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

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Rubi [A]  time = 0.07, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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 2637

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

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 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)^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*}

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Mathematica [A]  time = 0.09, size = 75, normalized size = 0.90 \[ \frac {6 \sqrt {\frac {1-a x}{a x+1}} (a x+1)-\text {sech}^{-1}(a x)^3+3 \sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)^2-6 \text {sech}^{-1}(a x)}{x} \]

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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fricas [A]  time = 0.71, size = 155, normalized size = 1.87 \[ \frac {3 \, a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} \log \left (\frac {a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )^{2} - \log \left (\frac {a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )^{3} + 6 \, a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} - 6 \, \log \left (\frac {a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )}{x} \]

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

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)

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maple [A]  time = 0.13, size = 98, normalized size = 1.18 \[ a \left (-\frac {\mathrm {arcsech}\left (a x \right )^{3}}{a x}+3 \mathrm {arcsech}\left (a x \right )^{2} \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}-\frac {6 \,\mathrm {arcsech}\left (a x \right )}{a x}+6 \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\right ) \]

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 = 0.32, size = 55, normalized size = 0.66 \[ 3 \, a \sqrt {\frac {1}{a^{2} x^{2}} - 1} \operatorname {arsech}\left (a x\right )^{2} - \frac {\operatorname {arsech}\left (a x\right )^{3}}{x} + 6 \, a \sqrt {\frac {1}{a^{2} x^{2}} - 1} - \frac {6 \, \operatorname {arsech}\left (a x\right )}{x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acosh(1/(a*x))^3/x^2, x)

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

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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