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

Optimal. Leaf size=67 \[ \frac{\csc (x) (b-a \cos (x))}{a^2-b^2}-\frac{2 b^2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0949208, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {2696, 12, 2659, 205} \[ \frac{\csc (x) (b-a \cos (x))}{a^2-b^2}-\frac{2 b^2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^2/(a + b*Cos[x]),x]

[Out]

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

Rule 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 12

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 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 (x)} \, dx &=\frac{(b-a \cos (x)) \csc (x)}{a^2-b^2}+\frac{\int \frac{b^2}{a+b \cos (x)} \, dx}{-a^2+b^2}\\ &=\frac{(b-a \cos (x)) \csc (x)}{a^2-b^2}-\frac{b^2 \int \frac{1}{a+b \cos (x)} \, dx}{a^2-b^2}\\ &=\frac{(b-a \cos (x)) \csc (x)}{a^2-b^2}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{a^2-b^2}\\ &=-\frac{2 b^2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}}+\frac{(b-a \cos (x)) \csc (x)}{a^2-b^2}\\ \end{align*}

Mathematica [A]  time = 0.311402, size = 66, normalized size = 0.99 \[ \frac{\csc (x) (b-a \cos (x))}{a^2-b^2}-\frac{2 b^2 \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^2/(a + b*Cos[x]),x]

[Out]

(-2*b^2*ArcTanh[((a - b)*Tan[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + ((b - a*Cos[x])*Csc[x])/(a^2 - b^2)

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Maple [A]  time = 0.08, size = 78, normalized size = 1.2 \begin{align*}{\frac{1}{2\,a-2\,b}\tan \left ({\frac{x}{2}} \right ) }-{\frac{1}{2\,a+2\,b} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-2\,{\frac{{b}^{2}}{ \left ( a-b \right ) \left ( a+b \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( x/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.77301, size = 544, normalized size = 8.12 \begin{align*} \left [\frac{\sqrt{-a^{2} + b^{2}} b^{2} \log \left (\frac{2 \, a b \cos \left (x\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) \sin \left (x\right ) + 2 \, a^{2} b - 2 \, b^{3} - 2 \,{\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )}, -\frac{\sqrt{a^{2} - b^{2}} b^{2} \arctan \left (-\frac{a \cos \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (x\right )}\right ) \sin \left (x\right ) - a^{2} b + b^{3} +{\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\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)),x, algorithm="fricas")

[Out]

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

[Out]

Integral(csc(x)**2/(a + b*cos(x)), x)

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Giac [A]  time = 1.15796, size = 123, normalized size = 1.84 \begin{align*} \frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x\right ) - b \tan \left (\frac{1}{2} \, x\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )} b^{2}}{{\left (a^{2} - b^{2}\right )}^{\frac{3}{2}}} + \frac{\tan \left (\frac{1}{2} \, x\right )}{2 \,{\left (a - b\right )}} - \frac{1}{2 \,{\left (a + b\right )} \tan \left (\frac{1}{2} \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*tan(1/2*x))/sqrt(a^2 - b^2)))*b^2/(a^2
- b^2)^(3/2) + 1/2*tan(1/2*x)/(a - b) - 1/2/((a + b)*tan(1/2*x))