3.122 \(\int \frac{1}{a+b \sin (x)} \, dx\)

Optimal. Leaf size=40 \[ \frac{2 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}} \]

[Out]

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

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Rubi [A]  time = 0.0399785, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {2660, 618, 204} \[ \frac{2 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[x])^(-1),x]

[Out]

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

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

Mathematica [A]  time = 0.0374633, size = 40, normalized size = 1. \[ \frac{2 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[x])^(-1),x]

[Out]

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

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Maple [A]  time = 0.014, size = 39, normalized size = 1. \begin{align*} 2\,{\frac{1}{\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(1/(a+b*sin(x)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.8461, size = 344, normalized size = 8.6 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}} \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 )}}, -\frac{\arctan \left (-\frac{a \sin \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (x\right )}\right )}{\sqrt{a^{2} - b^{2}}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 8.44262, size = 114, normalized size = 2.85 \begin{align*} \begin{cases} - \frac{2}{b - \sqrt{b^{2}} \tan{\left (\frac{x}{2} \right )}} & \text{for}\: a = - \sqrt{b^{2}} \\- \frac{2}{b + \sqrt{b^{2}} \tan{\left (\frac{x}{2} \right )}} & \text{for}\: a = \sqrt{b^{2}} \\\frac{\log{\left (\tan{\left (\frac{x}{2} \right )} \right )}}{b} & \text{for}\: a = 0 \\\frac{\log{\left (\tan{\left (\frac{x}{2} \right )} + \frac{b}{a} - \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{\sqrt{- a^{2} + b^{2}}} - \frac{\log{\left (\tan{\left (\frac{x}{2} \right )} + \frac{b}{a} + \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{\sqrt{- a^{2} + b^{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x)),x)

[Out]

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

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Giac [A]  time = 1.09591, size = 65, normalized size = 1.62 \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 )}}{\sqrt{a^{2} - b^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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)))/sqrt(a^2 - b^2)