3.839 \(\int \frac{\sin ^3(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=178 \[ -\frac{4 \tan ^9(c+d x)}{9 a^3 d}+\frac{\tan ^7(c+d x)}{7 a^3 d}-\frac{\tan ^5(c+d x)}{5 a^3 d}+\frac{\tan ^3(c+d x)}{3 a^3 d}-\frac{\tan (c+d x)}{a^3 d}+\frac{4 \sec ^9(c+d x)}{9 a^3 d}-\frac{15 \sec ^7(c+d x)}{7 a^3 d}+\frac{21 \sec ^5(c+d x)}{5 a^3 d}-\frac{13 \sec ^3(c+d x)}{3 a^3 d}+\frac{3 \sec (c+d x)}{a^3 d}+\frac{x}{a^3} \]

[Out]

x/a^3 + (3*Sec[c + d*x])/(a^3*d) - (13*Sec[c + d*x]^3)/(3*a^3*d) + (21*Sec[c + d*x]^5)/(5*a^3*d) - (15*Sec[c +
d*x]^7)/(7*a^3*d) + (4*Sec[c + d*x]^9)/(9*a^3*d) - Tan[c + d*x]/(a^3*d) + Tan[c + d*x]^3/(3*a^3*d) - Tan[c +
d*x]^5/(5*a^3*d) + Tan[c + d*x]^7/(7*a^3*d) - (4*Tan[c + d*x]^9)/(9*a^3*d)

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Rubi [A]  time = 0.367597, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {2875, 2873, 2606, 270, 2607, 30, 194, 3473, 8} \[ -\frac{4 \tan ^9(c+d x)}{9 a^3 d}+\frac{\tan ^7(c+d x)}{7 a^3 d}-\frac{\tan ^5(c+d x)}{5 a^3 d}+\frac{\tan ^3(c+d x)}{3 a^3 d}-\frac{\tan (c+d x)}{a^3 d}+\frac{4 \sec ^9(c+d x)}{9 a^3 d}-\frac{15 \sec ^7(c+d x)}{7 a^3 d}+\frac{21 \sec ^5(c+d x)}{5 a^3 d}-\frac{13 \sec ^3(c+d x)}{3 a^3 d}+\frac{3 \sec (c+d x)}{a^3 d}+\frac{x}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^3*Tan[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]

[Out]

x/a^3 + (3*Sec[c + d*x])/(a^3*d) - (13*Sec[c + d*x]^3)/(3*a^3*d) + (21*Sec[c + d*x]^5)/(5*a^3*d) - (15*Sec[c +
d*x]^7)/(7*a^3*d) + (4*Sec[c + d*x]^9)/(9*a^3*d) - Tan[c + d*x]/(a^3*d) + Tan[c + d*x]^3/(3*a^3*d) - Tan[c +
d*x]^5/(5*a^3*d) + Tan[c + d*x]^7/(7*a^3*d) - (4*Tan[c + d*x]^9)/(9*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 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

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]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 194

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

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{\sin ^3(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \sec ^3(c+d x) (a-a \sin (c+d x))^3 \tan ^7(c+d x) \, dx}{a^6}\\ &=\frac{\int \left (a^3 \sec ^3(c+d x) \tan ^7(c+d x)-3 a^3 \sec ^2(c+d x) \tan ^8(c+d x)+3 a^3 \sec (c+d x) \tan ^9(c+d x)-a^3 \tan ^{10}(c+d x)\right ) \, dx}{a^6}\\ &=\frac{\int \sec ^3(c+d x) \tan ^7(c+d x) \, dx}{a^3}-\frac{\int \tan ^{10}(c+d x) \, dx}{a^3}-\frac{3 \int \sec ^2(c+d x) \tan ^8(c+d x) \, dx}{a^3}+\frac{3 \int \sec (c+d x) \tan ^9(c+d x) \, dx}{a^3}\\ &=-\frac{\tan ^9(c+d x)}{9 a^3 d}+\frac{\int \tan ^8(c+d x) \, dx}{a^3}+\frac{\operatorname{Subst}\left (\int x^2 \left (-1+x^2\right )^3 \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int x^8 \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int \left (-1+x^2\right )^4 \, dx,x,\sec (c+d x)\right )}{a^3 d}\\ &=\frac{\tan ^7(c+d x)}{7 a^3 d}-\frac{4 \tan ^9(c+d x)}{9 a^3 d}-\frac{\int \tan ^6(c+d x) \, dx}{a^3}+\frac{\operatorname{Subst}\left (\int \left (-x^2+3 x^4-3 x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int \left (1-4 x^2+6 x^4-4 x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}\\ &=\frac{3 \sec (c+d x)}{a^3 d}-\frac{13 \sec ^3(c+d x)}{3 a^3 d}+\frac{21 \sec ^5(c+d x)}{5 a^3 d}-\frac{15 \sec ^7(c+d x)}{7 a^3 d}+\frac{4 \sec ^9(c+d x)}{9 a^3 d}-\frac{\tan ^5(c+d x)}{5 a^3 d}+\frac{\tan ^7(c+d x)}{7 a^3 d}-\frac{4 \tan ^9(c+d x)}{9 a^3 d}+\frac{\int \tan ^4(c+d x) \, dx}{a^3}\\ &=\frac{3 \sec (c+d x)}{a^3 d}-\frac{13 \sec ^3(c+d x)}{3 a^3 d}+\frac{21 \sec ^5(c+d x)}{5 a^3 d}-\frac{15 \sec ^7(c+d x)}{7 a^3 d}+\frac{4 \sec ^9(c+d x)}{9 a^3 d}+\frac{\tan ^3(c+d x)}{3 a^3 d}-\frac{\tan ^5(c+d x)}{5 a^3 d}+\frac{\tan ^7(c+d x)}{7 a^3 d}-\frac{4 \tan ^9(c+d x)}{9 a^3 d}-\frac{\int \tan ^2(c+d x) \, dx}{a^3}\\ &=\frac{3 \sec (c+d x)}{a^3 d}-\frac{13 \sec ^3(c+d x)}{3 a^3 d}+\frac{21 \sec ^5(c+d x)}{5 a^3 d}-\frac{15 \sec ^7(c+d x)}{7 a^3 d}+\frac{4 \sec ^9(c+d x)}{9 a^3 d}-\frac{\tan (c+d x)}{a^3 d}+\frac{\tan ^3(c+d x)}{3 a^3 d}-\frac{\tan ^5(c+d x)}{5 a^3 d}+\frac{\tan ^7(c+d x)}{7 a^3 d}-\frac{4 \tan ^9(c+d x)}{9 a^3 d}+\frac{\int 1 \, dx}{a^3}\\ &=\frac{x}{a^3}+\frac{3 \sec (c+d x)}{a^3 d}-\frac{13 \sec ^3(c+d x)}{3 a^3 d}+\frac{21 \sec ^5(c+d x)}{5 a^3 d}-\frac{15 \sec ^7(c+d x)}{7 a^3 d}+\frac{4 \sec ^9(c+d x)}{9 a^3 d}-\frac{\tan (c+d x)}{a^3 d}+\frac{\tan ^3(c+d x)}{3 a^3 d}-\frac{\tan ^5(c+d x)}{5 a^3 d}+\frac{\tan ^7(c+d x)}{7 a^3 d}-\frac{4 \tan ^9(c+d x)}{9 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.51058, size = 273, normalized size = 1.53 \[ \frac{93312 \sin (c+d x)+272160 (c+d x) \sin (2 (c+d x))-506277 \sin (2 (c+d x))+125248 \sin (3 (c+d x))+120960 (c+d x) \sin (4 (c+d x))-225012 \sin (4 (c+d x))+67776 \sin (5 (c+d x))-10080 (c+d x) \sin (6 (c+d x))+18751 \sin (6 (c+d x))+362880 (c+d x) \cos (c+d x)-675036 \cos (c+d x)+173952 \cos (2 (c+d x))+20160 (c+d x) \cos (3 (c+d x))-37502 \cos (3 (c+d x))+54912 \cos (4 (c+d x))-60480 (c+d x) \cos (5 (c+d x))+112506 \cos (5 (c+d x))-21376 \cos (6 (c+d x))+169344}{322560 d (a \sin (c+d x)+a)^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^3*Tan[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]

[Out]

(169344 - 675036*Cos[c + d*x] + 362880*(c + d*x)*Cos[c + d*x] + 173952*Cos[2*(c + d*x)] - 37502*Cos[3*(c + d*x
)] + 20160*(c + d*x)*Cos[3*(c + d*x)] + 54912*Cos[4*(c + d*x)] + 112506*Cos[5*(c + d*x)] - 60480*(c + d*x)*Cos
[5*(c + d*x)] - 21376*Cos[6*(c + d*x)] + 93312*Sin[c + d*x] - 506277*Sin[2*(c + d*x)] + 272160*(c + d*x)*Sin[2
*(c + d*x)] + 125248*Sin[3*(c + d*x)] - 225012*Sin[4*(c + d*x)] + 120960*(c + d*x)*Sin[4*(c + d*x)] + 67776*Si
n[5*(c + d*x)] + 18751*Sin[6*(c + d*x)] - 10080*(c + d*x)*Sin[6*(c + d*x)])/(322560*d*(Cos[(c + d*x)/2] - Sin[
(c + d*x)/2])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(a + a*Sin[c + d*x])^3)

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Maple [A]  time = 0.16, size = 272, normalized size = 1.5 \begin{align*} -{\frac{1}{24\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{16\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{7}{32\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}+{\frac{8}{9\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-9}}-4\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{8}}}+{\frac{40}{7\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-7}}-{\frac{4}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-6}}-{\frac{21}{10\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}-{\frac{3}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-4}}+{\frac{3}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{13}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{57}{32\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*sin(d*x+c)^7/(a+a*sin(d*x+c))^3,x)

[Out]

-1/24/d/a^3/(tan(1/2*d*x+1/2*c)-1)^3-1/16/d/a^3/(tan(1/2*d*x+1/2*c)-1)^2+7/32/d/a^3/(tan(1/2*d*x+1/2*c)-1)+2/d
/a^3*arctan(tan(1/2*d*x+1/2*c))+8/9/d/a^3/(tan(1/2*d*x+1/2*c)+1)^9-4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^8+40/7/d/a^3
/(tan(1/2*d*x+1/2*c)+1)^7-4/3/d/a^3/(tan(1/2*d*x+1/2*c)+1)^6-21/10/d/a^3/(tan(1/2*d*x+1/2*c)+1)^5-3/4/d/a^3/(t
an(1/2*d*x+1/2*c)+1)^4+3/4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^3+13/8/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2+57/32/d/a^3/(tan
(1/2*d*x+1/2*c)+1)

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Maxima [B]  time = 1.7886, size = 657, normalized size = 3.69 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^7/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

2/315*((1893*sin(d*x + c)/(cos(d*x + c) + 1) + 2526*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2939*sin(d*x + c)^3/
(cos(d*x + c) + 1)^3 - 9936*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 3546*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 1
1172*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 9702*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 3675*sin(d*x + c)^9/(cos
(d*x + c) + 1)^9 - 1890*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 315*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 36
8)/(a^3 + 6*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 12*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2*a^3*sin(d*x +
c)^3/(cos(d*x + c) + 1)^3 - 27*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 36*a^3*sin(d*x + c)^5/(cos(d*x + c)
+ 1)^5 + 36*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 27*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 2*a^3*sin(d
*x + c)^9/(cos(d*x + c) + 1)^9 - 12*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 6*a^3*sin(d*x + c)^11/(cos(d*x
+ c) + 1)^11 - a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) + 315*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/
d

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Fricas [A]  time = 1.8831, size = 478, normalized size = 2.69 \begin{align*} \frac{945 \, d x \cos \left (d x + c\right )^{5} + 668 \, \cos \left (d x + c\right )^{6} - 1260 \, d x \cos \left (d x + c\right )^{3} - 1431 \, \cos \left (d x + c\right )^{4} + 465 \, \cos \left (d x + c\right )^{2} +{\left (315 \, d x \cos \left (d x + c\right )^{5} - 1260 \, d x \cos \left (d x + c\right )^{3} - 1059 \, \cos \left (d x + c\right )^{4} + 305 \, \cos \left (d x + c\right )^{2} - 35\right )} \sin \left (d x + c\right ) - 70}{315 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3} +{\left (a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^7/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/315*(945*d*x*cos(d*x + c)^5 + 668*cos(d*x + c)^6 - 1260*d*x*cos(d*x + c)^3 - 1431*cos(d*x + c)^4 + 465*cos(d
*x + c)^2 + (315*d*x*cos(d*x + c)^5 - 1260*d*x*cos(d*x + c)^3 - 1059*cos(d*x + c)^4 + 305*cos(d*x + c)^2 - 35)
*sin(d*x + c) - 70)/(3*a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^3 + (a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x
+ c)^3)*sin(d*x + c))

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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(sec(d*x+c)**4*sin(d*x+c)**7/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.29692, size = 244, normalized size = 1.37 \begin{align*} \frac{\frac{10080 \,{\left (d x + c\right )}}{a^{3}} + \frac{105 \,{\left (21 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 48 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 23\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} + \frac{17955 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 160020 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 624960 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 1387260 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1884582 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 1556268 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 774792 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 215748 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 25967}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{9}}}{10080 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^7/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/10080*(10080*(d*x + c)/a^3 + 105*(21*tan(1/2*d*x + 1/2*c)^2 - 48*tan(1/2*d*x + 1/2*c) + 23)/(a^3*(tan(1/2*d*
x + 1/2*c) - 1)^3) + (17955*tan(1/2*d*x + 1/2*c)^8 + 160020*tan(1/2*d*x + 1/2*c)^7 + 624960*tan(1/2*d*x + 1/2*
c)^6 + 1387260*tan(1/2*d*x + 1/2*c)^5 + 1884582*tan(1/2*d*x + 1/2*c)^4 + 1556268*tan(1/2*d*x + 1/2*c)^3 + 7747
92*tan(1/2*d*x + 1/2*c)^2 + 215748*tan(1/2*d*x + 1/2*c) + 25967)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^9))/d