3.26 \(\int \frac {\text {sech}^{-1}(\sqrt {x})}{x^3} \, dx\)

Optimal. Leaf size=136 \[ \frac {3 (1-x)}{16 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} x^{3/2}}+\frac {1-x}{8 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} x^{5/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {3 \sqrt {1-x} \tanh ^{-1}\left (\sqrt {1-x}\right )}{16 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}} \]

[Out]

-1/2*arcsech(x^(1/2))/x^2+1/8*(1-x)/x^(5/2)/(-1+1/x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)+3/16*(1-x)/x^(3/2)/(-1+1/
x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)+3/16*arctanh((1-x)^(1/2))*(1-x)^(1/2)/x^(1/2)/(-1+1/x^(1/2))^(1/2)/(1+1/x^(
1/2))^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 136, 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 = {6345, 12, 51, 63, 206} \[ \frac {3 (1-x)}{16 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} x^{3/2}}+\frac {1-x}{8 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} x^{5/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {3 \sqrt {1-x} \tanh ^{-1}\left (\sqrt {1-x}\right )}{16 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - x)/(8*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*x^(5/2)) + (3*(1 - x))/(16*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1
/Sqrt[x]]*x^(3/2)) - ArcSech[Sqrt[x]]/(2*x^2) + (3*Sqrt[1 - x]*ArcTanh[Sqrt[1 - x]])/(16*Sqrt[-1 + 1/Sqrt[x]]*
Sqrt[1 + 1/Sqrt[x]]*Sqrt[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 206

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

Rule 6345

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

Rubi steps

\begin {align*} \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx &=-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{2 x^2}-\frac {\sqrt {1-x} \int \frac {1}{2 \sqrt {1-x} x^3} \, dx}{2 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{2 x^2}-\frac {\sqrt {1-x} \int \frac {1}{\sqrt {1-x} x^3} \, dx}{4 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1-x}{8 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{5/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{2 x^2}-\frac {\left (3 \sqrt {1-x}\right ) \int \frac {1}{\sqrt {1-x} x^2} \, dx}{16 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1-x}{8 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{5/2}}+\frac {3 (1-x)}{16 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{3/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{2 x^2}-\frac {\left (3 \sqrt {1-x}\right ) \int \frac {1}{\sqrt {1-x} x} \, dx}{32 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1-x}{8 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{5/2}}+\frac {3 (1-x)}{16 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{3/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {\left (3 \sqrt {1-x}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-x}\right )}{16 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1-x}{8 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{5/2}}+\frac {3 (1-x)}{16 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{3/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {3 \sqrt {1-x} \tanh ^{-1}\left (\sqrt {1-x}\right )}{16 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 125, normalized size = 0.92 \[ \frac {1}{16} \left (-\frac {8 \text {sech}^{-1}\left (\sqrt {x}\right )}{x^2}+\frac {\sqrt {\frac {1-\sqrt {x}}{\sqrt {x}+1}} \left (3 x^{3/2}+3 x+2 \sqrt {x}+2\right )}{x^2}+3 \log \left (\sqrt {x} \sqrt {\frac {1-\sqrt {x}}{\sqrt {x}+1}}+\sqrt {\frac {1-\sqrt {x}}{\sqrt {x}+1}}+1\right )-\frac {3 \log (x)}{2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.09, size = 54, normalized size = 0.40 \[ \frac {{\left (3 \, x + 2\right )} \sqrt {x} \sqrt {-\frac {x - 1}{x}} + {\left (3 \, x^{2} - 8\right )} \log \left (\frac {x \sqrt {-\frac {x - 1}{x}} + \sqrt {x}}{x}\right )}{16 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsech}\left (\sqrt {x}\right )}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(sqrt(x))/x^3, x)

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maple [A]  time = 0.07, size = 79, normalized size = 0.58 \[ -\frac {\mathrm {arcsech}\left (\sqrt {x}\right )}{2 x^{2}}+\frac {\sqrt {-\frac {-1+\sqrt {x}}{\sqrt {x}}}\, \sqrt {\frac {1+\sqrt {x}}{\sqrt {x}}}\, \left (3 \arctanh \left (\frac {1}{\sqrt {1-x}}\right ) x^{2}+3 x \sqrt {1-x}+2 \sqrt {1-x}\right )}{16 x^{\frac {3}{2}} \sqrt {1-x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arcsech(x^(1/2))/x^2+1/16*(-(-1+x^(1/2))/x^(1/2))^(1/2)/x^(3/2)*((1+x^(1/2))/x^(1/2))^(1/2)*(3*arctanh(1/
(1-x)^(1/2))*x^2+3*x*(1-x)^(1/2)+2*(1-x)^(1/2))/(1-x)^(1/2)

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maxima [A]  time = 0.34, size = 92, normalized size = 0.68 \[ -\frac {3 \, x^{\frac {3}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {3}{2}} - 5 \, \sqrt {x} \sqrt {\frac {1}{x} - 1}}{16 \, {\left (x^{2} {\left (\frac {1}{x} - 1\right )}^{2} - 2 \, x {\left (\frac {1}{x} - 1\right )} + 1\right )}} - \frac {\operatorname {arsech}\left (\sqrt {x}\right )}{2 \, x^{2}} + \frac {3}{32} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} - 1} + 1\right ) - \frac {3}{32} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} - 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(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*arcsec
h(sqrt(x))/x^2 + 3/32*log(sqrt(x)*sqrt(1/x - 1) + 1) - 3/32*log(sqrt(x)*sqrt(1/x - 1) - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acosh}\left (\frac {1}{\sqrt {x}}\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acosh(1/x^(1/2))/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asech}{\left (\sqrt {x} \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(sqrt(x))/x**3, x)

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