\(\int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [835]

Optimal result

Integrand size = 27, antiderivative size = 93 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sec ^3(c+d x)}{7 d (a+a \sin (c+d x))^2}-\frac {2 \sec ^3(c+d x)}{35 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {8 \tan (c+d x)}{35 a^2 d}+\frac {8 \tan ^3(c+d x)}{105 a^2 d} \]

[Out]

1/7*sec(d*x+c)^3/d/(a+a*sin(d*x+c))^2-2/35*sec(d*x+c)^3/d/(a^2+a^2*sin(d*x+c))+8/35*tan(d*x+c)/a^2/d+8/105*tan
(d*x+c)^3/a^2/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2938, 2751, 3852} \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {8 \tan ^3(c+d x)}{105 a^2 d}+\frac {8 \tan (c+d x)}{35 a^2 d}-\frac {2 \sec ^3(c+d x)}{35 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2} \]

[In]

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

[Out]

Sec[c + d*x]^3/(7*d*(a + a*Sin[c + d*x])^2) - (2*Sec[c + d*x]^3)/(35*d*(a^2 + a^2*Sin[c + d*x])) + (8*Tan[c +
d*x])/(35*a^2*d) + (8*Tan[c + d*x]^3)/(105*a^2*d)

Rule 2751

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*Simplify[2*m + p + 1])), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 2938

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p
+ 1))), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sec ^3(c+d x)}{7 d (a+a \sin (c+d x))^2}+\frac {2 \int \frac {\sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx}{7 a} \\ & = \frac {\sec ^3(c+d x)}{7 d (a+a \sin (c+d x))^2}-\frac {2 \sec ^3(c+d x)}{35 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {8 \int \sec ^4(c+d x) \, dx}{35 a^2} \\ & = \frac {\sec ^3(c+d x)}{7 d (a+a \sin (c+d x))^2}-\frac {2 \sec ^3(c+d x)}{35 d \left (a^2+a^2 \sin (c+d x)\right )}-\frac {8 \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{35 a^2 d} \\ & = \frac {\sec ^3(c+d x)}{7 d (a+a \sin (c+d x))^2}-\frac {2 \sec ^3(c+d x)}{35 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {8 \tan (c+d x)}{35 a^2 d}+\frac {8 \tan ^3(c+d x)}{105 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.44 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\sec ^3(c+d x) \left (-84+\frac {21}{4} \cos (c+d x)+32 \cos (2 (c+d x))+\frac {9}{8} \cos (3 (c+d x))+16 \cos (4 (c+d x))-\frac {3}{8} \cos (5 (c+d x))-56 \sin (c+d x)+3 \sin (2 (c+d x))-12 \sin (3 (c+d x))+\frac {3}{2} \sin (4 (c+d x))+4 \sin (5 (c+d x))\right )}{420 a^2 d (1+\sin (c+d x))^2} \]

[In]

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

[Out]

-1/420*(Sec[c + d*x]^3*(-84 + (21*Cos[c + d*x])/4 + 32*Cos[2*(c + d*x)] + (9*Cos[3*(c + d*x)])/8 + 16*Cos[4*(c
+ d*x)] - (3*Cos[5*(c + d*x)])/8 - 56*Sin[c + d*x] + 3*Sin[2*(c + d*x)] - 12*Sin[3*(c + d*x)] + (3*Sin[4*(c +
d*x)])/2 + 4*Sin[5*(c + d*x)]))/(a^2*d*(1 + Sin[c + d*x])^2)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.46 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04

method	result	size
risch	\(-\frac {32 \left (14 i {\mathrm e}^{4 i \left (d x +c \right )}+21 \,{\mathrm e}^{5 i \left (d x +c \right )}+3 i {\mathrm e}^{2 i \left (d x +c \right )}-8 \,{\mathrm e}^{3 i \left (d x +c \right )}-i-4 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{105 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} d \,a^{2}}\)	\(97\)
parallelrisch	\(-\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (9 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+36 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+132 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+68 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+14 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-84 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+140 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+140 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+105\right )}{105 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}\)	\(150\)
derivativedivides	\(\frac {-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {18}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {35}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {11}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16}}{d \,a^{2}}\)	\(160\)
default	\(\frac {-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {18}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {35}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {11}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16}}{d \,a^{2}}\)	\(160\)
norman	\(\frac {\frac {8 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}-\frac {6}{35 a d}-\frac {2 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {8 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {8 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {24 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{35 d a}-\frac {136 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d a}-\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}-\frac {88 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} a}\)	\(190\)

[In]

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

[Out]

-32/105*(14*I*exp(4*I*(d*x+c))+21*exp(5*I*(d*x+c))+3*I*exp(2*I*(d*x+c))-8*exp(3*I*(d*x+c))-I-4*exp(I*(d*x+c)))
/(exp(I*(d*x+c))-I)^3/(exp(I*(d*x+c))+I)^7/d/a^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.12 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {32 \, \cos \left (d x + c\right )^{4} - 16 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, \cos \left (d x + c\right )^{4} - 12 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) - 25}{105 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]

[In]

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

[Out]

1/105*(32*cos(d*x + c)^4 - 16*cos(d*x + c)^2 + 2*(8*cos(d*x + c)^4 - 12*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]

\[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sin {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[In]

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

[Out]

Integral(sin(c + d*x)*sec(c + d*x)**4/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (85) = 170\).

Time = 0.23 (sec) , antiderivative size = 376, normalized size of antiderivative = 4.04 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {36 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {132 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {68 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {14 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {84 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {140 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {140 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {105 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 9\right )}}{105 \, {\left (a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {8 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {14 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}\right )} d} \]

[In]

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

[Out]

2/105*(36*sin(d*x + c)/(cos(d*x + c) + 1) + 132*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 68*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 14*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 84*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 140*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 + 140*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 105*sin(d*x + c)^8/(cos(d*x + c) + 1)
^8 + 9)/((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*sin(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*si
n(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 = 0.34 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.57 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {35 \, {\left (6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {210 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 175 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 910 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 756 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 427 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 31}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \]

[In]

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

[Out]

-1/840*(35*(6*tan(1/2*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) + 5)/(a^2*(tan(1/2*d*x + 1/2*c) - 1)^3) - (210*t
an(1/2*d*x + 1/2*c)^6 + 105*tan(1/2*d*x + 1/2*c)^5 - 175*tan(1/2*d*x + 1/2*c)^4 - 910*tan(1/2*d*x + 1/2*c)^3 -
756*tan(1/2*d*x + 1/2*c)^2 - 427*tan(1/2*d*x + 1/2*c) - 31)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^7))/d

Mupad [B] (verification not implemented)

Time = 16.04 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.73 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (9\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+36\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+132\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+68\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+14\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-84\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+140\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+140\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+105\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\right )}{105\,a^2\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^7} \]

[In]

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

[Out]

(2*cos(c/2 + (d*x)/2)^2*(9*cos(c/2 + (d*x)/2)^8 + 105*sin(c/2 + (d*x)/2)^8 + 140*cos(c/2 + (d*x)/2)*sin(c/2 +
(d*x)/2)^7 + 36*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2) + 140*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^6 - 84*c
os(c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/2)^5 + 14*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^4 + 68*cos(c/2 + (d*x)/2
)^5*sin(c/2 + (d*x)/2)^3 + 132*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^2))/(105*a^2*d*(cos(c/2 + (d*x)/2) - si
n(c/2 + (d*x)/2))^3*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^7)