Integrand size = 15, antiderivative size = 82 \[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\frac {x}{6 \sqrt {x^2} \left (1-x^2\right )}-\frac {\sec ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}-\frac {\sec ^{-1}(x)}{\sqrt {-1+x^2}}-\frac {2 x \log (x)}{3 \sqrt {x^2}}+\frac {x \log \left (-1+x^2\right )}{3 \sqrt {x^2}} \] Output:
-1/3*arcsec(x)/(x^2-1)^(3/2)+1/6*x/(-x^2+1)/(x^2)^(1/2)-2/3*x*ln(x)/(x^2)^ (1/2)+1/3*x*ln(x^2-1)/(x^2)^(1/2)-arcsec(x)/(x^2-1)^(1/2)
Time = 0.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\frac {-2 \left (-2+3 x^2\right ) \sec ^{-1}(x)+\sqrt {1-\frac {1}{x^2}} x \left (-1+2 \left (-1+x^2\right ) \log (-1+x)-4 \left (-1+x^2\right ) \log (x)-2 \log (1+x)+2 x^2 \log (1+x)\right )}{6 \left (-1+x^2\right )^{3/2}} \] Input:
Integrate[(x^3*ArcSec[x])/(-1 + x^2)^(5/2),x]
Output:
(-2*(-2 + 3*x^2)*ArcSec[x] + Sqrt[1 - x^(-2)]*x*(-1 + 2*(-1 + x^2)*Log[-1 + x] - 4*(-1 + x^2)*Log[x] - 2*Log[1 + x] + 2*x^2*Log[1 + x]))/(6*(-1 + x^ 2)^(3/2))
Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5761, 27, 354, 86, 2009}
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.
\(\displaystyle \int \frac {x^3 \sec ^{-1}(x)}{\left (x^2-1\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5761 |
\(\displaystyle -\frac {x \int \frac {2-3 x^2}{3 x \left (1-x^2\right )^2}dx}{\sqrt {x^2}}-\frac {\sec ^{-1}(x)}{\sqrt {x^2-1}}-\frac {\sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {x \int \frac {2-3 x^2}{x \left (1-x^2\right )^2}dx}{3 \sqrt {x^2}}-\frac {\sec ^{-1}(x)}{\sqrt {x^2-1}}-\frac {\sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {x \int \frac {2-3 x^2}{x^2 \left (1-x^2\right )^2}dx^2}{6 \sqrt {x^2}}-\frac {\sec ^{-1}(x)}{\sqrt {x^2-1}}-\frac {\sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {x \int \left (\frac {2}{x^2}-\frac {2}{x^2-1}-\frac {1}{\left (x^2-1\right )^2}\right )dx^2}{6 \sqrt {x^2}}-\frac {\sec ^{-1}(x)}{\sqrt {x^2-1}}-\frac {\sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {x \left (\frac {1}{x^2-1}+2 \log \left (x^2\right )-2 \log \left (1-x^2\right )\right )}{6 \sqrt {x^2}}-\frac {\sec ^{-1}(x)}{\sqrt {x^2-1}}-\frac {\sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}}\) |
Input:
Int[(x^3*ArcSec[x])/(-1 + x^2)^(5/2),x]
Output:
-1/3*ArcSec[x]/(-1 + x^2)^(3/2) - ArcSec[x]/Sqrt[-1 + x^2] - (x*((-1 + x^2 )^(-1) + 2*Log[x^2] - 2*Log[1 - x^2]))/(6*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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(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] && IntegerQ [(m - 1)/2]
Int[((a_.) + ArcSec[(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*ArcSec[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 = 0.40 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.23
method | result | size |
default | \(\frac {\operatorname {csgn}\left (x \sqrt {1-\frac {1}{x^{2}}}\right ) \left (2 \ln \left (1-\frac {1}{x^{2}}\right ) x^{4}-6 \,\operatorname {arcsec}\left (x \right ) x^{3} \sqrt {\frac {x^{2}-1}{x^{2}}}-x^{4}-4 \ln \left (1-\frac {1}{x^{2}}\right ) x^{2}+4 \,\operatorname {arcsec}\left (x \right ) \sqrt {\frac {x^{2}-1}{x^{2}}}\, x +x^{2}+2 \ln \left (1-\frac {1}{x^{2}}\right )\right )}{6 \left (x^{2}-1\right )^{2}}\) | \(101\) |
Input:
int(x^3*arcsec(x)/(x^2-1)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/6*csgn(x*(1-1/x^2)^(1/2))*(2*ln(1-1/x^2)*x^4-6*arcsec(x)*x^3*((x^2-1)/x^ 2)^(1/2)-x^4-4*ln(1-1/x^2)*x^2+4*arcsec(x)*((x^2-1)/x^2)^(1/2)*x+x^2+2*ln( 1-1/x^2))/(x^2-1)^2
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84 \[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, x^{2} - 2\right )} \sqrt {x^{2} - 1} \operatorname {arcsec}\left (x\right ) + x^{2} - 2 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x^{2} - 1\right ) + 4 \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \log \left (x\right ) - 1}{6 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} \] Input:
integrate(x^3*arcsec(x)/(x^2-1)^(5/2),x, algorithm="fricas")
Output:
-1/6*(2*(3*x^2 - 2)*sqrt(x^2 - 1)*arcsec(x) + x^2 - 2*(x^4 - 2*x^2 + 1)*lo g(x^2 - 1) + 4*(x^4 - 2*x^2 + 1)*log(x) - 1)/(x^4 - 2*x^2 + 1)
Timed out. \[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(x**3*asec(x)/(x**2-1)**(5/2),x)
Output:
Timed out
\[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\int { \frac {x^{3} \operatorname {arcsec}\left (x\right )}{{\left (x^{2} - 1\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(x^3*arcsec(x)/(x^2-1)^(5/2),x, algorithm="maxima")
Output:
integrate(x^3*arcsec(x)/(x^2 - 1)^(5/2), x)
Time = 0.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.78 \[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=-\frac {{\left (3 \, x^{2} - 2\right )} \arccos \left (\frac {1}{x}\right )}{3 \, {\left (x^{2} - 1\right )}^{\frac {3}{2}}} - \frac {\log \left (x^{2}\right )}{3 \, \mathrm {sgn}\left (x\right )} + \frac {\log \left ({\left | x^{2} - 1 \right |}\right )}{3 \, \mathrm {sgn}\left (x\right )} - \frac {2 \, x^{2} - 1}{6 \, {\left (x^{2} - 1\right )} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^3*arcsec(x)/(x^2-1)^(5/2),x, algorithm="giac")
Output:
-1/3*(3*x^2 - 2)*arccos(1/x)/(x^2 - 1)^(3/2) - 1/3*log(x^2)/sgn(x) + 1/3*l og(abs(x^2 - 1))/sgn(x) - 1/6*(2*x^2 - 1)/((x^2 - 1)*sgn(x))
Timed out. \[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\mathrm {acos}\left (\frac {1}{x}\right )}{{\left (x^2-1\right )}^{5/2}} \,d x \] Input:
int((x^3*acos(1/x))/(x^2 - 1)^(5/2),x)
Output:
int((x^3*acos(1/x))/(x^2 - 1)^(5/2), x)
\[ \int \frac {x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx=\int \frac {\mathit {asec} \left (x \right ) x^{3}}{\sqrt {x^{2}-1}\, x^{4}-2 \sqrt {x^{2}-1}\, x^{2}+\sqrt {x^{2}-1}}d x \] Input:
int(x^3*asec(x)/(x^2-1)^(5/2),x)
Output:
int((asec(x)*x**3)/(sqrt(x**2 - 1)*x**4 - 2*sqrt(x**2 - 1)*x**2 + sqrt(x** 2 - 1)),x)