Integrand size = 15, antiderivative size = 78 \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2}}-\frac {1}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac {1}{(a-b)^2 \sqrt {a+b \cot ^2(x)}} \] Output:
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)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.60 \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \cot ^2(x)}{a-b}\right )}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}} \] Input:
Integrate[Cot[x]/(a + b*Cot[x]^2)^(5/2),x]
Output:
-1/3*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Cot[x]^2)/(a - b)]/((a - b)*( a + b*Cot[x]^2)^(3/2))
Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.19, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 25, 4153, 25, 353, 61, 61, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\tan \left (x+\frac {\pi }{2}\right )}{\left (a+b \tan \left (x+\frac {\pi }{2}\right )^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\tan \left (x+\frac {\pi }{2}\right )}{\left (b \tan \left (x+\frac {\pi }{2}\right )^2+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \int -\frac {\cot (x)}{\left (\cot ^2(x)+1\right ) \left (a+b \cot ^2(x)\right )^{5/2}}d\cot (x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cot (x)}{\left (\cot ^2(x)+1\right ) \left (b \cot ^2(x)+a\right )^{5/2}}d\cot (x)\) |
\(\Big \downarrow \) 353 |
\(\displaystyle -\frac {1}{2} \int \frac {1}{\left (\cot ^2(x)+1\right ) \left (b \cot ^2(x)+a\right )^{5/2}}d\cot ^2(x)\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {1}{\left (\cot ^2(x)+1\right ) \left (b \cot ^2(x)+a\right )^{3/2}}d\cot ^2(x)}{a-b}-\frac {2}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {\int \frac {1}{\left (\cot ^2(x)+1\right ) \sqrt {b \cot ^2(x)+a}}d\cot ^2(x)}{a-b}+\frac {2}{(a-b) \sqrt {a+b \cot ^2(x)}}}{a-b}-\frac {2}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {2 \int \frac {1}{\frac {\cot ^4(x)}{b}-\frac {a}{b}+1}d\sqrt {b \cot ^2(x)+a}}{b (a-b)}+\frac {2}{(a-b) \sqrt {a+b \cot ^2(x)}}}{a-b}-\frac {2}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {2}{(a-b) \sqrt {a+b \cot ^2(x)}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \cot ^2(x)}}{\sqrt {a-b}}\right )}{(a-b)^{3/2}}}{a-b}-\frac {2}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}\right )\) |
Input:
Int[Cot[x]/(a + b*Cot[x]^2)^(5/2),x]
Output:
(-2/(3*(a - b)*(a + b*Cot[x]^2)^(3/2)) - ((-2*ArcTanh[Sqrt[a + b*Cot[x]^2] /Sqrt[a - b]])/(a - b)^(3/2) + 2/((a - b)*Sqrt[a + b*Cot[x]^2]))/(a - b))/ 2
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] - Simp[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] && 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]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ {a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
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]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^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] && Ratio nalQ[n]))
Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(-\frac {\arctan \left (\frac {\sqrt {a +b \cot \left (x \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right )^{2} \sqrt {-a +b}}-\frac {1}{3 \left (a -b \right ) \left (a +b \cot \left (x \right )^{2}\right )^{\frac {3}{2}}}-\frac {1}{\left (a -b \right )^{2} \sqrt {a +b \cot \left (x \right )^{2}}}\) | \(75\) |
default | \(-\frac {\arctan \left (\frac {\sqrt {a +b \cot \left (x \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right )^{2} \sqrt {-a +b}}-\frac {1}{3 \left (a -b \right ) \left (a +b \cot \left (x \right )^{2}\right )^{\frac {3}{2}}}-\frac {1}{\left (a -b \right )^{2} \sqrt {a +b \cot \left (x \right )^{2}}}\) | \(75\) |
Input:
int(cot(x)/(a+b*cot(x)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/(a-b)^2/(-a+b)^(1/2)*arctan((a+b*cot(x)^2)^(1/2)/(-a+b)^(1/2))-1/3/(a-b )/(a+b*cot(x)^2)^(3/2)-1/(a-b)^2/(a+b*cot(x)^2)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (66) = 132\).
Time = 0.10 (sec) , antiderivative size = 645, normalized size of antiderivative = 8.27 \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate(cot(x)/(a+b*cot(x)^2)^(5/2),x, algorithm="fricas")
Output:
[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))*sq rt(((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))*(cos(2*x) - 1)/((a - b)*cos(2*x) - 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))]
Time = 7.77 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.41 \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=- \begin {cases} \frac {2 \left (\frac {b}{6 \left (a - b\right ) \left (a + b \cot ^{2}{\left (x \right )}\right )^{\frac {3}{2}}} + \frac {b}{2 \left (a - b\right )^{2} \sqrt {a + b \cot ^{2}{\left (x \right )}}} + \frac {b \operatorname {atan}{\left (\frac {\sqrt {a + b \cot ^{2}{\left (x \right )}}}{\sqrt {- a + b}} \right )}}{2 \sqrt {- a + b} \left (a - b\right )^{2}}\right )}{b} & \text {for}\: b \neq 0 \\\begin {cases} \tilde {\infty } \cot ^{2}{\left (x \right )} & \text {for}\: a^{\frac {5}{2}} = 0 \\\frac {\log {\left (2 a^{\frac {5}{2}} \cot ^{2}{\left (x \right )} + 2 a^{\frac {5}{2}} \right )}}{2 a^{\frac {5}{2}}} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \] Input:
integrate(cot(x)/(a+b*cot(x)**2)**(5/2),x)
Output:
-Piecewise((2*(b/(6*(a - b)*(a + b*cot(x)**2)**(3/2)) + b/(2*(a - b)**2*sq rt(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))
Exception generated. \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cot(x)/(a+b*cot(x)^2)^(5/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (66) = 132\).
Time = 0.16 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.76 \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=\frac {\log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (\sin \left (x\right )\right )}{2 \, {\left (\sqrt {a - b} a^{2} - 2 \, \sqrt {a - b} a b + \sqrt {a - b} b^{2}\right )}} - \frac {\frac {{\left (\frac {4 \, {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \sin \left (x\right )^{2}}{a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}} + \frac {3 \, {\left (a b^{2} - b^{3}\right )}}{a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}}\right )} \sin \left (x\right )}{{\left (a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b\right )}^{\frac {3}{2}}} + \frac {3 \, \log \left ({\left | -\sqrt {a - b} \sin \left (x\right ) + \sqrt {a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b} \right |}\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a - b}}}{3 \, \mathrm {sgn}\left (\sin \left (x\right )\right )} \] Input:
integrate(cot(x)/(a+b*cot(x)^2)^(5/2),x, algorithm="giac")
Output:
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))*sin( x)/(a*sin(x)^2 - b*sin(x)^2 + b)^(3/2) + 3*log(abs(-sqrt(a - b)*sin(x) + s qrt(a*sin(x)^2 - b*sin(x)^2 + b)))/((a^2 - 2*a*b + b^2)*sqrt(a - b)))/sgn( sin(x))
Time = 13.79 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.05 \[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {cot}\left (x\right )}^2+a}\,\left (2\,a^2-4\,a\,b+2\,b^2\right )}{2\,{\left (a-b\right )}^{5/2}}\right )}{{\left (a-b\right )}^{5/2}}-\frac {\frac {1}{3\,\left (a-b\right )}+\frac {b\,{\mathrm {cot}\left (x\right )}^2+a}{{\left (a-b\right )}^2}}{{\left (b\,{\mathrm {cot}\left (x\right )}^2+a\right )}^{3/2}} \] Input:
int(cot(x)/(a + b*cot(x)^2)^(5/2),x)
Output:
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)
\[ \int \frac {\cot (x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\cot \left (x \right )^{2} b +a}\, \cot \left (x \right )}{\cot \left (x \right )^{6} b^{3}+3 \cot \left (x \right )^{4} a \,b^{2}+3 \cot \left (x \right )^{2} a^{2} b +a^{3}}d x \] Input:
int(cot(x)/(a+b*cot(x)^2)^(5/2),x)
Output:
int((sqrt(cot(x)**2*b + a)*cot(x))/(cot(x)**6*b**3 + 3*cot(x)**4*a*b**2 + 3*cot(x)**2*a**2*b + a**3),x)