3.8 \(\int \frac{\sec ^{-1}(\sqrt{x})}{x^3} \, dx\)

Optimal. Leaf size=54 \[ \frac{\sqrt{x-1}}{8 x^2}-\frac{\sec ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{3 \sqrt{x-1}}{16 x}+\frac{3}{16} \tan ^{-1}\left (\sqrt{x-1}\right ) \]

[Out]

Sqrt[-1 + x]/(8*x^2) + (3*Sqrt[-1 + x])/(16*x) - ArcSec[Sqrt[x]]/(2*x^2) + (3*ArcTan[Sqrt[-1 + x]])/16

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Rubi [A]  time = 0.018859, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5270, 12, 51, 63, 203} \[ \frac{\sqrt{x-1}}{8 x^2}-\frac{\sec ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{3 \sqrt{x-1}}{16 x}+\frac{3}{16} \tan ^{-1}\left (\sqrt{x-1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[Sqrt[x]]/x^3,x]

[Out]

Sqrt[-1 + x]/(8*x^2) + (3*Sqrt[-1 + x])/(16*x) - ArcSec[Sqrt[x]]/(2*x^2) + (3*ArcTan[Sqrt[-1 + x]])/16

Rule 5270

Int[((a_.) + ArcSec[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSec[
u]))/(d*(m + 1)), x] - Dist[(b*u)/(d*(m + 1)*Sqrt[u^2]), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/(u*
Sqrt[u^2 - 1]), x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !Funct
ionOfQ[(c + d*x)^(m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^{-1}\left (\sqrt{x}\right )}{x^3} \, dx &=-\frac{\sec ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{1}{2} \int \frac{1}{2 \sqrt{-1+x} x^3} \, dx\\ &=-\frac{\sec ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{1}{4} \int \frac{1}{\sqrt{-1+x} x^3} \, dx\\ &=\frac{\sqrt{-1+x}}{8 x^2}-\frac{\sec ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{3}{16} \int \frac{1}{\sqrt{-1+x} x^2} \, dx\\ &=\frac{\sqrt{-1+x}}{8 x^2}+\frac{3 \sqrt{-1+x}}{16 x}-\frac{\sec ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{3}{32} \int \frac{1}{\sqrt{-1+x} x} \, dx\\ &=\frac{\sqrt{-1+x}}{8 x^2}+\frac{3 \sqrt{-1+x}}{16 x}-\frac{\sec ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{3}{16} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{-1+x}\right )\\ &=\frac{\sqrt{-1+x}}{8 x^2}+\frac{3 \sqrt{-1+x}}{16 x}-\frac{\sec ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{3}{16} \tan ^{-1}\left (\sqrt{-1+x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0315698, size = 55, normalized size = 1.02 \[ \sqrt{\frac{x-1}{x}} \left (\frac{1}{8 x^{3/2}}+\frac{3}{16 \sqrt{x}}\right )-\frac{\sec ^{-1}\left (\sqrt{x}\right )}{2 x^2}-\frac{3}{16} \sin ^{-1}\left (\frac{1}{\sqrt{x}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[Sqrt[x]]/x^3,x]

[Out]

(1/(8*x^(3/2)) + 3/(16*Sqrt[x]))*Sqrt[(-1 + x)/x] - ArcSec[Sqrt[x]]/(2*x^2) - (3*ArcSin[1/Sqrt[x]])/16

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Maple [A]  time = 0.111, size = 57, normalized size = 1.1 \begin{align*} -{\frac{1}{2\,{x}^{2}}{\rm arcsec} \left (\sqrt{x}\right )}+{\frac{1}{16}\sqrt{x-1} \left ( -3\,\arctan \left ({\frac{1}{\sqrt{x-1}}} \right ){x}^{2}+3\,x\sqrt{x-1}+2\,\sqrt{x-1} \right ){\frac{1}{\sqrt{{\frac{x-1}{x}}}}}{x}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(x^(1/2))/x^3,x)

[Out]

-1/2*arcsec(x^(1/2))/x^2+1/16*(x-1)^(1/2)*(-3*arctan(1/(x-1)^(1/2))*x^2+3*x*(x-1)^(1/2)+2*(x-1)^(1/2))/((x-1)/
x)^(1/2)/x^(5/2)

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Maxima [B]  time = 1.51742, size = 108, normalized size = 2. \begin{align*} \frac{3 \, x^{\frac{3}{2}}{\left (-\frac{1}{x} + 1\right )}^{\frac{3}{2}} + 5 \, \sqrt{x} \sqrt{-\frac{1}{x} + 1}}{16 \,{\left (x^{2}{\left (\frac{1}{x} - 1\right )}^{2} - 2 \, x{\left (\frac{1}{x} - 1\right )} + 1\right )}} - \frac{\operatorname{arcsec}\left (\sqrt{x}\right )}{2 \, x^{2}} + \frac{3}{16} \, \arctan \left (\sqrt{x} \sqrt{-\frac{1}{x} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*x^(3/2)*(-1/x + 1)^(3/2) + 5*sqrt(x)*sqrt(-1/x + 1))/(x^2*(1/x - 1)^2 - 2*x*(1/x - 1) + 1) - 1/2*arcse
c(sqrt(x))/x^2 + 3/16*arctan(sqrt(x)*sqrt(-1/x + 1))

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Fricas [A]  time = 2.24137, size = 86, normalized size = 1.59 \begin{align*} \frac{{\left (3 \, x^{2} - 8\right )} \operatorname{arcsec}\left (\sqrt{x}\right ) +{\left (3 \, x + 2\right )} \sqrt{x - 1}}{16 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((3*x^2 - 8)*arcsec(sqrt(x)) + (3*x + 2)*sqrt(x - 1))/x^2

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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**(1/2))/x**3,x)

[Out]

Timed out

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Giac [A]  time = 1.11598, size = 51, normalized size = 0.94 \begin{align*} \frac{3 \,{\left (x - 1\right )}^{\frac{3}{2}} + 5 \, \sqrt{x - 1}}{16 \, x^{2}} - \frac{\arccos \left (\frac{1}{\sqrt{x}}\right )}{2 \, x^{2}} + \frac{3}{16} \, \arctan \left (\sqrt{x - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*(x - 1)^(3/2) + 5*sqrt(x - 1))/x^2 - 1/2*arccos(1/sqrt(x))/x^2 + 3/16*arctan(sqrt(x - 1))