### 3.3 $$\int \frac{\sin (x)}{a+b \cos (x)+c \cos ^2(x)} \, dx$$

Optimal. Leaf size=35 $\frac{2 \tanh ^{-1}\left (\frac{b+2 c \cos (x)}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c}}$

[Out]

(2*ArcTanh[(b + 2*c*Cos[x])/Sqrt[b^2 - 4*a*c]])/Sqrt[b^2 - 4*a*c]

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Rubi [A]  time = 0.0456803, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.176, Rules used = {3259, 618, 206} $\frac{2 \tanh ^{-1}\left (\frac{b+2 c \cos (x)}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*ArcTanh[(b + 2*c*Cos[x])/Sqrt[b^2 - 4*a*c]])/Sqrt[b^2 - 4*a*c]

Rule 3259

Int[((a_.) + (b_.)*(cos[(d_.) + (e_.)*(x_)]*(f_.))^(n_.) + (c_.)*(cos[(d_.) + (e_.)*(x_)]*(f_.))^(n2_.))^(p_.)
*sin[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol] :> Module[{g = FreeFactors[Cos[d + e*x], x]}, -Dist[g/e, Subst[Int[(
1 - g^2*x^2)^((m - 1)/2)*(a + b*(f*g*x)^n + c*(f*g*x)^(2*n))^p, x], x, Cos[d + e*x]/g], x]] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] && EqQ[n2, 2*n] && IntegerQ[(m - 1)/2]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sin (x)}{a+b \cos (x)+c \cos ^2(x)} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,\cos (x)\right )\\ &=2 \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c \cos (x)\right )\\ &=\frac{2 \tanh ^{-1}\left (\frac{b+2 c \cos (x)}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c}}\\ \end{align*}

Mathematica [A]  time = 0.0329729, size = 39, normalized size = 1.11 $-\frac{2 \tan ^{-1}\left (\frac{b+2 c \cos (x)}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*ArcTan[(b + 2*c*Cos[x])/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c]

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Maple [A]  time = 0.013, size = 36, normalized size = 1. \begin{align*} -2\,{\frac{1}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{b+2\,c\cos \left ( x \right ) }{\sqrt{4\,ac-{b}^{2}}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(a+b*cos(x)+c*cos(x)^2),x)

[Out]

-2/(4*a*c-b^2)^(1/2)*arctan((b+2*c*cos(x))/(4*a*c-b^2)^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.70433, size = 317, normalized size = 9.06 \begin{align*} \left [\frac{\log \left (-\frac{2 \, c^{2} \cos \left (x\right )^{2} + 2 \, b c \cos \left (x\right ) + b^{2} - 2 \, a c + \sqrt{b^{2} - 4 \, a c}{\left (2 \, c \cos \left (x\right ) + b\right )}}{c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a}\right )}{\sqrt{b^{2} - 4 \, a c}}, \frac{2 \, \sqrt{-b^{2} + 4 \, a c} \arctan \left (-\frac{\sqrt{-b^{2} + 4 \, a c}{\left (2 \, c \cos \left (x\right ) + b\right )}}{b^{2} - 4 \, a c}\right )}{b^{2} - 4 \, a c}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[log(-(2*c^2*cos(x)^2 + 2*b*c*cos(x) + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*cos(x) + b))/(c*cos(x)^2 + b*cos(x
) + a))/sqrt(b^2 - 4*a*c), 2*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*cos(x) + b)/(b^2 - 4*a*c))/(b^
2 - 4*a*c)]

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Sympy [A]  time = 6.01091, size = 99, normalized size = 2.83 \begin{align*} \begin{cases} - \frac{\log{\left (\frac{a}{b} + \cos{\left (x \right )} \right )}}{b} & \text{for}\: c = 0 \\\frac{2}{b + 2 c \cos{\left (x \right )}} & \text{for}\: a = \frac{b^{2}}{4 c} \\- \frac{\log{\left (\frac{b}{2 c} + \cos{\left (x \right )} - \frac{\sqrt{- 4 a c + b^{2}}}{2 c} \right )}}{\sqrt{- 4 a c + b^{2}}} + \frac{\log{\left (\frac{b}{2 c} + \cos{\left (x \right )} + \frac{\sqrt{- 4 a c + b^{2}}}{2 c} \right )}}{\sqrt{- 4 a c + b^{2}}} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(a+b*cos(x)+c*cos(x)**2),x)

[Out]

Piecewise((-log(a/b + cos(x))/b, Eq(c, 0)), (2/(b + 2*c*cos(x)), Eq(a, b**2/(4*c))), (-log(b/(2*c) + cos(x) -
sqrt(-4*a*c + b**2)/(2*c))/sqrt(-4*a*c + b**2) + log(b/(2*c) + cos(x) + sqrt(-4*a*c + b**2)/(2*c))/sqrt(-4*a*c
+ b**2), True))

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Giac [A]  time = 1.1887, size = 47, normalized size = 1.34 \begin{align*} -\frac{2 \, \arctan \left (\frac{2 \, c \cos \left (x\right ) + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\sqrt{-b^{2} + 4 \, a c}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*arctan((2*c*cos(x) + b)/sqrt(-b^2 + 4*a*c))/sqrt(-b^2 + 4*a*c)