Integrand size = 12, antiderivative size = 65 \[ \int \frac {\sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\frac {\sqrt {x^2}}{6 \left (1-x^2\right )}+\frac {5}{6} \coth ^{-1}\left (\sqrt {x^2}\right )-\frac {x \sec ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}+\frac {2 x \sec ^{-1}(x)}{3 \sqrt {-1+x^2}} \] Output:
5/6*arccoth((x^2)^(1/2))-1/3*x*arcsec(x)/(x^2-1)^(3/2)+1/6*(x^2)^(1/2)/(-x ^2+1)+2/3*x*arcsec(x)/(x^2-1)^(1/2)
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03 \[ \int \frac {\sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\frac {4 x \left (-3+2 x^2\right ) \sec ^{-1}(x)+\sqrt {1-\frac {1}{x^2}} x \left (-2 x-5 \left (-1+x^2\right ) \log (1-x)+5 \left (-1+x^2\right ) \log (1+x)\right )}{12 \left (-1+x^2\right )^{3/2}} \] Input:
Integrate[ArcSec[x]/(-1 + x^2)^(5/2),x]
Output:
(4*x*(-3 + 2*x^2)*ArcSec[x] + Sqrt[1 - x^(-2)]*x*(-2*x - 5*(-1 + x^2)*Log[ 1 - x] + 5*(-1 + x^2)*Log[1 + x]))/(12*(-1 + x^2)^(3/2))
Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5751, 27, 298, 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.
\(\displaystyle \int \frac {\sec ^{-1}(x)}{\left (x^2-1\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5751 |
\(\displaystyle -\frac {x \int -\frac {3-2 x^2}{3 \left (1-x^2\right )^2}dx}{\sqrt {x^2}}+\frac {2 x \sec ^{-1}(x)}{3 \sqrt {x^2-1}}-\frac {x \sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x \int \frac {3-2 x^2}{\left (1-x^2\right )^2}dx}{3 \sqrt {x^2}}+\frac {2 x \sec ^{-1}(x)}{3 \sqrt {x^2-1}}-\frac {x \sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {x \left (\frac {5}{2} \int \frac {1}{1-x^2}dx+\frac {x}{2 \left (1-x^2\right )}\right )}{3 \sqrt {x^2}}+\frac {2 x \sec ^{-1}(x)}{3 \sqrt {x^2-1}}-\frac {x \sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {x \left (\frac {5 \text {arctanh}(x)}{2}+\frac {x}{2 \left (1-x^2\right )}\right )}{3 \sqrt {x^2}}+\frac {2 x \sec ^{-1}(x)}{3 \sqrt {x^2-1}}-\frac {x \sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}\) |
Input:
Int[ArcSec[x]/(-1 + x^2)^(5/2),x]
Output:
-1/3*(x*ArcSec[x])/(-1 + x^2)^(3/2) + (2*x*ArcSec[x])/(3*Sqrt[-1 + x^2]) + (x*(x/(2*(1 - x^2)) + (5*ArcTanh[x])/2))/(3*Sqrt[x^2])
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symb ol] :> With[{u = IntHide[(d + e*x^2)^p, x]}, Simp[(a + b*ArcSec[c*x]) u, x] - Simp[b*c*(x/Sqrt[c^2*x^2]) Int[SimplifyIntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && (IGtQ[p, 0] || ILtQ[p + 1 /2, 0])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.76 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.95
method | result | size |
default | \(\frac {\sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{2} \operatorname {csgn}\left (x \sqrt {1-\frac {1}{x^{2}}}\right ) \left (4 x^{2} \operatorname {arcsec}\left (x \right )-x \sqrt {\frac {x^{2}-1}{x^{2}}}-6 \,\operatorname {arcsec}\left (x \right )\right )}{6 \left (x^{2}-1\right )^{2}}+\frac {5 \,\operatorname {csgn}\left (x \sqrt {1-\frac {1}{x^{2}}}\right ) \ln \left (\frac {1}{x}+i \sqrt {1-\frac {1}{x^{2}}}+1\right )}{6}-\frac {5 \,\operatorname {csgn}\left (x \sqrt {1-\frac {1}{x^{2}}}\right ) \ln \left (\frac {1}{x}+i \sqrt {1-\frac {1}{x^{2}}}-1\right )}{6}\) | \(127\) |
Input:
int(arcsec(x)/(x^2-1)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/6*((x^2-1)/x^2)^(1/2)*x^2/(x^2-1)^2*csgn(x*(1-1/x^2)^(1/2))*(4*x^2*arcse c(x)-x*((x^2-1)/x^2)^(1/2)-6*arcsec(x))+5/6*csgn(x*(1-1/x^2)^(1/2))*ln(1/x +I*(1-1/x^2)^(1/2)+1)-5/6*csgn(x*(1-1/x^2)^(1/2))*ln(1/x+I*(1-1/x^2)^(1/2) -1)
Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.15 \[ \int \frac {\sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=-\frac {2 \, x^{3} - 4 \, {\left (2 \, x^{3} - 3 \, x\right )} \sqrt {x^{2} - 1} \operatorname {arcsec}\left (x\right ) - 5 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x + 1\right ) + 5 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x - 1\right ) - 2 \, x}{12 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} \] Input:
integrate(arcsec(x)/(x^2-1)^(5/2),x, algorithm="fricas")
Output:
-1/12*(2*x^3 - 4*(2*x^3 - 3*x)*sqrt(x^2 - 1)*arcsec(x) - 5*(x^4 - 2*x^2 + 1)*log(x + 1) + 5*(x^4 - 2*x^2 + 1)*log(x - 1) - 2*x)/(x^4 - 2*x^2 + 1)
Timed out. \[ \int \frac {\sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(asec(x)/(x**2-1)**(5/2),x)
Output:
Timed out
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \frac {\sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\frac {1}{3} \, {\left (\frac {2 \, x}{\sqrt {x^{2} - 1}} - \frac {x}{{\left (x^{2} - 1\right )}^{\frac {3}{2}}}\right )} \operatorname {arcsec}\left (x\right ) - \frac {x}{6 \, {\left (x^{2} - 1\right )}} + \frac {5}{12} \, \log \left (x + 1\right ) - \frac {5}{12} \, \log \left (x - 1\right ) \] Input:
integrate(arcsec(x)/(x^2-1)^(5/2),x, algorithm="maxima")
Output:
1/3*(2*x/sqrt(x^2 - 1) - x/(x^2 - 1)^(3/2))*arcsec(x) - 1/6*x/(x^2 - 1) + 5/12*log(x + 1) - 5/12*log(x - 1)
Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, x^{2} - 3\right )} x \arccos \left (\frac {1}{x}\right )}{3 \, {\left (x^{2} - 1\right )}^{\frac {3}{2}}} + \frac {5 \, \log \left ({\left | x + 1 \right |}\right )}{12 \, \mathrm {sgn}\left (x\right )} - \frac {5 \, \log \left ({\left | x - 1 \right |}\right )}{12 \, \mathrm {sgn}\left (x\right )} - \frac {x}{6 \, {\left (x^{2} - 1\right )} \mathrm {sgn}\left (x\right )} \] Input:
integrate(arcsec(x)/(x^2-1)^(5/2),x, algorithm="giac")
Output:
1/3*(2*x^2 - 3)*x*arccos(1/x)/(x^2 - 1)^(3/2) + 5/12*log(abs(x + 1))/sgn(x ) - 5/12*log(abs(x - 1))/sgn(x) - 1/6*x/((x^2 - 1)*sgn(x))
Timed out. \[ \int \frac {\sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\int \frac {\mathrm {acos}\left (\frac {1}{x}\right )}{{\left (x^2-1\right )}^{5/2}} \,d x \] Input:
int(acos(1/x)/(x^2 - 1)^(5/2),x)
Output:
int(acos(1/x)/(x^2 - 1)^(5/2), x)
\[ \int \frac {\sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\int \frac {\mathit {asec} \left (x \right )}{\sqrt {x^{2}-1}\, x^{4}-2 \sqrt {x^{2}-1}\, x^{2}+\sqrt {x^{2}-1}}d x \] Input:
int(asec(x)/(x^2-1)^(5/2),x)
Output:
int(asec(x)/(sqrt(x**2 - 1)*x**4 - 2*sqrt(x**2 - 1)*x**2 + sqrt(x**2 - 1)) ,x)