3.1.37 \(\int \frac {\cos ^2(x)}{a+b \cos ^2(x)} \, dx\) [37]

Optimal. Leaf size=38 \[ \frac {x}{b}+\frac {\sqrt {a} \text {ArcTan}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{b \sqrt {a+b}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3250, 3260, 211} \begin {gather*} \frac {\sqrt {a} \text {ArcTan}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{b \sqrt {a+b}}+\frac {x}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x]^2/(a + b*Cos[x]^2),x]

[Out]

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

Rule 211

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

Rule 3250

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[B*(x
/b), x] + Dist[(A*b - a*B)/b, Int[1/(a + b*Sin[e + f*x]^2), x], x] /; FreeQ[{a, b, e, f, A, B}, x]

Rule 3260

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 36, normalized size = 0.95 \begin {gather*} \frac {x-\frac {\sqrt {a} \text {ArcTan}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 34, normalized size = 0.89

method	result	size
default	\(\frac {\arctan \left (\tan \left (x \right )\right )}{b}-\frac {a \arctan \left (\frac {a \tan \left (x \right )}{\sqrt {\left (a +b \right ) a}}\right )}{b \sqrt {\left (a +b \right ) a}}\)	\(34\)
risch	\(\frac {x}{b}+\frac {\sqrt {-\left (a +b \right ) a}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-\left (a +b \right ) a}+2 a +b}{b}\right )}{2 \left (a +b \right ) b}-\frac {\sqrt {-\left (a +b \right ) a}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-\left (a +b \right ) a}-2 a -b}{b}\right )}{2 \left (a +b \right ) b}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^2/(a+b*cos(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.48, size = 31, normalized size = 0.82 \begin {gather*} -\frac {a \arctan \left (\frac {a \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} b} + \frac {x}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.42, size = 183, normalized size = 4.82 \begin {gather*} \left [\frac {\sqrt {-\frac {a}{a + b}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \, {\left (4 \, a^{2} + 3 \, a b\right )} \cos \left (x\right )^{2} + 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (x\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (x\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (x\right ) + a^{2}}{b^{2} \cos \left (x\right )^{4} + 2 \, a b \cos \left (x\right )^{2} + a^{2}}\right ) + 4 \, x}{4 \, b}, \frac {\sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (x\right ) \sin \left (x\right )}\right ) + 2 \, x}{2 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.41, size = 48, normalized size = 1.26 \begin {gather*} -\frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )\right )} a}{\sqrt {a^{2} + a b} b} + \frac {x}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.55, size = 425, normalized size = 11.18 \begin {gather*} \frac {x}{b}-\frac {\mathrm {atan}\left (\frac {\frac {\left (2\,a^3\,\mathrm {tan}\left (x\right )-\frac {\left (2\,a^2\,b^2-\frac {\mathrm {tan}\left (x\right )\,\left (16\,a^3\,b^2+8\,a^2\,b^3\right )\,\sqrt {-a\,\left (a+b\right )}}{4\,\left (b^2+a\,b\right )}\right )\,\sqrt {-a\,\left (a+b\right )}}{2\,\left (b^2+a\,b\right )}\right )\,\sqrt {-a\,\left (a+b\right )}\,1{}\mathrm {i}}{b^2+a\,b}+\frac {\left (2\,a^3\,\mathrm {tan}\left (x\right )+\frac {\left (2\,a^2\,b^2+\frac {\mathrm {tan}\left (x\right )\,\left (16\,a^3\,b^2+8\,a^2\,b^3\right )\,\sqrt {-a\,\left (a+b\right )}}{4\,\left (b^2+a\,b\right )}\right )\,\sqrt {-a\,\left (a+b\right )}}{2\,\left (b^2+a\,b\right )}\right )\,\sqrt {-a\,\left (a+b\right )}\,1{}\mathrm {i}}{b^2+a\,b}}{\frac {\left (2\,a^3\,\mathrm {tan}\left (x\right )-\frac {\left (2\,a^2\,b^2-\frac {\mathrm {tan}\left (x\right )\,\left (16\,a^3\,b^2+8\,a^2\,b^3\right )\,\sqrt {-a\,\left (a+b\right )}}{4\,\left (b^2+a\,b\right )}\right )\,\sqrt {-a\,\left (a+b\right )}}{2\,\left (b^2+a\,b\right )}\right )\,\sqrt {-a\,\left (a+b\right )}}{b^2+a\,b}-\frac {\left (2\,a^3\,\mathrm {tan}\left (x\right )+\frac {\left (2\,a^2\,b^2+\frac {\mathrm {tan}\left (x\right )\,\left (16\,a^3\,b^2+8\,a^2\,b^3\right )\,\sqrt {-a\,\left (a+b\right )}}{4\,\left (b^2+a\,b\right )}\right )\,\sqrt {-a\,\left (a+b\right )}}{2\,\left (b^2+a\,b\right )}\right )\,\sqrt {-a\,\left (a+b\right )}}{b^2+a\,b}}\right )\,\sqrt {-a\,\left (a+b\right )}\,1{}\mathrm {i}}{b^2+a\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^2/(a + b*cos(x)^2),x)

[Out]

x/b - (atan((((2*a^3*tan(x) - ((2*a^2*b^2 - (tan(x)*(8*a^2*b^3 + 16*a^3*b^2)*(-a*(a + b))^(1/2))/(4*(a*b + b^2
)))*(-a*(a + b))^(1/2))/(2*(a*b + b^2)))*(-a*(a + b))^(1/2)*1i)/(a*b + b^2) + ((2*a^3*tan(x) + ((2*a^2*b^2 + (
tan(x)*(8*a^2*b^3 + 16*a^3*b^2)*(-a*(a + b))^(1/2))/(4*(a*b + b^2)))*(-a*(a + b))^(1/2))/(2*(a*b + b^2)))*(-a*
(a + b))^(1/2)*1i)/(a*b + b^2))/(((2*a^3*tan(x) - ((2*a^2*b^2 - (tan(x)*(8*a^2*b^3 + 16*a^3*b^2)*(-a*(a + b))^
(1/2))/(4*(a*b + b^2)))*(-a*(a + b))^(1/2))/(2*(a*b + b^2)))*(-a*(a + b))^(1/2))/(a*b + b^2) - ((2*a^3*tan(x)
+ ((2*a^2*b^2 + (tan(x)*(8*a^2*b^3 + 16*a^3*b^2)*(-a*(a + b))^(1/2))/(4*(a*b + b^2)))*(-a*(a + b))^(1/2))/(2*(
a*b + b^2)))*(-a*(a + b))^(1/2))/(a*b + b^2)))*(-a*(a + b))^(1/2)*1i)/(a*b + b^2)

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