3.9 \(\int \frac{\text{sech}^{-1}(a x)^2}{x^4} \, dx\)

Optimal. Leaf size=102 \[ -\frac{4 a^2}{9 x}+\frac{4 a^2 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)}{9 x}-\frac{\text{sech}^{-1}(a x)^2}{3 x^3}+\frac{2 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)}{9 x^3}-\frac{2}{27 x^3} \]

[Out]

-2/(27*x^3) - (4*a^2)/(9*x) + (2*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x])/(9*x^3) + (4*a^2*Sqrt[(1 -
a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x])/(9*x) - ArcSech[a*x]^2/(3*x^3)

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Rubi [A]  time = 0.084085, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6285, 5373, 3310, 3296, 2638} \[ -\frac{4 a^2}{9 x}+\frac{4 a^2 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)}{9 x}-\frac{\text{sech}^{-1}(a x)^2}{3 x^3}+\frac{2 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)}{9 x^3}-\frac{2}{27 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcSech[a*x]^2/x^4,x]

[Out]

-2/(27*x^3) - (4*a^2)/(9*x) + (2*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x])/(9*x^3) + (4*a^2*Sqrt[(1 -
a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x])/(9*x) - ArcSech[a*x]^2/(3*x^3)

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 5373

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[(x^(m -
n + 1)*Cosh[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Cosh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -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 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 2638

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

Rubi steps

\begin{align*} \int \frac{\text{sech}^{-1}(a x)^2}{x^4} \, dx &=-\left (a^3 \operatorname{Subst}\left (\int x^2 \cosh ^2(x) \sinh (x) \, dx,x,\text{sech}^{-1}(a x)\right )\right )\\ &=-\frac{\text{sech}^{-1}(a x)^2}{3 x^3}+\frac{1}{3} \left (2 a^3\right ) \operatorname{Subst}\left (\int x \cosh ^3(x) \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=-\frac{2}{27 x^3}+\frac{2 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)}{9 x^3}-\frac{\text{sech}^{-1}(a x)^2}{3 x^3}+\frac{1}{9} \left (4 a^3\right ) \operatorname{Subst}\left (\int x \cosh (x) \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=-\frac{2}{27 x^3}+\frac{2 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)}{9 x^3}+\frac{4 a^2 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)}{9 x}-\frac{\text{sech}^{-1}(a x)^2}{3 x^3}-\frac{1}{9} \left (4 a^3\right ) \operatorname{Subst}\left (\int \sinh (x) \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=-\frac{2}{27 x^3}-\frac{4 a^2}{9 x}+\frac{2 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)}{9 x^3}+\frac{4 a^2 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)}{9 x}-\frac{\text{sech}^{-1}(a x)^2}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0949477, size = 73, normalized size = 0.72 \[ \frac{-2 \left (6 a^2 x^2+1\right )+6 \sqrt{\frac{1-a x}{a x+1}} \left (2 a^3 x^3+2 a^2 x^2+a x+1\right ) \text{sech}^{-1}(a x)-9 \text{sech}^{-1}(a x)^2}{27 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSech[a*x]^2/x^4,x]

[Out]

(-2*(1 + 6*a^2*x^2) + 6*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x + 2*a^2*x^2 + 2*a^3*x^3)*ArcSech[a*x] - 9*ArcSech[a
*x]^2)/(27*x^3)

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Maple [A]  time = 0.237, size = 146, normalized size = 1.4 \begin{align*}{a}^{3} \left ({\frac{ \left ({\rm arcsech} \left (ax\right ) \right ) ^{2} \left ( ax-1 \right ) \left ( ax+1 \right ) }{3\,{x}^{3}{a}^{3}}}-{\frac{ \left ({\rm arcsech} \left (ax\right ) \right ) ^{2}}{3\,ax}}+{\frac{2\,{\rm arcsech} \left (ax\right )}{9\,{a}^{2}{x}^{2}}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}}}+{\frac{4\,{\rm arcsech} \left (ax\right )}{9}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}}}+{\frac{ \left ( 2\,ax-2 \right ) \left ( ax+1 \right ) }{27\,{x}^{3}{a}^{3}}}-{\frac{14}{27\,ax}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsech(a*x)^2/x^4,x)

[Out]

a^3*(1/3*arcsech(a*x)^2/a^3/x^3*(a*x-1)*(a*x+1)-1/3*arcsech(a*x)^2/a/x+2/9*arcsech(a*x)/a^2/x^2*(-(a*x-1)/a/x)
^(1/2)*((a*x+1)/a/x)^(1/2)+4/9*arcsech(a*x)*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)+2/27*(a*x-1)/a^3/x^3*(a*x
+1)-14/27/a/x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arsech}\left (a x\right )^{2}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(a*x)^2/x^4,x, algorithm="maxima")

[Out]

integrate(arcsech(a*x)^2/x^4, x)

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Fricas [A]  time = 1.69588, size = 258, normalized size = 2.53 \begin{align*} -\frac{12 \, a^{2} x^{2} - 6 \,{\left (2 \, a^{3} x^{3} + a x\right )} \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 ) + 9 \, \log \left (\frac{a x \sqrt{-\frac{a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )^{2} + 2}{27 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*(12*a^2*x^2 - 6*(2*a^3*x^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)) + 9*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x))^2 + 2)/x^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asech}^{2}{\left (a x \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(a*x)**2/x**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arsech}\left (a x\right )^{2}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(a*x)^2/x^4, x)