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

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

[Out]

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

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Rubi [A]  time = 0.0449551, 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 = {3258, 618, 206} \[ -\frac{2 \tanh ^{-1}\left (\frac{b+2 c \sin (x)}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3258

Int[cos[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*sin[(d_.) + (e_.)*(x_)])^(n_.) + (c_.)*((f_.)*sin[(d_.
) + (e_.)*(x_)])^(n2_.))^(p_.), x_Symbol] :> Module[{g = FreeFactors[Sin[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, Sin[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{\cos (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,\sin (x)\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c \sin (x)\right )\right )\\ &=-\frac{2 \tanh ^{-1}\left (\frac{b+2 c \sin (x)}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c}}\\ \end{align*}

Mathematica [A]  time = 0.0143901, size = 35, normalized size = 1. \[ -\frac{2 \tanh ^{-1}\left (\frac{b+2 c \sin (x)}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.121, size = 36, normalized size = 1. \begin{align*} 2\,{\frac{1}{\sqrt{4\,ca-{b}^{2}}}\arctan \left ({\frac{b+2\,c\sin \left ( x \right ) }{\sqrt{4\,ca-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(4*a*c-b^2)^(1/2)*arctan((b+2*c*sin(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(cos(x)/(a+b*sin(x)+c*sin(x)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.13063, size = 335, normalized size = 9.57 \begin{align*} \left [\frac{\log \left (-\frac{2 \, c^{2} \cos \left (x\right )^{2} - 2 \, b c \sin \left (x\right ) - b^{2} + 2 \, a c - 2 \, c^{2} + \sqrt{b^{2} - 4 \, a c}{\left (2 \, c \sin \left (x\right ) + b\right )}}{c \cos \left (x\right )^{2} - b \sin \left (x\right ) - a - c}\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 \sin \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(cos(x)/(a+b*sin(x)+c*sin(x)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.10367, size = 99, normalized size = 2.83 \begin{align*} \begin{cases} \frac{\log{\left (\frac{a}{b} + \sin{\left (x \right )} \right )}}{b} & \text{for}\: c = 0 \\- \frac{2}{b + 2 c \sin{\left (x \right )}} & \text{for}\: a = \frac{b^{2}}{4 c} \\\frac{\log{\left (\frac{b}{2 c} + \sin{\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} + \sin{\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(cos(x)/(a+b*sin(x)+c*sin(x)**2),x)

[Out]

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

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Giac [A]  time = 1.16669, size = 47, normalized size = 1.34 \begin{align*} \frac{2 \, \arctan \left (\frac{2 \, c \sin \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(cos(x)/(a+b*sin(x)+c*sin(x)^2),x, algorithm="giac")

[Out]

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