3.1.87 \(\int \frac {\tan ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx\) [87]

Optimal. Leaf size=127 \[ -\frac {4 \sec ^5(c+d x)}{5 a^4 d}+\frac {12 \sec ^7(c+d x)}{7 a^4 d}-\frac {8 \sec ^9(c+d x)}{9 a^4 d}+\frac {\tan ^3(c+d x)}{3 a^4 d}+\frac {9 \tan ^5(c+d x)}{5 a^4 d}+\frac {16 \tan ^7(c+d x)}{7 a^4 d}+\frac {8 \tan ^9(c+d x)}{9 a^4 d} \]

[Out]

-4/5*sec(d*x+c)^5/a^4/d+12/7*sec(d*x+c)^7/a^4/d-8/9*sec(d*x+c)^9/a^4/d+1/3*tan(d*x+c)^3/a^4/d+9/5*tan(d*x+c)^5
/a^4/d+16/7*tan(d*x+c)^7/a^4/d+8/9*tan(d*x+c)^9/a^4/d

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Rubi [A]
time = 0.23, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2790, 2687, 276, 2686, 14} \begin {gather*} \frac {8 \tan ^9(c+d x)}{9 a^4 d}+\frac {16 \tan ^7(c+d x)}{7 a^4 d}+\frac {9 \tan ^5(c+d x)}{5 a^4 d}+\frac {\tan ^3(c+d x)}{3 a^4 d}-\frac {8 \sec ^9(c+d x)}{9 a^4 d}+\frac {12 \sec ^7(c+d x)}{7 a^4 d}-\frac {4 \sec ^5(c+d x)}{5 a^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sec[c + d*x]^5)/(5*a^4*d) + (12*Sec[c + d*x]^7)/(7*a^4*d) - (8*Sec[c + d*x]^9)/(9*a^4*d) + Tan[c + d*x]^3/
(3*a^4*d) + (9*Tan[c + d*x]^5)/(5*a^4*d) + (16*Tan[c + d*x]^7)/(7*a^4*d) + (8*Tan[c + d*x]^9)/(9*a^4*d)

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 276

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 2686

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 2687

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 2790

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[a^(2*
m), Int[ExpandIntegrand[(g*Tan[e + f*x])^p/Sec[e + f*x]^m, (a*Sec[e + f*x] - b*Tan[e + f*x])^(-m), x], x], x]
/; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {\tan ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx &=\frac {\int \left (a^4 \sec ^8(c+d x) \tan ^2(c+d x)-4 a^4 \sec ^7(c+d x) \tan ^3(c+d x)+6 a^4 \sec ^6(c+d x) \tan ^4(c+d x)-4 a^4 \sec ^5(c+d x) \tan ^5(c+d x)+a^4 \sec ^4(c+d x) \tan ^6(c+d x)\right ) \, dx}{a^8}\\ &=\frac {\int \sec ^8(c+d x) \tan ^2(c+d x) \, dx}{a^4}+\frac {\int \sec ^4(c+d x) \tan ^6(c+d x) \, dx}{a^4}-\frac {4 \int \sec ^7(c+d x) \tan ^3(c+d x) \, dx}{a^4}-\frac {4 \int \sec ^5(c+d x) \tan ^5(c+d x) \, dx}{a^4}+\frac {6 \int \sec ^6(c+d x) \tan ^4(c+d x) \, dx}{a^4}\\ &=\frac {\text {Subst}\left (\int x^6 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a^4 d}+\frac {\text {Subst}\left (\int x^2 \left (1+x^2\right )^3 \, dx,x,\tan (c+d x)\right )}{a^4 d}-\frac {4 \text {Subst}\left (\int x^6 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d}-\frac {4 \text {Subst}\left (\int x^4 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{a^4 d}+\frac {6 \text {Subst}\left (\int x^4 \left (1+x^2\right )^2 \, dx,x,\tan (c+d x)\right )}{a^4 d}\\ &=\frac {\text {Subst}\left (\int \left (x^6+x^8\right ) \, dx,x,\tan (c+d x)\right )}{a^4 d}+\frac {\text {Subst}\left (\int \left (x^2+3 x^4+3 x^6+x^8\right ) \, dx,x,\tan (c+d x)\right )}{a^4 d}-\frac {4 \text {Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d}-\frac {4 \text {Subst}\left (\int \left (-x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d}+\frac {6 \text {Subst}\left (\int \left (x^4+2 x^6+x^8\right ) \, dx,x,\tan (c+d x)\right )}{a^4 d}\\ &=-\frac {4 \sec ^5(c+d x)}{5 a^4 d}+\frac {12 \sec ^7(c+d x)}{7 a^4 d}-\frac {8 \sec ^9(c+d x)}{9 a^4 d}+\frac {\tan ^3(c+d x)}{3 a^4 d}+\frac {9 \tan ^5(c+d x)}{5 a^4 d}+\frac {16 \tan ^7(c+d x)}{7 a^4 d}+\frac {8 \tan ^9(c+d x)}{9 a^4 d}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 124, normalized size = 0.98 \begin {gather*} \frac {\sec (c+d x) (16128+1554 \cos (c+d x)-16896 \cos (2 (c+d x))-999 \cos (3 (c+d x))+2816 \cos (4 (c+d x))+37 \cos (5 (c+d x))+34944 \sin (c+d x)+1776 \sin (2 (c+d x))-9504 \sin (3 (c+d x))-296 \sin (4 (c+d x))+352 \sin (5 (c+d x)))}{80640 a^4 d (1+\sin (c+d x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(16128 + 1554*Cos[c + d*x] - 16896*Cos[2*(c + d*x)] - 999*Cos[3*(c + d*x)] + 2816*Cos[4*(c + d*x
)] + 37*Cos[5*(c + d*x)] + 34944*Sin[c + d*x] + 1776*Sin[2*(c + d*x)] - 9504*Sin[3*(c + d*x)] - 296*Sin[4*(c +
d*x)] + 352*Sin[5*(c + d*x)]))/(80640*a^4*d*(1 + Sin[c + d*x])^4)

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Maple [A]
time = 0.27, size = 158, normalized size = 1.24

method	result	size
risch	\(-\frac {4 i \left (504 i {\mathrm e}^{5 i \left (d x +c \right )}+315 \,{\mathrm e}^{6 i \left (d x +c \right )}-528 i {\mathrm e}^{3 i \left (d x +c \right )}-777 \,{\mathrm e}^{4 i \left (d x +c \right )}+88 i {\mathrm e}^{i \left (d x +c \right )}+297 \,{\mathrm e}^{2 i \left (d x +c \right )}-11\right )}{315 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{9} d \,a^{4}}\)	\(109\)
derivativedivides	\(\frac {-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {16}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}+\frac {8}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {116}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {62}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {83}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {17}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {29}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+128}}{d \,a^{4}}\)	\(158\)
default	\(\frac {-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {16}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}+\frac {8}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {116}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {62}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {83}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {17}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {29}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+128}}{d \,a^{4}}\)	\(158\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

8/d/a^4*(-1/128/(tan(1/2*d*x+1/2*c)-1)-2/9/(tan(1/2*d*x+1/2*c)+1)^9+1/(tan(1/2*d*x+1/2*c)+1)^8-29/14/(tan(1/2*
d*x+1/2*c)+1)^7+31/12/(tan(1/2*d*x+1/2*c)+1)^6-83/40/(tan(1/2*d*x+1/2*c)+1)^5+17/16/(tan(1/2*d*x+1/2*c)+1)^4-2
9/96/(tan(1/2*d*x+1/2*c)+1)^3+1/64/(tan(1/2*d*x+1/2*c)+1)^2+1/128/(tan(1/2*d*x+1/2*c)+1))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 356 vs. \(2 (113) = 226\).
time = 0.29, size = 356, normalized size = 2.80 \begin {gather*} \frac {8 \, {\left (\frac {16 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {54 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {201 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {294 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {378 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {210 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {105 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 2\right )}}{315 \, {\left (a^{4} + \frac {8 \, a^{4} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {27 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {48 \, a^{4} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {42 \, a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {42 \, a^{4} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {48 \, a^{4} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {27 \, a^{4} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {8 \, a^{4} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {a^{4} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

8/315*(16*sin(d*x + c)/(cos(d*x + c) + 1) + 54*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 201*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 294*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 378*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 210*sin(d
*x + c)^6/(cos(d*x + c) + 1)^6 + 105*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 2)/((a^4 + 8*a^4*sin(d*x + c)/(cos(
d*x + c) + 1) + 27*a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 48*a^4*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 42*a
^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 42*a^4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 48*a^4*sin(d*x + c)^7/(c
os(d*x + c) + 1)^7 - 27*a^4*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 8*a^4*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 -
a^4*sin(d*x + c)^10/(cos(d*x + c) + 1)^10)*d)

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Fricas [A]
time = 0.40, size = 129, normalized size = 1.02 \begin {gather*} \frac {88 \, \cos \left (d x + c\right )^{4} - 220 \, \cos \left (d x + c\right )^{2} + {\left (22 \, \cos \left (d x + c\right )^{4} - 165 \, \cos \left (d x + c\right )^{2} + 175\right )} \sin \left (d x + c\right ) + 140}{315 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} - 8 \, a^{4} d \cos \left (d x + c\right )^{3} + 8 \, a^{4} d \cos \left (d x + c\right ) - 4 \, {\left (a^{4} d \cos \left (d x + c\right )^{3} - 2 \, a^{4} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/315*(88*cos(d*x + c)^4 - 220*cos(d*x + c)^2 + (22*cos(d*x + c)^4 - 165*cos(d*x + c)^2 + 175)*sin(d*x + c) +
140)/(a^4*d*cos(d*x + c)^5 - 8*a^4*d*cos(d*x + c)^3 + 8*a^4*d*cos(d*x + c) - 4*(a^4*d*cos(d*x + c)^3 - 2*a^4*d
*cos(d*x + c))*sin(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\tan ^{2}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2/(a+a*sin(d*x+c))**4,x)

[Out]

Integral(tan(c + d*x)**2/(sin(c + d*x)**4 + 4*sin(c + d*x)**3 + 6*sin(c + d*x)**2 + 4*sin(c + d*x) + 1), x)/a*
*4

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Giac [A]
time = 14.26, size = 146, normalized size = 1.15 \begin {gather*} -\frac {\frac {315}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} - \frac {315 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 3150 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1050 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 630 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8064 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 6006 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5274 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 846 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 59}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{9}}}{5040 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

-1/5040*(315/(a^4*(tan(1/2*d*x + 1/2*c) - 1)) - (315*tan(1/2*d*x + 1/2*c)^8 + 3150*tan(1/2*d*x + 1/2*c)^7 + 10
50*tan(1/2*d*x + 1/2*c)^6 + 630*tan(1/2*d*x + 1/2*c)^5 - 8064*tan(1/2*d*x + 1/2*c)^4 - 6006*tan(1/2*d*x + 1/2*
c)^3 - 5274*tan(1/2*d*x + 1/2*c)^2 - 846*tan(1/2*d*x + 1/2*c) - 59)/(a^4*(tan(1/2*d*x + 1/2*c) + 1)^9))/d

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Mupad [B]
time = 7.58, size = 231, normalized size = 1.82 \begin {gather*} \frac {\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{315}+\frac {128\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{315}+\frac {48\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}+\frac {536\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{105}+\frac {112\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}+\frac {48\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}}{a^4\,d\,\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2/(a + a*sin(c + d*x))^4,x)

[Out]

((16*cos(c/2 + (d*x)/2)^10)/315 + (128*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2))/315 + (8*cos(c/2 + (d*x)/2)^3*
sin(c/2 + (d*x)/2)^7)/3 + (16*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6)/3 + (48*cos(c/2 + (d*x)/2)^5*sin(c/2
+ (d*x)/2)^5)/5 + (112*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4)/15 + (536*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*
x)/2)^3)/105 + (48*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2)/35)/(a^4*d*(cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)
/2))*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^9)

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