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

Optimal. Leaf size=66 \[ -\frac{2 b \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac{a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))} \]

[Out]

(-2*b*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) - (a*Cos[x])/((a^2 - b^2)*(a + b*Sin[x]))

________________________________________________________________________________________

Rubi [A]  time = 0.0633352, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {2754, 12, 2660, 618, 204} \[ -\frac{2 b \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac{a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]/(a + b*Sin[x])^2,x]

[Out]

(-2*b*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) - (a*Cos[x])/((a^2 - b^2)*(a + b*Sin[x]))

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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))^2} \, dx &=-\frac{a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}+\frac{\int \frac{b}{a+b \sin (x)} \, dx}{-a^2+b^2}\\ &=-\frac{a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}-\frac{b \int \frac{1}{a+b \sin (x)} \, dx}{a^2-b^2}\\ &=-\frac{a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{a^2-b^2}\\ &=-\frac{a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}+\frac{(4 b) \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 )}{a^2-b^2}\\ &=-\frac{2 b \tan ^{-1}\left (\frac{b+a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac{a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}\\ \end{align*}

Mathematica [A]  time = 0.105275, size = 67, normalized size = 1.02 \[ -\frac{2 b \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac{a \cos (x)}{(a-b) (a+b) (a+b \sin (x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]/(a + b*Sin[x])^2,x]

[Out]

(-2*b*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) - (a*Cos[x])/((a - b)*(a + b)*(a + b*Sin[x])
)

________________________________________________________________________________________

Maple [A]  time = 0.039, size = 99, normalized size = 1.5 \begin{align*} 4\,{\frac{-2\,\tan \left ( x/2 \right ) b-2\,a}{ \left ( 4\,{a}^{2}-4\,{b}^{2} \right ) \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) }}-8\,{\frac{b}{ \left ( 4\,{a}^{2}-4\,{b}^{2} \right ) \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))^2,x)

[Out]

4*(-2*tan(1/2*x)*b-2*a)/(4*a^2-4*b^2)/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)-8*b/(4*a^2-4*b^2)/(a^2-b^2)^(1/2)*arct
an(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))

________________________________________________________________________________________

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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.72511, size = 612, normalized size = 9.27 \begin{align*} \left [\frac{{\left (b^{2} \sin \left (x\right ) + a b\right )} \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^{3} - a b^{2}\right )} \cos \left (x\right )}{2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )}}, \frac{{\left (b^{2} \sin \left (x\right ) + a b\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (x\right )}\right ) -{\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(a+b*sin(x))^2,x, algorithm="fricas")

[Out]

[1/2*((b^2*sin(x) + a*b)*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)) - 2*(a^3 - a*b^2)*cos(x))/(a^
5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*sin(x)), ((b^2*sin(x) + a*b)*sqrt(a^2 - b^2)*arctan(-(a*sin(
x) + b)/(sqrt(a^2 - b^2)*cos(x))) - (a^3 - a*b^2)*cos(x))/(a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)
*sin(x))]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(a+b*sin(x))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.84589, size = 122, normalized size = 1.85 \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 )} b}{{\left (a^{2} - b^{2}\right )}^{\frac{3}{2}}} - \frac{2 \,{\left (b \tan \left (\frac{1}{2} \, x\right ) + a\right )}}{{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\right )}{\left (a^{2} - b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(a+b*sin(x))^2,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)))*b/(a^2 - b^2)^(3/2) - 2*(b*t
an(1/2*x) + a)/((a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)*(a^2 - b^2))