\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx\) [68]

Optimal result

Integrand size = 31, antiderivative size = 136 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=-\frac {2 (A+27 B) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}+\frac {(13 A+36 B) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))}+\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(A-8 B) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3} \]

[Out]

-2/105*(A+27*B)*sin(d*x+c)/a^4/d/(1+cos(d*x+c))^2+1/105*(13*A+36*B)*sin(d*x+c)/a^4/d/(1+cos(d*x+c))+1/7*(A-B)*
cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^4-1/35*(A-8*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^3

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 136, 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, 2829, 2727} \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=\frac {(13 A+36 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac {2 (A+27 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {(A-8 B) \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3} \]

[In]

Int[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^4,x]

[Out]

(-2*(A + 27*B)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) + ((13*A + 36*B)*Sin[c + d*x])/(105*a^4*d*(1 + C
os[c + d*x])) + ((A - B)*Cos[c + d*x]^2*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - ((A - 8*B)*Sin[c + d*x])/
(35*a*d*(a + a*Cos[c + d*x])^3)

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 2829

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

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)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {\cos (c+d x) (2 a (A-B)+a (A+6 B) \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx}{7 a^2} \\ & = \frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {2 a (A-B) \cos (c+d x)+a (A+6 B) \cos ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2} \\ & = \frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(A-8 B) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {\int \frac {-3 a^2 (A-8 B)-5 a^2 (A+6 B) \cos (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4} \\ & = -\frac {2 (A+27 B) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}+\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(A-8 B) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(13 A+36 B) \int \frac {1}{a+a \cos (c+d x)} \, dx}{105 a^3} \\ & = -\frac {2 (A+27 B) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}+\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(A-8 B) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(13 A+36 B) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.09 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.42 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (70 (4 A+9 B) \sin \left (\frac {d x}{2}\right )-35 (5 A+18 B) \sin \left (c+\frac {d x}{2}\right )+168 A \sin \left (c+\frac {3 d x}{2}\right )+441 B \sin \left (c+\frac {3 d x}{2}\right )-105 A \sin \left (2 c+\frac {3 d x}{2}\right )-315 B \sin \left (2 c+\frac {3 d x}{2}\right )+91 A \sin \left (2 c+\frac {5 d x}{2}\right )+147 B \sin \left (2 c+\frac {5 d x}{2}\right )-105 B \sin \left (3 c+\frac {5 d x}{2}\right )+13 A \sin \left (3 c+\frac {7 d x}{2}\right )+36 B \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{420 a^4 d (1+\cos (c+d x))^4} \]

[In]

Integrate[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^4,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(70*(4*A + 9*B)*Sin[(d*x)/2] - 35*(5*A + 18*B)*Sin[c + (d*x)/2] + 168*A*Sin[c + (3*
d*x)/2] + 441*B*Sin[c + (3*d*x)/2] - 105*A*Sin[2*c + (3*d*x)/2] - 315*B*Sin[2*c + (3*d*x)/2] + 91*A*Sin[2*c +
(5*d*x)/2] + 147*B*Sin[2*c + (5*d*x)/2] - 105*B*Sin[3*c + (5*d*x)/2] + 13*A*Sin[3*c + (7*d*x)/2] + 36*B*Sin[3*
c + (7*d*x)/2]))/(420*a^4*d*(1 + Cos[c + d*x])^4)

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.60

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A -B \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {7 \left (-A +3 B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+7 \left (-\frac {A}{3}-B \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 A +7 B \right )}{56 a^{4} d}\)	\(82\)
derivativedivides	\(\frac {\frac {\left (A -B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (-A +3 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (-A -3 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(90\)
default	\(\frac {\frac {\left (A -B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (-A +3 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (-A -3 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(90\)
risch	\(\frac {2 i \left (105 B \,{\mathrm e}^{6 i \left (d x +c \right )}+105 A \,{\mathrm e}^{5 i \left (d x +c \right )}+315 B \,{\mathrm e}^{5 i \left (d x +c \right )}+175 A \,{\mathrm e}^{4 i \left (d x +c \right )}+630 B \,{\mathrm e}^{4 i \left (d x +c \right )}+280 A \,{\mathrm e}^{3 i \left (d x +c \right )}+630 B \,{\mathrm e}^{3 i \left (d x +c \right )}+168 A \,{\mathrm e}^{2 i \left (d x +c \right )}+441 B \,{\mathrm e}^{2 i \left (d x +c \right )}+91 A \,{\mathrm e}^{i \left (d x +c \right )}+147 B \,{\mathrm e}^{i \left (d x +c \right )}+13 A +36 B \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(162\)
norman	\(\frac {\frac {\left (A -B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 a d}+\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {3 \left (3 A +B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 a d}+\frac {\left (4 A +3 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}-\frac {\left (4 A +3 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{70 a d}+\frac {\left (4 A +3 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{140 a d}-\frac {\left (53 A -39 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{840 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a^{3}}\)	\(193\)

[In]

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^4,x,method=_RETURNVERBOSE)

[Out]

1/56*tan(1/2*d*x+1/2*c)*((A-B)*tan(1/2*d*x+1/2*c)^6+7/5*(-A+3*B)*tan(1/2*d*x+1/2*c)^4+7*(-1/3*A-B)*tan(1/2*d*x
+1/2*c)^2+7*A+7*B)/a^4/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=\frac {{\left ({\left (13 \, A + 36 \, B\right )} \cos \left (d x + c\right )^{3} + 13 \, {\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 8 \, A + 6 \, B\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/105*((13*A + 36*B)*cos(d*x + c)^3 + 13*(4*A + 3*B)*cos(d*x + c)^2 + 8*(4*A + 3*B)*cos(d*x + c) + 8*A + 6*B)*
sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) +
a^4*d)

Sympy [A] (verification not implemented)

Time = 2.20 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.34 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=\begin {cases} \frac {A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} - \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} - \frac {B \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} + \frac {3 B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} - \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right ) \cos ^{2}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**4,x)

[Out]

Piecewise((A*tan(c/2 + d*x/2)**7/(56*a**4*d) - A*tan(c/2 + d*x/2)**5/(40*a**4*d) - A*tan(c/2 + d*x/2)**3/(24*a
**4*d) + A*tan(c/2 + d*x/2)/(8*a**4*d) - B*tan(c/2 + d*x/2)**7/(56*a**4*d) + 3*B*tan(c/2 + d*x/2)**5/(40*a**4*
d) - B*tan(c/2 + d*x/2)**3/(8*a**4*d) + B*tan(c/2 + d*x/2)/(8*a**4*d), Ne(d, 0)), (x*(A + B*cos(c))*cos(c)**2/
(a*cos(c) + a)**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.29 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {A {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {3 \, B {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

1/840*(A*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 + 3*B*(35*sin(d*x + c)/(cos(d*x + c) + 1) - 35
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7)/a^4)/d

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=\frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 63 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \]

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 - 15*B*tan(1/2*d*x + 1/2*c)^7 - 21*A*tan(1/2*d*x + 1/2*c)^5 + 63*B*tan(1/2*
d*x + 1/2*c)^5 - 35*A*tan(1/2*d*x + 1/2*c)^3 - 105*B*tan(1/2*d*x + 1/2*c)^3 + 105*A*tan(1/2*d*x + 1/2*c) + 105
*B*tan(1/2*d*x + 1/2*c))/(a^4*d)

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.63 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx=-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+3\,B\right )}{24\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-3\,B\right )}{40\,a^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B\right )}{56\,a^4}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B\right )}{8\,a^4}}{d} \]

[In]

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

[Out]

-((tan(c/2 + (d*x)/2)^3*(A + 3*B))/(24*a^4) + (tan(c/2 + (d*x)/2)^5*(A - 3*B))/(40*a^4) - (tan(c/2 + (d*x)/2)^
7*(A - B))/(56*a^4) - (tan(c/2 + (d*x)/2)*(A + B))/(8*a^4))/d