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

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

[Out]

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

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Rubi [A]  time = 0.109112, 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 = {391, 203, 205} \[ \frac{x}{a-b}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (x)}{\sqrt{a}}\right )}{\sqrt{a} (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 391

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

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

Mathematica [A]  time = 0.0583285, size = 36, normalized size = 0.84 \[ \frac{x-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (x)}{\sqrt{a}}\right )}{\sqrt{a}}}{a-b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 36, normalized size = 0.8 \begin{align*} -{\frac{b}{a-b}\arctan \left ({b\tan \left ( x \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{x}{a-b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b/(a-b)/(a*b)^(1/2)*arctan(tan(x)*b/(a*b)^(1/2))+x/(a-b)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.94024, size = 432, normalized size = 10.05 \begin{align*} \left [-\frac{\sqrt{-\frac{b}{a}} \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^{2} + a b\right )} \cos \left (x\right )^{3} - a b \cos \left (x\right )\right )} \sqrt{-\frac{b}{a}} \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{b}{a}} \arctan \left (\frac{{\left ({\left (a + b\right )} \cos \left (x\right )^{2} - b\right )} \sqrt{\frac{b}{a}}}{2 \, b \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(cos(x)^2/(a*cos(x)^2+b*sin(x)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.5537, size = 267, normalized size = 6.21 \begin{align*} \begin{cases} \tilde{\infty } \left (- x - \frac{\cos{\left (x \right )}}{\sin{\left (x \right )}}\right ) & \text{for}\: a = 0 \wedge b = 0 \\\frac{- x - \frac{\cos{\left (x \right )}}{\sin{\left (x \right )}}}{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}{a} & \text{for}\: b = 0 \\\frac{2 i a \sqrt{b} x \sqrt{\frac{1}{a}}}{2 i a^{2} \sqrt{b} \sqrt{\frac{1}{a}} - 2 i a b^{\frac{3}{2}} \sqrt{\frac{1}{a}}} + \frac{b \log{\left (- i \sqrt{b} \sqrt{\frac{1}{a}} \sin{\left (x \right )} + \cos{\left (x \right )} \right )}}{2 i a^{2} \sqrt{b} \sqrt{\frac{1}{a}} - 2 i a b^{\frac{3}{2}} \sqrt{\frac{1}{a}}} - \frac{b \log{\left (i \sqrt{b} \sqrt{\frac{1}{a}} \sin{\left (x \right )} + \cos{\left (x \right )} \right )}}{2 i a^{2} \sqrt{b} \sqrt{\frac{1}{a}} - 2 i a 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(cos(x)**2/(a*cos(x)**2+b*sin(x)**2),x)

[Out]

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

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Giac [B]  time = 1.12483, size = 200, normalized size = 4.65 \begin{align*} -\frac{2 \, \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 |}}{{\left (a - b\right )}^{2} b -{\left (a b + b^{2}\right )}{\left | -a + b \right |}} - \frac{2 \,{\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 )} b}{{\left (a - b\right )}^{2} + a{\left | -a + b \right |} + b{\left | -a + b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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