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

Optimal. Leaf size=87 \[ -\frac{2 a \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2}}+\frac{a^2 \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}+\frac{x}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.125101, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2790, 2735, 2660, 618, 204} \[ -\frac{2 a \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2}}+\frac{a^2 \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}+\frac{x}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a + b*Sin[x])^2,x]

[Out]

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

Rule 2790

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 - b^2)), x] - Di
st[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*(2*b*c*d - a*(c^2 + d^2)) + (a^2
*d^2 - 2*a*b*c*d*(m + 2) + b^2*(d^2*(m + 1) + c^2*(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]

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

Mathematica [A]  time = 0.214287, size = 83, normalized size = 0.95 \[ \frac{-\frac{2 a \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{a^2 b \cos (x)}{(a-b) (a+b) (a+b \sin (x))}+x}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(a + b*Sin[x])^2,x]

[Out]

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

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Maple [B]  time = 0.045, size = 170, normalized size = 2. \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) }{{b}^{2}}}+2\,{\frac{a\tan \left ( x/2 \right ) }{ \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}+2\,{\frac{{a}^{2}}{b \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}-2\,{\frac{{a}^{3}}{{b}^{2} \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+4\,{\frac{a}{ \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\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))^2,x)

[Out]

2/b^2*arctan(tan(1/2*x))+2*a/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)/(a^2-b^2)*tan(1/2*x)+2*a^2/b/(tan(1/2*x)^2*a+2*
tan(1/2*x)*b+a)/(a^2-b^2)-2*a^3/b^2/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))+4*a/(a^2-
b^2)^(3/2)*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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.85131, size = 896, normalized size = 10.3 \begin{align*} \left [\frac{2 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} x \sin \left (x\right ) -{\left (a^{4} - 2 \, a^{2} b^{2} +{\left (a^{3} b - 2 \, a b^{3}\right )} \sin \left (x\right )\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^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} x + 2 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (x\right )}{2 \,{\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6} +{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sin \left (x\right )\right )}}, \frac{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} x \sin \left (x\right ) +{\left (a^{4} - 2 \, a^{2} b^{2} +{\left (a^{3} b - 2 \, a b^{3}\right )} \sin \left (x\right )\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^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} x +{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (x\right )}{a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6} +{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sin \left (x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(a^4*b - 2*a^2*b^3 + b^5)*x*sin(x) - (a^4 - 2*a^2*b^2 + (a^3*b - 2*a*b^3)*sin(x))*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*c
os(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + 2*(a^5 - 2*a^3*b^2 + a*b^4)*x + 2*(a^4*b - a^2*b^3)*cos(x))/(a^5*b^2 -
2*a^3*b^4 + a*b^6 + (a^4*b^3 - 2*a^2*b^5 + b^7)*sin(x)), ((a^4*b - 2*a^2*b^3 + b^5)*x*sin(x) + (a^4 - 2*a^2*b^
2 + (a^3*b - 2*a*b^3)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x))) + (a^5 - 2*a^3*
b^2 + a*b^4)*x + (a^4*b - a^2*b^3)*cos(x))/(a^5*b^2 - 2*a^3*b^4 + a*b^6 + (a^4*b^3 - 2*a^2*b^5 + b^7)*sin(x))]

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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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.93779, size = 167, normalized size = 1.92 \begin{align*} -\frac{2 \,{\left (a^{3} - 2 \, a b^{2}\right )}{\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 )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{2 \,{\left (a b \tan \left (\frac{1}{2} \, x\right ) + a^{2}\right )}}{{\left (a^{2} b - b^{3}\right )}{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\right )}} + \frac{x}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(a^3 - 2*a*b^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*b^2 -
b^4)*sqrt(a^2 - b^2)) + 2*(a*b*tan(1/2*x) + a^2)/((a^2*b - b^3)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)) + x/b^2