3.702 \(\int \frac{1}{a+b \sin (e+f x)} \, dx\)

Optimal. Leaf size=47 \[ \frac{2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \sqrt{a^2-b^2}} \]

[Out]

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

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Rubi [A]  time = 0.0341297, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2660, 618, 204} \[ \frac{2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^(-1),x]

[Out]

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

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

Mathematica [A]  time = 0.031157, size = 47, normalized size = 1. \[ \frac{2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x])^(-1),x]

[Out]

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

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Maple [A]  time = 0.033, size = 47, normalized size = 1. \begin{align*} 2\,{\frac{1}{f\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/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(f*x+e)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.3701, size = 428, normalized size = 9.11 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right )}{2 \,{\left (a^{2} - b^{2}\right )} f}, -\frac{\arctan \left (-\frac{a \sin \left (f x + e\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (f x + e\right )}\right )}{\sqrt{a^{2} - b^{2}} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 7.72019, size = 162, normalized size = 3.45 \begin{align*} \begin{cases} \frac{\tilde{\infty } x}{\sin{\left (e \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac{2}{b f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - b f} & \text{for}\: a = - b \\- \frac{2}{b f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + b f} & \text{for}\: a = b \\\frac{x}{a + b \sin{\left (e \right )}} & \text{for}\: f = 0 \\\frac{\log{\left (\tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} \right )}}{b f} & \text{for}\: a = 0 \\- \frac{\sqrt{- a^{2} + b^{2}} \log{\left (\tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + \frac{b}{a} - \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{a^{2} f - b^{2} f} + \frac{\sqrt{- a^{2} + b^{2}} \log{\left (\tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + \frac{b}{a} + \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{a^{2} f - b^{2} f} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.27913, size = 84, normalized size = 1.79 \begin{align*} \frac{2 \,{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

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