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

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Rubi [A]  time = 0.11, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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 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]

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]

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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 1.49, size = 181, normalized size = 4.21 \[ \left [-\frac {\sqrt {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \relax (x)^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \relax (x)^{2} - 4 \, {\left ({\left (a^{2} + a b\right )} \cos \relax (x)^{3} - a b \cos \relax (x)\right )} \sqrt {-\frac {b}{a}} \sin \relax (x) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \relax (x)^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \relax (x)^{2} + b^{2}}\right ) - 4 \, x}{4 \, {\left (a - b\right )}}, \frac {\sqrt {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \relax (x)^{2} - b\right )} \sqrt {\frac {b}{a}}}{2 \, b \cos \relax (x) \sin \relax (x)}\right ) + 2 \, x}{2 \, {\left (a - b\right )}}\right ] \]

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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giac [A]  time = 0.14, size = 48, normalized size = 1.12 \[ -\frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \relax (x)}{\sqrt {a b}}\right )\right )} b}{\sqrt {a b} {\left (a - b\right )}} + \frac {x}{a - b} \]

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]

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

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maple [A]  time = 0.10, size = 36, normalized size = 0.84 \[ -\frac {b \arctan \left (\frac {\tan \relax (x ) b}{\sqrt {a b}}\right )}{\left (a -b \right ) \sqrt {a b}}+\frac {x}{a -b} \]

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 [A]  time = 0.41, size = 35, normalized size = 0.81 \[ -\frac {b \arctan \left (\frac {b \tan \relax (x)}{\sqrt {a b}}\right )}{\sqrt {a b} {\left (a - b\right )}} + \frac {x}{a - b} \]

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]

-b*arctan(b*tan(x)/sqrt(a*b))/(sqrt(a*b)*(a - b)) + x/(a - b)

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mupad [B]  time = 2.67, size = 48, normalized size = 1.12 \[ \left \{\begin {array}{cl} \frac {2\,x+\sin \left (2\,x\right )}{4\,b} & \text {\ if\ \ }a=b\\ \frac {x-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\relax (x)}{\sqrt {a}}\right )}{\sqrt {a}}}{a-b} & \text {\ if\ \ }a\neq b \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(a == b, (2*x + sin(2*x))/(4*b), a ~= b, (x - (b^(1/2)*atan((b^(1/2)*tan(x))/a^(1/2)))/a^(1/2))/(a -
b))

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sympy [A]  time = 1.62, size = 267, normalized size = 6.21 \[ \begin {cases} \tilde {\infty } \left (- x - \frac {\cos {\relax (x )}}{\sin {\relax (x )}}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {x \sin ^{2}{\relax (x )}}{2 b \sin ^{2}{\relax (x )} + 2 b \cos ^{2}{\relax (x )}} + \frac {x \cos ^{2}{\relax (x )}}{2 b \sin ^{2}{\relax (x )} + 2 b \cos ^{2}{\relax (x )}} + \frac {\sin {\relax (x )} \cos {\relax (x )}}{2 b \sin ^{2}{\relax (x )} + 2 b \cos ^{2}{\relax (x )}} & \text {for}\: a = b \\\frac {x}{a} & \text {for}\: b = 0 \\\frac {- x - \frac {\cos {\relax (x )}}{\sin {\relax (x )}}}{b} & \text {for}\: a = 0 \\\frac {2 i a \sqrt {b} x \sqrt {\frac {1}{a}}}{2 i a^{2} \sqrt {b} \sqrt {\frac {1}{a}} - 2 i a b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} + \frac {b \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} \sin {\relax (x )} + \cos {\relax (x )} \right )}}{2 i a^{2} \sqrt {b} \sqrt {\frac {1}{a}} - 2 i a b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} - \frac {b \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} \sin {\relax (x )} + \cos {\relax (x )} \right )}}{2 i a^{2} \sqrt {b} \sqrt {\frac {1}{a}} - 2 i a b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} & \text {otherwise} \end {cases} \]

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*sin(x)**2/(2*b*sin(x)**2 + 2*b*cos(x)**2) + x*co
s(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)), ((-x - cos(x)/sin(x))/b, Eq(a, 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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