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

Optimal. Leaf size=59 \[ \frac{a x}{b^2}-\frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b^2}-\frac{\sin (x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.108881, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {2695, 2735, 2659, 205} \[ \frac{a x}{b^2}-\frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b^2}-\frac{\sin (x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a + b*Cos[x]),x]

[Out]

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

Rule 2695

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(b*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

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 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0788361, size = 54, normalized size = 0.92 \[ \frac{-2 \sqrt{b^2-a^2} \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )+a x-b \sin (x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(a + b*Cos[x]),x]

[Out]

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

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Maple [B]  time = 0.076, size = 108, normalized size = 1.8 \begin{align*} -2\,{\frac{\tan \left ( x/2 \right ) }{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{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( x/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }+2\,{\frac{1}{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( x/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(a+b*cos(x)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.76365, size = 378, normalized size = 6.41 \begin{align*} \left [\frac{2 \, a x - 2 \, b \sin \left (x\right ) + \sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (x\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right )}{2 \, b^{2}}, \frac{a x - b \sin \left (x\right ) - \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (x\right )}\right )}{b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*cos(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.15667, size = 122, normalized size = 2.07 \begin{align*} \frac{a x}{b^{2}} + \frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x\right ) - b \tan \left (\frac{1}{2} \, x\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )} \sqrt{a^{2} - b^{2}}}{b^{2}} - \frac{2 \, \tan \left (\frac{1}{2} \, x\right )}{{\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*cos(x)),x, algorithm="giac")

[Out]

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