3.688 \(\int \frac{x^3 \sec ^{-1}(x)}{(-1+x^2)^{5/2}} \, dx\)

Optimal. Leaf size=82 \[ \frac{x}{6 \sqrt{x^2} \left (1-x^2\right )}-\frac{2 x \log (x)}{3 \sqrt{x^2}}+\frac{x \log \left (x^2-1\right )}{3 \sqrt{x^2}}-\frac{\sec ^{-1}(x)}{\sqrt{x^2-1}}-\frac{\sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}} \]

[Out]

x/(6*Sqrt[x^2]*(1 - x^2)) - ArcSec[x]/(3*(-1 + x^2)^(3/2)) - ArcSec[x]/Sqrt[-1 + x^2] - (2*x*Log[x])/(3*Sqrt[x
^2]) + (x*Log[-1 + x^2])/(3*Sqrt[x^2])

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Rubi [A]  time = 0.0897121, antiderivative size = 84, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {266, 43, 5238, 12, 446, 77} \[ \frac{x}{6 \sqrt{x^2} \left (1-x^2\right )}-\frac{2 x \log (x)}{3 \sqrt{x^2}}+\frac{x \log \left (1-x^2\right )}{3 \sqrt{x^2}}-\frac{\sec ^{-1}(x)}{\sqrt{x^2-1}}-\frac{\sec ^{-1}(x)}{3 \left (x^2-1\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Int[(x^3*ArcSec[x])/(-1 + x^2)^(5/2),x]

[Out]

x/(6*Sqrt[x^2]*(1 - x^2)) - ArcSec[x]/(3*(-1 + x^2)^(3/2)) - ArcSec[x]/Sqrt[-1 + x^2] - (2*x*Log[x])/(3*Sqrt[x
^2]) + (x*Log[1 - x^2])/(3*Sqrt[x^2])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 5238

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]}, Dist[a + b*ArcSec[c*x], u, x] - Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyI
ntegrand[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])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((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]))))

Rubi steps

\begin{align*} \int \frac{x^3 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx &=-\frac{\sec ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}-\frac{\sec ^{-1}(x)}{\sqrt{-1+x^2}}-\frac{x \int \frac{2-3 x^2}{3 x \left (1-x^2\right )^2} \, dx}{\sqrt{x^2}}\\ &=-\frac{\sec ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}-\frac{\sec ^{-1}(x)}{\sqrt{-1+x^2}}-\frac{x \int \frac{2-3 x^2}{x \left (1-x^2\right )^2} \, dx}{3 \sqrt{x^2}}\\ &=-\frac{\sec ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}-\frac{\sec ^{-1}(x)}{\sqrt{-1+x^2}}-\frac{x \operatorname{Subst}\left (\int \frac{2-3 x}{(1-x)^2 x} \, dx,x,x^2\right )}{6 \sqrt{x^2}}\\ &=-\frac{\sec ^{-1}(x)}{3 \left (-1+x^2\right )^{3/2}}-\frac{\sec ^{-1}(x)}{\sqrt{-1+x^2}}-\frac{x \operatorname{Subst}\left (\int \left (-\frac{1}{(-1+x)^2}-\frac{2}{-1+x}+\frac{2}{x}\right ) \, dx,x,x^2\right )}{6 \sqrt{x^2}}\\ &=\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}}\\ \end{align*}

Mathematica [A]  time = 0.174818, size = 72, normalized size = 0.88 \[ \frac{-\frac{\left (x^2-1\right ) \left (4 \left (x^2-1\right ) \log (x)-2 \left (x^2-1\right ) \log \left (1-x^2\right )+1\right )}{\sqrt{1-\frac{1}{x^2}} x}-2 \left (3 x^2-2\right ) \sec ^{-1}(x)}{6 \left (x^2-1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*ArcSec[x])/(-1 + x^2)^(5/2),x]

[Out]

(-2*(-2 + 3*x^2)*ArcSec[x] - ((-1 + x^2)*(1 + 4*(-1 + x^2)*Log[x] - 2*(-1 + x^2)*Log[1 - x^2]))/(Sqrt[1 - x^(-
2)]*x))/(6*(-1 + x^2)^(3/2))

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Maple [C]  time = 0.403, size = 197, normalized size = 2.4 \begin{align*}{-{\frac{4\,i}{3}}x{\rm arcsec} \left (x\right )\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{\frac{1}{\sqrt{{x}^{2}-1}}}}+{\frac{1}{6\,{x}^{2} \left ( 4\,{x}^{6}-11\,{x}^{4}+10\,{x}^{2}-3 \right ) }\sqrt{{x}^{2}-1} \left ( 2\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}-2\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x-3\,{x}^{2}+2 \right ) \left ( 8\,{\rm arcsec} \left (x\right ){x}^{4}+2\,i{x}^{4}+3\,\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}-6\,{\rm arcsec} \left (x\right ){x}^{2}-4\,i{x}^{2}-2\,\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x+2\,i \right ) }+{\frac{2\,x}{3}\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}\ln \left ( \left ({x}^{-1}+i\sqrt{1-{x}^{-2}} \right ) ^{2}-1 \right ){\frac{1}{\sqrt{{x}^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsec(x)/(x^2-1)^(5/2),x)

[Out]

-4/3*I/(x^2-1)^(1/2)*((x^2-1)/x^2)^(1/2)*x*arcsec(x)+1/6*(x^2-1)^(1/2)/x^2*(2*I*((x^2-1)/x^2)^(1/2)*x^3-2*I*((
x^2-1)/x^2)^(1/2)*x-3*x^2+2)*(8*arcsec(x)*x^4+2*I*x^4+3*((x^2-1)/x^2)^(1/2)*x^3-6*arcsec(x)*x^2-4*I*x^2-2*((x^
2-1)/x^2)^(1/2)*x+2*I)/(4*x^6-11*x^4+10*x^2-3)+2/3/(x^2-1)^(1/2)*((x^2-1)/x^2)^(1/2)*x*ln((1/x+I*(1-1/x^2)^(1/
2))^2-1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \operatorname{arcsec}\left (x\right )}{{\left (x^{2} - 1\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsec(x)/(x^2-1)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^3*arcsec(x)/(x^2 - 1)^(5/2), x)

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Fricas [A]  time = 2.62936, size = 186, normalized size = 2.27 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsec(x)/(x^2-1)^(5/2),x, algorithm="fricas")

[Out]

-1/6*(2*(3*x^2 - 2)*sqrt(x^2 - 1)*arcsec(x) + x^2 - 2*(x^4 - 2*x^2 + 1)*log(x^2 - 1) + 4*(x^4 - 2*x^2 + 1)*log
(x) - 1)/(x^4 - 2*x^2 + 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asec(x)/(x**2-1)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.11892, size = 86, normalized size = 1.05 \begin{align*} -\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsec(x)/(x^2-1)^(5/2),x, algorithm="giac")

[Out]

-1/3*(3*x^2 - 2)*arccos(1/x)/(x^2 - 1)^(3/2) - 1/3*log(x^2)/sgn(x) + 1/3*log(abs(x^2 - 1))/sgn(x) - 1/6*(2*x^2
- 1)/((x^2 - 1)*sgn(x))