cos \left (\frac{9 c}{2}+4 d x\right )-63 \cos \left (\frac{9 c}{2}+5 d x\right )-63 \cos \left (\frac{11 c}{2}+5 d x\right )+21 \cos \left (\frac{11 c}{2}+6 d x\right )-21 \cos \left (\frac{13 c}{2}+6 d x\right )+3 \cos \left (\frac{13 c}{2}+7 d x\right )+3 \cos \left (\frac{15 c}{2}+7 d x\right )-16632 c \sin \left (\frac{c}{2}\right )+16996 \sin \left (\frac{c}{2}\right )}{2688 a^3 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-168*(99*c - 5*d*x)*Cos[c/2] + 609*Cos[c/2 + d*x] + 609*Cos[(3*c)/2 + d*x] - 63*Cos[(3*c)/2 + 2*d*x] + 63*Cos
[(5*c)/2 + 2*d*x] + 91*Cos[(5*c)/2 + 3*d*x] + 91*Cos[(7*c)/2 + 3*d*x] - 105*Cos[(7*c)/2 + 4*d*x] + 105*Cos[(9*
c)/2 + 4*d*x] - 63*Cos[(9*c)/2 + 5*d*x] - 63*Cos[(11*c)/2 + 5*d*x] + 21*Cos[(11*c)/2 + 6*d*x] - 21*Cos[(13*c)/
2 + 6*d*x] + 3*Cos[(13*c)/2 + 7*d*x] + 3*Cos[(15*c)/2 + 7*d*x] + 16996*Sin[c/2] - 16632*c*Sin[c/2] + 840*d*x*S
in[c/2] - 609*Sin[c/2 + d*x] + 609*Sin[(3*c)/2 + d*x] - 63*Sin[(3*c)/2 + 2*d*x] - 63*Sin[(5*c)/2 + 2*d*x] - 91
*Sin[(5*c)/2 + 3*d*x] + 91*Sin[(7*c)/2 + 3*d*x] - 105*Sin[(7*c)/2 + 4*d*x] - 105*Sin[(9*c)/2 + 4*d*x] + 63*Sin
[(9*c)/2 + 5*d*x] - 63*Sin[(11*c)/2 + 5*d*x] + 21*Sin[(11*c)/2 + 6*d*x] + 21*Sin[(13*c)/2 + 6*d*x] - 3*Sin[(13
*c)/2 + 7*d*x] + 3*Sin[(15*c)/2 + 7*d*x])/(2688*a^3*d*(Cos[c/2] + Sin[c/2]))

________________________________________________________________________________________

Maple [B]  time = 0.102, size = 415, normalized size = 3.1 \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)^2/(a+a*sin(d*x+c))^3,x)

[Out]

5/8/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+1/2*c)^13+3/2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+1/
2*c)^11+12/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+1/2*c)^10-119/8/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1
/2*d*x+1/2*c)^9+92/3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+1/2*c)^8+8/3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^
7*tan(1/2*d*x+1/2*c)^6+119/8/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+1/2*c)^5+8/d/a^3/(1+tan(1/2*d*x+1/2*
c)^2)^7*tan(1/2*d*x+1/2*c)^4-3/2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+1/2*c)^3+20/3/d/a^3/(1+tan(1/2*d
*x+1/2*c)^2)^7*tan(1/2*d*x+1/2*c)^2-5/8/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+1/2*c)+20/21/d/a^3/(1+tan
(1/2*d*x+1/2*c)^2)^7+5/8/d/a^3*arctan(tan(1/2*d*x+1/2*c))

________________________________________________________________________________________

Maxima [B]  time = 1.56745, size = 562, normalized size = 4.23 \begin{align*} -\frac{\frac{\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{1120 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{252 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{1344 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{2499 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{448 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{5152 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac{2499 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac{2016 \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac{252 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac{105 \, \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - 160}{a^{3} + \frac{7 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{21 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{35 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{35 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac{21 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac{7 \, a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + \frac{a^{3} \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}}} - \frac{105 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{168 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/168*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 1120*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 252*sin(d*x + c)^3/(
cos(d*x + c) + 1)^3 - 1344*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2499*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 44
8*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 5152*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 2499*sin(d*x + c)^9/(cos(d*
x + c) + 1)^9 - 2016*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 252*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 105*s
in(d*x + c)^13/(cos(d*x + c) + 1)^13 - 160)/(a^3 + 7*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 21*a^3*sin(d*x
+ c)^4/(cos(d*x + c) + 1)^4 + 35*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 35*a^3*sin(d*x + c)^8/(cos(d*x + c)
+ 1)^8 + 21*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 7*a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + a^3*sin
(d*x + c)^14/(cos(d*x + c) + 1)^14) - 105*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

________________________________________________________________________________________

Fricas [A]  time = 1.11327, size = 217, normalized size = 1.63 \begin{align*} \frac{48 \, \cos \left (d x + c\right )^{7} - 336 \, \cos \left (d x + c\right )^{5} + 448 \, \cos \left (d x + c\right )^{3} + 105 \, d x + 21 \,{\left (8 \, \cos \left (d x + c\right )^{5} - 18 \, \cos \left (d x + c\right )^{3} + 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{336 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/336*(48*cos(d*x + c)^7 - 336*cos(d*x + c)^5 + 448*cos(d*x + c)^3 + 105*d*x + 21*(8*cos(d*x + c)^5 - 18*cos(d
*x + c)^3 + 5*cos(d*x + c))*sin(d*x + c))/(a^3*d)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.28112, size = 242, normalized size = 1.82 \begin{align*} \frac{\frac{105 \,{\left (d x + c\right )}}{a^{3}} + \frac{2 \,{\left (105 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} + 252 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 2016 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} - 2499 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 5152 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 448 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 2499 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1344 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 252 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1120 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 105 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 160\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{7} a^{3}}}{336 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/336*(105*(d*x + c)/a^3 + 2*(105*tan(1/2*d*x + 1/2*c)^13 + 252*tan(1/2*d*x + 1/2*c)^11 + 2016*tan(1/2*d*x + 1
/2*c)^10 - 2499*tan(1/2*d*x + 1/2*c)^9 + 5152*tan(1/2*d*x + 1/2*c)^8 + 448*tan(1/2*d*x + 1/2*c)^6 + 2499*tan(1
/2*d*x + 1/2*c)^5 + 1344*tan(1/2*d*x + 1/2*c)^4 - 252*tan(1/2*d*x + 1/2*c)^3 + 1120*tan(1/2*d*x + 1/2*c)^2 - 1
05*tan(1/2*d*x + 1/2*c) + 160)/((tan(1/2*d*x + 1/2*c)^2 + 1)^7*a^3))/d