3.100 \(\int \frac{\sin ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=157 \[ \frac{3 \sin ^7(c+d x)}{7 a^3 d}-\frac{7 \sin ^5(c+d x)}{5 a^3 d}+\frac{4 \sin ^3(c+d x)}{3 a^3 d}+\frac{\sin (c+d x) \cos ^7(c+d x)}{8 a^3 d}+\frac{23 \sin (c+d x) \cos ^5(c+d x)}{48 a^3 d}-\frac{29 \sin (c+d x) \cos ^3(c+d x)}{192 a^3 d}-\frac{29 \sin (c+d x) \cos (c+d x)}{128 a^3 d}-\frac{29 x}{128 a^3} \]

[Out]

(-29*x)/(128*a^3) - (29*Cos[c + d*x]*Sin[c + d*x])/(128*a^3*d) - (29*Cos[c + d*x]^3*Sin[c + d*x])/(192*a^3*d)
+ (23*Cos[c + d*x]^5*Sin[c + d*x])/(48*a^3*d) + (Cos[c + d*x]^7*Sin[c + d*x])/(8*a^3*d) + (4*Sin[c + d*x]^3)/(
3*a^3*d) - (7*Sin[c + d*x]^5)/(5*a^3*d) + (3*Sin[c + d*x]^7)/(7*a^3*d)

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Rubi [A]  time = 0.460523, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3872, 2875, 2873, 2564, 14, 2568, 2635, 8, 270} \[ \frac{3 \sin ^7(c+d x)}{7 a^3 d}-\frac{7 \sin ^5(c+d x)}{5 a^3 d}+\frac{4 \sin ^3(c+d x)}{3 a^3 d}+\frac{\sin (c+d x) \cos ^7(c+d x)}{8 a^3 d}+\frac{23 \sin (c+d x) \cos ^5(c+d x)}{48 a^3 d}-\frac{29 \sin (c+d x) \cos ^3(c+d x)}{192 a^3 d}-\frac{29 \sin (c+d x) \cos (c+d x)}{128 a^3 d}-\frac{29 x}{128 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-29*x)/(128*a^3) - (29*Cos[c + d*x]*Sin[c + d*x])/(128*a^3*d) - (29*Cos[c + d*x]^3*Sin[c + d*x])/(192*a^3*d)
+ (23*Cos[c + d*x]^5*Sin[c + d*x])/(48*a^3*d) + (Cos[c + d*x]^7*Sin[c + d*x])/(8*a^3*d) + (4*Sin[c + d*x]^3)/(
3*a^3*d) - (7*Sin[c + d*x]^5)/(5*a^3*d) + (3*Sin[c + d*x]^7)/(7*a^3*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

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 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, 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{\sin ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\int \frac{\cos ^3(c+d x) \sin ^8(c+d x)}{(-a-a \cos (c+d x))^3} \, dx\\ &=-\frac{\int \cos ^3(c+d x) (-a+a \cos (c+d x))^3 \sin ^2(c+d x) \, dx}{a^6}\\ &=-\frac{\int \left (-a^3 \cos ^3(c+d x) \sin ^2(c+d x)+3 a^3 \cos ^4(c+d x) \sin ^2(c+d x)-3 a^3 \cos ^5(c+d x) \sin ^2(c+d x)+a^3 \cos ^6(c+d x) \sin ^2(c+d x)\right ) \, dx}{a^6}\\ &=\frac{\int \cos ^3(c+d x) \sin ^2(c+d x) \, dx}{a^3}-\frac{\int \cos ^6(c+d x) \sin ^2(c+d x) \, dx}{a^3}-\frac{3 \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx}{a^3}+\frac{3 \int \cos ^5(c+d x) \sin ^2(c+d x) \, dx}{a^3}\\ &=\frac{\cos ^5(c+d x) \sin (c+d x)}{2 a^3 d}+\frac{\cos ^7(c+d x) \sin (c+d x)}{8 a^3 d}-\frac{\int \cos ^6(c+d x) \, dx}{8 a^3}-\frac{\int \cos ^4(c+d x) \, dx}{2 a^3}+\frac{\operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\sin (c+d x)\right )}{a^3 d}\\ &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{8 a^3 d}+\frac{23 \cos ^5(c+d x) \sin (c+d x)}{48 a^3 d}+\frac{\cos ^7(c+d x) \sin (c+d x)}{8 a^3 d}-\frac{5 \int \cos ^4(c+d x) \, dx}{48 a^3}-\frac{3 \int \cos ^2(c+d x) \, dx}{8 a^3}+\frac{\operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sin (c+d x)\right )}{a^3 d}\\ &=-\frac{3 \cos (c+d x) \sin (c+d x)}{16 a^3 d}-\frac{29 \cos ^3(c+d x) \sin (c+d x)}{192 a^3 d}+\frac{23 \cos ^5(c+d x) \sin (c+d x)}{48 a^3 d}+\frac{\cos ^7(c+d x) \sin (c+d x)}{8 a^3 d}+\frac{4 \sin ^3(c+d x)}{3 a^3 d}-\frac{7 \sin ^5(c+d x)}{5 a^3 d}+\frac{3 \sin ^7(c+d x)}{7 a^3 d}-\frac{5 \int \cos ^2(c+d x) \, dx}{64 a^3}-\frac{3 \int 1 \, dx}{16 a^3}\\ &=-\frac{3 x}{16 a^3}-\frac{29 \cos (c+d x) \sin (c+d x)}{128 a^3 d}-\frac{29 \cos ^3(c+d x) \sin (c+d x)}{192 a^3 d}+\frac{23 \cos ^5(c+d x) \sin (c+d x)}{48 a^3 d}+\frac{\cos ^7(c+d x) \sin (c+d x)}{8 a^3 d}+\frac{4 \sin ^3(c+d x)}{3 a^3 d}-\frac{7 \sin ^5(c+d x)}{5 a^3 d}+\frac{3 \sin ^7(c+d x)}{7 a^3 d}-\frac{5 \int 1 \, dx}{128 a^3}\\ &=-\frac{29 x}{128 a^3}-\frac{29 \cos (c+d x) \sin (c+d x)}{128 a^3 d}-\frac{29 \cos ^3(c+d x) \sin (c+d x)}{192 a^3 d}+\frac{23 \cos ^5(c+d x) \sin (c+d x)}{48 a^3 d}+\frac{\cos ^7(c+d x) \sin (c+d x)}{8 a^3 d}+\frac{4 \sin ^3(c+d x)}{3 a^3 d}-\frac{7 \sin ^5(c+d x)}{5 a^3 d}+\frac{3 \sin ^7(c+d x)}{7 a^3 d}\\ \end{align*}

Mathematica [A]  time = 4.53154, size = 131, normalized size = 0.83 \[ \frac{\cos ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (38640 \sin (c+d x)-6720 \sin (2 (c+d x))-3920 \sin (3 (c+d x))+5880 \sin (4 (c+d x))-4368 \sin (5 (c+d x))+2240 \sin (6 (c+d x))-720 \sin (7 (c+d x))+105 \sin (8 (c+d x))+294 \tan \left (\frac{c}{2}\right )-24360 d x\right )}{13440 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^6*Sec[c + d*x]^3*(-24360*d*x + 38640*Sin[c + d*x] - 6720*Sin[2*(c + d*x)] - 3920*Sin[3*(c +
d*x)] + 5880*Sin[4*(c + d*x)] - 4368*Sin[5*(c + d*x)] + 2240*Sin[6*(c + d*x)] - 720*Sin[7*(c + d*x)] + 105*Sin
[8*(c + d*x)] + 294*Tan[c/2]))/(13440*a^3*d*(1 + Sec[c + d*x])^3)

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Maple [B]  time = 0.105, size = 290, normalized size = 1.9 \begin{align*}{\frac{29}{64\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}+{\frac{667}{192\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}+{\frac{11107}{960\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}+{\frac{146537}{6720\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}+{\frac{72669}{2240\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{9} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}+{\frac{1759}{320\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{11} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}+{\frac{1143}{64\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{13} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}-{\frac{29}{64\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{15} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}-{\frac{29}{64\,d{a}^{3}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

29/64/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)+667/192/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x
+1/2*c)^3+11107/960/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^5+146537/6720/d/a^3/(1+tan(1/2*d*x+1/2
*c)^2)^8*tan(1/2*d*x+1/2*c)^7+72669/2240/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^9+1759/320/d/a^3/
(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^11+1143/64/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^1
3-29/64/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^15-29/64/d/a^3*arctan(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.53776, size = 510, normalized size = 3.25 \begin{align*} \frac{\frac{\frac{3045 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{23345 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{77749 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{146537 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{218007 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac{36939 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} + \frac{120015 \, \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - \frac{3045 \, \sin \left (d x + c\right )^{15}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{15}}}{a^{3} + \frac{8 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{28 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{56 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{70 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac{56 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac{28 \, a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + \frac{8 \, a^{3} \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}} + \frac{a^{3} \sin \left (d x + c\right )^{16}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{16}}} - \frac{3045 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{6720 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6720*((3045*sin(d*x + c)/(cos(d*x + c) + 1) + 23345*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 77749*sin(d*x + c)
^5/(cos(d*x + c) + 1)^5 + 146537*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 218007*sin(d*x + c)^9/(cos(d*x + c) + 1
)^9 + 36939*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 120015*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 3045*sin(d*
x + c)^15/(cos(d*x + c) + 1)^15)/(a^3 + 8*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*a^3*sin(d*x + c)^4/(cos
(d*x + c) + 1)^4 + 56*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 70*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 5
6*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 28*a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 8*a^3*sin(d*x + c
)^14/(cos(d*x + c) + 1)^14 + a^3*sin(d*x + c)^16/(cos(d*x + c) + 1)^16) - 3045*arctan(sin(d*x + c)/(cos(d*x +
c) + 1))/a^3)/d

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Fricas [A]  time = 1.80264, size = 274, normalized size = 1.75 \begin{align*} -\frac{3045 \, d x -{\left (1680 \, \cos \left (d x + c\right )^{7} - 5760 \, \cos \left (d x + c\right )^{6} + 6440 \, \cos \left (d x + c\right )^{5} - 1536 \, \cos \left (d x + c\right )^{4} - 2030 \, \cos \left (d x + c\right )^{3} + 2432 \, \cos \left (d x + c\right )^{2} - 3045 \, \cos \left (d x + c\right ) + 4864\right )} \sin \left (d x + c\right )}{13440 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/13440*(3045*d*x - (1680*cos(d*x + c)^7 - 5760*cos(d*x + c)^6 + 6440*cos(d*x + c)^5 - 1536*cos(d*x + c)^4 -
2030*cos(d*x + c)^3 + 2432*cos(d*x + c)^2 - 3045*cos(d*x + c) + 4864)*sin(d*x + c))/(a^3*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(sin(d*x+c)**8/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.28195, size = 188, normalized size = 1.2 \begin{align*} -\frac{\frac{3045 \,{\left (d x + c\right )}}{a^{3}} + \frac{2 \,{\left (3045 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{15} - 120015 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} - 36939 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 218007 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 146537 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 77749 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 23345 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3045 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{8} a^{3}}}{13440 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/13440*(3045*(d*x + c)/a^3 + 2*(3045*tan(1/2*d*x + 1/2*c)^15 - 120015*tan(1/2*d*x + 1/2*c)^13 - 36939*tan(1/
2*d*x + 1/2*c)^11 - 218007*tan(1/2*d*x + 1/2*c)^9 - 146537*tan(1/2*d*x + 1/2*c)^7 - 77749*tan(1/2*d*x + 1/2*c)
^5 - 23345*tan(1/2*d*x + 1/2*c)^3 - 3045*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^8*a^3))/d