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

Optimal. Leaf size=83 \[ \frac{6 \sqrt{\frac{1-a x}{a x+1}} (a x+1)}{x}-\frac{\text{sech}^{-1}(a x)^3}{x}+\frac{3 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2}{x}-\frac{6 \text{sech}^{-1}(a x)}{x} \]

[Out]

(6*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x))/x - (6*ArcSech[a*x])/x + (3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSec
h[a*x]^2)/x - ArcSech[a*x]^3/x

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Rubi [A]  time = 0.0711182, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {6285, 3296, 2637} \[ \frac{6 \sqrt{\frac{1-a x}{a x+1}} (a x+1)}{x}-\frac{\text{sech}^{-1}(a x)^3}{x}+\frac{3 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2}{x}-\frac{6 \text{sech}^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x))/x - (6*ArcSech[a*x])/x + (3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSec
h[a*x]^2)/x - ArcSech[a*x]^3/x

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])

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\text{sech}^{-1}(a x)^3}{x^2} \, dx &=-\left (a \operatorname{Subst}\left (\int x^3 \sinh (x) \, dx,x,\text{sech}^{-1}(a x)\right )\right )\\ &=-\frac{\text{sech}^{-1}(a x)^3}{x}+(3 a) \operatorname{Subst}\left (\int x^2 \cosh (x) \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=\frac{3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{x}-\frac{\text{sech}^{-1}(a x)^3}{x}-(6 a) \operatorname{Subst}\left (\int x \sinh (x) \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=-\frac{6 \text{sech}^{-1}(a x)}{x}+\frac{3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{x}-\frac{\text{sech}^{-1}(a x)^3}{x}+(6 a) \operatorname{Subst}\left (\int \cosh (x) \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=\frac{6 \sqrt{\frac{1-a x}{1+a x}} (1+a x)}{x}-\frac{6 \text{sech}^{-1}(a x)}{x}+\frac{3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{x}-\frac{\text{sech}^{-1}(a x)^3}{x}\\ \end{align*}

Mathematica [A]  time = 0.0718714, size = 75, normalized size = 0.9 \[ \frac{6 \sqrt{\frac{1-a x}{a x+1}} (a x+1)-\text{sech}^{-1}(a x)^3+3 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2-6 \text{sech}^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x) - 6*ArcSech[a*x] + 3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x]^2
- ArcSech[a*x]^3)/x

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Maple [A]  time = 0.211, size = 98, normalized size = 1.2 \begin{align*} a \left ( -{\frac{ \left ({\rm arcsech} \left (ax\right ) \right ) ^{3}}{ax}}+3\, \left ({\rm arcsech} \left (ax\right ) \right ) ^{2}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}}-6\,{\frac{{\rm arcsech} \left (ax\right )}{ax}}+6\,\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(-arcsech(a*x)^3/a/x+3*arcsech(a*x)^2*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)-6/a/x*arcsech(a*x)+6*(-(a*x-1
)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2))

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Maxima [A]  time = 1.03425, size = 74, normalized size = 0.89 \begin{align*} 3 \, a \sqrt{\frac{1}{a^{2} x^{2}} - 1} \operatorname{arsech}\left (a x\right )^{2} - \frac{\operatorname{arsech}\left (a x\right )^{3}}{x} + 6 \, a \sqrt{\frac{1}{a^{2} x^{2}} - 1} - \frac{6 \, \operatorname{arsech}\left (a x\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a*sqrt(1/(a^2*x^2) - 1)*arcsech(a*x)^2 - arcsech(a*x)^3/x + 6*a*sqrt(1/(a^2*x^2) - 1) - 6*arcsech(a*x)/x

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Fricas [A]  time = 1.93658, size = 333, normalized size = 4.01 \begin{align*} \frac{3 \, 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 )^{2} - \log \left (\frac{a x \sqrt{-\frac{a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )^{3} + 6 \, a x \sqrt{-\frac{a^{2} x^{2} - 1}{a^{2} x^{2}}} - 6 \, \log \left (\frac{a x \sqrt{-\frac{a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*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))^2 - log((a*x*sqrt(-(
a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x))^3 + 6*a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) - 6*log((a*x*sqrt(-(a^2*x^2 - 1)/
(a^2*x^2)) + 1)/(a*x)))/x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asech}^{3}{\left (a x \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(a*x)**3/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arsech}\left (a x\right )^{3}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(a*x)^3/x^2, x)