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

Optimal. Leaf size=43 \[ \frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} \tan (x)}{\sqrt{a}}\right )}{\sqrt{b} (a-b)}-\frac{x}{a-b} \]

[Out]

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

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Rubi [A]  time = 0.149493, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {481, 203, 205} \[ \frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} \tan (x)}{\sqrt{a}}\right )}{\sqrt{b} (a-b)}-\frac{x}{a-b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 481

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> -Dist[(a*e^n)/(b*c -
a*d), Int[(e*x)^(m - n)/(a + b*x^n), x], x] + Dist[(c*e^n)/(b*c - a*d), Int[(e*x)^(m - n)/(c + d*x^n), x], x]
/; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 \cos ^2(x)+b \sin ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (x)\right )\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (x)\right )}{a-b}+\frac{a \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (x)\right )}{a-b}\\ &=-\frac{x}{a-b}+\frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} \tan (x)}{\sqrt{a}}\right )}{(a-b) \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0989602, size = 36, normalized size = 0.84 \[ \frac{x-\frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} \tan (x)}{\sqrt{a}}\right )}{\sqrt{b}}}{b-a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.051, size = 38, normalized size = 0.9 \begin{align*} -{\frac{\arctan \left ( \tan \left ( x \right ) \right ) }{a-b}}+{\frac{a}{a-b}\arctan \left ({b\tan \left ( x \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.99184, size = 433, normalized size = 10.07 \begin{align*} \left [-\frac{\sqrt{-\frac{a}{b}} \log \left (\frac{{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \,{\left (3 \, a b + b^{2}\right )} \cos \left (x\right )^{2} + 4 \,{\left ({\left (a b + b^{2}\right )} \cos \left (x\right )^{3} - b^{2} \cos \left (x\right )\right )} \sqrt{-\frac{a}{b}} \sin \left (x\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (x\right )^{4} + 2 \,{\left (a b - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}}\right ) + 4 \, x}{4 \,{\left (a - b\right )}}, -\frac{\sqrt{\frac{a}{b}} \arctan \left (\frac{{\left ({\left (a + b\right )} \cos \left (x\right )^{2} - b\right )} \sqrt{\frac{a}{b}}}{2 \, a \cos \left (x\right ) \sin \left (x\right )}\right ) + 2 \, x}{2 \,{\left (a - b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.53662, size = 241, normalized size = 5.6 \begin{align*} \begin{cases} \tilde{\infty } x & \text{for}\: a = 0 \wedge b = 0 \\\frac{x}{b} & \text{for}\: a = 0 \\\frac{x \sin ^{2}{\left (x \right )}}{2 b \sin ^{2}{\left (x \right )} + 2 b \cos ^{2}{\left (x \right )}} + \frac{x \cos ^{2}{\left (x \right )}}{2 b \sin ^{2}{\left (x \right )} + 2 b \cos ^{2}{\left (x \right )}} - \frac{\sin{\left (x \right )} \cos{\left (x \right )}}{2 b \sin ^{2}{\left (x \right )} + 2 b \cos ^{2}{\left (x \right )}} & \text{for}\: a = b \\\frac{- x + \frac{\sin{\left (x \right )}}{\cos{\left (x \right )}}}{a} & \text{for}\: b = 0 \\- \frac{2 i \sqrt{b} x \sqrt{\frac{1}{a}}}{2 i a \sqrt{b} \sqrt{\frac{1}{a}} - 2 i b^{\frac{3}{2}} \sqrt{\frac{1}{a}}} - \frac{\log{\left (- i \sqrt{b} \sqrt{\frac{1}{a}} \sin{\left (x \right )} + \cos{\left (x \right )} \right )}}{2 i a \sqrt{b} \sqrt{\frac{1}{a}} - 2 i b^{\frac{3}{2}} \sqrt{\frac{1}{a}}} + \frac{\log{\left (i \sqrt{b} \sqrt{\frac{1}{a}} \sin{\left (x \right )} + \cos{\left (x \right )} \right )}}{2 i a \sqrt{b} \sqrt{\frac{1}{a}} - 2 i b^{\frac{3}{2}} \sqrt{\frac{1}{a}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0)), (x/b, Eq(a, 0)), (x*sin(x)**2/(2*b*sin(x)**2 + 2*b*cos(x)**2) + x*cos(
x)**2/(2*b*sin(x)**2 + 2*b*cos(x)**2) - sin(x)*cos(x)/(2*b*sin(x)**2 + 2*b*cos(x)**2), Eq(a, b)), ((-x + sin(x
)/cos(x))/a, Eq(b, 0)), (-2*I*sqrt(b)*x*sqrt(1/a)/(2*I*a*sqrt(b)*sqrt(1/a) - 2*I*b**(3/2)*sqrt(1/a)) - log(-I*
sqrt(b)*sqrt(1/a)*sin(x) + cos(x))/(2*I*a*sqrt(b)*sqrt(1/a) - 2*I*b**(3/2)*sqrt(1/a)) + log(I*sqrt(b)*sqrt(1/a
)*sin(x) + cos(x))/(2*I*a*sqrt(b)*sqrt(1/a) - 2*I*b**(3/2)*sqrt(1/a)), True))

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Giac [B]  time = 1.21473, size = 151, normalized size = 3.51 \begin{align*} \frac{\pi \left \lfloor \frac{x}{\pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{2 \, \sqrt{\frac{1}{2}} \tan \left (x\right )}{\sqrt{\frac{a + b + \sqrt{{\left (a + b\right )}^{2} - 4 \, a b}}{b}}}\right )}{{\left | -a + b \right |}} - \frac{\sqrt{a b}{\left (\pi \left \lfloor \frac{x}{\pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{2 \, \sqrt{\frac{1}{2}} \tan \left (x\right )}{\sqrt{\frac{a + b - \sqrt{{\left (a + b\right )}^{2} - 4 \, a b}}{b}}}\right )\right )}{\left | b \right |}}{b^{2}{\left | -a + b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(pi*floor(x/pi + 1/2) + arctan(2*sqrt(1/2)*tan(x)/sqrt((a + b + sqrt((a + b)^2 - 4*a*b))/b)))/abs(-a + b) - sq
rt(a*b)*(pi*floor(x/pi + 1/2) + arctan(2*sqrt(1/2)*tan(x)/sqrt((a + b - sqrt((a + b)^2 - 4*a*b))/b)))*abs(b)/(
b^2*abs(-a + b))