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+sin(x)/a/(b+a*cos(x))+2*b*arctanh((a-b)^(1/2)*tan(1/2*x)/(a+b)^(1/2))/a^2/(a-b)^(1/2)/(a+b)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 67, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.27, 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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fricas [B]  time = 1.05, size = 307, normalized size = 4.58 \[ \left [-\frac {2 \, {\left (a^{3} - a b^{2}\right )} x \cos \relax (x) - {\left (a b \cos \relax (x) + b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \relax (x) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \relax (x) + a\right )} \sin \relax (x) + 2 \, a^{2} - b^{2}}{a^{2} \cos \relax (x)^{2} + 2 \, a b \cos \relax (x) + b^{2}}\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} x - 2 \, {\left (a^{3} - a b^{2}\right )} \sin \relax (x)}{2 \, {\left (a^{4} b - a^{2} b^{3} + {\left (a^{5} - a^{3} b^{2}\right )} \cos \relax (x)\right )}}, -\frac {{\left (a^{3} - a b^{2}\right )} x \cos \relax (x) - {\left (a b \cos \relax (x) + b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \relax (x)}\right ) + {\left (a^{2} b - b^{3}\right )} x - {\left (a^{3} - a b^{2}\right )} \sin \relax (x)}{a^{4} b - a^{2} b^{3} + {\left (a^{5} - a^{3} b^{2}\right )} \cos \relax (x)}\right ] \]

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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giac [A]  time = 0.14, size = 107, normalized size = 1.60 \[ \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b}{\sqrt {-a^{2} + b^{2}} a^{2}} - \frac {x}{a^{2}} - \frac {2 \, \tan \left (\frac {1}{2} \, x\right )}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - b \tan \left (\frac {1}{2} \, x\right )^{2} - a - b\right )} a} \]

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)

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maple [A]  time = 0.12, size = 86, normalized size = 1.28 \[ -\frac {2 \tan \left (\frac {x}{2}\right )}{a \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-a -b \right )}+\frac {2 b \arctanh \left (\frac {\tan \left (\frac {x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> 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 details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 3.04, size = 440, normalized size = 6.57 \[ \frac {a^3\,\sin \relax (x)+b^2\,\left (-a\,\sin \relax (x)+\mathrm {atan}\left (\frac {-a^5\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}+b^3\,\sin \left (\frac {x}{2}\right )\,{\left (a^2-b^2\right )}^{3/2}\,2{}\mathrm {i}+b^5\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,2{}\mathrm {i}+a^4\,b\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}-a^2\,b^3\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,3{}\mathrm {i}+a^3\,b^2\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}}{\cos \left (\frac {x}{2}\right )\,a^6-2\,\cos \left (\frac {x}{2}\right )\,a^4\,b^2+\cos \left (\frac {x}{2}\right )\,a^2\,b^4}\right )\,\sqrt {a^2-b^2}\,2{}\mathrm {i}\right )+a\,b\,\mathrm {atan}\left (\frac {-a^5\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}+b^3\,\sin \left (\frac {x}{2}\right )\,{\left (a^2-b^2\right )}^{3/2}\,2{}\mathrm {i}+b^5\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,2{}\mathrm {i}+a^4\,b\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}-a^2\,b^3\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,3{}\mathrm {i}+a^3\,b^2\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}}{\cos \left (\frac {x}{2}\right )\,a^6-2\,\cos \left (\frac {x}{2}\right )\,a^4\,b^2+\cos \left (\frac {x}{2}\right )\,a^2\,b^4}\right )\,\cos \relax (x)\,\sqrt {a^2-b^2}\,2{}\mathrm {i}}{\cos \relax (x)\,a^5+a^4\,b-\cos \relax (x)\,a^3\,b^2-a^2\,b^3}-\frac {2\,\mathrm {atan}\left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/sin(x) + a*cot(x))^2,x)

[Out]

(a^3*sin(x) + b^2*(atan((b^3*sin(x/2)*(a^2 - b^2)^(3/2)*2i - a^5*sin(x/2)*(a^2 - b^2)^(1/2)*1i + b^5*sin(x/2)*
(a^2 - b^2)^(1/2)*2i + a^4*b*sin(x/2)*(a^2 - b^2)^(1/2)*1i - a^2*b^3*sin(x/2)*(a^2 - b^2)^(1/2)*3i + a^3*b^2*s
in(x/2)*(a^2 - b^2)^(1/2)*1i)/(a^6*cos(x/2) + a^2*b^4*cos(x/2) - 2*a^4*b^2*cos(x/2)))*(a^2 - b^2)^(1/2)*2i - a
*sin(x)) + a*b*atan((b^3*sin(x/2)*(a^2 - b^2)^(3/2)*2i - a^5*sin(x/2)*(a^2 - b^2)^(1/2)*1i + b^5*sin(x/2)*(a^2
- b^2)^(1/2)*2i + a^4*b*sin(x/2)*(a^2 - b^2)^(1/2)*1i - a^2*b^3*sin(x/2)*(a^2 - b^2)^(1/2)*3i + a^3*b^2*sin(x
/2)*(a^2 - b^2)^(1/2)*1i)/(a^6*cos(x/2) + a^2*b^4*cos(x/2) - 2*a^4*b^2*cos(x/2)))*cos(x)*(a^2 - b^2)^(1/2)*2i)
/(a^4*b - a^2*b^3 + a^5*cos(x) - a^3*b^2*cos(x)) - (2*atan(sin(x/2)/cos(x/2)))/a^2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \cot {\relax (x )} + b \csc {\relax (x )}\right )^{2}}\, dx \]

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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