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

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

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 52, normalized size of antiderivative = 1.00, 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 = {1093, 205} \[ \frac {x}{2 b}-\frac {\sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (x)}{\sqrt {a+b}}\right )}{2 b \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[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*b*Sqrt[a + b])

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 1093

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 50, normalized size = 0.96 \[ \frac {\frac {(a-b) \tanh ^{-1}\left (\frac {(a-b) \tan (x)}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+x}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[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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fricas [A]  time = 1.11, size = 224, normalized size = 4.31 \[ \left [\frac {\sqrt {-\frac {a - b}{a + b}} \log \left (\frac {4 \, {\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{4} - 4 \, {\left (2 \, a^{2} - a b - b^{2}\right )} \cos \relax (x)^{2} + 4 \, {\left (2 \, {\left (a^{2} + a b\right )} \cos \relax (x)^{3} - {\left (a^{2} - b^{2}\right )} \cos \relax (x)\right )} \sqrt {-\frac {a - b}{a + b}} \sin \relax (x) + a^{2} - 2 \, a b + b^{2}}{4 \, b^{2} \cos \relax (x)^{4} + 4 \, {\left (a b - b^{2}\right )} \cos \relax (x)^{2} + a^{2} - 2 \, a b + b^{2}}\right ) + 4 \, x}{8 \, b}, -\frac {\sqrt {\frac {a - b}{a + b}} \arctan \left (-\frac {{\left (2 \, a \cos \relax (x)^{2} - a + b\right )} \sqrt {\frac {a - b}{a + b}}}{2 \, {\left (a - b\right )} \cos \relax (x) \sin \relax (x)}\right ) - 2 \, x}{4 \, b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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 - 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)*s
qrt((a - b)/(a + b))/((a - b)*cos(x)*sin(x))) - 2*x)/b]

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giac [B]  time = 0.15, size = 159, normalized size = 3.06 \[ -\frac {\sqrt {a^{2} - b^{2}} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \relax (x)}{\sqrt {\frac {4 \, a + \sqrt {-16 \, {\left (a + b\right )} {\left (a - b\right )} + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{2 \, {\left ({\left (a - b\right )} b^{2} + {\left (a^{2} - a b\right )} {\left | b \right |}\right )}} - \frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \relax (x)}{\sqrt {\frac {4 \, a - \sqrt {-16 \, {\left (a + b\right )} {\left (a - b\right )} + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left (a - b\right )}}{2 \, {\left (b^{2} - a {\left | b \right |}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(a^2 - b^2)*(pi*floor(x/pi + 1/2) + arctan(2*tan(x)/sqrt((4*a + sqrt(-16*(a + b)*(a - b) + 16*a^2))/(
a - b))))*abs(a - b)/((a - b)*b^2 + (a^2 - a*b)*abs(b)) - 1/2*(pi*floor(x/pi + 1/2) + arctan(2*tan(x)/sqrt((4*
a - sqrt(-16*(a + b)*(a - b) + 16*a^2))/(a - b))))*(a - b)/(b^2 - a*abs(b))

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maple [A]  time = 0.13, size = 80, normalized size = 1.54 \[ -\frac {\arctan \left (\frac {\tan \relax (x ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) a}{2 b \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\arctan \left (\frac {\tan \relax (x ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\arctan \left (\tan \relax (x )\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 3.32, size = 684, normalized size = 13.15 \[ \frac {\mathrm {atan}\left (\frac {2\,a^2\,\mathrm {tan}\relax (x)}{2\,a^2-4\,a\,b+2\,b^2}+\frac {2\,b^2\,\mathrm {tan}\relax (x)}{2\,a^2-4\,a\,b+2\,b^2}-\frac {4\,a\,b\,\mathrm {tan}\relax (x)}{2\,a^2-4\,a\,b+2\,b^2}\right )}{2\,b}+\frac {\mathrm {atan}\left (\frac {\frac {\left (\frac {\mathrm {tan}\relax (x)\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )}{4}+\frac {\sqrt {b^2-a^2}\,\left (4\,b^4-8\,a\,b^3+4\,a^2\,b^2+\frac {\mathrm {tan}\relax (x)\,\sqrt {b^2-a^2}\,\left (64\,a^3\,b^2-128\,a^2\,b^3+64\,a\,b^4\right )}{16\,\left (b^2+a\,b\right )}\right )}{4\,\left (b^2+a\,b\right )}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}}{b^2+a\,b}+\frac {\left (\frac {\mathrm {tan}\relax (x)\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )}{4}+\frac {\sqrt {b^2-a^2}\,\left (8\,a\,b^3-4\,b^4-4\,a^2\,b^2+\frac {\mathrm {tan}\relax (x)\,\sqrt {b^2-a^2}\,\left (64\,a^3\,b^2-128\,a^2\,b^3+64\,a\,b^4\right )}{16\,\left (b^2+a\,b\right )}\right )}{4\,\left (b^2+a\,b\right )}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}}{b^2+a\,b}}{\frac {\left (\frac {\mathrm {tan}\relax (x)\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )}{4}+\frac {\sqrt {b^2-a^2}\,\left (4\,b^4-8\,a\,b^3+4\,a^2\,b^2+\frac {\mathrm {tan}\relax (x)\,\sqrt {b^2-a^2}\,\left (64\,a^3\,b^2-128\,a^2\,b^3+64\,a\,b^4\right )}{16\,\left (b^2+a\,b\right )}\right )}{4\,\left (b^2+a\,b\right )}\right )\,\sqrt {b^2-a^2}}{b^2+a\,b}-\frac {\left (\frac {\mathrm {tan}\relax (x)\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )}{4}+\frac {\sqrt {b^2-a^2}\,\left (8\,a\,b^3-4\,b^4-4\,a^2\,b^2+\frac {\mathrm {tan}\relax (x)\,\sqrt {b^2-a^2}\,\left (64\,a^3\,b^2-128\,a^2\,b^3+64\,a\,b^4\right )}{16\,\left (b^2+a\,b\right )}\right )}{4\,\left (b^2+a\,b\right )}\right )\,\sqrt {b^2-a^2}}{b^2+a\,b}}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}}{2\,\left (b^2+a\,b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^2/(a + b*cos(2*x)),x)

[Out]

atan((2*a^2*tan(x))/(2*a^2 - 4*a*b + 2*b^2) + (2*b^2*tan(x))/(2*a^2 - 4*a*b + 2*b^2) - (4*a*b*tan(x))/(2*a^2 -
4*a*b + 2*b^2))/(2*b) + (atan(((((tan(x)*(12*a*b^2 - 12*a^2*b + 4*a^3 - 4*b^3))/4 + ((b^2 - a^2)^(1/2)*(4*b^4
- 8*a*b^3 + 4*a^2*b^2 + (tan(x)*(b^2 - a^2)^(1/2)*(64*a*b^4 - 128*a^2*b^3 + 64*a^3*b^2))/(16*(a*b + b^2))))/(
4*(a*b + b^2)))*(b^2 - a^2)^(1/2)*1i)/(a*b + b^2) + (((tan(x)*(12*a*b^2 - 12*a^2*b + 4*a^3 - 4*b^3))/4 + ((b^2
- a^2)^(1/2)*(8*a*b^3 - 4*b^4 - 4*a^2*b^2 + (tan(x)*(b^2 - a^2)^(1/2)*(64*a*b^4 - 128*a^2*b^3 + 64*a^3*b^2))/
(16*(a*b + b^2))))/(4*(a*b + b^2)))*(b^2 - a^2)^(1/2)*1i)/(a*b + b^2))/((((tan(x)*(12*a*b^2 - 12*a^2*b + 4*a^3
- 4*b^3))/4 + ((b^2 - a^2)^(1/2)*(4*b^4 - 8*a*b^3 + 4*a^2*b^2 + (tan(x)*(b^2 - a^2)^(1/2)*(64*a*b^4 - 128*a^2
*b^3 + 64*a^3*b^2))/(16*(a*b + b^2))))/(4*(a*b + b^2)))*(b^2 - a^2)^(1/2))/(a*b + b^2) - (((tan(x)*(12*a*b^2 -
12*a^2*b + 4*a^3 - 4*b^3))/4 + ((b^2 - a^2)^(1/2)*(8*a*b^3 - 4*b^4 - 4*a^2*b^2 + (tan(x)*(b^2 - a^2)^(1/2)*(6
4*a*b^4 - 128*a^2*b^3 + 64*a^3*b^2))/(16*(a*b + b^2))))/(4*(a*b + b^2)))*(b^2 - a^2)^(1/2))/(a*b + b^2)))*(b^2
- a^2)^(1/2)*1i)/(2*(a*b + b^2))

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sympy [B]  time = 35.33, size = 432, normalized size = 8.31 \[ \begin {cases} \tilde {\infty } \left (- \frac {\log {\left (\tan {\relax (x )} - 1 \right )}}{2} + \frac {\log {\left (\tan {\relax (x )} + 1 \right )}}{2}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {1}{4 b \tan {\relax (x )}} & \text {for}\: a = - b \\\frac {\tan {\relax (x )}}{4 b} & \text {for}\: a = b \\\frac {\log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\relax (x )} \right )}}{4 a \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - 4 b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {\log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\relax (x )} \right )}}{4 a \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - 4 b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} & \text {otherwise} \end {cases} + \begin {cases} \tilde {\infty } x & \text {for}\: a = 0 \wedge b = 0 \\\frac {x}{2 b} - \frac {\tan {\relax (x )}}{4 b} & \text {for}\: a = b \\\frac {x}{2 b} + \frac {1}{4 b \tan {\relax (x )}} & \text {for}\: a = - b \\\frac {\sin {\left (2 x \right )}}{4 a} & \text {for}\: b = 0 \\\frac {2 a x \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}}{4 a b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - 4 b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {a \log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\relax (x )} \right )}}{4 a b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - 4 b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} + \frac {a \log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\relax (x )} \right )}}{4 a b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - 4 b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {2 b x \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}}{4 a b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - 4 b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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)), (1/(4*b*tan(x)), Eq(a, -b)), (t
an(x)/(4*b), Eq(a, b)), (log(-sqrt(-a/(a - b) - b/(a - b)) + tan(x))/(4*a*sqrt(-a/(a - b) - b/(a - b)) - 4*b*s
qrt(-a/(a - b) - b/(a - b))) - log(sqrt(-a/(a - b) - b/(a - b)) + tan(x))/(4*a*sqrt(-a/(a - b) - b/(a - b)) -
4*b*sqrt(-a/(a - b) - b/(a - b))), True)) + Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0)), (x/(2*b) - tan(x)/(4*b), E
q(a, b)), (x/(2*b) + 1/(4*b*tan(x)), Eq(a, -b)), (sin(2*x)/(4*a), Eq(b, 0)), (2*a*x*sqrt(-a/(a - b) - b/(a - b
))/(4*a*b*sqrt(-a/(a - b) - b/(a - b)) - 4*b**2*sqrt(-a/(a - b) - b/(a - b))) - a*log(-sqrt(-a/(a - b) - b/(a
- b)) + tan(x))/(4*a*b*sqrt(-a/(a - b) - b/(a - b)) - 4*b**2*sqrt(-a/(a - b) - b/(a - b))) + a*log(sqrt(-a/(a
- b) - b/(a - b)) + tan(x))/(4*a*b*sqrt(-a/(a - b) - b/(a - b)) - 4*b**2*sqrt(-a/(a - b) - b/(a - b))) - 2*b*x
*sqrt(-a/(a - b) - b/(a - b))/(4*a*b*sqrt(-a/(a - b) - b/(a - b)) - 4*b**2*sqrt(-a/(a - b) - b/(a - b))), True
))

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