3.787 \(\int \frac{\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=151 \[ \frac{\cos (c+d x)}{a^3 d}+\frac{4 \tan ^7(c+d x)}{7 a^3 d}-\frac{3 \tan ^5(c+d x)}{5 a^3 d}+\frac{\tan ^3(c+d x)}{a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}-\frac{4 \sec ^7(c+d x)}{7 a^3 d}+\frac{13 \sec ^5(c+d x)}{5 a^3 d}-\frac{5 \sec ^3(c+d x)}{a^3 d}+\frac{7 \sec (c+d x)}{a^3 d}+\frac{3 x}{a^3} \]

[Out]

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

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Rubi [A]  time = 0.34146, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 10, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.345, Rules used = {2875, 2873, 2607, 30, 2606, 194, 3473, 8, 2590, 270} \[ \frac{\cos (c+d x)}{a^3 d}+\frac{4 \tan ^7(c+d x)}{7 a^3 d}-\frac{3 \tan ^5(c+d x)}{5 a^3 d}+\frac{\tan ^3(c+d x)}{a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}-\frac{4 \sec ^7(c+d x)}{7 a^3 d}+\frac{13 \sec ^5(c+d x)}{5 a^3 d}-\frac{5 \sec ^3(c+d x)}{a^3 d}+\frac{7 \sec (c+d x)}{a^3 d}+\frac{3 x}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^4*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

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

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

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

Mathematica [A]  time = 0.650508, size = 224, normalized size = 1.48 \[ \frac{8008 \sin (c+d x)+11760 c \sin (2 (c+d x))+11760 d x \sin (2 (c+d x))-20762 \sin (2 (c+d x))+6588 \sin (3 (c+d x))-840 c \sin (4 (c+d x))-840 d x \sin (4 (c+d x))+1483 \sin (4 (c+d x))-140 \sin (5 (c+d x))+14 (840 c+840 d x-1483) \cos (c+d x)+5152 \cos (2 (c+d x))-5040 c \cos (3 (c+d x))-5040 d x \cos (3 (c+d x))+8898 \cos (3 (c+d x))-2288 \cos (4 (c+d x))+8400}{2240 a^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^4*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

(8400 + 14*(-1483 + 840*c + 840*d*x)*Cos[c + d*x] + 5152*Cos[2*(c + d*x)] + 8898*Cos[3*(c + d*x)] - 5040*c*Cos
[3*(c + d*x)] - 5040*d*x*Cos[3*(c + d*x)] - 2288*Cos[4*(c + d*x)] + 8008*Sin[c + d*x] - 20762*Sin[2*(c + d*x)]
+ 11760*c*Sin[2*(c + d*x)] + 11760*d*x*Sin[2*(c + d*x)] + 6588*Sin[3*(c + d*x)] + 1483*Sin[4*(c + d*x)] - 840
*c*Sin[4*(c + d*x)] - 840*d*x*Sin[4*(c + d*x)] - 140*Sin[5*(c + d*x)])/(2240*a^3*d*(Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^7)

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Maple [A]  time = 0.12, size = 211, normalized size = 1.4 \begin{align*} -{\frac{1}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{1}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}-{\frac{8}{7\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-7}}+4\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{6}}}-{\frac{14}{5\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}-3\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}+{\frac{1}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{17}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{49}{8\,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)^2*sin(d*x+c)^6/(a+a*sin(d*x+c))^3,x)

[Out]

-1/8/d/a^3/(tan(1/2*d*x+1/2*c)-1)+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)+6/d/a^3*arctan(tan(1/2*d*x+1/2*c))-8/7/d/a^
3/(tan(1/2*d*x+1/2*c)+1)^7+4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^6-14/5/d/a^3/(tan(1/2*d*x+1/2*c)+1)^5-3/d/a^3/(tan(1
/2*d*x+1/2*c)+1)^4+1/2/d/a^3/(tan(1/2*d*x+1/2*c)+1)^3+17/4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2+49/8/d/a^3/(tan(1/2*
d*x+1/2*c)+1)

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Maxima [B]  time = 1.56908, size = 568, normalized size = 3.76 \begin{align*} \frac{2 \,{\left (\frac{\frac{951 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{2010 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{1980 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{574 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{966 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{1890 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{1540 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{630 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac{105 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 176}{a^{3} + \frac{6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{15 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{20 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{14 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{20 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{15 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac{6 \, a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac{a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac{105 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{35 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

2/35*((951*sin(d*x + c)/(cos(d*x + c) + 1) + 2010*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1980*sin(d*x + c)^3/(c
os(d*x + c) + 1)^3 + 574*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 966*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1890*
sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 1540*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 630*sin(d*x + c)^8/(cos(d*x +
c) + 1)^8 - 105*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 176)/(a^3 + 6*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 15*
a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 20*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 14*a^3*sin(d*x + c)^4/(
cos(d*x + c) + 1)^4 - 14*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 20*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7
- 15*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 6*a^3*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - a^3*sin(d*x + c)^10
/(cos(d*x + c) + 1)^10) + 105*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [A]  time = 1.14304, size = 431, normalized size = 2.85 \begin{align*} \frac{315 \, d x \cos \left (d x + c\right )^{3} + 286 \, \cos \left (d x + c\right )^{4} - 420 \, d x \cos \left (d x + c\right ) - 447 \, \cos \left (d x + c\right )^{2} +{\left (105 \, d x \cos \left (d x + c\right )^{3} + 35 \, \cos \left (d x + c\right )^{4} - 420 \, d x \cos \left (d x + c\right ) - 438 \, \cos \left (d x + c\right )^{2} - 20\right )} \sin \left (d x + c\right ) - 15}{35 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) +{\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\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)^2*sin(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/35*(315*d*x*cos(d*x + c)^3 + 286*cos(d*x + c)^4 - 420*d*x*cos(d*x + c) - 447*cos(d*x + c)^2 + (105*d*x*cos(d
*x + c)^3 + 35*cos(d*x + c)^4 - 420*d*x*cos(d*x + c) - 438*cos(d*x + c)^2 - 20)*sin(d*x + c) - 15)/(3*a^3*d*co
s(d*x + c)^3 - 4*a^3*d*cos(d*x + c) + (a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d*x + c))*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)**2*sin(d*x+c)**6/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.1958, size = 239, normalized size = 1.58 \begin{align*} \frac{\frac{840 \,{\left (d x + c\right )}}{a^{3}} - \frac{35 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 16 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 17\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )} a^{3}} + \frac{1715 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 11480 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 31815 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 45920 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 35161 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 13832 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2221}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{7}}}{280 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/280*(840*(d*x + c)/a^3 - 35*(tan(1/2*d*x + 1/2*c)^2 - 16*tan(1/2*d*x + 1/2*c) + 17)/((tan(1/2*d*x + 1/2*c)^3
- tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) - 1)*a^3) + (1715*tan(1/2*d*x + 1/2*c)^6 + 11480*tan(1/2*d*x
+ 1/2*c)^5 + 31815*tan(1/2*d*x + 1/2*c)^4 + 45920*tan(1/2*d*x + 1/2*c)^3 + 35161*tan(1/2*d*x + 1/2*c)^2 + 1383
2*tan(1/2*d*x + 1/2*c) + 2221)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d