3.179 \(\int \frac{\sin (x)}{a+b \sin (x)} \, dx\)

Optimal. Leaf size=50 \[ \frac{x}{b}-\frac{2 a \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2}} \]

[Out]

x/b - (2*a*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(b*Sqrt[a^2 - b^2])

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Rubi [A]  time = 0.0552835, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {2735, 2660, 618, 204} \[ \frac{x}{b}-\frac{2 a \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b \sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]/(a + b*Sin[x]),x]

[Out]

x/b - (2*a*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(b*Sqrt[a^2 - b^2])

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 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 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 204

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

Rubi steps

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

Mathematica [A]  time = 0.0398464, size = 47, normalized size = 0.94 \[ \frac{x-\frac{2 a \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]/(a + b*Sin[x]),x]

[Out]

(x - (2*a*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2])/b

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Maple [A]  time = 0.029, size = 54, normalized size = 1.1 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) }{b}}-2\,{\frac{a}{b\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(a+b*sin(x)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80871, size = 423, normalized size = 8.46 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}} a \log \left (-\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} - 2 \,{\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - 2 \,{\left (a^{2} - b^{2}\right )} x}{2 \,{\left (a^{2} b - b^{3}\right )}}, \frac{\sqrt{a^{2} - b^{2}} a \arctan \left (-\frac{a \sin \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (x\right )}\right ) +{\left (a^{2} - b^{2}\right )} x}{a^{2} b - b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(a+b*sin(x)),x, algorithm="fricas")

[Out]

[-1/2*(sqrt(-a^2 + b^2)*a*log(-((2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2 - 2*(a*cos(x)*sin(x) + b*cos
(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) - 2*(a^2 - b^2)*x)/(a^2*b - b^3), (sqrt(a^2
- b^2)*a*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x))) + (a^2 - b^2)*x)/(a^2*b - b^3)]

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Sympy [A]  time = 99.1242, size = 202, normalized size = 4.04 \begin{align*} \begin{cases} \tilde{\infty } x & \text{for}\: a = 0 \wedge b = 0 \\\frac{x}{b} & \text{for}\: a = 0 \\\frac{x \tan{\left (\frac{x}{2} \right )}}{b \tan{\left (\frac{x}{2} \right )} - b} - \frac{x}{b \tan{\left (\frac{x}{2} \right )} - b} + \frac{2 \tan{\left (\frac{x}{2} \right )}}{b \tan{\left (\frac{x}{2} \right )} - b} & \text{for}\: a = - b \\\frac{x \tan{\left (\frac{x}{2} \right )}}{b \tan{\left (\frac{x}{2} \right )} + b} + \frac{x}{b \tan{\left (\frac{x}{2} \right )} + b} - \frac{2 \tan{\left (\frac{x}{2} \right )}}{b \tan{\left (\frac{x}{2} \right )} + b} & \text{for}\: a = b \\- \frac{\cos{\left (x \right )}}{a} & \text{for}\: b = 0 \\\frac{a^{2} x}{a^{2} b - b^{3}} + \frac{a \sqrt{- a^{2} + b^{2}} \log{\left (\tan{\left (\frac{x}{2} \right )} + \frac{b}{a} - \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} - \frac{a \sqrt{- a^{2} + b^{2}} \log{\left (\tan{\left (\frac{x}{2} \right )} + \frac{b}{a} + \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} - \frac{b^{2} x}{a^{2} b - b^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(a+b*sin(x)),x)

[Out]

Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0)), (x/b, Eq(a, 0)), (x*tan(x/2)/(b*tan(x/2) - b) - x/(b*tan(x/2) - b) + 2
*tan(x/2)/(b*tan(x/2) - b), Eq(a, -b)), (x*tan(x/2)/(b*tan(x/2) + b) + x/(b*tan(x/2) + b) - 2*tan(x/2)/(b*tan(
x/2) + b), Eq(a, b)), (-cos(x)/a, Eq(b, 0)), (a**2*x/(a**2*b - b**3) + a*sqrt(-a**2 + b**2)*log(tan(x/2) + b/a
- sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) - a*sqrt(-a**2 + b**2)*log(tan(x/2) + b/a + sqrt(-a**2 + b**2)/a)/(a*
*2*b - b**3) - b**2*x/(a**2*b - b**3), True))

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Giac [A]  time = 1.85884, size = 78, normalized size = 1.56 \begin{align*} -\frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )} a}{\sqrt{a^{2} - b^{2}} b} + \frac{x}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(a+b*sin(x)),x, algorithm="giac")

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))*a/(sqrt(a^2 - b^2)*b) + x/b