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

Optimal. Leaf size=55 \[ \frac{\tan ^{-1}\left (\frac{a \tan (x)+b}{\sqrt{a^2-b^2}}\right )}{2 \sqrt{a^2-b^2}}+\frac{\log (a+b \sin (2 x))}{4 b} \]

[Out]

ArcTan[(b + a*Tan[x])/Sqrt[a^2 - b^2]]/(2*Sqrt[a^2 - b^2]) + Log[a + b*Sin[2*x]]/(4*b)

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Rubi [A]  time = 0.132885, antiderivative size = 70, normalized size of antiderivative = 1.27, number of steps used = 8, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {981, 634, 618, 204, 628, 12, 260} \[ \frac{\tan ^{-1}\left (\frac{a \tan (x)+b}{\sqrt{a^2-b^2}}\right )}{2 \sqrt{a^2-b^2}}+\frac{\log \left (a \tan ^2(x)+a+2 b \tan (x)\right )}{4 b}+\frac{\log (\cos (x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^2/(a + b*Sin[2*x]),x]

[Out]

ArcTan[(b + a*Tan[x])/Sqrt[a^2 - b^2]]/(2*Sqrt[a^2 - b^2]) + Log[Cos[x]]/(2*b) + Log[a + 2*b*Tan[x] + a*Tan[x]
^2]/(4*b)

Rule 981

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

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.062303, size = 54, normalized size = 0.98 \[ \frac{1}{4} \left (\frac{2 \tan ^{-1}\left (\frac{a \tan (x)+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{\log (a+b \sin (2 x))}{b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*ArcTan[(b + a*Tan[x])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + Log[a + b*Sin[2*x]]/b)/4

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Maple [A]  time = 0.061, size = 69, normalized size = 1.3 \begin{align*} -{\frac{\ln \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) }{4\,b}}+{\frac{\ln \left ( a+2\,b\tan \left ( x \right ) +a \left ( \tan \left ( x \right ) \right ) ^{2} \right ) }{4\,b}}+{\frac{1}{2}\arctan \left ({\frac{2\,a\tan \left ( x \right ) +2\,b}{2}{\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/b*ln(1+tan(x)^2)+1/4*ln(a+2*b*tan(x)+a*tan(x)^2)/b+1/2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(x)+2*b)/(a^2-b
^2)^(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(cos(x)^2/(a+b*sin(2*x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.4104, size = 767, normalized size = 13.95 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}} b \log \left (-\frac{4 \,{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{4} - 4 \, a b \cos \left (x\right ) \sin \left (x\right ) - 4 \,{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2} - 2 \, b^{2} + 2 \,{\left (2 \, b \cos \left (x\right )^{2} + 2 \,{\left (2 \, a \cos \left (x\right )^{3} - a \cos \left (x\right )\right )} \sin \left (x\right ) - b\right )} \sqrt{-a^{2} + b^{2}}}{4 \, b^{2} \cos \left (x\right )^{4} - 4 \, b^{2} \cos \left (x\right )^{2} - 4 \, a b \cos \left (x\right ) \sin \left (x\right ) - a^{2}}\right ) -{\left (a^{2} - b^{2}\right )} \log \left (-4 \, b^{2} \cos \left (x\right )^{4} + 4 \, b^{2} \cos \left (x\right )^{2} + 4 \, a b \cos \left (x\right ) \sin \left (x\right ) + a^{2}\right )}{8 \,{\left (a^{2} b - b^{3}\right )}}, -\frac{2 \, \sqrt{a^{2} - b^{2}} b \arctan \left (-\frac{{\left (2 \, a \cos \left (x\right ) \sin \left (x\right ) + b\right )} \sqrt{a^{2} - b^{2}}}{2 \,{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} + b^{2}}\right ) -{\left (a^{2} - b^{2}\right )} \log \left (-4 \, b^{2} \cos \left (x\right )^{4} + 4 \, b^{2} \cos \left (x\right )^{2} + 4 \, a b \cos \left (x\right ) \sin \left (x\right ) + a^{2}\right )}{8 \,{\left (a^{2} b - b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 10.0253, size = 136, normalized size = 2.47 \begin{align*} \frac{\begin{cases} \frac{\log{\left (\frac{a}{b} + \sin{\left (2 x \right )} \right )}}{2 b} & \text{for}\: b \neq 0 \\\frac{\sin{\left (2 x \right )}}{2 a} & \text{otherwise} \end{cases}}{2} + \frac{\begin{cases} \frac{\log{\left (\tan{\left (x \right )} \right )}}{2 b} & \text{for}\: a = 0 \\- \frac{1}{b - \sqrt{b^{2}} \tan{\left (x \right )}} & \text{for}\: a = - \sqrt{b^{2}} \\- \frac{1}{b + \sqrt{b^{2}} \tan{\left (x \right )}} & \text{for}\: a = \sqrt{b^{2}} \\\frac{\log{\left (\tan{\left (x \right )} + \frac{b}{a} - \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{2 \sqrt{- a^{2} + b^{2}}} - \frac{\log{\left (\tan{\left (x \right )} + \frac{b}{a} + \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{2 \sqrt{- a^{2} + b^{2}}} & \text{otherwise} \end{cases}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((log(a/b + sin(2*x))/(2*b), Ne(b, 0)), (sin(2*x)/(2*a), True))/2 + Piecewise((log(tan(x))/(2*b), Eq(
a, 0)), (-1/(b - sqrt(b**2)*tan(x)), Eq(a, -sqrt(b**2))), (-1/(b + sqrt(b**2)*tan(x)), Eq(a, sqrt(b**2))), (lo
g(tan(x) + b/a - sqrt(-a**2 + b**2)/a)/(2*sqrt(-a**2 + b**2)) - log(tan(x) + b/a + sqrt(-a**2 + b**2)/a)/(2*sq
rt(-a**2 + b**2)), True))/2

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Giac [A]  time = 1.08591, size = 104, normalized size = 1.89 \begin{align*} \frac{\pi \left \lfloor \frac{x}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (x\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )}{2 \, \sqrt{a^{2} - b^{2}}} + \frac{\log \left (a \tan \left (x\right )^{2} + 2 \, b \tan \left (x\right ) + a\right )}{4 \, b} - \frac{\log \left (\tan \left (x\right )^{2} + 1\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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