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

Optimal. Leaf size=41 \[ -\frac{\cot (x)}{a+b}-\frac{b \tan ^{-1}\left (\frac{\sqrt{a+b} \cot (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^{3/2}} \]

[Out]

-((b*ArcTan[(Sqrt[a + b]*Cot[x])/Sqrt[a]])/(Sqrt[a]*(a + b)^(3/2))) - Cot[x]/(a + b)

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Rubi [A]  time = 0.0560492, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3191, 388, 205} \[ -\frac{\cot (x)}{a+b}-\frac{b \tan ^{-1}\left (\frac{\sqrt{a+b} \cot (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^2/(a + b*Cos[x]^2),x]

[Out]

-((b*ArcTan[(Sqrt[a + b]*Cot[x])/Sqrt[a]])/(Sqrt[a]*(a + b)^(3/2))) - Cot[x]/(a + b)

Rule 3191

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]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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]

Rubi steps

\begin{align*} \int \frac{\csc ^2(x)}{a+b \cos ^2(x)} \, dx &=-\operatorname{Subst}\left (\int \frac{1+x^2}{a+(a+b) x^2} \, dx,x,\cot (x)\right )\\ &=-\frac{\cot (x)}{a+b}-\frac{b \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\cot (x)\right )}{a+b}\\ &=-\frac{b \tan ^{-1}\left (\frac{\sqrt{a+b} \cot (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^{3/2}}-\frac{\cot (x)}{a+b}\\ \end{align*}

Mathematica [A]  time = 0.0894784, size = 40, normalized size = 0.98 \[ \frac{b \tan ^{-1}\left (\frac{\sqrt{a} \tan (x)}{\sqrt{a+b}}\right )}{\sqrt{a} (a+b)^{3/2}}-\frac{\cot (x)}{a+b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ArcTan[(Sqrt[a]*Tan[x])/Sqrt[a + b]])/(Sqrt[a]*(a + b)^(3/2)) - Cot[x]/(a + b)

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Maple [A]  time = 0.03, size = 39, normalized size = 1. \begin{align*}{\frac{b}{a+b}\arctan \left ({a\tan \left ( x \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}}-{\frac{1}{ \left ( a+b \right ) \tan \left ( x \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.91062, size = 571, normalized size = 13.93 \begin{align*} \left [-\frac{\sqrt{-a^{2} - a b} 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 + b\right )} \cos \left (x\right )^{3} - a \cos \left (x\right )\right )} \sqrt{-a^{2} - 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 ) \sin \left (x\right ) + 4 \,{\left (a^{2} + a b\right )} \cos \left (x\right )}{4 \,{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sin \left (x\right )}, -\frac{\sqrt{a^{2} + a b} b \arctan \left (\frac{{\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a}{2 \, \sqrt{a^{2} + a b} \cos \left (x\right ) \sin \left (x\right )}\right ) \sin \left (x\right ) + 2 \,{\left (a^{2} + a b\right )} \cos \left (x\right )}{2 \,{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sin \left (x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{2}{\left (x \right )}}{a + b \cos ^{2}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(x)**2/(a + b*cos(x)**2), x)

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Giac [A]  time = 1.17665, size = 74, normalized size = 1.8 \begin{align*} \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 )} b}{\sqrt{a^{2} + a b}{\left (a + b\right )}} - \frac{1}{{\left (a + b\right )} \tan \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(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)))*b/(sqrt(a^2 + a*b)*(a + b)) - 1/((a + b)*tan(
x))