\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx\) [49]

Optimal result

Integrand size = 31, antiderivative size = 99 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {(A-2 B) x}{a^2}-\frac {(A-4 B) \sin (c+d x)}{3 a^2 d}-\frac {(A-2 B) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}+\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3056, 3047, 3102, 12, 2814, 2727} \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=-\frac {(A-4 B) \sin (c+d x)}{3 a^2 d}-\frac {(A-2 B) \sin (c+d x)}{a^2 d (\cos (c+d x)+1)}+\frac {x (A-2 B)}{a^2}+\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 3102

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

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

Mathematica [A] (verified)

Time = 1.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.38 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left ((A-B) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-2 (5 A-8 B) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+6 \cos ^3\left (\frac {1}{2} (c+d x)\right ) ((A-2 B) d x+B \sin (c+d x))+(A-B) \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{3 a^2 d (1+\cos (c+d x))^2} \]

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*((A - B)*Sec[c/2]*Sin[(d*x)/2] - 2*(5*A - 8*B)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] +
6*Cos[(c + d*x)/2]^3*((A - 2*B)*d*x + B*Sin[c + d*x]) + (A - B)*Cos[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Cos[
c + d*x])^2)

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.74

method	result	size
parallelrisch	\(\frac {\left (-20 A +\left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (3 B \cos \left (2 d x +2 c \right )+28 B \cos \left (d x +c \right )+2 A +23 B \right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+12 d x \left (A -2 B \right )}{12 a^{2} d}\)	\(73\)
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}-3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {4 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+4 \left (A -2 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(106\)
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}-3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {4 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+4 \left (A -2 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(106\)
risch	\(\frac {x A}{a^{2}}-\frac {2 B x}{a^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{2 a^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 a^{2} d}-\frac {2 i \left (6 A \,{\mathrm e}^{2 i \left (d x +c \right )}-9 B \,{\mathrm e}^{2 i \left (d x +c \right )}+9 A \,{\mathrm e}^{i \left (d x +c \right )}-15 B \,{\mathrm e}^{i \left (d x +c \right )}+5 A -8 B \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(130\)
norman	\(\frac {\frac {\left (A -2 B \right ) x}{a}+\frac {\left (A -2 B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {3 \left (A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (A -2 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {3 \left (A -2 B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {3 \left (A -2 B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {\left (A -B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (4 A -9 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (13 A -34 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a}\)	\(218\)

[In]

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

[Out]

1/12*((-20*A+sec(1/2*d*x+1/2*c)^2*(3*B*cos(2*d*x+2*c)+28*B*cos(d*x+c)+2*A+23*B))*tan(1/2*d*x+1/2*c)+12*d*x*(A-
2*B))/a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {3 \, {\left (A - 2 \, B\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (A - 2 \, B\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (A - 2 \, B\right )} d x + {\left (3 \, B \cos \left (d x + c\right )^{2} - {\left (5 \, A - 14 \, B\right )} \cos \left (d x + c\right ) - 4 \, A + 10 \, B\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

1/3*(3*(A - 2*B)*d*x*cos(d*x + c)^2 + 6*(A - 2*B)*d*x*cos(d*x + c) + 3*(A - 2*B)*d*x + (3*B*cos(d*x + c)^2 - (
5*A - 14*B)*cos(d*x + c) - 4*A + 10*B)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 411 vs. \(2 (90) = 180\).

Time = 1.09 (sec) , antiderivative size = 411, normalized size of antiderivative = 4.15 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\begin {cases} \frac {6 A d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {6 A d x}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {8 A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {9 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {12 B d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {12 B d x}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {14 B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {27 B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right ) \cos ^{2}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((6*A*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 6*A*d*x/(6*a**2*d*tan(c/2 +
d*x/2)**2 + 6*a**2*d) + A*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 8*A*tan(c/2 + d*x/2
)**3/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 9*A*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d
) - 12*B*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 12*B*d*x/(6*a**2*d*tan(c/2 + d*x/
2)**2 + 6*a**2*d) - B*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 14*B*tan(c/2 + d*x/2)**3
/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 27*B*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d),
Ne(d, 0)), (x*(A + B*cos(c))*cos(c)**2/(a*cos(c) + a)**2, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (95) = 190\).

Time = 0.32 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.93 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {B {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \]

[In]

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

[Out]

1/6*(B*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1
))) - A*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {6 \, {\left (d x + c\right )} {\left (A - 2 \, B\right )}}{a^{2}} + \frac {12 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

[In]

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

[Out]

1/6*(6*(d*x + c)*(A - 2*B)/a^2 + 12*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) + (A*a^4*tan(1/2
*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*A*a^4*tan(1/2*d*x + 1/2*c) + 15*B*a^4*tan(1/2*d*x + 1/2*c))
/a^6)/d

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.06 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx=\frac {x\,\left (A-2\,B\right )}{a^2}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A-B}{a^2}+\frac {A-3\,B}{2\,a^2}\right )}{d}+\frac {2\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d} \]

[In]

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

[Out]

(x*(A - 2*B))/a^2 - (tan(c/2 + (d*x)/2)*((A - B)/a^2 + (A - 3*B)/(2*a^2)))/d + (2*B*tan(c/2 + (d*x)/2))/(d*(a^
2*tan(c/2 + (d*x)/2)^2 + a^2)) + (tan(c/2 + (d*x)/2)^3*(A - B))/(6*a^2*d)