\(\int \frac {\sin ^2(x)}{a+b \cos (2 x)} \, dx\) [848]

Optimal result

Integrand size = 15, antiderivative size = 52 \[ \int \frac {\sin ^2(x)}{a+b \cos (2 x)} \, dx=-\frac {x}{2 b}+\frac {\sqrt {a+b} \arctan \left (\frac {\sqrt {a-b} \tan (x)}{\sqrt {a+b}}\right )}{2 \sqrt {a-b} b} \]

[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)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1144, 211} \[ \int \frac {\sin ^2(x)}{a+b \cos (2 x)} \, dx=\frac {\sqrt {a+b} \arctan \left (\frac {\sqrt {a-b} \tan (x)}{\sqrt {a+b}}\right )}{2 b \sqrt {a-b}}-\frac {x}{2 b} \]

[In]

Int[Sin[x]^2/(a + b*Cos[2*x]),x]

[Out]

-1/2*x/b + (Sqrt[a + b]*ArcTan[(Sqrt[a - b]*Tan[x])/Sqrt[a + b]])/(2*Sqrt[a - b]*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 1144

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

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.92 \[ \int \frac {\sin ^2(x)}{a+b \cos (2 x)} \, dx=-\frac {x+\frac {(a+b) \text {arctanh}\left (\frac {(a-b) \tan (x)}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}}{2 b} \]

[In]

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

[Out]

-1/2*(x + ((a + b)*ArcTanh[((a - b)*Tan[x])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2])/b

Maple [A] (verified)

Time = 2.23 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {\arctan \left (\tan \left (x \right )\right )}{2 b}+\frac {\left (a +b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (x \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 b \sqrt {\left (a +b \right ) \left (a -b \right )}}\)	\(49\)
risch	\(-\frac {x}{2 b}-\frac {\sqrt {-\left (a +b \right ) \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i x}+\frac {i \sqrt {-\left (a +b \right ) \left (a -b \right )}+a}{b}\right )}{4 \left (a -b \right ) b}+\frac {\sqrt {-\left (a +b \right ) \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i x}-\frac {i \sqrt {-\left (a +b \right ) \left (a -b \right )}-a}{b}\right )}{4 \left (a -b \right ) b}\)	\(115\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 225, normalized size of antiderivative = 4.33 \[ \int \frac {\sin ^2(x)}{a+b \cos (2 x)} \, dx=\left [\frac {\sqrt {-\frac {a + b}{a - b}} \log \left (\frac {4 \, {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{4} - 4 \, {\left (2 \, a^{2} - a b - b^{2}\right )} \cos \left (x\right )^{2} - 4 \, {\left (2 \, {\left (a^{2} - a b\right )} \cos \left (x\right )^{3} - {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (x\right )\right )} \sqrt {-\frac {a + b}{a - b}} \sin \left (x\right ) + a^{2} - 2 \, a b + b^{2}}{4 \, b^{2} \cos \left (x\right )^{4} + 4 \, {\left (a b - b^{2}\right )} \cos \left (x\right )^{2} + a^{2} - 2 \, a b + b^{2}}\right ) - 4 \, x}{8 \, b}, -\frac {\sqrt {\frac {a + b}{a - b}} \arctan \left (\frac {{\left (2 \, a \cos \left (x\right )^{2} - a + b\right )} \sqrt {\frac {a + b}{a - b}}}{2 \, {\left (a + b\right )} \cos \left (x\right ) \sin \left (x\right )}\right ) + 2 \, x}{4 \, b}\right ] \]

[In]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (39) = 78\).

Time = 16.32 (sec) , antiderivative size = 432, normalized size of antiderivative = 8.31 \[ \int \frac {\sin ^2(x)}{a+b \cos (2 x)} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {\log {\left (\tan {\left (x \right )} - 1 \right )}}{2} + \frac {\log {\left (\tan {\left (x \right )} + 1 \right )}}{2}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\tan {\left (x \right )}}{4 b} & \text {for}\: a = b \\\frac {1}{4 b \tan {\left (x \right )}} & \text {for}\: a = - b \\\frac {\log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (x \right )} \right )}}{4 a \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - 4 b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {\log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (x \right )} \right )}}{4 a \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - 4 b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} & \text {otherwise} \end {cases} - \begin {cases} \tilde {\infty } x & \text {for}\: a = 0 \wedge b = 0 \\\frac {x}{2 b} - \frac {\tan {\left (x \right )}}{4 b} & \text {for}\: a = b \\\frac {x}{2 b} + \frac {1}{4 b \tan {\left (x \right )}} & \text {for}\: a = - b \\\frac {\sin {\left (2 x \right )}}{4 a} & \text {for}\: b = 0 \\\frac {2 a x \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}}{4 a b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - 4 b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {a \log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (x \right )} \right )}}{4 a b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - 4 b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} + \frac {a \log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (x \right )} \right )}}{4 a b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - 4 b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {2 b x \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}}{4 a b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - 4 b^{2} \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} & \text {otherwise} \end {cases} \]

[In]

integrate(sin(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)), (tan(x)/(4*b), Eq(a, b)), (1/(4
*b*tan(x)), 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
))

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sin ^2(x)}{a+b \cos (2 x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(sin(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 de

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (40) = 80\).

Time = 0.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.71 \[ \int \frac {\sin ^2(x)}{a+b \cos (2 x)} \, dx=\frac {\sqrt {a^{2} - b^{2}} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (x\right )}{\sqrt {\frac {4 \, a + \sqrt {-16 \, {\left (a + b\right )} {\left (a - b\right )} + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} {\left | b \right |}} - \frac {\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (x\right )}{\sqrt {\frac {4 \, a - \sqrt {-16 \, {\left (a + b\right )} {\left (a - b\right )} + 16 \, a^{2}}}{a - b}}}\right )}{2 \, {\left | b \right |}} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 26.85 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.08 \[ \int \frac {\sin ^2(x)}{a+b \cos (2 x)} \, dx=\frac {\mathrm {atan}\left (\frac {2\,b^3\,\mathrm {tan}\left (x\right )}{2\,a^2\,b-2\,b^3}-\frac {2\,a^2\,b\,\mathrm {tan}\left (x\right )}{2\,a^2\,b-2\,b^3}\right )}{2\,b}+\frac {\mathrm {atanh}\left (\frac {a\,\mathrm {tan}\left (x\right )}{\sqrt {b^2-a^2}}-\frac {b\,\mathrm {tan}\left (x\right )}{\sqrt {b^2-a^2}}\right )\,\sqrt {b^2-a^2}}{2\,\left (a\,b-b^2\right )} \]

[In]

int(sin(x)^2/(a + b*cos(2*x)),x)

[Out]

atan((2*b^3*tan(x))/(2*a^2*b - 2*b^3) - (2*a^2*b*tan(x))/(2*a^2*b - 2*b^3))/(2*b) + (atanh((a*tan(x))/(b^2 - a
^2)^(1/2) - (b*tan(x))/(b^2 - a^2)^(1/2))*(b^2 - a^2)^(1/2))/(2*(a*b - b^2))