\(\int \frac {\cot ^3(x)}{\sqrt {a+b \cot ^2(x)}} \, dx\) [44]

Optimal result

Integrand size = 17, antiderivative size = 52 \[ \int \frac {\cot ^3(x)}{\sqrt {a+b \cot ^2(x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}-\frac {\sqrt {a+b \cot ^2(x)}}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3751, 457, 81, 65, 214} \[ \int \frac {\cot ^3(x)}{\sqrt {a+b \cot ^2(x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}-\frac {\sqrt {a+b \cot ^2(x)}}{b} \]

[In]

Int[Cot[x]^3/Sqrt[a + b*Cot[x]^2],x]

[Out]

-(ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]/Sqrt[a - b]) - Sqrt[a + b*Cot[x]^2]/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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

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^3}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\cot (x)\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {x}{(1+x) \sqrt {a+b x}} \, dx,x,\cot ^2(x)\right )\right ) \\ & = -\frac {\sqrt {a+b \cot ^2(x)}}{b}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\cot ^2(x)\right ) \\ & = -\frac {\sqrt {a+b \cot ^2(x)}}{b}+\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}}-\frac {\sqrt {a+b \cot ^2(x)}}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {\cot ^3(x)}{\sqrt {a+b \cot ^2(x)}} \, dx=-\frac {\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\sqrt {a+b \cot ^2(x)}}{b} \]

[In]

Integrate[Cot[x]^3/Sqrt[a + b*Cot[x]^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(-\frac {\sqrt {a +b \cot \left (x \right )^{2}}}{b}+\frac {\arctan \left (\frac {\sqrt {a +b \cot \left (x \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}\)	\(44\)
default	\(-\frac {\sqrt {a +b \cot \left (x \right )^{2}}}{b}+\frac {\arctan \left (\frac {\sqrt {a +b \cot \left (x \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}\)	\(44\)

[In]

int(cot(x)^3/(a+b*cot(x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-(a+b*cot(x)^2)^(1/2)/b+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. 118 vs. \(2 (44) = 88\).

Time = 0.35 (sec) , antiderivative size = 284, normalized size of antiderivative = 5.46 \[ \int \frac {\cot ^3(x)}{\sqrt {a+b \cot ^2(x)}} \, dx=\left [\frac {\sqrt {a - b} b \log \left (-2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (2 \, x\right )^{2} - 2 \, a^{2} + b^{2} + 2 \, {\left ({\left (a - b\right )} \cos \left (2 \, x\right )^{2} - {\left (2 \, a - b\right )} \cos \left (2 \, x\right ) + a\right )} \sqrt {a - b} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}} + 4 \, {\left (a^{2} - a b\right )} \cos \left (2 \, x\right )\right ) - 4 \, {\left (a - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}}{4 \, {\left (a b - b^{2}\right )}}, -\frac {\sqrt {-a + b} b \arctan \left (-\frac {\sqrt {-a + b} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}} {\left (\cos \left (2 \, x\right ) - 1\right )}}{{\left (a - b\right )} \cos \left (2 \, x\right ) - a}\right ) + 2 \, {\left (a - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}}{2 \, {\left (a b - b^{2}\right )}}\right ] \]

[In]

integrate(cot(x)^3/(a+b*cot(x)^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\cot ^3(x)}{\sqrt {a+b \cot ^2(x)}} \, dx=\int \frac {\cot ^{3}{\left (x \right )}}{\sqrt {a + b \cot ^{2}{\left (x \right )}}}\, dx \]

[In]

integrate(cot(x)**3/(a+b*cot(x)**2)**(1/2),x)

[Out]

Integral(cot(x)**3/sqrt(a + b*cot(x)**2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cot ^3(x)}{\sqrt {a+b \cot ^2(x)}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(cot(x)^3/(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. 96 vs. \(2 (44) = 88\).

Time = 0.36 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.85 \[ \int \frac {\cot ^3(x)}{\sqrt {a+b \cot ^2(x)}} \, dx=\frac {\frac {\log \left ({\left (\sqrt {a - b} \sin \left (x\right ) - \sqrt {a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b}\right )}^{2}\right )}{\sqrt {a - b}} + \frac {4 \, \sqrt {a - b}}{{\left (\sqrt {a - b} \sin \left (x\right ) - \sqrt {a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b}\right )}^{2} - b}}{2 \, \mathrm {sgn}\left (\sin \left (x\right )\right )} \]

[In]

integrate(cot(x)^3/(a+b*cot(x)^2)^(1/2),x, algorithm="giac")

[Out]

1/2*(log((sqrt(a - b)*sin(x) - sqrt(a*sin(x)^2 - b*sin(x)^2 + b))^2)/sqrt(a - b) + 4*sqrt(a - b)/((sqrt(a - b)
*sin(x) - sqrt(a*sin(x)^2 - b*sin(x)^2 + b))^2 - b))/sgn(sin(x))

Mupad [B] (verification not implemented)

Time = 14.61 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85 \[ \int \frac {\cot ^3(x)}{\sqrt {a+b \cot ^2(x)}} \, dx=-\frac {\sqrt {b\,{\mathrm {cot}\left (x\right )}^2+a}}{b}-\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {cot}\left (x\right )}^2+a}}{\sqrt {a-b}}\right )}{\sqrt {a-b}} \]

[In]

int(cot(x)^3/(a + b*cot(x)^2)^(1/2),x)

[Out]

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