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

Optimal. Leaf size=68 \[ \frac {\sqrt {x-1}}{18 x^3}-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {5 \sqrt {x-1}}{72 x^2}+\frac {5 \sqrt {x-1}}{48 x}+\frac {5}{48} \tan ^{-1}\left (\sqrt {x-1}\right ) \]

[Out]

-1/3*arcsec(x^(1/2))/x^3+5/48*arctan((-1+x)^(1/2))+1/18*(-1+x)^(1/2)/x^3+5/72*(-1+x)^(1/2)/x^2+5/48*(-1+x)^(1/
2)/x

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Rubi [A]  time = 0.02, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {5 \sqrt {x-1}}{72 x^2}+\frac {\sqrt {x-1}}{18 x^3}-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {5 \sqrt {x-1}}{48 x}+\frac {5}{48} \tan ^{-1}\left (\sqrt {x-1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[Sqrt[x]]/x^4,x]

[Out]

Sqrt[-1 + x]/(18*x^3) + (5*Sqrt[-1 + x])/(72*x^2) + (5*Sqrt[-1 + x])/(48*x) - ArcSec[Sqrt[x]]/(3*x^3) + (5*Arc
Tan[Sqrt[-1 + x]])/48

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^4} \, dx &=-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {1}{3} \int \frac {1}{2 \sqrt {-1+x} x^4} \, dx\\ &=-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {1}{6} \int \frac {1}{\sqrt {-1+x} x^4} \, dx\\ &=\frac {\sqrt {-1+x}}{18 x^3}-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {5}{36} \int \frac {1}{\sqrt {-1+x} x^3} \, dx\\ &=\frac {\sqrt {-1+x}}{18 x^3}+\frac {5 \sqrt {-1+x}}{72 x^2}-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {5}{48} \int \frac {1}{\sqrt {-1+x} x^2} \, dx\\ &=\frac {\sqrt {-1+x}}{18 x^3}+\frac {5 \sqrt {-1+x}}{72 x^2}+\frac {5 \sqrt {-1+x}}{48 x}-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {5}{96} \int \frac {1}{\sqrt {-1+x} x} \, dx\\ &=\frac {\sqrt {-1+x}}{18 x^3}+\frac {5 \sqrt {-1+x}}{72 x^2}+\frac {5 \sqrt {-1+x}}{48 x}-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {5}{48} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {-1+x}\right )\\ &=\frac {\sqrt {-1+x}}{18 x^3}+\frac {5 \sqrt {-1+x}}{72 x^2}+\frac {5 \sqrt {-1+x}}{48 x}-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {5}{48} \tan ^{-1}\left (\sqrt {-1+x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 45, normalized size = 0.66 \[ \frac {-15 x^3 \sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )+\sqrt {x-1} \left (15 x^2+10 x+8\right )-48 \sec ^{-1}\left (\sqrt {x}\right )}{144 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[Sqrt[x]]/x^4,x]

[Out]

(Sqrt[-1 + x]*(8 + 10*x + 15*x^2) - 48*ArcSec[Sqrt[x]] - 15*x^3*ArcSin[1/Sqrt[x]])/(144*x^3)

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fricas [A]  time = 0.54, size = 35, normalized size = 0.51 \[ \frac {3 \, {\left (5 \, x^{3} - 16\right )} \operatorname {arcsec}\left (\sqrt {x}\right ) + {\left (15 \, x^{2} + 10 \, x + 8\right )} \sqrt {x - 1}}{144 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*(3*(5*x^3 - 16)*arcsec(sqrt(x)) + (15*x^2 + 10*x + 8)*sqrt(x - 1))/x^3

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giac [A]  time = 0.15, size = 58, normalized size = 0.85 \[ \frac {5 \, \sqrt {-\frac {1}{x} + 1}}{48 \, \sqrt {x}} + \frac {5 \, \sqrt {-\frac {1}{x} + 1}}{72 \, x^{\frac {3}{2}}} + \frac {\sqrt {-\frac {1}{x} + 1}}{18 \, x^{\frac {5}{2}}} - \frac {\arccos \left (\frac {1}{\sqrt {x}}\right )}{3 \, x^{3}} + \frac {5}{48} \, \arccos \left (\frac {1}{\sqrt {x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/48*sqrt(-1/x + 1)/sqrt(x) + 5/72*sqrt(-1/x + 1)/x^(3/2) + 1/18*sqrt(-1/x + 1)/x^(5/2) - 1/3*arccos(1/sqrt(x)
)/x^3 + 5/48*arccos(1/sqrt(x))

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maple [A]  time = 0.05, size = 67, normalized size = 0.99 \[ -\frac {\mathrm {arcsec}\left (\sqrt {x}\right )}{3 x^{3}}-\frac {\sqrt {x -1}\, \left (15 \arctan \left (\frac {1}{\sqrt {x -1}}\right ) x^{3}-15 x^{2} \sqrt {x -1}-10 x \sqrt {x -1}-8 \sqrt {x -1}\right )}{144 \sqrt {\frac {x -1}{x}}\, x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*arcsec(x^(1/2))/x^3-1/144*(x-1)^(1/2)*(15*arctan(1/(x-1)^(1/2))*x^3-15*x^2*(x-1)^(1/2)-10*x*(x-1)^(1/2)-8
*(x-1)^(1/2))/((x-1)/x)^(1/2)/x^(7/2)

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maxima [B]  time = 0.43, size = 106, normalized size = 1.56 \[ -\frac {15 \, x^{\frac {5}{2}} {\left (-\frac {1}{x} + 1\right )}^{\frac {5}{2}} + 40 \, x^{\frac {3}{2}} {\left (-\frac {1}{x} + 1\right )}^{\frac {3}{2}} + 33 \, \sqrt {x} \sqrt {-\frac {1}{x} + 1}}{144 \, {\left (x^{3} {\left (\frac {1}{x} - 1\right )}^{3} - 3 \, x^{2} {\left (\frac {1}{x} - 1\right )}^{2} + 3 \, x {\left (\frac {1}{x} - 1\right )} - 1\right )}} - \frac {\operatorname {arcsec}\left (\sqrt {x}\right )}{3 \, x^{3}} + \frac {5}{48} \, \arctan \left (\sqrt {x} \sqrt {-\frac {1}{x} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/144*(15*x^(5/2)*(-1/x + 1)^(5/2) + 40*x^(3/2)*(-1/x + 1)^(3/2) + 33*sqrt(x)*sqrt(-1/x + 1))/(x^3*(1/x - 1)^
3 - 3*x^2*(1/x - 1)^2 + 3*x*(1/x - 1) - 1) - 1/3*arcsec(sqrt(x))/x^3 + 5/48*arctan(sqrt(x)*sqrt(-1/x + 1))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acos}\left (\frac {1}{\sqrt {x}}\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(1/x^(1/2))/x^4,x)

[Out]

int(acos(1/x^(1/2))/x^4, x)

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sympy [C]  time = 132.62, size = 180, normalized size = 2.65 \[ \frac {\begin {cases} \frac {5 i \operatorname {acosh}{\left (\frac {1}{\sqrt {x}} \right )}}{8} - \frac {5 i}{8 \sqrt {x} \sqrt {-1 + \frac {1}{x}}} + \frac {5 i}{24 x^{\frac {3}{2}} \sqrt {-1 + \frac {1}{x}}} + \frac {i}{12 x^{\frac {5}{2}} \sqrt {-1 + \frac {1}{x}}} + \frac {i}{3 x^{\frac {7}{2}} \sqrt {-1 + \frac {1}{x}}} & \text {for}\: \frac {1}{\left |{x}\right |} > 1 \\- \frac {5 \operatorname {asin}{\left (\frac {1}{\sqrt {x}} \right )}}{8} + \frac {5}{8 \sqrt {x} \sqrt {1 - \frac {1}{x}}} - \frac {5}{24 x^{\frac {3}{2}} \sqrt {1 - \frac {1}{x}}} - \frac {1}{12 x^{\frac {5}{2}} \sqrt {1 - \frac {1}{x}}} - \frac {1}{3 x^{\frac {7}{2}} \sqrt {1 - \frac {1}{x}}} & \text {otherwise} \end {cases}}{6} - \frac {\operatorname {asec}{\left (\sqrt {x} \right )}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((5*I*acosh(1/sqrt(x))/8 - 5*I/(8*sqrt(x)*sqrt(-1 + 1/x)) + 5*I/(24*x**(3/2)*sqrt(-1 + 1/x)) + I/(12*
x**(5/2)*sqrt(-1 + 1/x)) + I/(3*x**(7/2)*sqrt(-1 + 1/x)), 1/Abs(x) > 1), (-5*asin(1/sqrt(x))/8 + 5/(8*sqrt(x)*
sqrt(1 - 1/x)) - 5/(24*x**(3/2)*sqrt(1 - 1/x)) - 1/(12*x**(5/2)*sqrt(1 - 1/x)) - 1/(3*x**(7/2)*sqrt(1 - 1/x)),
True))/6 - asec(sqrt(x))/(3*x**3)

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