\(\int \frac {\cot (x)}{(a+b \cot ^2(x))^{5/2}} \, dx\) [56]

Optimal result

Integrand size = 15, antiderivative size = 78 \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2}}-\frac {1}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac {1}{(a-b)^2 \sqrt {a+b \cot ^2(x)}} \]

[Out]

arctanh((a+b*cot(x)^2)^(1/2)/(a-b)^(1/2))/(a-b)^(5/2)-1/3/(a-b)/(a+b*cot(x)^2)^(3/2)-1/(a-b)^2/(a+b*cot(x)^2)^
(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3751, 455, 53, 65, 214} \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2}}-\frac {1}{(a-b)^2 \sqrt {a+b \cot ^2(x)}}-\frac {1}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}} \]

[In]

Int[Cot[x]/(a + b*Cot[x]^2)^(5/2),x]

[Out]

ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]/(a - b)^(5/2) - 1/(3*(a - b)*(a + b*Cot[x]^2)^(3/2)) - 1/((a - b)^2*
Sqrt[a + b*Cot[x]^2])

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 ) \left (a+b x^2\right )^{5/2}} \, dx,x,\cot (x)\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{(1+x) (a+b x)^{5/2}} \, dx,x,\cot ^2(x)\right )\right ) \\ & = -\frac {1}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac {\text {Subst}\left (\int \frac {1}{(1+x) (a+b x)^{3/2}} \, dx,x,\cot ^2(x)\right )}{2 (a-b)} \\ & = -\frac {1}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac {1}{(a-b)^2 \sqrt {a+b \cot ^2(x)}}-\frac {\text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\cot ^2(x)\right )}{2 (a-b)^2} \\ & = -\frac {1}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac {1}{(a-b)^2 \sqrt {a+b \cot ^2(x)}}-\frac {\text {Subst}\left (\int \frac {1}{1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \cot ^2(x)}\right )}{(a-b)^2 b} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2}}-\frac {1}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac {1}{(a-b)^2 \sqrt {a+b \cot ^2(x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.60 \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \cot ^2(x)}{a-b}\right )}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}} \]

[In]

Integrate[Cot[x]/(a + b*Cot[x]^2)^(5/2),x]

[Out]

-1/3*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Cot[x]^2)/(a - b)]/((a - b)*(a + b*Cot[x]^2)^(3/2))

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(-\frac {1}{3 \left (a -b \right ) \left (a +b \cot \left (x \right )^{2}\right )^{\frac {3}{2}}}-\frac {1}{\left (a -b \right )^{2} \sqrt {a +b \cot \left (x \right )^{2}}}-\frac {\arctan \left (\frac {\sqrt {a +b \cot \left (x \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right )^{2} \sqrt {-a +b}}\)	\(75\)
default	\(-\frac {1}{3 \left (a -b \right ) \left (a +b \cot \left (x \right )^{2}\right )^{\frac {3}{2}}}-\frac {1}{\left (a -b \right )^{2} \sqrt {a +b \cot \left (x \right )^{2}}}-\frac {\arctan \left (\frac {\sqrt {a +b \cot \left (x \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right )^{2} \sqrt {-a +b}}\)	\(75\)

[In]

int(cot(x)/(a+b*cot(x)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/(a-b)/(a+b*cot(x)^2)^(3/2)-1/(a-b)^2/(a+b*cot(x)^2)^(1/2)-1/(a-b)^2/(-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. 307 vs. \(2 (66) = 132\).

Time = 0.34 (sec) , antiderivative size = 627, normalized size of antiderivative = 8.04 \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (2 \, x\right )^{2} + a^{2} + 2 \, a b + b^{2} - 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, x\right )\right )} \sqrt {a - b} \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 ) - 4 \, {\left (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (2 \, x\right )^{2} + 2 \, a^{2} - a b - b^{2} - {\left (4 \, a^{2} - 5 \, a b + b^{2}\right )} \cos \left (2 \, x\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}}{6 \, {\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5} + {\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} \cos \left (2 \, x\right )^{2} - 2 \, {\left (a^{5} - 3 \, a^{4} b + 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 3 \, a b^{4} + b^{5}\right )} \cos \left (2 \, x\right )\right )}}, \frac {3 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (2 \, x\right )^{2} + a^{2} + 2 \, a b + b^{2} - 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, x\right )\right )} \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 ) - 2 \, {\left (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (2 \, x\right )^{2} + 2 \, a^{2} - a b - b^{2} - {\left (4 \, a^{2} - 5 \, a b + b^{2}\right )} \cos \left (2 \, x\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}}{3 \, {\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5} + {\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} \cos \left (2 \, x\right )^{2} - 2 \, {\left (a^{5} - 3 \, a^{4} b + 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 3 \, a b^{4} + b^{5}\right )} \cos \left (2 \, x\right )\right )}}\right ] \]

[In]

integrate(cot(x)/(a+b*cot(x)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*((a^2 - 2*a*b + b^2)*cos(2*x)^2 + a^2 + 2*a*b + b^2 - 2*(a^2 - b^2)*cos(2*x))*sqrt(a - b)*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) - 4*(2*(a^2 - 2*a*
b + b^2)*cos(2*x)^2 + 2*a^2 - a*b - b^2 - (4*a^2 - 5*a*b + b^2)*cos(2*x))*sqrt(((a - b)*cos(2*x) - a - b)/(cos
(2*x) - 1)))/(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5
*a*b^4 - b^5)*cos(2*x)^2 - 2*(a^5 - 3*a^4*b + 2*a^3*b^2 + 2*a^2*b^3 - 3*a*b^4 + b^5)*cos(2*x)), 1/3*(3*((a^2 -
2*a*b + b^2)*cos(2*x)^2 + a^2 + 2*a*b + b^2 - 2*(a^2 - b^2)*cos(2*x))*sqrt(-a + b)*arctan(-sqrt(-a + b)*sqrt(
((a - b)*cos(2*x) - a - b)/(cos(2*x) - 1))/(a - b)) - 2*(2*(a^2 - 2*a*b + b^2)*cos(2*x)^2 + 2*a^2 - a*b - b^2
- (4*a^2 - 5*a*b + b^2)*cos(2*x))*sqrt(((a - b)*cos(2*x) - a - b)/(cos(2*x) - 1)))/(a^5 - a^4*b - 2*a^3*b^2 +
2*a^2*b^3 + a*b^4 - b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*cos(2*x)^2 - 2*(a^5 - 3*a^
4*b + 2*a^3*b^2 + 2*a^2*b^3 - 3*a*b^4 + b^5)*cos(2*x))]

Sympy [A] (verification not implemented)

Time = 7.39 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.41 \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=- \begin {cases} \frac {2 \left (\frac {b}{6 \left (a - b\right ) \left (a + b \cot ^{2}{\left (x \right )}\right )^{\frac {3}{2}}} + \frac {b}{2 \left (a - b\right )^{2} \sqrt {a + b \cot ^{2}{\left (x \right )}}} + \frac {b \operatorname {atan}{\left (\frac {\sqrt {a + b \cot ^{2}{\left (x \right )}}}{\sqrt {- a + b}} \right )}}{2 \sqrt {- a + b} \left (a - b\right )^{2}}\right )}{b} & \text {for}\: b \neq 0 \\\begin {cases} \tilde {\infty } \cot ^{2}{\left (x \right )} & \text {for}\: a^{\frac {5}{2}} = 0 \\\frac {\log {\left (2 a^{\frac {5}{2}} \cot ^{2}{\left (x \right )} + 2 a^{\frac {5}{2}} \right )}}{2 a^{\frac {5}{2}}} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \]

[In]

integrate(cot(x)/(a+b*cot(x)**2)**(5/2),x)

[Out]

-Piecewise((2*(b/(6*(a - b)*(a + b*cot(x)**2)**(3/2)) + b/(2*(a - b)**2*sqrt(a + b*cot(x)**2)) + b*atan(sqrt(a
+ b*cot(x)**2)/sqrt(-a + b))/(2*sqrt(-a + b)*(a - b)**2))/b, Ne(b, 0)), (Piecewise((zoo*cot(x)**2, Eq(a**(5/2
), 0)), (log(2*a**(5/2)*cot(x)**2 + 2*a**(5/2))/(2*a**(5/2)), True)), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(cot(x)/(a+b*cot(x)^2)^(5/2),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. 215 vs. \(2 (66) = 132\).

Time = 0.32 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.76 \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=\frac {\log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (\sin \left (x\right )\right )}{2 \, {\left (\sqrt {a - b} a^{2} - 2 \, \sqrt {a - b} a b + \sqrt {a - b} b^{2}\right )}} - \frac {\frac {{\left (\frac {4 \, {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \sin \left (x\right )^{2}}{a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}} + \frac {3 \, {\left (a b^{2} - b^{3}\right )}}{a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}}\right )} \sin \left (x\right )}{{\left (a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b\right )}^{\frac {3}{2}}} + \frac {3 \, \log \left ({\left | -\sqrt {a - b} \sin \left (x\right ) + \sqrt {a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b} \right |}\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a - b}}}{3 \, \mathrm {sgn}\left (\sin \left (x\right )\right )} \]

[In]

integrate(cot(x)/(a+b*cot(x)^2)^(5/2),x, algorithm="giac")

[Out]

1/2*log(abs(b))*sgn(sin(x))/(sqrt(a - b)*a^2 - 2*sqrt(a - b)*a*b + sqrt(a - b)*b^2) - 1/3*((4*(a^2*b - 2*a*b^2
+ b^3)*sin(x)^2/(a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4) + 3*(a*b^2 - b^3)/(a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4))*si
n(x)/(a*sin(x)^2 - b*sin(x)^2 + b)^(3/2) + 3*log(abs(-sqrt(a - b)*sin(x) + sqrt(a*sin(x)^2 - b*sin(x)^2 + b)))
/((a^2 - 2*a*b + b^2)*sqrt(a - b)))/sgn(sin(x))

Mupad [B] (verification not implemented)

Time = 17.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.05 \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {cot}\left (x\right )}^2+a}\,\left (2\,a^2-4\,a\,b+2\,b^2\right )}{2\,{\left (a-b\right )}^{5/2}}\right )}{{\left (a-b\right )}^{5/2}}-\frac {\frac {1}{3\,\left (a-b\right )}+\frac {b\,{\mathrm {cot}\left (x\right )}^2+a}{{\left (a-b\right )}^2}}{{\left (b\,{\mathrm {cot}\left (x\right )}^2+a\right )}^{3/2}} \]

[In]

int(cot(x)/(a + b*cot(x)^2)^(5/2),x)

[Out]

atanh(((a + b*cot(x)^2)^(1/2)*(2*a^2 - 4*a*b + 2*b^2))/(2*(a - b)^(5/2)))/(a - b)^(5/2) - (1/(3*(a - b)) + (a
+ b*cot(x)^2)/(a - b)^2)/(a + b*cot(x)^2)^(3/2)