3.27 \(\int \frac {\text {sech}^{-1}(\sqrt {x})}{x^4} \, dx\)

Optimal. Leaf size=172 \[ \frac {5 (1-x)}{48 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} x^{3/2}}+\frac {5 (1-x)}{72 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} x^{5/2}}+\frac {1-x}{18 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} x^{7/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {5 \sqrt {1-x} \tanh ^{-1}\left (\sqrt {1-x}\right )}{48 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}} \]

[Out]

-1/3*arcsech(x^(1/2))/x^3+1/18*(1-x)/x^(7/2)/(-1+1/x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)+5/72*(1-x)/x^(5/2)/(-1+1
/x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)+5/48*(1-x)/x^(3/2)/(-1+1/x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)+5/48*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 = 0.04, antiderivative size = 172, 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 = {6345, 12, 51, 63, 206} \[ \frac {5 (1-x)}{48 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} x^{3/2}}+\frac {5 (1-x)}{72 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} x^{5/2}}+\frac {1-x}{18 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} x^{7/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {5 \sqrt {1-x} \tanh ^{-1}\left (\sqrt {1-x}\right )}{48 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - x)/(18*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*x^(7/2)) + (5*(1 - x))/(72*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 +
1/Sqrt[x]]*x^(5/2)) + (5*(1 - x))/(48*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*x^(3/2)) - ArcSech[Sqrt[x]]/(3*
x^3) + (5*Sqrt[1 - x]*ArcTanh[Sqrt[1 - x]])/(48*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^4} \, dx &=-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}-\frac {\sqrt {1-x} \int \frac {1}{2 \sqrt {1-x} x^4} \, dx}{3 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}-\frac {\sqrt {1-x} \int \frac {1}{\sqrt {1-x} x^4} \, dx}{6 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1-x}{18 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{7/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}-\frac {\left (5 \sqrt {1-x}\right ) \int \frac {1}{\sqrt {1-x} x^3} \, dx}{36 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1-x}{18 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{7/2}}+\frac {5 (1-x)}{72 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{5/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}-\frac {\left (5 \sqrt {1-x}\right ) \int \frac {1}{\sqrt {1-x} x^2} \, dx}{48 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1-x}{18 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{7/2}}+\frac {5 (1-x)}{72 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{5/2}}+\frac {5 (1-x)}{48 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{3/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}-\frac {\left (5 \sqrt {1-x}\right ) \int \frac {1}{\sqrt {1-x} x} \, dx}{96 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1-x}{18 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{7/2}}+\frac {5 (1-x)}{72 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{5/2}}+\frac {5 (1-x)}{48 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{3/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {\left (5 \sqrt {1-x}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-x}\right )}{48 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1-x}{18 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{7/2}}+\frac {5 (1-x)}{72 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{5/2}}+\frac {5 (1-x)}{48 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{3/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {5 \sqrt {1-x} \tanh ^{-1}\left (\sqrt {1-x}\right )}{48 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 140, normalized size = 0.81 \[ \frac {15 x^3 \log \left (\sqrt {x} \sqrt {\frac {1-\sqrt {x}}{\sqrt {x}+1}}+\sqrt {\frac {1-\sqrt {x}}{\sqrt {x}+1}}+1\right )-\frac {15}{2} x^3 \log (x)+\sqrt {\frac {1-\sqrt {x}}{\sqrt {x}+1}} \left (15 x^{5/2}+10 x^{3/2}+15 x^2+10 x+8 \sqrt {x}+8\right )-48 \text {sech}^{-1}\left (\sqrt {x}\right )}{144 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])]*(8 + 8*Sqrt[x] + 10*x + 10*x^(3/2) + 15*x^2 + 15*x^(5/2)) - 48*ArcSech[Sqrt
[x]] + 15*x^3*Log[1 + Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])] + Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])]*Sqrt[x]] - (15*x^3
*Log[x])/2)/(144*x^3)

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fricas [A]  time = 1.39, size = 60, normalized size = 0.35 \[ \frac {{\left (15 \, x^{2} + 10 \, x + 8\right )} \sqrt {x} \sqrt {-\frac {x - 1}{x}} + 3 \, {\left (5 \, x^{3} - 16\right )} \log \left (\frac {x \sqrt {-\frac {x - 1}{x}} + \sqrt {x}}{x}\right )}{144 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsech}\left (\sqrt {x}\right )}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(sqrt(x))/x^4, x)

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maple [A]  time = 0.07, size = 91, normalized size = 0.53 \[ -\frac {\mathrm {arcsech}\left (\sqrt {x}\right )}{3 x^{3}}+\frac {\sqrt {-\frac {-1+\sqrt {x}}{\sqrt {x}}}\, \sqrt {\frac {1+\sqrt {x}}{\sqrt {x}}}\, \left (15 \arctanh \left (\frac {1}{\sqrt {1-x}}\right ) x^{3}+15 \sqrt {1-x}\, x^{2}+10 x \sqrt {1-x}+8 \sqrt {1-x}\right )}{144 x^{\frac {5}{2}} \sqrt {1-x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*arcsech(x^(1/2))/x^3+1/144*(-(-1+x^(1/2))/x^(1/2))^(1/2)/x^(5/2)*((1+x^(1/2))/x^(1/2))^(1/2)*(15*arctanh(
1/(1-x)^(1/2))*x^3+15*(1-x)^(1/2)*x^2+10*x*(1-x)^(1/2)+8*(1-x)^(1/2))/(1-x)^(1/2)

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maxima [A]  time = 0.32, size = 116, normalized size = 0.67 \[ -\frac {15 \, x^{\frac {5}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {5}{2}} - 40 \, x^{\frac {3}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {3}{2}} + 33 \, \sqrt {x} \sqrt {\frac {1}{x} - 1}}{144 \, {\left (x^{3} {\left (\frac {1}{x} - 1\right )}^{3} - 3 \, x^{2} {\left (\frac {1}{x} - 1\right )}^{2} + 3 \, x {\left (\frac {1}{x} - 1\right )} - 1\right )}} - \frac {\operatorname {arsech}\left (\sqrt {x}\right )}{3 \, x^{3}} + \frac {5}{96} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} - 1} + 1\right ) - \frac {5}{96} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} - 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(x^(1/2))/x^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*arcsech(sqrt(x))/x^3 + 5/96*log(sqrt(x)*sqrt(1/x - 1) + 1) - 5/9
6*log(sqrt(x)*sqrt(1/x - 1) - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acosh(1/x^(1/2))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asech}{\left (\sqrt {x} \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(sqrt(x))/x**4, x)

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