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

Optimal. Leaf size=303 \[ \frac{\sin (c+d x) \left (2 a^2 b B-3 a^3 C-a b^2 (A-2 C)-b^3 B\right )}{b^3 d \left (a^2-b^2\right )}-\frac{2 a \left (a^2 A b^2-4 a^2 b^2 C-2 a^3 b B+3 a^4 C+3 a b^3 B-2 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\sin (c+d x) \cos ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\sin (c+d x) \cos (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right )}{2 b^2 d \left (a^2-b^2\right )}+\frac{x \left (6 a^2 C-4 a b B+2 A b^2+b^2 C\right )}{2 b^4} \]

[Out]

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

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Rubi [A]  time = 1.11738, antiderivative size = 303, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {3047, 3049, 3023, 2735, 2659, 205} \[ \frac{\sin (c+d x) \left (2 a^2 b B-3 a^3 C-a b^2 (A-2 C)-b^3 B\right )}{b^3 d \left (a^2-b^2\right )}-\frac{2 a \left (a^2 A b^2-4 a^2 b^2 C-2 a^3 b B+3 a^4 C+3 a b^3 B-2 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\sin (c+d x) \cos ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\sin (c+d x) \cos (c+d x) \left (3 a^2 C-2 a b B+2 A b^2-b^2 C\right )}{2 b^2 d \left (a^2-b^2\right )}+\frac{x \left (6 a^2 C-4 a b B+2 A b^2+b^2 C\right )}{2 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3047

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

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

Mathematica [A]  time = 1.3755, size = 208, normalized size = 0.69 \[ \frac{2 (c+d x) \left (6 a^2 C-4 a b B+2 A b^2+b^2 C\right )-\frac{4 a^2 b \sin (c+d x) \left (a (a C-b B)+A b^2\right )}{(a-b) (a+b) (a+b \cos (c+d x))}-\frac{8 a \left (a^2 b^2 (A-4 C)-2 a^3 b B+3 a^4 C+3 a b^3 B-2 A b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+4 b (b B-2 a C) \sin (c+d x)+b^2 C \sin (2 (c+d x))}{4 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(2*A*b^2 - 4*a*b*B + 6*a^2*C + b^2*C)*(c + d*x) - (8*a*(-2*A*b^4 - 2*a^3*b*B + 3*a*b^3*B + a^2*b^2*(A - 4*C
) + 3*a^4*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + 4*b*(b*B - 2*a*C)*Sin[
c + d*x] - (4*a^2*b*(A*b^2 + a*(-(b*B) + a*C))*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])) + b^2*C*Si
n[2*(c + d*x)])/(4*b^4*d)

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Maple [B]  time = 0.043, size = 845, normalized size = 2.8 \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*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x)

[Out]

2/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c)^3*B-4/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c
)^3*a*C-1/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c)^3*C+2/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d
*x+1/2*c)*B-4/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c)*a*C+1/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1
/2*d*x+1/2*c)*C+2/d/b^2*arctan(tan(1/2*d*x+1/2*c))*A-4/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a*B+6/d/b^4*arctan(tan
(1/2*d*x+1/2*c))*a^2*C+1/d/b^2*arctan(tan(1/2*d*x+1/2*c))*C-2/d*a^2/b/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*
d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)*A+2/d*a^3/b^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-ta
n(1/2*d*x+1/2*c)^2*b+a+b)*B-2/d*a^4/b^3/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c
)^2*b+a+b)*C-2/d*a^3/b^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*
A+4/d*a/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+4/d*a^4/b^3/(a+
b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B-6/d*a^2/b/(a+b)/(a-b)/((a+
b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B-6/d*a^5/b^4/(a+b)/(a-b)/((a+b)*(a-b))^(
1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C+8/d*a^3/b^2/(a+b)/(a-b)/((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*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.54595, size = 2381, normalized size = 7.86 \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)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.24799, size = 508, normalized size = 1.68 \begin{align*} \frac{\frac{4 \,{\left (3 \, C a^{5} - 2 \, B a^{4} b + A a^{3} b^{2} - 4 \, C a^{3} b^{2} + 3 \, B a^{2} b^{3} - 2 \, A a b^{4}\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^{2} b^{4} - b^{6}\right )} \sqrt{a^{2} - b^{2}}} - \frac{4 \,{\left (C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\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 )}} + \frac{{\left (6 \, C a^{2} - 4 \, B a b + 2 \, A b^{2} + C b^{2}\right )}{\left (d x + c\right )}}{b^{4}} - \frac{2 \,{\left (4 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C b \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 )}^{2} b^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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