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

Optimal. Leaf size=136 \[ \frac{(2 A+27 C) \sin (c+d x)}{15 a^3 d}+\frac{3 C \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{3 C x}{a^3}-\frac{(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(A-9 C) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[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]))

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Rubi [A]  time = 0.462583, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {3042, 2977, 2968, 3023, 12, 2735, 2648} \[ \frac{(2 A+27 C) \sin (c+d x)}{15 a^3 d}+\frac{3 C \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{3 C x}{a^3}-\frac{(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(A-9 C) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[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 3042

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)]

Rule 2977

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 2968

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 3023

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 12

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

Rule 2735

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 2648

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]

Rubi steps

\begin{align*} \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{(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]  time = 0.905447, size = 283, normalized size = 2.08 \[ -\frac{\sec \left (\frac{c}{2}\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) \left (120 A \sin \left (c+\frac{d x}{2}\right )-80 A \sin \left (c+\frac{3 d x}{2}\right )+60 A \sin \left (2 c+\frac{3 d x}{2}\right )-28 A \sin \left (2 c+\frac{5 d x}{2}\right )-160 A \sin \left (\frac{d x}{2}\right )+1125 C \sin \left (c+\frac{d x}{2}\right )-1215 C \sin \left (c+\frac{3 d x}{2}\right )+225 C \sin \left (2 c+\frac{3 d x}{2}\right )-363 C \sin \left (2 c+\frac{5 d x}{2}\right )-75 C \sin \left (3 c+\frac{5 d x}{2}\right )-15 C \sin \left (3 c+\frac{7 d x}{2}\right )-15 C \sin \left (4 c+\frac{7 d x}{2}\right )+900 C d x \cos \left (c+\frac{d x}{2}\right )+450 C d x \cos \left (c+\frac{3 d x}{2}\right )+450 C d x \cos \left (2 c+\frac{3 d x}{2}\right )+90 C d x \cos \left (2 c+\frac{5 d x}{2}\right )+90 C d x \cos \left (3 c+\frac{5 d x}{2}\right )-1755 C \sin \left (\frac{d x}{2}\right )+900 C d x \cos \left (\frac{d x}{2}\right )\right )}{960 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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] - 160*A*S
in[(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]))/(960*a^3*d)

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Maple [A]  time = 0.031, size = 170, normalized size = 1.3 \begin{align*}{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{A}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{17\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5+1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5-1/6/d/a^3*tan(1/2*d*x+1/2*c)^3*A-1/2/d/a^3
*C*tan(1/2*d*x+1/2*c)^3+1/4/d/a^3*A*tan(1/2*d*x+1/2*c)+17/4/d/a^3*C*tan(1/2*d*x+1/2*c)+2/d/a^3*C*tan(1/2*d*x+1
/2*c)/(tan(1/2*d*x+1/2*c)^2+1)-6/d/a^3*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [A]  time = 1.49045, size = 277, normalized size = 2.04 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.64396, size = 386, normalized size = 2.84 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 16.1421, size = 422, normalized size = 3.1 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.29864, size = 204, normalized size = 1.5 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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