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

Optimal. Leaf size=168 \[ -\frac{2 \left (3 a^2 b^2 C+a^3 b B-2 a^4 C-2 a b^3 B+A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{x (b B-2 a C)}{b^3}+\frac{C \sin (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.461326, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {3031, 3023, 2735, 2659, 205} \[ -\frac{2 \left (3 a^2 b^2 C+a^3 b B-2 a^4 C-2 a b^3 B+A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{x (b B-2 a C)}{b^3}+\frac{C \sin (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*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] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

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

Mathematica [A]  time = 0.959694, size = 159, normalized size = 0.95 \[ \frac{-\frac{2 \left (a \left (a^2 b B-2 a^3 C+3 a b^2 C-2 b^3 B\right )+A b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac{a b \sin (c+d x) \left (a (a C-b B)+A b^2\right )}{(a-b) (a+b) (a+b \cos (c+d x))}+(c+d x) (b B-2 a C)+b C \sin (c+d x)}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.042, size = 561, normalized size = 3.3 \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)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x)

[Out]

2/d*C/b^2*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2+1)+2/d/b^2*B*arctan(tan(1/2*d*x+1/2*c))-4/d/b^3*C*arctan(ta
n(1/2*d*x+1/2*c))*a+2/d*a/(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^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)*B+2/d/b^2*a^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*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))*A-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))*B+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))*B+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))*C-6/d/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)
)*a^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)*(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 [B]  time = 2.31104, size = 1782, normalized size = 10.61 \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)*(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*(2*(2*C*a^5*b - B*a^4*b^2 - 4*C*a^3*b^3 + 2*B*a^2*b^4 + 2*C*a*b^5 - B*b^6)*d*x*cos(d*x + c) + 2*(2*C*a^6
- B*a^5*b - 4*C*a^4*b^2 + 2*B*a^3*b^3 + 2*C*a^2*b^4 - B*a*b^5)*d*x + (2*C*a^5 - B*a^4*b - 3*C*a^3*b^2 + 2*B*a
^2*b^3 - A*a*b^4 + (2*C*a^4*b - B*a^3*b^2 - 3*C*a^2*b^3 + 2*B*a*b^4 - A*b^5)*cos(d*x + c))*sqrt(-a^2 + b^2)*lo
g((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^5*b - B*a^4*b^2 + (A - 3*C)*a^3*b^3 +
B*a^2*b^4 - (A - C)*a*b^5 + (C*a^4*b^2 - 2*C*a^2*b^4 + C*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4*b^4 - 2*a^2*b
^6 + b^8)*d*cos(d*x + c) + (a^5*b^3 - 2*a^3*b^5 + a*b^7)*d), -((2*C*a^5*b - B*a^4*b^2 - 4*C*a^3*b^3 + 2*B*a^2*
b^4 + 2*C*a*b^5 - B*b^6)*d*x*cos(d*x + c) + (2*C*a^6 - B*a^5*b - 4*C*a^4*b^2 + 2*B*a^3*b^3 + 2*C*a^2*b^4 - B*a
*b^5)*d*x - (2*C*a^5 - B*a^4*b - 3*C*a^3*b^2 + 2*B*a^2*b^3 - A*a*b^4 + (2*C*a^4*b - B*a^3*b^2 - 3*C*a^2*b^3 +
2*B*a*b^4 - 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)))
- (2*C*a^5*b - B*a^4*b^2 + (A - 3*C)*a^3*b^3 + B*a^2*b^4 - (A - C)*a*b^5 + (C*a^4*b^2 - 2*C*a^2*b^4 + C*b^6)*c
os(d*x + c))*sin(d*x + c))/((a^4*b^4 - 2*a^2*b^6 + b^8)*d*cos(d*x + c) + (a^5*b^3 - 2*a^3*b^5 + a*b^7)*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)*(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 [B]  time = 1.21048, size = 554, normalized size = 3.3 \begin{align*} -\frac{\frac{2 \,{\left (2 \, C a^{4} - B a^{3} b - 3 \, C a^{2} b^{2} + 2 \, B a b^{3} - 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^{3} - b^{5}\right )} \sqrt{a^{2} - b^{2}}} - \frac{2 \,{\left (2 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b\right )}{\left (a^{2} b^{2} - b^{4}\right )}} + \frac{{\left (2 \, C a - B b\right )}{\left (d x + c\right )}}{b^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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