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

Optimal. Leaf size=77 \[ -\frac {a \cot (x)}{a^2-b^2}+\frac {b \csc (x)}{a^2-b^2}-\frac {2 a^2 \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}} \]

[Out]

-2*a^2*arctan((a-b)^(1/2)*tan(1/2*x)/(a+b)^(1/2))/(a-b)^(3/2)/(a+b)^(3/2)-a*cot(x)/(a^2-b^2)+b*csc(x)/(a^2-b^2
)

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Rubi [A]  time = 0.10, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2727, 3767, 8, 2606, 2659, 205} \[ -\frac {a \cot (x)}{a^2-b^2}+\frac {b \csc (x)}{a^2-b^2}-\frac {2 a^2 \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^2/(a + b*Cos[x]),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

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 2727

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.33, size = 67, normalized size = 0.87 \[ \frac {b \csc (x)-a \cot (x)}{a^2-b^2}-\frac {2 a^2 \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^2/(a + b*Cos[x]),x]

[Out]

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

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fricas [A]  time = 0.85, size = 230, normalized size = 2.99 \[ \left [\frac {\sqrt {-a^{2} + b^{2}} a^{2} \log \left (\frac {2 \, a b \cos \relax (x) + {\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \relax (x) + b\right )} \sin \relax (x) - a^{2} + 2 \, b^{2}}{b^{2} \cos \relax (x)^{2} + 2 \, a b \cos \relax (x) + a^{2}}\right ) \sin \relax (x) + 2 \, a^{2} b - 2 \, b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \relax (x)}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \relax (x)}, -\frac {\sqrt {a^{2} - b^{2}} a^{2} \arctan \left (-\frac {a \cos \relax (x) + b}{\sqrt {a^{2} - b^{2}} \sin \relax (x)}\right ) \sin \relax (x) - a^{2} b + b^{3} + {\left (a^{3} - a b^{2}\right )} \cos \relax (x)}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \relax (x)}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(-a^2 + b^2)*a^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)*a^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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giac [A]  time = 0.49, size = 91, normalized size = 1.18 \[ \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{2}}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {\tan \left (\frac {1}{2} \, x\right )}{2 \, {\left (a - b\right )}} - \frac {1}{2 \, {\left (a + b\right )} \tan \left (\frac {1}{2} \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(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)))*a^2/(a^2
- b^2)^(3/2) + 1/2*tan(1/2*x)/(a - b) - 1/2/((a + b)*tan(1/2*x))

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maple [A]  time = 0.06, size = 78, normalized size = 1.01 \[ \frac {\tan \left (\frac {x}{2}\right )}{2 a -2 b}-\frac {2 a^{2} \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{2 \left (a +b \right ) \tan \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(a-b)*tan(1/2*x)-2/(a-b)/(a+b)*a^2/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*x)*(a-b)/((a-b)*(a+b))^(1/2))-1/2/(a
+b)/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(cot(x)^2/(a+b*cos(x)),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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 0.54, size = 86, normalized size = 1.12 \[ \frac {\mathrm {tan}\left (\frac {x}{2}\right )}{2\,a-2\,b}-\frac {2\,a^2\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2-b^2\right )}{{\left (a+b\right )}^{3/2}\,\sqrt {a-b}}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {a-b}{\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a+b\right )\,\left (2\,a-2\,b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^2/(a + b*cos(x)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cot(x)**2/(a + b*cos(x)), x)

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