3.685 \(\int \frac{\sqrt{-1+x^2} \sec ^{-1}(x)}{x^4} \, dx\)

Optimal. Leaf size=41 \[ \frac{1}{3 \sqrt{x^2}}-\frac{1}{9 \left (x^2\right )^{3/2}}+\frac{\left (x^2-1\right )^{3/2} \sec ^{-1}(x)}{3 x^3} \]

[Out]

-1/(9*(x^2)^(3/2)) + 1/(3*Sqrt[x^2]) + ((-1 + x^2)^(3/2)*ArcSec[x])/(3*x^3)

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Rubi [A]  time = 0.0519343, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {264, 5238, 12, 14} \[ \frac{1}{3 \sqrt{x^2}}-\frac{1}{9 \left (x^2\right )^{3/2}}+\frac{\left (x^2-1\right )^{3/2} \sec ^{-1}(x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[-1 + x^2]*ArcSec[x])/x^4,x]

[Out]

-1/(9*(x^2)^(3/2)) + 1/(3*Sqrt[x^2]) + ((-1 + x^2)^(3/2)*ArcSec[x])/(3*x^3)

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{\sqrt{-1+x^2} \sec ^{-1}(x)}{x^4} \, dx &=\frac{\left (-1+x^2\right )^{3/2} \sec ^{-1}(x)}{3 x^3}-\frac{x \int \frac{-1+x^2}{3 x^4} \, dx}{\sqrt{x^2}}\\ &=\frac{\left (-1+x^2\right )^{3/2} \sec ^{-1}(x)}{3 x^3}-\frac{x \int \frac{-1+x^2}{x^4} \, dx}{3 \sqrt{x^2}}\\ &=\frac{\left (-1+x^2\right )^{3/2} \sec ^{-1}(x)}{3 x^3}-\frac{x \int \left (-\frac{1}{x^4}+\frac{1}{x^2}\right ) \, dx}{3 \sqrt{x^2}}\\ &=-\frac{1}{9 \left (x^2\right )^{3/2}}+\frac{1}{3 \sqrt{x^2}}+\frac{\left (-1+x^2\right )^{3/2} \sec ^{-1}(x)}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0526947, size = 48, normalized size = 1.17 \[ \frac{\sqrt{1-\frac{1}{x^2}} x \left (3 x^2-1\right )+3 \left (x^2-1\right )^2 \sec ^{-1}(x)}{9 x^3 \sqrt{x^2-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[-1 + x^2]*ArcSec[x])/x^4,x]

[Out]

(Sqrt[1 - x^(-2)]*x*(-1 + 3*x^2) + 3*(-1 + x^2)^2*ArcSec[x])/(9*x^3*Sqrt[-1 + x^2])

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Maple [C]  time = 0.383, size = 602, normalized size = 14.7 \begin{align*} -{\frac{1}{144\,{x}^{3}}\sqrt{{x}^{2}-1} \left ( \sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{5}-5\,i{x}^{4}-12\,\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}+20\,i{x}^{2}+16\,\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x-16\,i \right ) \left ( -i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x+{x}^{2}-1 \right ) ^{-1}}-{\frac{1}{18\,x}\sqrt{{x}^{2}-1} \left ( \sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}-3\,i{x}^{2}-4\,\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x+4\,i \right ) \left ( -i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x+{x}^{2}-1 \right ) ^{-1}}+{\frac{{\rm arcsec} \left (x\right )}{48\,{x}^{3}} \left ( i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{5}-8\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}+4\,{x}^{4}+8\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x-12\,{x}^{2}+8 \right ){\frac{1}{\sqrt{{x}^{2}-1}}}}+{\frac{{\rm arcsec} \left (x\right )}{24\,x} \left ( i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}-2\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x+2\,{x}^{2}-2 \right ){\frac{1}{\sqrt{{x}^{2}-1}}}}+{\frac{x}{8}\sqrt{{x}^{2}-1} \left ( \sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x-i \right ) \left ( -i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x+{x}^{2}-1 \right ) ^{-1}}-{\frac{{\rm arcsec} \left (x\right )}{24\,x} \left ( i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}-2\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x-2\,{x}^{2}+2 \right ){\frac{1}{\sqrt{{x}^{2}-1}}}}-{i{x}^{3}\sqrt{{x}^{2}-1} \left ( 18\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}-36\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x+36\,{x}^{2}-36 \right ) ^{-1}}-{i{x}^{5}\sqrt{{x}^{2}-1} \left ( 144\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{5}-1152\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}+576\,{x}^{4}+1152\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x-1728\,{x}^{2}+1152 \right ) ^{-1}}-{\frac{{\rm arcsec} \left (x\right )}{48\,{x}^{3}} \left ( i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{5}-8\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}{x}^{3}-4\,{x}^{4}+8\,i\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}x+12\,{x}^{2}-8 \right ){\frac{1}{\sqrt{{x}^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(x)*(x^2-1)^(1/2)/x^4,x)

[Out]

-1/144*(x^2-1)^(1/2)*(((x^2-1)/x^2)^(1/2)*x^5-5*I*x^4-12*((x^2-1)/x^2)^(1/2)*x^3+20*I*x^2+16*((x^2-1)/x^2)^(1/
2)*x-16*I)/(-I*((x^2-1)/x^2)^(1/2)*x+x^2-1)/x^3-1/18*(x^2-1)^(1/2)*(((x^2-1)/x^2)^(1/2)*x^3-3*I*x^2-4*((x^2-1)
/x^2)^(1/2)*x+4*I)/(-I*((x^2-1)/x^2)^(1/2)*x+x^2-1)/x+1/48/(x^2-1)^(1/2)/x^3*(I*((x^2-1)/x^2)^(1/2)*x^5-8*I*((
x^2-1)/x^2)^(1/2)*x^3+4*x^4+8*I*((x^2-1)/x^2)^(1/2)*x-12*x^2+8)*arcsec(x)+1/24/(x^2-1)^(1/2)/x*(I*((x^2-1)/x^2
)^(1/2)*x^3-2*I*((x^2-1)/x^2)^(1/2)*x+2*x^2-2)*arcsec(x)+1/8*(x^2-1)^(1/2)*(((x^2-1)/x^2)^(1/2)*x-I)*x/(-I*((x
^2-1)/x^2)^(1/2)*x+x^2-1)-1/24/(x^2-1)^(1/2)*(I*((x^2-1)/x^2)^(1/2)*x^3-2*I*((x^2-1)/x^2)^(1/2)*x-2*x^2+2)*arc
sec(x)/x-I*(x^2-1)^(1/2)*x^3/(18*I*((x^2-1)/x^2)^(1/2)*x^3-36*I*((x^2-1)/x^2)^(1/2)*x+36*x^2-36)-I*(x^2-1)^(1/
2)*x^5/(144*I*((x^2-1)/x^2)^(1/2)*x^5-1152*I*((x^2-1)/x^2)^(1/2)*x^3+576*x^4+1152*I*((x^2-1)/x^2)^(1/2)*x-1728
*x^2+1152)-1/48/(x^2-1)^(1/2)*(I*((x^2-1)/x^2)^(1/2)*x^5-8*I*((x^2-1)/x^2)^(1/2)*x^3-4*x^4+8*I*((x^2-1)/x^2)^(
1/2)*x+12*x^2-8)*arcsec(x)/x^3

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Maxima [A]  time = 1.4774, size = 36, normalized size = 0.88 \begin{align*} \frac{{\left (x^{2} - 1\right )}^{\frac{3}{2}} \operatorname{arcsec}\left (x\right )}{3 \, x^{3}} + \frac{3 \, x^{2} - 1}{9 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)*(x^2-1)^(1/2)/x^4,x, algorithm="maxima")

[Out]

1/3*(x^2 - 1)^(3/2)*arcsec(x)/x^3 + 1/9*(3*x^2 - 1)/x^3

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Fricas [A]  time = 2.3381, size = 69, normalized size = 1.68 \begin{align*} \frac{3 \,{\left (x^{2} - 1\right )}^{\frac{3}{2}} \operatorname{arcsec}\left (x\right ) + 3 \, x^{2} - 1}{9 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)*(x^2-1)^(1/2)/x^4,x, algorithm="fricas")

[Out]

1/9*(3*(x^2 - 1)^(3/2)*arcsec(x) + 3*x^2 - 1)/x^3

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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(asec(x)*(x**2-1)**(1/2)/x**4,x)

[Out]

Timed out

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Giac [B]  time = 1.10548, size = 101, normalized size = 2.46 \begin{align*} -\frac{2 \, \arctan \left (-x + \sqrt{x^{2} - 1}\right )}{3 \, \mathrm{sgn}\left (x\right )} + \frac{2 \,{\left (3 \,{\left (x - \sqrt{x^{2} - 1}\right )}^{4} + 1\right )} \arccos \left (\frac{1}{x}\right )}{3 \,{\left ({\left (x - \sqrt{x^{2} - 1}\right )}^{2} + 1\right )}^{3}} + \frac{3 \, x^{2} - 1}{9 \, x^{3} \mathrm{sgn}\left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)*(x^2-1)^(1/2)/x^4,x, algorithm="giac")

[Out]

-2/3*arctan(-x + sqrt(x^2 - 1))/sgn(x) + 2/3*(3*(x - sqrt(x^2 - 1))^4 + 1)*arccos(1/x)/((x - sqrt(x^2 - 1))^2
+ 1)^3 + 1/9*(3*x^2 - 1)/(x^3*sgn(x))