\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\) [59]

Optimal result

Integrand size = 31, antiderivative size = 116 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {B x}{a^3}+\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-7 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(4 A-29 B) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

B*x/a^3+1/5*(A-B)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15*(2*A-7*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^
2+1/15*(4*A-29*B)*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3056, 3047, 3098, 2814, 2727} \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {(4 A-29 B) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {B x}{a^3}+\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {(2 A-7 B) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[In]

Int[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

(B*x)/a^3 + ((A - B)*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((2*A - 7*B)*Sin[c + d*x])/(1
5*a*d*(a + a*Cos[c + d*x])^2) + ((4*A - 29*B)*Sin[c + d*x])/(15*d*(a^3 + a^3*Cos[c + d*x]))

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3056

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 3098

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a*B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos (c+d x) (2 a (A-B)+5 a B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = \frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {2 a (A-B) \cos (c+d x)+5 a B \cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = \frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-7 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {\int \frac {-2 a^2 (2 A-7 B)-15 a^2 B \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4} \\ & = \frac {B x}{a^3}+\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-7 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(4 A-29 B) \int \frac {1}{a+a \cos (c+d x)} \, dx}{15 a^2} \\ & = \frac {B x}{a^3}+\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(2 A-7 B) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(4 A-29 B) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(241\) vs. \(2(116)=232\).

Time = 1.21 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.08 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (150 B d x \cos \left (\frac {d x}{2}\right )+150 B d x \cos \left (c+\frac {d x}{2}\right )+75 B d x \cos \left (c+\frac {3 d x}{2}\right )+75 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+15 B d x \cos \left (2 c+\frac {5 d x}{2}\right )+15 B d x \cos \left (3 c+\frac {5 d x}{2}\right )+80 A \sin \left (\frac {d x}{2}\right )-370 B \sin \left (\frac {d x}{2}\right )-60 A \sin \left (c+\frac {d x}{2}\right )+270 B \sin \left (c+\frac {d x}{2}\right )+40 A \sin \left (c+\frac {3 d x}{2}\right )-230 B \sin \left (c+\frac {3 d x}{2}\right )-30 A \sin \left (2 c+\frac {3 d x}{2}\right )+90 B \sin \left (2 c+\frac {3 d x}{2}\right )+14 A \sin \left (2 c+\frac {5 d x}{2}\right )-64 B \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{480 a^3 d} \]

[In]

Integrate[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(150*B*d*x*Cos[(d*x)/2] + 150*B*d*x*Cos[c + (d*x)/2] + 75*B*d*x*Cos[c + (3*d*x)/2
] + 75*B*d*x*Cos[2*c + (3*d*x)/2] + 15*B*d*x*Cos[2*c + (5*d*x)/2] + 15*B*d*x*Cos[3*c + (5*d*x)/2] + 80*A*Sin[(
d*x)/2] - 370*B*Sin[(d*x)/2] - 60*A*Sin[c + (d*x)/2] + 270*B*Sin[c + (d*x)/2] + 40*A*Sin[c + (3*d*x)/2] - 230*
B*Sin[c + (3*d*x)/2] - 30*A*Sin[2*c + (3*d*x)/2] + 90*B*Sin[2*c + (3*d*x)/2] + 14*A*Sin[2*c + (5*d*x)/2] - 64*
B*Sin[2*c + (5*d*x)/2]))/(480*a^3*d)

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.59

method	result	size
parallelrisch	\(\frac {3 \left (A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \left (2 B -A \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \left (A -7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+60 d x B}{60 a^{3} d}\)	\(69\)
derivativedivides	\(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(102\)
default	\(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(102\)
risch	\(\frac {B x}{a^{3}}+\frac {2 i \left (15 A \,{\mathrm e}^{4 i \left (d x +c \right )}-45 B \,{\mathrm e}^{4 i \left (d x +c \right )}+30 A \,{\mathrm e}^{3 i \left (d x +c \right )}-135 B \,{\mathrm e}^{3 i \left (d x +c \right )}+40 A \,{\mathrm e}^{2 i \left (d x +c \right )}-185 B \,{\mathrm e}^{2 i \left (d x +c \right )}+20 A \,{\mathrm e}^{i \left (d x +c \right )}-115 B \,{\mathrm e}^{i \left (d x +c \right )}+7 A -32 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(133\)
norman	\(\frac {\frac {B x}{a}+\frac {B x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {3 B x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {3 B x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {\left (A -11 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}+\frac {\left (A -7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (A -B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}-\frac {\left (A +9 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10 a d}+\frac {\left (3 A -43 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10 a d}+\frac {\left (7 A -59 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a^{2}}\)	\(226\)

[In]

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^3,x,method=_RETURNVERBOSE)

[Out]

1/60*(3*(A-B)*tan(1/2*d*x+1/2*c)^5+10*(2*B-A)*tan(1/2*d*x+1/2*c)^3+15*(A-7*B)*tan(1/2*d*x+1/2*c)+60*d*x*B)/a^3
/d

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, B d x \cos \left (d x + c\right )^{3} + 45 \, B d x \cos \left (d x + c\right )^{2} + 45 \, B d x \cos \left (d x + c\right ) + 15 \, B d x + {\left ({\left (7 \, A - 32 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A - 17 \, B\right )} \cos \left (d x + c\right ) + 2 \, A - 22 \, B\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(15*B*d*x*cos(d*x + c)^3 + 45*B*d*x*cos(d*x + c)^2 + 45*B*d*x*cos(d*x + c) + 15*B*d*x + ((7*A - 32*B)*cos
(d*x + c)^2 + 3*(2*A - 17*B)*cos(d*x + c) + 2*A - 22*B)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x
+ c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [A] (verification not implemented)

Time = 1.34 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.28 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} \frac {A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} - \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} + \frac {B x}{a^{3}} - \frac {B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{3} d} - \frac {7 B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right ) \cos ^{2}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**3,x)

[Out]

Piecewise((A*tan(c/2 + d*x/2)**5/(20*a**3*d) - A*tan(c/2 + d*x/2)**3/(6*a**3*d) + A*tan(c/2 + d*x/2)/(4*a**3*d
) + B*x/a**3 - B*tan(c/2 + d*x/2)**5/(20*a**3*d) + B*tan(c/2 + d*x/2)**3/(3*a**3*d) - 7*B*tan(c/2 + d*x/2)/(4*
a**3*d), Ne(d, 0)), (x*(A + B*cos(c))*cos(c)**2/(a*cos(c) + a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.38 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=-\frac {B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - \frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(B*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3) - A*(15*sin(d*x + c)/(cos(d*x + c) +
1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {60 \, {\left (d x + c\right )} B}{a^{3}} + \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(60*(d*x + c)*B/a^3 + (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 10*A*a^12*tan(
1/2*d*x + 1/2*c)^3 + 20*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 15*A*a^12*tan(1/2*d*x + 1/2*c) - 105*B*a^12*tan(1/2*d*
x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.16 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {B\,x}{a^3}-\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}\right )-{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {7\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )-\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]

[In]

int((cos(c + d*x)^2*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^3,x)

[Out]

(B*x)/a^3 - (cos(c/2 + (d*x)/2)^2*((A*sin(c/2 + (d*x)/2)^3)/6 - (B*sin(c/2 + (d*x)/2)^3)/3) - cos(c/2 + (d*x)/
2)^4*((A*sin(c/2 + (d*x)/2))/4 - (7*B*sin(c/2 + (d*x)/2))/4) - (A*sin(c/2 + (d*x)/2)^5)/20 + (B*sin(c/2 + (d*x
)/2)^5)/20)/(a^3*d*cos(c/2 + (d*x)/2)^5)