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

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

[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

________________________________________________________________________________________

Rubi [A]  time = 0.0236281, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{5 \sqrt{x-1}}{72 x^2}+\frac{\sqrt{x-1}}{18 x^3}-\frac{\sec ^{-1}\left (\sqrt{x}\right )}{3 x^3}+\frac{5 \sqrt{x-1}}{48 x}+\frac{5}{48} \tan ^{-1}\left (\sqrt{x-1}\right ) \]

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 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^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*}

Mathematica [A]  time = 0.0494117, size = 45, normalized size = 0.66 \[ \frac{\sqrt{x-1} \left (15 x^2+10 x+8\right )-15 x^3 \sin ^{-1}\left (\frac{1}{\sqrt{x}}\right )-48 \sec ^{-1}\left (\sqrt{x}\right )}{144 x^3} \]

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)

________________________________________________________________________________________

Maple [A]  time = 0.108, size = 67, normalized size = 1. \begin{align*} -{\frac{1}{3\,{x}^{3}}{\rm arcsec} \left (\sqrt{x}\right )}+{\frac{1}{144}\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 ){\frac{1}{\sqrt{{\frac{x-1}{x}}}}}{x}^{-{\frac{7}{2}}}} \end{align*}

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)

________________________________________________________________________________________

Maxima [B]  time = 1.50427, size = 143, normalized size = 2.1 \begin{align*} -\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 ) \end{align*}

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))

________________________________________________________________________________________

Fricas [A]  time = 2.39751, size = 105, normalized size = 1.54 \begin{align*} \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}} \end{align*}

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

________________________________________________________________________________________

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**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.12156, size = 61, normalized size = 0.9 \begin{align*} \frac{15 \,{\left (x - 1\right )}^{\frac{5}{2}} + 40 \,{\left (x - 1\right )}^{\frac{3}{2}} + 33 \, \sqrt{x - 1}}{144 \, x^{3}} - \frac{\arccos \left (\frac{1}{\sqrt{x}}\right )}{3 \, x^{3}} + \frac{5}{48} \, \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^4,x, algorithm="giac")

[Out]

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