3.8 \(\int \frac {\text {sech}^{-1}(a x)^2}{x^3} \, dx\)

Optimal. Leaf size=90 \[ -\frac {1}{4} a^2 \text {sech}^{-1}(a x)^2-\frac {(1-a x) (a x+1)}{4 x^2}-\frac {(1-a x) (a x+1) \text {sech}^{-1}(a x)^2}{2 x^2}+\frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)}{2 x^2} \]

[Out]

-1/4*(-a*x+1)*(a*x+1)/x^2-1/4*a^2*arcsech(a*x)^2-1/2*(-a*x+1)*(a*x+1)*arcsech(a*x)^2/x^2+1/2*(a*x+1)*arcsech(a
*x)*((-a*x+1)/(a*x+1))^(1/2)/x^2

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Rubi [A]  time = 0.06, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6285, 5372, 3310, 30} \[ -\frac {1}{4} a^2 \text {sech}^{-1}(a x)^2-\frac {(1-a x) (a x+1)}{4 x^2}-\frac {(1-a x) (a x+1) \text {sech}^{-1}(a x)^2}{2 x^2}+\frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcSech[a*x]^2/x^3,x]

[Out]

-((1 - a*x)*(1 + a*x))/(4*x^2) + (Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x])/(2*x^2) - (a^2*ArcSech[a*x
]^2)/4 - ((1 - a*x)*(1 + a*x)*ArcSech[a*x]^2)/(2*x^2)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 5372

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m -
n + 1)*Sinh[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sinh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 6285

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b
*x)^n*Sech[x]^(m + 1)*Tanh[x], x], x, ArcSech[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] &
& (GtQ[n, 0] || LtQ[m, -1])

Rubi steps

\begin {align*} \int \frac {\text {sech}^{-1}(a x)^2}{x^3} \, dx &=-\left (a^2 \operatorname {Subst}\left (\int x^2 \cosh (x) \sinh (x) \, dx,x,\text {sech}^{-1}(a x)\right )\right )\\ &=-\frac {(1-a x) (1+a x) \text {sech}^{-1}(a x)^2}{2 x^2}+a^2 \operatorname {Subst}\left (\int x \sinh ^2(x) \, dx,x,\text {sech}^{-1}(a x)\right )\\ &=-\frac {(1-a x) (1+a x)}{4 x^2}+\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{2 x^2}-\frac {(1-a x) (1+a x) \text {sech}^{-1}(a x)^2}{2 x^2}-\frac {1}{2} a^2 \operatorname {Subst}\left (\int x \, dx,x,\text {sech}^{-1}(a x)\right )\\ &=-\frac {(1-a x) (1+a x)}{4 x^2}+\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{2 x^2}-\frac {1}{4} a^2 \text {sech}^{-1}(a x)^2-\frac {(1-a x) (1+a x) \text {sech}^{-1}(a x)^2}{2 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 54, normalized size = 0.60 \[ \frac {\left (a^2 x^2-2\right ) \text {sech}^{-1}(a x)^2+2 \sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)-1}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSech[a*x]^2/x^3,x]

[Out]

(-1 + 2*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x] + (-2 + a^2*x^2)*ArcSech[a*x]^2)/(4*x^2)

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fricas [A]  time = 0.56, size = 106, normalized size = 1.18 \[ \frac {2 \, a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} \log \left (\frac {a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right ) + {\left (a^{2} x^{2} - 2\right )} \log \left (\frac {a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )^{2} - 1}{4 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2))*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x)) + (a^2*x^2 - 2)*
log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x))^2 - 1)/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsech}\left (a x\right )^{2}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(a*x)^2/x^3, x)

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maple [A]  time = 0.12, size = 77, normalized size = 0.86 \[ a^{2} \left (-\frac {\mathrm {arcsech}\left (a x \right )^{2}}{2 a^{2} x^{2}}+\frac {\mathrm {arcsech}\left (a x \right ) \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}}{2 a x}+\frac {\mathrm {arcsech}\left (a x \right )^{2}}{4}-\frac {1}{4 x^{2} a^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsech(a*x)^2/x^3,x)

[Out]

a^2*(-1/2*arcsech(a*x)^2/a^2/x^2+1/2*arcsech(a*x)/a/x*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)+1/4*arcsech(a*x
)^2-1/4/x^2/a^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsech}\left (a x\right )^{2}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(a*x)^2/x^3,x, algorithm="maxima")

[Out]

integrate(arcsech(a*x)^2/x^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {acosh}\left (\frac {1}{a\,x}\right )}^2}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acosh(1/(a*x))^2/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asech}^{2}{\left (a x \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(a*x)**2/x**3, x)

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