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(1/2*x))/(a^2-b^2)^(1/2))/b/(a^2-b^2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.78, size = 192, normalized size = 3.84 \[ \left [-\frac {\sqrt {-a^{2} + b^{2}} a \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2} - 2 \, {\left (a \cos \relax (x) \sin \relax (x) + b \cos \relax (x)\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{2} - b^{2}\right )} x}{2 \, {\left (a^{2} b - b^{3}\right )}}, \frac {\sqrt {a^{2} - b^{2}} a \arctan \left (-\frac {a \sin \relax (x) + b}{\sqrt {a^{2} - b^{2}} \cos \relax (x)}\right ) + {\left (a^{2} - b^{2}\right )} x}{a^{2} b - b^{3}}\right ] \]

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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giac [A]  time = 0.32, size = 58, normalized size = 1.16 \[ -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a}{\sqrt {a^{2} - b^{2}} b} + \frac {x}{b} \]

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

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maple [A]  time = 0.06, size = 54, normalized size = 1.08 \[ \frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{b}-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b \sqrt {a^{2}-b^{2}}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 6.70, size = 101, normalized size = 2.02 \[ \frac {x}{b}-\frac {2\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {x}{2}\right )\,a^4-\cos \left (\frac {x}{2}\right )\,a^3\,b-3\,\sin \left (\frac {x}{2}\right )\,a^2\,b^2+\cos \left (\frac {x}{2}\right )\,a\,b^3+2\,\sin \left (\frac {x}{2}\right )\,b^4}{{\left (b^2-a^2\right )}^{3/2}\,\left (2\,b\,\sin \left (\frac {x}{2}\right )+a\,\cos \left (\frac {x}{2}\right )\right )}\right )}{b\,\sqrt {b^2-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(a + b*sin(x)),x)

[Out]

x/b - (2*a*atanh((a^4*sin(x/2) + 2*b^4*sin(x/2) - 3*a^2*b^2*sin(x/2) + a*b^3*cos(x/2) - a^3*b*cos(x/2))/((b^2
- a^2)^(3/2)*(2*b*sin(x/2) + a*cos(x/2)))))/(b*(b^2 - a^2)^(1/2))

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sympy [A]  time = 80.94, size = 236, normalized size = 4.72 \[ \begin {cases} \tilde {\infty } x & \text {for}\: a = 0 \wedge b = 0 \\\frac {b x \tan {\left (\frac {x}{2} \right )}}{b^{2} \tan {\left (\frac {x}{2} \right )} - b \sqrt {b^{2}}} + \frac {2 b}{b^{2} \tan {\left (\frac {x}{2} \right )} - b \sqrt {b^{2}}} - \frac {x \sqrt {b^{2}}}{b^{2} \tan {\left (\frac {x}{2} \right )} - b \sqrt {b^{2}}} & \text {for}\: a = - \sqrt {b^{2}} \\\frac {b x \tan {\left (\frac {x}{2} \right )}}{b^{2} \tan {\left (\frac {x}{2} \right )} + b \sqrt {b^{2}}} + \frac {2 b}{b^{2} \tan {\left (\frac {x}{2} \right )} + b \sqrt {b^{2}}} + \frac {x \sqrt {b^{2}}}{b^{2} \tan {\left (\frac {x}{2} \right )} + b \sqrt {b^{2}}} & \text {for}\: a = \sqrt {b^{2}} \\- \frac {\cos {\relax (x )}}{a} & \text {for}\: b = 0 \\\frac {x}{b} & \text {for}\: a = 0 \\- \frac {a \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{b \sqrt {- a^{2} + b^{2}}} + \frac {a \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{b \sqrt {- a^{2} + b^{2}}} + \frac {x}{b} & \text {otherwise} \end {cases} \]

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

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