Integrand size = 15, antiderivative size = 70 \[ \int \frac {\csc ^{-1}(x)}{x^2 \left (-1+x^2\right )^{5/2}} \, dx=-\frac {1}{\sqrt {x^2}}+\frac {\sqrt {x^2}}{6 \left (-1+x^2\right )}-\frac {11}{6} \coth ^{-1}\left (\sqrt {x^2}\right )+\frac {\left (3-12 x^2+8 x^4\right ) \csc ^{-1}(x)}{3 x \left (-1+x^2\right )^{3/2}} \]
-11/6*arccoth((x^2)^(1/2))+1/3*(8*x^4-12*x^2+3)*arccsc(x)/x/(x^2-1)^(3/2)- 1/(x^2)^(1/2)+1/6*(x^2)^(1/2)/(x^2-1)
Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.13 \[ \int \frac {\csc ^{-1}(x)}{x^2 \left (-1+x^2\right )^{5/2}} \, dx=\frac {4 \left (3-12 x^2+8 x^4\right ) \csc ^{-1}(x)+\sqrt {1-\frac {1}{x^2}} x \left (12-10 x^2+11 x \left (-1+x^2\right ) \log (1-x)-11 x \left (-1+x^2\right ) \log (1+x)\right )}{12 x \left (-1+x^2\right )^{3/2}} \]
(4*(3 - 12*x^2 + 8*x^4)*ArcCsc[x] + Sqrt[1 - x^(-2)]*x*(12 - 10*x^2 + 11*x *(-1 + x^2)*Log[1 - x] - 11*x*(-1 + x^2)*Log[1 + x]))/(12*x*(-1 + x^2)^(3/ 2))
Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.27, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5762, 27, 1582, 25, 359, 219}
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.
\(\displaystyle \int \frac {\csc ^{-1}(x)}{x^2 \left (x^2-1\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5762 |
\(\displaystyle \frac {x \int \frac {8 x^4-12 x^2+3}{3 x^2 \left (1-x^2\right )^2}dx}{\sqrt {x^2}}+\frac {8 x \csc ^{-1}(x)}{3 \sqrt {x^2-1}}-\frac {4 x \csc ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}+\frac {\csc ^{-1}(x)}{x \left (x^2-1\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x \int \frac {8 x^4-12 x^2+3}{x^2 \left (1-x^2\right )^2}dx}{3 \sqrt {x^2}}+\frac {8 x \csc ^{-1}(x)}{3 \sqrt {x^2-1}}-\frac {4 x \csc ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}+\frac {\csc ^{-1}(x)}{x \left (x^2-1\right )^{3/2}}\) |
\(\Big \downarrow \) 1582 |
\(\displaystyle \frac {x \left (-\frac {1}{2} \int -\frac {6-17 x^2}{x^2 \left (1-x^2\right )}dx-\frac {x}{2 \left (1-x^2\right )}\right )}{3 \sqrt {x^2}}+\frac {8 x \csc ^{-1}(x)}{3 \sqrt {x^2-1}}-\frac {4 x \csc ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}+\frac {\csc ^{-1}(x)}{x \left (x^2-1\right )^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {x \left (\frac {1}{2} \int \frac {6-17 x^2}{x^2 \left (1-x^2\right )}dx-\frac {x}{2 \left (1-x^2\right )}\right )}{3 \sqrt {x^2}}+\frac {8 x \csc ^{-1}(x)}{3 \sqrt {x^2-1}}-\frac {4 x \csc ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}+\frac {\csc ^{-1}(x)}{x \left (x^2-1\right )^{3/2}}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {x \left (\frac {1}{2} \left (-11 \int \frac {1}{1-x^2}dx-\frac {6}{x}\right )-\frac {x}{2 \left (1-x^2\right )}\right )}{3 \sqrt {x^2}}+\frac {8 x \csc ^{-1}(x)}{3 \sqrt {x^2-1}}-\frac {4 x \csc ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}+\frac {\csc ^{-1}(x)}{x \left (x^2-1\right )^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {x \left (\frac {1}{2} \left (-11 \text {arctanh}(x)-\frac {6}{x}\right )-\frac {x}{2 \left (1-x^2\right )}\right )}{3 \sqrt {x^2}}+\frac {8 x \csc ^{-1}(x)}{3 \sqrt {x^2-1}}-\frac {4 x \csc ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}+\frac {\csc ^{-1}(x)}{x \left (x^2-1\right )^{3/2}}\) |
ArcCsc[x]/(x*(-1 + x^2)^(3/2)) - (4*x*ArcCsc[x])/(3*(-1 + x^2)^(3/2)) + (8 *x*ArcCsc[x])/(3*Sqrt[-1 + x^2]) + (x*(-1/2*x/(1 - x^2) + (-6/x - 11*ArcTa nh[x])/2))/(3*Sqrt[x^2])
3.7.91.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[(-d)^(m/2 - 1)/(2*e^ (2*p)*(q + 1)) Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e *x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]
Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x _)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Sim p[(a + b*ArcCsc[c*x]) u, x] + Simp[b*c*(x/Sqrt[c^2*x^2]) Int[SimplifyIn tegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] && !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) | | (IGtQ[(m + 1)/2, 0] && !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (ILtQ[(m + 2*p + 1)/2, 0] && !ILtQ[(m - 1)/2, 0]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.72 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.91
method | result | size |
default | \(\frac {\operatorname {csgn}\left (x \sqrt {1-\frac {1}{x^{2}}}\right ) \left (\operatorname {arccsc}\left (x \right )+i\right ) \left (\sqrt {\frac {x^{2}-1}{x^{2}}}\, x +i\right )}{2 x}+\frac {\left (\sqrt {\frac {x^{2}-1}{x^{2}}}\, x -i\right ) \operatorname {csgn}\left (x \sqrt {1-\frac {1}{x^{2}}}\right ) \left (\operatorname {arccsc}\left (x \right )-i\right )}{2 x}+\frac {\sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{2} \operatorname {csgn}\left (x \sqrt {1-\frac {1}{x^{2}}}\right ) \left (10 \,\operatorname {arccsc}\left (x \right ) x^{2}+\sqrt {\frac {x^{2}-1}{x^{2}}}\, x -12 \,\operatorname {arccsc}\left (x \right )\right )}{6 \left (x^{2}-1\right )^{2}}-\frac {11 \,\operatorname {csgn}\left (x \sqrt {1-\frac {1}{x^{2}}}\right ) \ln \left (\frac {i}{x}+\sqrt {1-\frac {1}{x^{2}}}+i\right )}{6}+\frac {11 \,\operatorname {csgn}\left (x \sqrt {1-\frac {1}{x^{2}}}\right ) \ln \left (\frac {i}{x}+\sqrt {1-\frac {1}{x^{2}}}-i\right )}{6}\) | \(204\) |
1/2*csgn(x*(1-1/x^2)^(1/2))*(arccsc(x)+I)*(((x^2-1)/x^2)^(1/2)*x+I)/x+1/2* (((x^2-1)/x^2)^(1/2)*x-I)/x*csgn(x*(1-1/x^2)^(1/2))*(arccsc(x)-I)+1/6*((x^ 2-1)/x^2)^(1/2)*x^2/(x^2-1)^2*csgn(x*(1-1/x^2)^(1/2))*(10*arccsc(x)*x^2+(( x^2-1)/x^2)^(1/2)*x-12*arccsc(x))-11/6*csgn(x*(1-1/x^2)^(1/2))*ln(I/x+(1-1 /x^2)^(1/2)+I)+11/6*csgn(x*(1-1/x^2)^(1/2))*ln(I/x+(1-1/x^2)^(1/2)-I)
Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.16 \[ \int \frac {\csc ^{-1}(x)}{x^2 \left (-1+x^2\right )^{5/2}} \, dx=-\frac {10 \, x^{4} - 4 \, {\left (8 \, x^{4} - 12 \, x^{2} + 3\right )} \sqrt {x^{2} - 1} \operatorname {arccsc}\left (x\right ) - 22 \, x^{2} + 11 \, {\left (x^{5} - 2 \, x^{3} + x\right )} \log \left (x + 1\right ) - 11 \, {\left (x^{5} - 2 \, x^{3} + x\right )} \log \left (x - 1\right ) + 12}{12 \, {\left (x^{5} - 2 \, x^{3} + x\right )}} \]
-1/12*(10*x^4 - 4*(8*x^4 - 12*x^2 + 3)*sqrt(x^2 - 1)*arccsc(x) - 22*x^2 + 11*(x^5 - 2*x^3 + x)*log(x + 1) - 11*(x^5 - 2*x^3 + x)*log(x - 1) + 12)/(x ^5 - 2*x^3 + x)
Timed out. \[ \int \frac {\csc ^{-1}(x)}{x^2 \left (-1+x^2\right )^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (56) = 112\).
Time = 0.55 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.76 \[ \int \frac {\csc ^{-1}(x)}{x^2 \left (-1+x^2\right )^{5/2}} \, dx=\frac {32 \, x^{4} \arctan \left (1, \sqrt {x + 1} \sqrt {x - 1}\right ) - {\left (x^{3} - x\right )} \sqrt {x + 1} \sqrt {x - 1} {\left (\frac {2 \, {\left (5 \, x^{2} - 6\right )}}{x^{3} - x} + 11 \, \log \left (x + 1\right ) - 11 \, \log \left (x - 1\right )\right )} - 48 \, x^{2} \arctan \left (1, \sqrt {x + 1} \sqrt {x - 1}\right ) + 12 \, \arctan \left (1, \sqrt {x + 1} \sqrt {x - 1}\right )}{12 \, {\left (x^{3} - x\right )} \sqrt {x + 1} \sqrt {x - 1}} \]
1/12*(32*x^4*arctan2(1, sqrt(x + 1)*sqrt(x - 1)) - (x^3 - x)*sqrt(x + 1)*s qrt(x - 1)*(2*(5*x^2 - 6)/(x^3 - x) + 11*log(x + 1) - 11*log(x - 1)) - 48* x^2*arctan2(1, sqrt(x + 1)*sqrt(x - 1)) + 12*arctan2(1, sqrt(x + 1)*sqrt(x - 1)))/((x^3 - x)*sqrt(x + 1)*sqrt(x - 1))
Time = 0.35 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.50 \[ \int \frac {\csc ^{-1}(x)}{x^2 \left (-1+x^2\right )^{5/2}} \, dx=\frac {1}{3} \, {\left (\frac {{\left (5 \, x^{2} - 6\right )} x}{{\left (x^{2} - 1\right )}^{\frac {3}{2}}} + \frac {6}{{\left (x - \sqrt {x^{2} - 1}\right )}^{2} + 1}\right )} \arcsin \left (\frac {1}{x}\right ) + \frac {2 \, \arctan \left (-x + \sqrt {x^{2} - 1}\right )}{\mathrm {sgn}\left (x\right )} - \frac {11 \, \log \left ({\left | x + 1 \right |}\right )}{12 \, \mathrm {sgn}\left (x\right )} + \frac {11 \, \log \left ({\left | x - 1 \right |}\right )}{12 \, \mathrm {sgn}\left (x\right )} - \frac {5 \, x^{2} - 6}{6 \, {\left (x^{3} - x\right )} \mathrm {sgn}\left (x\right )} \]
1/3*((5*x^2 - 6)*x/(x^2 - 1)^(3/2) + 6/((x - sqrt(x^2 - 1))^2 + 1))*arcsin (1/x) + 2*arctan(-x + sqrt(x^2 - 1))/sgn(x) - 11/12*log(abs(x + 1))/sgn(x) + 11/12*log(abs(x - 1))/sgn(x) - 1/6*(5*x^2 - 6)/((x^3 - x)*sgn(x))
Timed out. \[ \int \frac {\csc ^{-1}(x)}{x^2 \left (-1+x^2\right )^{5/2}} \, dx=\int \frac {\mathrm {asin}\left (\frac {1}{x}\right )}{x^2\,{\left (x^2-1\right )}^{5/2}} \,d x \]