3.1.19 \(\int \frac {\sin ^2(x)}{a+b \cos ^2(x)} \, dx\) [19]

Optimal. Leaf size=40 \[ -\frac {x}{b}-\frac {\sqrt {a+b} \text {ArcTan}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{\sqrt {a} b} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3270, 400, 209, 211} \begin {gather*} -\frac {\sqrt {a+b} \text {ArcTan}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{\sqrt {a} b}-\frac {x}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a + b*Cos[x]^2),x]

[Out]

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

Rule 209

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

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 400

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]

Rule 3270

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 37, normalized size = 0.92 \begin {gather*} \frac {-x+\frac {\sqrt {a+b} \text {ArcTan}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a+b}}\right )}{\sqrt {a}}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 36, normalized size = 0.90

method	result	size
default	\(-\frac {\arctan \left (\tan \left (x \right )\right )}{b}+\frac {\left (a +b \right ) \arctan \left (\frac {a \tan \left (x \right )}{\sqrt {\left (a +b \right ) a}}\right )}{b \sqrt {\left (a +b \right ) a}}\)	\(36\)
risch	\(-\frac {x}{b}-\frac {\sqrt {-\left (a +b \right ) a}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-\left (a +b \right ) a}+2 a +b}{b}\right )}{2 a b}+\frac {\sqrt {-\left (a +b \right ) a}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-\left (a +b \right ) a}-2 a -b}{b}\right )}{2 a b}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 33, normalized size = 0.82 \begin {gather*} \frac {{\left (a + b\right )} \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(sin(x)^2/(a+b*cos(x)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.45, size = 177, normalized size = 4.42 \begin {gather*} \left [\frac {\sqrt {-\frac {a + b}{a}} \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} + a b\right )} \cos \left (x\right )^{3} - a^{2} \cos \left (x\right )\right )} \sqrt {-\frac {a + b}{a}} \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 + b}{a}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a\right )} \sqrt {\frac {a + b}{a}}}{2 \, {\left (a + b\right )} \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(sin(x)^2/(a+b*cos(x)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]
time = 0.40, size = 50, normalized size = 1.25 \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 )} {\left (a + b\right )}}{\sqrt {a^{2} + a b} b} - \frac {x}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [B]
time = 2.37, size = 108, normalized size = 2.70 \begin {gather*} -\frac {\mathrm {atan}\left (\frac {2\,a\,b^2\,\mathrm {tan}\left (x\right )}{2\,a^2\,b+2\,a\,b^2}+\frac {2\,a^2\,b\,\mathrm {tan}\left (x\right )}{2\,a^2\,b+2\,a\,b^2}\right )}{b}-\frac {\mathrm {atanh}\left (\frac {2\,a^2\,b\,\mathrm {tan}\left (x\right )\,\sqrt {-a^2-b\,a}}{2\,a^3\,b+2\,a^2\,b^2}\right )\,\sqrt {-a\,\left (a+b\right )}}{a\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(a + b*cos(x)^2),x)

[Out]

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

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