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]

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

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Rubi [A]  time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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])

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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 1.69, size = 29, normalized size = 0.54 \[ \frac {{\left (3 \, x^{2} - 8\right )} \operatorname {arcsec}\left (\sqrt {x}\right ) + {\left (3 \, x + 2\right )} \sqrt {x - 1}}{16 \, x^{2}} \]

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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giac [A]  time = 0.14, size = 44, normalized size = 0.81 \[ \frac {3 \, \sqrt {-\frac {1}{x} + 1}}{16 \, \sqrt {x}} + \frac {\sqrt {-\frac {1}{x} + 1}}{8 \, x^{\frac {3}{2}}} - \frac {\arccos \left (\frac {1}{\sqrt {x}}\right )}{2 \, x^{2}} + \frac {3}{16} \, \arccos \left (\frac {1}{\sqrt {x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 57, normalized size = 1.06 \[ -\frac {\mathrm {arcsec}\left (\sqrt {x}\right )}{2 x^{2}}+\frac {\sqrt {x -1}\, \left (-3 \arctan \left (\frac {1}{\sqrt {x -1}}\right ) x^{2}+3 x \sqrt {x -1}+2 \sqrt {x -1}\right )}{16 \sqrt {\frac {x -1}{x}}\, x^{\frac {5}{2}}} \]

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 = 0.44, size = 80, normalized size = 1.48 \[ \frac {3 \, x^{\frac {3}{2}} {\left (-\frac {1}{x} + 1\right )}^{\frac {3}{2}} + 5 \, \sqrt {x} \sqrt {-\frac {1}{x} + 1}}{16 \, {\left (x^{2} {\left (\frac {1}{x} - 1\right )}^{2} - 2 \, x {\left (\frac {1}{x} - 1\right )} + 1\right )}} - \frac {\operatorname {arcsec}\left (\sqrt {x}\right )}{2 \, x^{2}} + \frac {3}{16} \, \arctan \left (\sqrt {x} \sqrt {-\frac {1}{x} + 1}\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {acos}\left (\frac {1}{\sqrt {x}}\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acos(1/x^(1/2))/x^3, x)

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sympy [C]  time = 53.24, size = 144, normalized size = 2.67 \[ \frac {\begin {cases} \frac {3 i \operatorname {acosh}{\left (\frac {1}{\sqrt {x}} \right )}}{4} - \frac {3 i}{4 \sqrt {x} \sqrt {-1 + \frac {1}{x}}} + \frac {i}{4 x^{\frac {3}{2}} \sqrt {-1 + \frac {1}{x}}} + \frac {i}{2 x^{\frac {5}{2}} \sqrt {-1 + \frac {1}{x}}} & \text {for}\: \frac {1}{\left |{x}\right |} > 1 \\- \frac {3 \operatorname {asin}{\left (\frac {1}{\sqrt {x}} \right )}}{4} + \frac {3}{4 \sqrt {x} \sqrt {1 - \frac {1}{x}}} - \frac {1}{4 x^{\frac {3}{2}} \sqrt {1 - \frac {1}{x}}} - \frac {1}{2 x^{\frac {5}{2}} \sqrt {1 - \frac {1}{x}}} & \text {otherwise} \end {cases}}{4} - \frac {\operatorname {asec}{\left (\sqrt {x} \right )}}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*I*acosh(1/sqrt(x))/4 - 3*I/(4*sqrt(x)*sqrt(-1 + 1/x)) + I/(4*x**(3/2)*sqrt(-1 + 1/x)) + I/(2*x**(
5/2)*sqrt(-1 + 1/x)), 1/Abs(x) > 1), (-3*asin(1/sqrt(x))/4 + 3/(4*sqrt(x)*sqrt(1 - 1/x)) - 1/(4*x**(3/2)*sqrt(
1 - 1/x)) - 1/(2*x**(5/2)*sqrt(1 - 1/x)), True))/4 - asec(sqrt(x))/(2*x**2)

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