\(\int \frac {\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^3} \, dx\) [58]

Optimal result

Integrand size = 33, antiderivative size = 136 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=-\frac {3 C x}{a^3}+\frac {(2 A+27 C) \sin (c+d x)}{15 a^3 d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {3 C \sin (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

-3*C*x/a^3+1/15*(2*A+27*C)*sin(d*x+c)/a^3/d-1/5*(A+C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+1/15*(A-9*C
)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2+3*C*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3121, 3056, 3047, 3102, 12, 2814, 2727} \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {(2 A+27 C) \sin (c+d x)}{15 a^3 d}+\frac {3 C \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {3 C x}{a^3}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {(A-9 C) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

(-3*C*x)/a^3 + ((2*A + 27*C)*Sin[c + d*x])/(15*a^3*d) - ((A + C)*Cos[c + d*x]^3*Sin[c + d*x])/(5*d*(a + a*Cos[
c + d*x])^3) + ((A - 9*C)*Cos[c + d*x]^2*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) + (3*C*Sin[c + d*x])/(d
*(a^3 + a^3*Cos[c + d*x]))

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]

Rule 3121

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos ^2(c+d x) (a (2 A-3 C)+a (A+6 C) \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\cos (c+d x) \left (2 a^2 (A-9 C)+a^2 (2 A+27 C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {2 a^2 (A-9 C) \cos (c+d x)+a^2 (2 A+27 C) \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4} \\ & = \frac {(2 A+27 C) \sin (c+d x)}{15 a^3 d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int -\frac {45 a^3 C \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^5} \\ & = \frac {(2 A+27 C) \sin (c+d x)}{15 a^3 d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 C) \int \frac {\cos (c+d x)}{a+a \cos (c+d x)} \, dx}{a^2} \\ & = -\frac {3 C x}{a^3}+\frac {(2 A+27 C) \sin (c+d x)}{15 a^3 d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(3 C) \int \frac {1}{a+a \cos (c+d x)} \, dx}{a^2} \\ & = -\frac {3 C x}{a^3}+\frac {(2 A+27 C) \sin (c+d x)}{15 a^3 d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {3 C \sin (c+d x)}{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. \(283\) vs. \(2(136)=272\).

Time = 1.82 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.08 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=-\frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (900 C d x \cos \left (\frac {d x}{2}\right )+900 C d x \cos \left (c+\frac {d x}{2}\right )+450 C d x \cos \left (c+\frac {3 d x}{2}\right )+450 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+90 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+90 C d x \cos \left (3 c+\frac {5 d x}{2}\right )-160 A \sin \left (\frac {d x}{2}\right )-1755 C \sin \left (\frac {d x}{2}\right )+120 A \sin \left (c+\frac {d x}{2}\right )+1125 C \sin \left (c+\frac {d x}{2}\right )-80 A \sin \left (c+\frac {3 d x}{2}\right )-1215 C \sin \left (c+\frac {3 d x}{2}\right )+60 A \sin \left (2 c+\frac {3 d x}{2}\right )+225 C \sin \left (2 c+\frac {3 d x}{2}\right )-28 A \sin \left (2 c+\frac {5 d x}{2}\right )-363 C \sin \left (2 c+\frac {5 d x}{2}\right )-75 C \sin \left (3 c+\frac {5 d x}{2}\right )-15 C \sin \left (3 c+\frac {7 d x}{2}\right )-15 C \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{960 a^3 d} \]

[In]

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

[Out]

-1/960*(Sec[c/2]*Sec[(c + d*x)/2]^5*(900*C*d*x*Cos[(d*x)/2] + 900*C*d*x*Cos[c + (d*x)/2] + 450*C*d*x*Cos[c + (
3*d*x)/2] + 450*C*d*x*Cos[2*c + (3*d*x)/2] + 90*C*d*x*Cos[2*c + (5*d*x)/2] + 90*C*d*x*Cos[3*c + (5*d*x)/2] - 1
60*A*Sin[(d*x)/2] - 1755*C*Sin[(d*x)/2] + 120*A*Sin[c + (d*x)/2] + 1125*C*Sin[c + (d*x)/2] - 80*A*Sin[c + (3*d
*x)/2] - 1215*C*Sin[c + (3*d*x)/2] + 60*A*Sin[2*c + (3*d*x)/2] + 225*C*Sin[2*c + (3*d*x)/2] - 28*A*Sin[2*c + (
5*d*x)/2] - 363*C*Sin[2*c + (5*d*x)/2] - 75*C*Sin[3*c + (5*d*x)/2] - 15*C*Sin[3*c + (7*d*x)/2] - 15*C*Sin[4*c
+ (7*d*x)/2]))/(a^3*d)

Maple [A] (verified)

Time = 1.81 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.62

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\frac {\left (\frac {7 A}{3}+39 C \right ) \cos \left (2 d x +2 c \right )}{4}+\frac {5 C \cos \left (3 d x +3 c \right )}{8}+\left (A +\frac {243 C}{8}\right ) \cos \left (d x +c \right )+\frac {11 A}{12}+\frac {87 C}{4}\right ) \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-30 d x C}{10 a^{3} d}\)	\(85\)
derivativedivides	\(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C +A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -16 C \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{4 d \,a^{3}}\)	\(131\)
default	\(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C +A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -16 C \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{4 d \,a^{3}}\)	\(131\)
risch	\(-\frac {3 C x}{a^{3}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a^{3} d}+\frac {2 i \left (15 A \,{\mathrm e}^{4 i \left (d x +c \right )}+90 C \,{\mathrm e}^{4 i \left (d x +c \right )}+30 A \,{\mathrm e}^{3 i \left (d x +c \right )}+300 C \,{\mathrm e}^{3 i \left (d x +c \right )}+40 A \,{\mathrm e}^{2 i \left (d x +c \right )}+420 C \,{\mathrm e}^{2 i \left (d x +c \right )}+20 A \,{\mathrm e}^{i \left (d x +c \right )}+270 C \,{\mathrm e}^{i \left (d x +c \right )}+7 A +72 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(172\)
norman	\(\frac {-\frac {3 C x}{a}-\frac {12 C x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {18 C x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {12 C x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {3 C x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {\left (A -9 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}+\frac {\left (A +C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {\left (A +25 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {5 \left (A +27 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (A +81 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}-\frac {\left (7 A -153 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}+\frac {\left (53 A +1773 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} a^{2}}\)	\(268\)

[In]

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

[Out]

1/10*(tan(1/2*d*x+1/2*c)*(1/4*(7/3*A+39*C)*cos(2*d*x+2*c)+5/8*C*cos(3*d*x+3*c)+(A+243/8*C)*cos(d*x+c)+11/12*A+
87/4*C)*sec(1/2*d*x+1/2*c)^4-30*d*x*C)/a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=-\frac {45 \, C d x \cos \left (d x + c\right )^{3} + 135 \, C d x \cos \left (d x + c\right )^{2} + 135 \, C d x \cos \left (d x + c\right ) + 45 \, C d x - {\left (15 \, C \cos \left (d x + c\right )^{3} + {\left (7 \, A + 117 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + 57 \, C\right )} \cos \left (d x + c\right ) + 2 \, A + 72 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

-1/15*(45*C*d*x*cos(d*x + c)^3 + 135*C*d*x*cos(d*x + c)^2 + 135*C*d*x*cos(d*x + c) + 45*C*d*x - (15*C*cos(d*x
+ c)^3 + (7*A + 117*C)*cos(d*x + c)^2 + 3*(2*A + 57*C)*cos(d*x + c) + 2*A + 72*C)*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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (129) = 258\).

Time = 2.17 (sec) , antiderivative size = 422, normalized size of antiderivative = 3.10 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} \frac {3 A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {7 A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {5 A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {15 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {180 C d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {180 C d x}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {3 C \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {27 C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {225 C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {375 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + C \cos ^{2}{\left (c \right )}\right ) \cos ^{2}{\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((3*A*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 7*A*tan(c/2 + d*x/2)**5/(60*a
**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 5*A*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) +
15*A*tan(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 180*C*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d*
tan(c/2 + d*x/2)**2 + 60*a**3*d) - 180*C*d*x/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 3*C*tan(c/2 + d*x/2
)**7/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 27*C*tan(c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**2 + 6
0*a**3*d) + 225*C*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 375*C*tan(c/2 + d*x/2)/(60
*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d), Ne(d, 0)), (x*(A + C*cos(c)**2)*cos(c)**2/(a*cos(c) + a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.51 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \, C {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + \frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]

[In]

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

[Out]

1/60*(3*C*(40*sin(d*x + c)/((a^3 + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x
+ c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(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 = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=-\frac {\frac {180 \, {\left (d x + c\right )} C}{a^{3}} - \frac {120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

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

[Out]

-1/60*(180*(d*x + c)*C/a^3 - 120*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) - (3*A*a^12*tan(1/2
*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 10*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*C*a^12*tan(1/2*d*x +
1/2*c)^3 + 15*A*a^12*tan(1/2*d*x + 1/2*c) + 255*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 1.09 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {\left (\frac {7\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+\frac {24\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-\frac {4\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}-\frac {3\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{20}+\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}-\frac {3\,C\,x}{a^3}+\frac {2\,C\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3\,d} \]

[In]

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

[Out]

((A*sin(c/2 + (d*x)/2))/20 + (C*sin(c/2 + (d*x)/2))/20 - cos(c/2 + (d*x)/2)^2*((4*A*sin(c/2 + (d*x)/2))/15 + (
3*C*sin(c/2 + (d*x)/2))/5) + cos(c/2 + (d*x)/2)^4*((7*A*sin(c/2 + (d*x)/2))/15 + (24*C*sin(c/2 + (d*x)/2))/5))
/(a^3*d*cos(c/2 + (d*x)/2)^5) - (3*C*x)/a^3 + (2*C*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2))/(a^3*d)