3.809 \(\int \frac{\cos (c+d x) (B \cos (c+d x)+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=211 \[ \frac{\left (a^2 b^3 B+5 a^3 b^2 C-2 a^5 C-6 a b^4 C+2 b^5 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a^2 (b B-a C) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{a \left (a^2 b B-3 a^3 C+6 a b^2 C-4 b^3 B\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{C x}{b^3} \]

[Out]

(C*x)/b^3 + ((a^2*b^3*B + 2*b^5*B - 2*a^5*C + 5*a^3*b^2*C - 6*a*b^4*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/S
qrt[a + b]])/((a - b)^(5/2)*b^3*(a + b)^(5/2)*d) - (a^2*(b*B - a*C)*Sin[c + d*x])/(2*b^2*(a^2 - b^2)*d*(a + b*
Cos[c + d*x])^2) + (a*(a^2*b*B - 4*b^3*B - 3*a^3*C + 6*a*b^2*C)*Sin[c + d*x])/(2*b^2*(a^2 - b^2)^2*d*(a + b*Co
s[c + d*x]))

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Rubi [A]  time = 0.615256, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3029, 2988, 3021, 2735, 2659, 205} \[ \frac{\left (a^2 b^3 B+5 a^3 b^2 C-2 a^5 C-6 a b^4 C+2 b^5 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a^2 (b B-a C) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{a \left (a^2 b B-3 a^3 C+6 a b^2 C-4 b^3 B\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{C x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*x)/b^3 + ((a^2*b^3*B + 2*b^5*B - 2*a^5*C + 5*a^3*b^2*C - 6*a*b^4*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/S
qrt[a + b]])/((a - b)^(5/2)*b^3*(a + b)^(5/2)*d) - (a^2*(b*B - a*C)*Sin[c + d*x])/(2*b^2*(a^2 - b^2)*d*(a + b*
Cos[c + d*x])^2) + (a*(a^2*b*B - 4*b^3*B - 3*a^3*C + 6*a*b^2*C)*Sin[c + d*x])/(2*b^2*(a^2 - b^2)^2*d*(a + b*Co
s[c + d*x]))

Rule 3029

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

Rule 2988

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

Rule 3021

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^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

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 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx &=\int \frac{\cos ^2(c+d x) (B+C \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx\\ &=-\frac{a^2 (b B-a C) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\int \frac{2 a b (b B-a C)+\left (a^2-2 b^2\right ) (b B-a C) \cos (c+d x)+2 b \left (a^2-b^2\right ) C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=-\frac{a^2 (b B-a C) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{a \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 C\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\int \frac{-b^2 \left (a^2 b B+2 b^3 B+a^3 C-4 a b^2 C\right )-2 b \left (a^2-b^2\right )^2 C \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{C x}{b^3}-\frac{a^2 (b B-a C) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{a \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 C\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (a^2 b^3 B+2 b^5 B-2 a^5 C+5 a^3 b^2 C-6 a b^4 C\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{C x}{b^3}-\frac{a^2 (b B-a C) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{a \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 C\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (a^2 b^3 B+2 b^5 B-2 a^5 C+5 a^3 b^2 C-6 a b^4 C\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^3 \left (a^2-b^2\right )^2 d}\\ &=\frac{C x}{b^3}+\frac{\left (a^2 b^3 B+2 b^5 B-2 a^5 C+5 a^3 b^2 C-6 a b^4 C\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} b^3 (a+b)^{5/2} d}-\frac{a^2 (b B-a C) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{a \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 C\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.38075, size = 204, normalized size = 0.97 \[ \frac{\frac{a b \left (a^2 b B-3 a^3 C+6 a b^2 C-4 b^3 B\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}+\frac{2 \left (-a^2 b^3 B-5 a^3 b^2 C+2 a^5 C+6 a b^4 C-2 b^5 B\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac{a^2 b (a C-b B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+2 C (c+d x)}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*(c + d*x) + (2*(-(a^2*b^3*B) - 2*b^5*B + 2*a^5*C - 5*a^3*b^2*C + 6*a*b^4*C)*ArcTanh[((a - b)*Tan[(c + d*x
)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + (a^2*b*(-(b*B) + a*C)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[
c + d*x])^2) + (a*b*(a^2*b*B - 4*b^3*B - 3*a^3*C + 6*a*b^2*C)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c
+ d*x])))/(2*b^3*d)

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Maple [B]  time = 0.042, size = 1023, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/b^3*arctan(tan(1/2*d*x+1/2*c))*C-1/d*a^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+
2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B-4/d*b/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*a/(a-b)/(a^2+2*a
*b+b^2)*tan(1/2*d*x+1/2*c)^3*B-2/d*a^4/b^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*
a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C+1/d/b/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*a^3/(a-b)/(a^2+2*a
*b+b^2)*tan(1/2*d*x+1/2*c)^3*C+6/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)
*tan(1/2*d*x+1/2*c)^3*a^2*C+1/d*a^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a-b)^2*tan(1/
2*d*x+1/2*c)*B-4/d*b/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*a/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*
B-2/d*a^4/b^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*C-1/d/b/(
a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*a^3*C+6/d/(a*tan(1/2*d*x
+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*a^2*C+1/d*a^2/(a^4-2*a^2*b^2+b^4)/((a
+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B+2/d*b^2/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-
b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B-2/d*a^5/b^3/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))
^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C+5/d*a^3/b/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2
)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-6/d*b*a/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arcta
n((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.0715, size = 2462, normalized size = 11.67 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*(C*a^6*b^2 - 3*C*a^4*b^4 + 3*C*a^2*b^6 - C*b^8)*d*x*cos(d*x + c)^2 + 8*(C*a^7*b - 3*C*a^5*b^3 + 3*C*a^
3*b^5 - C*a*b^7)*d*x*cos(d*x + c) + 4*(C*a^8 - 3*C*a^6*b^2 + 3*C*a^4*b^4 - C*a^2*b^6)*d*x + (2*C*a^7 - 5*C*a^5
*b^2 - B*a^4*b^3 + 6*C*a^3*b^4 - 2*B*a^2*b^5 + (2*C*a^5*b^2 - 5*C*a^3*b^4 - B*a^2*b^5 + 6*C*a*b^6 - 2*B*b^7)*c
os(d*x + c)^2 + 2*(2*C*a^6*b - 5*C*a^4*b^3 - B*a^3*b^4 + 6*C*a^2*b^5 - 2*B*a*b^6)*cos(d*x + c))*sqrt(-a^2 + b^
2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x +
c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - 2*(2*C*a^7*b - 7*C*a^5*b^3 + 3*B*a^4*b^4
+ 5*C*a^3*b^5 - 3*B*a^2*b^6 + (3*C*a^6*b^2 - B*a^5*b^3 - 9*C*a^4*b^4 + 5*B*a^3*b^5 + 6*C*a^2*b^6 - 4*B*a*b^7)*
cos(d*x + c))*sin(d*x + c))/((a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d*cos(d*x + c)^2 + 2*(a^7*b^4 - 3*a^5*b^
6 + 3*a^3*b^8 - a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 3*a^6*b^5 + 3*a^4*b^7 - a^2*b^9)*d), 1/2*(2*(C*a^6*b^2 - 3
*C*a^4*b^4 + 3*C*a^2*b^6 - C*b^8)*d*x*cos(d*x + c)^2 + 4*(C*a^7*b - 3*C*a^5*b^3 + 3*C*a^3*b^5 - C*a*b^7)*d*x*c
os(d*x + c) + 2*(C*a^8 - 3*C*a^6*b^2 + 3*C*a^4*b^4 - C*a^2*b^6)*d*x - (2*C*a^7 - 5*C*a^5*b^2 - B*a^4*b^3 + 6*C
*a^3*b^4 - 2*B*a^2*b^5 + (2*C*a^5*b^2 - 5*C*a^3*b^4 - B*a^2*b^5 + 6*C*a*b^6 - 2*B*b^7)*cos(d*x + c)^2 + 2*(2*C
*a^6*b - 5*C*a^4*b^3 - B*a^3*b^4 + 6*C*a^2*b^5 - 2*B*a*b^6)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x +
c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (2*C*a^7*b - 7*C*a^5*b^3 + 3*B*a^4*b^4 + 5*C*a^3*b^5 - 3*B*a^2*b^6
+ (3*C*a^6*b^2 - B*a^5*b^3 - 9*C*a^4*b^4 + 5*B*a^3*b^5 + 6*C*a^2*b^6 - 4*B*a*b^7)*cos(d*x + c))*sin(d*x + c))/
((a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d*cos(d*x + c)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 3*a^3*b^8 - a*b^10)*d*co
s(d*x + c) + (a^8*b^3 - 3*a^6*b^5 + 3*a^4*b^7 - a^2*b^9)*d)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.32382, size = 614, normalized size = 2.91 \begin{align*} -\frac{\frac{{\left (2 \, C a^{5} - 5 \, C a^{3} b^{2} - B a^{2} b^{3} + 6 \, C a b^{4} - 2 \, B b^{5}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sqrt{a^{2} - b^{2}}} - \frac{{\left (d x + c\right )} C}{b^{3}} + \frac{2 \, C a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, C a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + B a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, C a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, B a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, C a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, B a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, C a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, C a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, C a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, B a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b\right )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((2*C*a^5 - 5*C*a^3*b^2 - B*a^2*b^3 + 6*C*a*b^4 - 2*B*b^5)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) +
arctan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^4*b^3 - 2*a^2*b^5 + b^7)*sqrt(
a^2 - b^2)) - (d*x + c)*C/b^3 + (2*C*a^5*tan(1/2*d*x + 1/2*c)^3 - 3*C*a^4*b*tan(1/2*d*x + 1/2*c)^3 + B*a^3*b^2
*tan(1/2*d*x + 1/2*c)^3 - 5*C*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^2*b^
3*tan(1/2*d*x + 1/2*c)^3 - 4*B*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 2*C*a^5*tan(1/2*d*x + 1/2*c) + 3*C*a^4*b*tan(1/2
*d*x + 1/2*c) - B*a^3*b^2*tan(1/2*d*x + 1/2*c) - 5*C*a^3*b^2*tan(1/2*d*x + 1/2*c) + 3*B*a^2*b^3*tan(1/2*d*x +
1/2*c) - 6*C*a^2*b^3*tan(1/2*d*x + 1/2*c) + 4*B*a*b^4*tan(1/2*d*x + 1/2*c))/((a^4*b^2 - 2*a^2*b^4 + b^6)*(a*ta
n(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2))/d