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

Optimal. Leaf size=61 \[ \frac{2 a^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 \sqrt{a^2-b^2}}-\frac{a x}{b^2}-\frac{\cos (x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.103534, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2746, 12, 2735, 2660, 618, 204} \[ \frac{2 a^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 \sqrt{a^2-b^2}}-\frac{a x}{b^2}-\frac{\cos (x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a + b*Sin[x]),x]

[Out]

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

Rule 2746

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

Rule 12

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

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

Mathematica [A]  time = 0.0890538, size = 56, normalized size = 0.92 \[ -\frac{-\frac{2 a^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+a x+b \cos (x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(a + b*Sin[x]),x]

[Out]

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

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Maple [A]  time = 0.033, size = 72, normalized size = 1.2 \begin{align*} -2\,{\frac{1}{b \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}-2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) a}{{b}^{2}}}+2\,{\frac{{a}^{2}}{{b}^{2}\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)^2/(a+b*sin(x)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80041, size = 504, normalized size = 8.26 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}} a^{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 )} x + 2 \,{\left (a^{2} b - b^{3}\right )} \cos \left (x\right )}{2 \,{\left (a^{2} b^{2} - b^{4}\right )}}, -\frac{\sqrt{a^{2} - b^{2}} a^{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 )} x +{\left (a^{2} b - b^{3}\right )} \cos \left (x\right )}{a^{2} b^{2} - b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.87935, size = 104, normalized size = 1.7 \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^{2}}{\sqrt{a^{2} - b^{2}} b^{2}} - \frac{a x}{b^{2}} - \frac{2}{{\left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(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^2/(sqrt(a^2 - b^2)*b^2) - a
*x/b^2 - 2/((tan(1/2*x)^2 + 1)*b)