3.289 \(\int \frac{1}{(a \cot (x)+b \csc (x))^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.117185, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {4392, 2693, 2735, 2659, 208} \[ \frac{2 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^2 \sqrt{a-b} \sqrt{a+b}}-\frac{x}{a^2}+\frac{\sin (x)}{a (a \cos (x)+b)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cot[x] + b*Csc[x])^(-2),x]

[Out]

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

Rule 4392

Int[(cot[(c_.) + (d_.)*(x_)]^(n_.)*(a_.) + csc[(c_.) + (d_.)*(x_)]^(n_.)*(b_.))^(p_)*(u_.), x_Symbol] :> Int[A
ctivateTrig[u]*Csc[c + d*x]^(n*p)*(b + a*Cos[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rule 2693

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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 208

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

Rubi steps

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

Mathematica [A]  time = 0.255074, size = 71, normalized size = 1.06 \[ -\frac{\frac{2 b \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{-a \sin (x)+a x \cos (x)+b x}{a \cos (x)+b}}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cot[x] + b*Csc[x])^(-2),x]

[Out]

-(((2*b*ArcTanh[((-a + b)*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (b*x + a*x*Cos[x] - a*Sin[x])/(b + a*C
os[x]))/a^2)

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Maple [A]  time = 0.051, size = 86, normalized size = 1.3 \begin{align*} -2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) }{{a}^{2}}}-2\,{\frac{\tan \left ( x/2 \right ) }{a \left ( a \left ( \tan \left ( x/2 \right ) \right ) ^{2}-b \left ( \tan \left ( x/2 \right ) \right ) ^{2}-a-b \right ) }}+2\,{\frac{b}{{a}^{2}\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( x/2 \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(1/(a*cot(x)+b*csc(x))^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.39142, size = 689, normalized size = 10.28 \begin{align*} \left [-\frac{2 \,{\left (a^{3} - a b^{2}\right )} x \cos \left (x\right ) -{\left (a b \cos \left (x\right ) + b^{2}\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (x\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (x\right ) + a\right )} \sin \left (x\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + b^{2}}\right ) + 2 \,{\left (a^{2} b - b^{3}\right )} x - 2 \,{\left (a^{3} - a b^{2}\right )} \sin \left (x\right )}{2 \,{\left (a^{4} b - a^{2} b^{3} +{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (x\right )\right )}}, -\frac{{\left (a^{3} - a b^{2}\right )} x \cos \left (x\right ) -{\left (a b \cos \left (x\right ) + b^{2}\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (x\right )}\right ) +{\left (a^{2} b - b^{3}\right )} x -{\left (a^{3} - a b^{2}\right )} \sin \left (x\right )}{a^{4} b - a^{2} b^{3} +{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))^2,x, algorithm="fricas")

[Out]

[-1/2*(2*(a^3 - a*b^2)*x*cos(x) - (a*b*cos(x) + b^2)*sqrt(a^2 - b^2)*log((2*a*b*cos(x) - (a^2 - 2*b^2)*cos(x)^
2 + 2*sqrt(a^2 - b^2)*(b*cos(x) + a)*sin(x) + 2*a^2 - b^2)/(a^2*cos(x)^2 + 2*a*b*cos(x) + b^2)) + 2*(a^2*b - b
^3)*x - 2*(a^3 - a*b^2)*sin(x))/(a^4*b - a^2*b^3 + (a^5 - a^3*b^2)*cos(x)), -((a^3 - a*b^2)*x*cos(x) - (a*b*co
s(x) + b^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(x) + a)/((a^2 - b^2)*sin(x))) + (a^2*b - b^3)*x -
(a^3 - a*b^2)*sin(x))/(a^4*b - a^2*b^3 + (a^5 - a^3*b^2)*cos(x))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \cot{\left (x \right )} + b \csc{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))**2,x)

[Out]

Integral((a*cot(x) + b*csc(x))**(-2), x)

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Giac [A]  time = 1.15058, size = 144, normalized size = 2.15 \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}{\sqrt{-a^{2} + b^{2}} a^{2}} - \frac{x}{a^{2}} - \frac{2 \, \tan \left (\frac{1}{2} \, x\right )}{{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} - b \tan \left (\frac{1}{2} \, x\right )^{2} - a - b\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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