\(\int \frac {\cot (x)}{\sqrt {a+b \cot ^2(x)}} \, dx\) [46]

Optimal result

Integrand size = 15, antiderivative size = 33 \[ \int \frac {\cot (x)}{\sqrt {a+b \cot ^2(x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3751, 455, 65, 214} \[ \int \frac {\cot (x)}{\sqrt {a+b \cot ^2(x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}} \]

[In]

Int[Cot[x]/Sqrt[a + b*Cot[x]^2],x]

[Out]

ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]/Sqrt[a - b]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 455

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

Rule 3751

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[c*(ff/f), Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2
+ ff^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {x}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\cot (x)\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\cot ^2(x)\right )\right ) \\ & = -\frac {\text {Subst}\left (\int \frac {1}{1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \cot ^2(x)}\right )}{b} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {\cot (x)}{\sqrt {a+b \cot ^2(x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}} \]

[In]

Integrate[Cot[x]/Sqrt[a + b*Cot[x]^2],x]

[Out]

ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]/Sqrt[a - b]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(-\frac {\arctan \left (\frac {\sqrt {a +b \cot \left (x \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}\)	\(29\)
default	\(-\frac {\arctan \left (\frac {\sqrt {a +b \cot \left (x \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}\)	\(29\)

[In]

int(cot(x)/(a+b*cot(x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (27) = 54\).

Time = 0.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 3.85 \[ \int \frac {\cot (x)}{\sqrt {a+b \cot ^2(x)}} \, dx=\left [\frac {\log \left (-\sqrt {a - b} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}} {\left (\cos \left (2 \, x\right ) - 1\right )} - {\left (a - b\right )} \cos \left (2 \, x\right ) + a\right )}{2 \, \sqrt {a - b}}, \frac {\sqrt {-a + b} \arctan \left (-\frac {\sqrt {-a + b} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}}{a - b}\right )}{a - b}\right ] \]

[In]

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

[Out]

[1/2*log(-sqrt(a - b)*sqrt(((a - b)*cos(2*x) - a - b)/(cos(2*x) - 1))*(cos(2*x) - 1) - (a - b)*cos(2*x) + a)/s
qrt(a - b), sqrt(-a + b)*arctan(-sqrt(-a + b)*sqrt(((a - b)*cos(2*x) - a - b)/(cos(2*x) - 1))/(a - b))/(a - b)
]

Sympy [F]

\[ \int \frac {\cot (x)}{\sqrt {a+b \cot ^2(x)}} \, dx=\int \frac {\cot {\left (x \right )}}{\sqrt {a + b \cot ^{2}{\left (x \right )}}}\, dx \]

[In]

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

[Out]

Integral(cot(x)/sqrt(a + b*cot(x)**2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cot (x)}{\sqrt {a+b \cot ^2(x)}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(cot(x)/(a+b*cot(x)^2)^(1/2),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-4*b>0)', see `assume?` for
more detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (27) = 54\).

Time = 0.33 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.85 \[ \int \frac {\cot (x)}{\sqrt {a+b \cot ^2(x)}} \, dx=\frac {\log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (\sin \left (x\right )\right )}{2 \, \sqrt {a - b}} - \frac {\log \left ({\left | -\sqrt {a - b} \sin \left (x\right ) + \sqrt {a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b} \right |}\right )}{\sqrt {a - b} \mathrm {sgn}\left (\sin \left (x\right )\right )} \]

[In]

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

[Out]

1/2*log(abs(b))*sgn(sin(x))/sqrt(a - b) - log(abs(-sqrt(a - b)*sin(x) + sqrt(a*sin(x)^2 - b*sin(x)^2 + b)))/(s
qrt(a - b)*sgn(sin(x)))

Mupad [B] (verification not implemented)

Time = 13.51 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {\cot (x)}{\sqrt {a+b \cot ^2(x)}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {cot}\left (x\right )}^2+a}}{\sqrt {a-b}}\right )}{\sqrt {a-b}} \]

[In]

int(cot(x)/(a + b*cot(x)^2)^(1/2),x)

[Out]

atanh((a + b*cot(x)^2)^(1/2)/(a - b)^(1/2))/(a - b)^(1/2)