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

Optimal. Leaf size=178 \[ -\frac{\left (-3 a^2 C+3 a b B-2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}-\frac{2 a^3 (b B-a C) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (2 a^2+b^2\right ) (b B-a C)}{2 b^4}+\frac{(b B-a C) \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{C \sin (c+d x) \cos ^2(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.569581, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {3029, 2990, 3049, 3023, 2735, 2659, 205} \[ -\frac{\left (-3 a^2 C+3 a b B-2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}-\frac{2 a^3 (b B-a C) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (2 a^2+b^2\right ) (b B-a C)}{2 b^4}+\frac{(b B-a C) \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{C \sin (c+d x) \cos ^2(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2990

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x
])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c -
b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m,
1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3049

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] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 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 ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=\int \frac{\cos ^3(c+d x) (B+C \cos (c+d x))}{a+b \cos (c+d x)} \, dx\\ &=\frac{C \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\int \frac{\cos (c+d x) \left (2 a C+2 b C \cos (c+d x)+3 (b B-a C) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{3 b}\\ &=\frac{(b B-a C) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{C \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\int \frac{3 a (b B-a C)+b (3 b B+a C) \cos (c+d x)-2 \left (3 a b B-3 a^2 C-2 b^2 C\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^2}\\ &=-\frac{\left (3 a b B-3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}+\frac{(b B-a C) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{C \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\int \frac{3 a b (b B-a C)+3 \left (2 a^2+b^2\right ) (b B-a C) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^3}\\ &=\frac{\left (2 a^2+b^2\right ) (b B-a C) x}{2 b^4}-\frac{\left (3 a b B-3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}+\frac{(b B-a C) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{C \cos ^2(c+d x) \sin (c+d x)}{3 b d}-\frac{\left (a^3 (b B-a C)\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^4}\\ &=\frac{\left (2 a^2+b^2\right ) (b B-a C) x}{2 b^4}-\frac{\left (3 a b B-3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}+\frac{(b B-a C) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{C \cos ^2(c+d x) \sin (c+d x)}{3 b d}-\frac{\left (2 a^3 (b B-a 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^4 d}\\ &=\frac{\left (2 a^2+b^2\right ) (b B-a C) x}{2 b^4}-\frac{2 a^3 (b B-a C) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} b^4 \sqrt{a+b} d}-\frac{\left (3 a b B-3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}+\frac{(b B-a C) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{C \cos ^2(c+d x) \sin (c+d x)}{3 b d}\\ \end{align*}

Mathematica [A]  time = 0.464254, size = 152, normalized size = 0.85 \[ \frac{6 \left (2 a^2+b^2\right ) (c+d x) (b B-a C)+3 b \left (4 a^2 C-4 a b B+3 b^2 C\right ) \sin (c+d x)-\frac{24 a^3 (a C-b B) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+3 b^2 (b B-a C) \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(2*a^2 + b^2)*(b*B - a*C)*(c + d*x) - (24*a^3*(-(b*B) + a*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 +
b^2]])/Sqrt[-a^2 + b^2] + 3*b*(-4*a*b*B + 4*a^2*C + 3*b^2*C)*Sin[c + d*x] + 3*b^2*(b*B - a*C)*Sin[2*(c + d*x)
] + b^3*C*Sin[3*(c + d*x)])/(12*b^4*d)

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Maple [B]  time = 0.037, size = 641, normalized size = 3.6 \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)^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64978, size = 1164, normalized size = 6.54 \begin{align*} \left [-\frac{3 \,{\left (2 \, C a^{5} - 2 \, B a^{4} b - C a^{3} b^{2} + B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} d x - 3 \,{\left (C a^{4} - B a^{3} b\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) -{\left (6 \, C a^{4} b - 6 \, B a^{3} b^{2} - 2 \, C a^{2} b^{3} + 6 \, B a b^{4} - 4 \, C b^{5} + 2 \,{\left (C a^{2} b^{3} - C b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (C a^{3} b^{2} - B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} b^{4} - b^{6}\right )} d}, -\frac{3 \,{\left (2 \, C a^{5} - 2 \, B a^{4} b - C a^{3} b^{2} + B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} d x - 6 \,{\left (C a^{4} - B a^{3} b\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) -{\left (6 \, C a^{4} b - 6 \, B a^{3} b^{2} - 2 \, C a^{2} b^{3} + 6 \, B a b^{4} - 4 \, C b^{5} + 2 \,{\left (C a^{2} b^{3} - C b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (C a^{3} b^{2} - B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} b^{4} - b^{6}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*(2*C*a^5 - 2*B*a^4*b - C*a^3*b^2 + B*a^2*b^3 - C*a*b^4 + B*b^5)*d*x - 3*(C*a^4 - B*a^3*b)*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)) - (6*C*a^4*b - 6*B*a^3*b^2 - 2*C*a^2*b^3
+ 6*B*a*b^4 - 4*C*b^5 + 2*(C*a^2*b^3 - C*b^5)*cos(d*x + c)^2 - 3*(C*a^3*b^2 - B*a^2*b^3 - C*a*b^4 + B*b^5)*co
s(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d), -1/6*(3*(2*C*a^5 - 2*B*a^4*b - C*a^3*b^2 + B*a^2*b^3 - C*a*b^4
+ B*b^5)*d*x - 6*(C*a^4 - B*a^3*b)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))
) - (6*C*a^4*b - 6*B*a^3*b^2 - 2*C*a^2*b^3 + 6*B*a*b^4 - 4*C*b^5 + 2*(C*a^2*b^3 - C*b^5)*cos(d*x + c)^2 - 3*(C
*a^3*b^2 - B*a^2*b^3 - C*a*b^4 + B*b^5)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*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)**2*(B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.8073, size = 486, normalized size = 2.73 \begin{align*} -\frac{\frac{3 \,{\left (2 \, C a^{3} - 2 \, B a^{2} b + C a b^{2} - B b^{3}\right )}{\left (d x + c\right )}}{b^{4}} + \frac{12 \,{\left (C a^{4} - B a^{3} b\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 )}}{\sqrt{a^{2} - b^{2}} b^{4}} - \frac{2 \,{\left (6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(2*C*a^3 - 2*B*a^2*b + C*a*b^2 - B*b^3)*(d*x + c)/b^4 + 12*(C*a^4 - B*a^3*b)*(pi*floor(1/2*(d*x + c)/p
i + 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)))/(sqrt(a
^2 - b^2)*b^4) - 2*(6*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 6*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 3*C*a*b*tan(1/2*d*x + 1/
2*c)^5 - 3*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*b^2*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 1
2*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 4*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^2*tan(1/2*d*x + 1/2*c) - 6*B*a*b*tan(1
/2*d*x + 1/2*c) - 3*C*a*b*tan(1/2*d*x + 1/2*c) + 3*B*b^2*tan(1/2*d*x + 1/2*c) + 6*C*b^2*tan(1/2*d*x + 1/2*c))/
((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^3))/d