\(\int \frac {\sec (c+d x) \tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [833]

Optimal result

Integrand size = 27, antiderivative size = 91 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sec ^3(c+d x)}{3 a^2 d}-\frac {3 \sec ^5(c+d x)}{5 a^2 d}+\frac {2 \sec ^7(c+d x)}{7 a^2 d}-\frac {2 \tan ^5(c+d x)}{5 a^2 d}-\frac {2 \tan ^7(c+d x)}{7 a^2 d} \]

[Out]

1/3*sec(d*x+c)^3/a^2/d-3/5*sec(d*x+c)^5/a^2/d+2/7*sec(d*x+c)^7/a^2/d-2/5*tan(d*x+c)^5/a^2/d-2/7*tan(d*x+c)^7/a
^2/d

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2954, 2952, 2686, 14, 2687, 276} \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 \tan ^7(c+d x)}{7 a^2 d}-\frac {2 \tan ^5(c+d x)}{5 a^2 d}+\frac {2 \sec ^7(c+d x)}{7 a^2 d}-\frac {3 \sec ^5(c+d x)}{5 a^2 d}+\frac {\sec ^3(c+d x)}{3 a^2 d} \]

[In]

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

[Out]

Sec[c + d*x]^3/(3*a^2*d) - (3*Sec[c + d*x]^5)/(5*a^2*d) + (2*Sec[c + d*x]^7)/(7*a^2*d) - (2*Tan[c + d*x]^5)/(5
*a^2*d) - (2*Tan[c + d*x]^7)/(7*a^2*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 2952

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 2954

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]

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

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.38 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sec ^3(c+d x) (672-182 \cos (c+d x)-736 \cos (2 (c+d x))-39 \cos (3 (c+d x))+192 \cos (4 (c+d x))+13 \cos (5 (c+d x))+448 \sin (c+d x)-104 \sin (2 (c+d x))-144 \sin (3 (c+d x))-52 \sin (4 (c+d x))+48 \sin (5 (c+d x)))}{6720 a^2 d (1+\sin (c+d x))^2} \]

[In]

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

[Out]

(Sec[c + d*x]^3*(672 - 182*Cos[c + d*x] - 736*Cos[2*(c + d*x)] - 39*Cos[3*(c + d*x)] + 192*Cos[4*(c + d*x)] +
13*Cos[5*(c + d*x)] + 448*Sin[c + d*x] - 104*Sin[2*(c + d*x)] - 144*Sin[3*(c + d*x)] - 52*Sin[4*(c + d*x)] + 4
8*Sin[5*(c + d*x)]))/(6720*a^2*d*(1 + Sin[c + d*x])^2)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.48 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.32

method	result	size
risch	\(\frac {\frac {8 i {\mathrm e}^{6 i \left (d x +c \right )}}{3}+\frac {8 \,{\mathrm e}^{7 i \left (d x +c \right )}}{3}+\frac {8 i {\mathrm e}^{4 i \left (d x +c \right )}}{15}-\frac {16 \,{\mathrm e}^{5 i \left (d x +c \right )}}{5}+\frac {24 i {\mathrm e}^{2 i \left (d x +c \right )}}{35}+\frac {88 \,{\mathrm e}^{3 i \left (d x +c \right )}}{105}-\frac {8 i}{35}-\frac {32 \,{\mathrm e}^{i \left (d x +c \right )}}{35}}{\left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d \,a^{2}}\)	\(120\)
parallelrisch	\(-\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )+4 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+91 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+84 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+105\right )}{105 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}\)	\(122\)
derivativedivides	\(\frac {-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {14}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {7}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{d \,a^{2}}\)	\(130\)
default	\(\frac {-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {14}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {7}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{d \,a^{2}}\)	\(130\)
norman	\(\frac {-\frac {16 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}-\frac {4}{105 a d}-\frac {4 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {52 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{105 d a}-\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d a}+\frac {32 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d a}}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}\)	\(152\)

[In]

int(sec(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

8/105*(35*I*exp(6*I*(d*x+c))+35*exp(7*I*(d*x+c))+7*I*exp(4*I*(d*x+c))-42*exp(5*I*(d*x+c))+9*I*exp(2*I*(d*x+c))
+11*exp(3*I*(d*x+c))-3*I-12*exp(I*(d*x+c)))/(exp(I*(d*x+c))+I)^7/(exp(I*(d*x+c))-I)^3/d/a^2

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.14 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {24 \, \cos \left (d x + c\right )^{4} - 47 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} + 5\right )} \sin \left (d x + c\right ) + 25}{105 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/105*(24*cos(d*x + c)^4 - 47*cos(d*x + c)^2 + 2*(6*cos(d*x + c)^4 - 9*cos(d*x + c)^2 + 5)*sin(d*x + c) + 25)
/(a^2*d*cos(d*x + c)^5 - 2*a^2*d*cos(d*x + c)^3*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**4*sin(d*x+c)**3/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (81) = 162\).

Time = 0.22 (sec) , antiderivative size = 336, normalized size of antiderivative = 3.69 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {4 \, {\left (\frac {4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {91 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {84 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {105 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}}{105 \, {\left (a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {8 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {14 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}\right )} d} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

4/105*(4*sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 8*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 91*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 84*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 105*sin(d*x + c
)^6/(cos(d*x + c) + 1)^6 + 1)/((a^2 + 4*a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 3*a^2*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 - 8*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 14*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 14*a^2*si
n(d*x + c)^6/(cos(d*x + c) + 1)^6 + 8*a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 3*a^2*sin(d*x + c)^8/(cos(d*x
+ c) + 1)^8 - 4*a^2*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10)*d)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.32 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {35 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1015 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 1330 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1302 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 469 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 67}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/840*(35*(3*tan(1/2*d*x + 1/2*c) - 1)/(a^2*(tan(1/2*d*x + 1/2*c) - 1)^3) - (105*tan(1/2*d*x + 1/2*c)^5 + 101
5*tan(1/2*d*x + 1/2*c)^4 + 1330*tan(1/2*d*x + 1/2*c)^3 + 1302*tan(1/2*d*x + 1/2*c)^2 + 469*tan(1/2*d*x + 1/2*c
) + 67)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^7))/d

Mupad [B] (verification not implemented)

Time = 14.89 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.27 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{105}+\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}+\frac {4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}-\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{105}+\frac {52\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}+\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^2\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^7} \]

[In]

int(sin(c + d*x)^3/(cos(c + d*x)^4*(a + a*sin(c + d*x))^2),x)

[Out]

((4*cos(c/2 + (d*x)/2)^10)/105 + (16*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2))/105 + 4*cos(c/2 + (d*x)/2)^4*sin
(c/2 + (d*x)/2)^6 + (16*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5)/5 + (52*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x
)/2)^4)/15 - (32*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^3)/105 + (4*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2
)/35)/(a^2*d*(cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2))^3*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^7)