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

Optimal. Leaf size=52 \[ \frac{\sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (x)}{\sqrt{a+b}}\right )}{2 b \sqrt{a-b}}-\frac{x}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.124984, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {1130, 205} \[ \frac{\sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (x)}{\sqrt{a+b}}\right )}{2 b \sqrt{a-b}}-\frac{x}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a + b*Cos[2*x]),x]

[Out]

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

Rule 1130

Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(
d^2*(b/q + 1))/2, Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(d^2*(b/q - 1))/2, Int[(d*x)^(m - 2)/(b
/2 - q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 2]

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

Mathematica [A]  time = 0.0895307, size = 48, normalized size = 0.92 \[ -\frac{\frac{(a+b) \tanh ^{-1}\left (\frac{(a-b) \tan (x)}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+x}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.045, size = 80, normalized size = 1.5 \begin{align*} -{\frac{\arctan \left ( \tan \left ( x \right ) \right ) }{2\,b}}+{\frac{a}{2\,b}\arctan \left ({\tan \left ( x \right ) \left ( a-b \right ){\frac{1}{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}}}+{\frac{1}{2}\arctan \left ({\tan \left ( x \right ) \left ( a-b \right ){\frac{1}{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.30778, size = 535, normalized size = 10.29 \begin{align*} \left [\frac{\sqrt{-\frac{a + b}{a - b}} \log \left (\frac{4 \,{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{4} - 4 \,{\left (2 \, a^{2} - a b - b^{2}\right )} \cos \left (x\right )^{2} - 4 \,{\left (2 \,{\left (a^{2} - a b\right )} \cos \left (x\right )^{3} -{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (x\right )\right )} \sqrt{-\frac{a + b}{a - b}} \sin \left (x\right ) + a^{2} - 2 \, a b + b^{2}}{4 \, b^{2} \cos \left (x\right )^{4} + 4 \,{\left (a b - b^{2}\right )} \cos \left (x\right )^{2} + a^{2} - 2 \, a b + b^{2}}\right ) - 4 \, x}{8 \, b}, -\frac{\sqrt{\frac{a + b}{a - b}} \arctan \left (\frac{{\left (2 \, a \cos \left (x\right )^{2} - a + b\right )} \sqrt{\frac{a + b}{a - b}}}{2 \,{\left (a + b\right )} \cos \left (x\right ) \sin \left (x\right )}\right ) + 2 \, x}{4 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(-(a + b)/(a - b))*log((4*(2*a^2 - b^2)*cos(x)^4 - 4*(2*a^2 - a*b - b^2)*cos(x)^2 - 4*(2*(a^2 - a*b)
*cos(x)^3 - (a^2 - 2*a*b + b^2)*cos(x))*sqrt(-(a + b)/(a - b))*sin(x) + a^2 - 2*a*b + b^2)/(4*b^2*cos(x)^4 + 4
*(a*b - b^2)*cos(x)^2 + a^2 - 2*a*b + b^2)) - 4*x)/b, -1/4*(sqrt((a + b)/(a - b))*arctan(1/2*(2*a*cos(x)^2 - a
+ b)*sqrt((a + b)/(a - b))/((a + b)*cos(x)*sin(x))) + 2*x)/b]

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Sympy [B]  time = 63.917, size = 432, normalized size = 8.31 \begin{align*} \frac{\begin{cases} \tilde{\infty } \left (- \frac{\log{\left (\tan{\left (x \right )} - 1 \right )}}{2} + \frac{\log{\left (\tan{\left (x \right )} + 1 \right )}}{2}\right ) & \text{for}\: a = 0 \wedge b = 0 \\\frac{\tan{\left (x \right )}}{2 b} & \text{for}\: a = b \\\frac{1}{2 b \tan{\left (x \right )}} & \text{for}\: a = - b \\\frac{\log{\left (- \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (x \right )} \right )}}{2 a \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - 2 b \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} - \frac{\log{\left (\sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (x \right )} \right )}}{2 a \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - 2 b \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} & \text{otherwise} \end{cases}}{2} - \frac{\begin{cases} \tilde{\infty } x & \text{for}\: a = 0 \wedge b = 0 \\\frac{x}{b} - \frac{\tan{\left (x \right )}}{2 b} & \text{for}\: a = b \\\frac{x}{b} + \frac{1}{2 b \tan{\left (x \right )}} & \text{for}\: a = - b \\\frac{\sin{\left (2 x \right )}}{2 a} & \text{for}\: b = 0 \\\frac{2 a x \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}}{2 a b \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - 2 b^{2} \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} - \frac{a \log{\left (- \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (x \right )} \right )}}{2 a b \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - 2 b^{2} \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} + \frac{a \log{\left (\sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} + \tan{\left (x \right )} \right )}}{2 a b \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - 2 b^{2} \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} - \frac{2 b x \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}}{2 a b \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}} - 2 b^{2} \sqrt{- \frac{a}{a - b} - \frac{b}{a - b}}} & \text{otherwise} \end{cases}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-log(tan(x) - 1)/2 + log(tan(x) + 1)/2), Eq(a, 0) & Eq(b, 0)), (tan(x)/(2*b), Eq(a, b)), (1/(2
*b*tan(x)), Eq(a, -b)), (log(-sqrt(-a/(a - b) - b/(a - b)) + tan(x))/(2*a*sqrt(-a/(a - b) - b/(a - b)) - 2*b*s
qrt(-a/(a - b) - b/(a - b))) - log(sqrt(-a/(a - b) - b/(a - b)) + tan(x))/(2*a*sqrt(-a/(a - b) - b/(a - b)) -
2*b*sqrt(-a/(a - b) - b/(a - b))), True))/2 - Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0)), (x/b - tan(x)/(2*b), Eq(
a, b)), (x/b + 1/(2*b*tan(x)), Eq(a, -b)), (sin(2*x)/(2*a), Eq(b, 0)), (2*a*x*sqrt(-a/(a - b) - b/(a - b))/(2*
a*b*sqrt(-a/(a - b) - b/(a - b)) - 2*b**2*sqrt(-a/(a - b) - b/(a - b))) - a*log(-sqrt(-a/(a - b) - b/(a - b))
+ tan(x))/(2*a*b*sqrt(-a/(a - b) - b/(a - b)) - 2*b**2*sqrt(-a/(a - b) - b/(a - b))) + a*log(sqrt(-a/(a - b) -
b/(a - b)) + tan(x))/(2*a*b*sqrt(-a/(a - b) - b/(a - b)) - 2*b**2*sqrt(-a/(a - b) - b/(a - b))) - 2*b*x*sqrt(
-a/(a - b) - b/(a - b))/(2*a*b*sqrt(-a/(a - b) - b/(a - b)) - 2*b**2*sqrt(-a/(a - b) - b/(a - b))), True))/2

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Giac [A]  time = 1.08859, size = 93, normalized size = 1.79 \begin{align*} -\frac{{\left (\pi \left \lfloor \frac{x}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (x\right ) - b \tan \left (x\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}{\left (a + b\right )}}{2 \, \sqrt{a^{2} - b^{2}} b} - \frac{x}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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