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

Optimal. Leaf size=137 \[ -\frac{1}{4} a^2 \text{sech}^{-1}(a x)^3-\frac{3}{8} a^2 \text{sech}^{-1}(a x)+\frac{3 \sqrt{\frac{1-a x}{a x+1}} (a x+1)}{8 x^2}-\frac{(1-a x) (a x+1) \text{sech}^{-1}(a x)^3}{2 x^2}+\frac{3 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2}{4 x^2}-\frac{3 (1-a x) (a x+1) \text{sech}^{-1}(a x)}{4 x^2} \]

[Out]

(3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x))/(8*x^2) - (3*a^2*ArcSech[a*x])/8 - (3*(1 - a*x)*(1 + a*x)*ArcSech[a*x]
)/(4*x^2) + (3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x]^2)/(4*x^2) - (a^2*ArcSech[a*x]^3)/4 - ((1 - a*
x)*(1 + a*x)*ArcSech[a*x]^3)/(2*x^2)

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Rubi [A]  time = 0.0862309, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6285, 5372, 3311, 30, 2635, 8} \[ -\frac{1}{4} a^2 \text{sech}^{-1}(a x)^3-\frac{3}{8} a^2 \text{sech}^{-1}(a x)+\frac{3 \sqrt{\frac{1-a x}{a x+1}} (a x+1)}{8 x^2}-\frac{(1-a x) (a x+1) \text{sech}^{-1}(a x)^3}{2 x^2}+\frac{3 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2}{4 x^2}-\frac{3 (1-a x) (a x+1) \text{sech}^{-1}(a x)}{4 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x))/(8*x^2) - (3*a^2*ArcSech[a*x])/8 - (3*(1 - a*x)*(1 + a*x)*ArcSech[a*x]
)/(4*x^2) + (3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x]^2)/(4*x^2) - (a^2*ArcSech[a*x]^3)/4 - ((1 - a*
x)*(1 + a*x)*ArcSech[a*x]^3)/(2*x^2)

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 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 3311

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

Rule 30

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

Rule 2635

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\text{sech}^{-1}(a x)^3}{x^3} \, dx &=-\left (a^2 \operatorname{Subst}\left (\int x^3 \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)^3}{2 x^2}+\frac{1}{2} \left (3 a^2\right ) \operatorname{Subst}\left (\int x^2 \sinh ^2(x) \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=-\frac{3 (1-a x) (1+a x) \text{sech}^{-1}(a x)}{4 x^2}+\frac{3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{4 x^2}-\frac{(1-a x) (1+a x) \text{sech}^{-1}(a x)^3}{2 x^2}-\frac{1}{4} \left (3 a^2\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,\text{sech}^{-1}(a x)\right )+\frac{1}{4} \left (3 a^2\right ) \operatorname{Subst}\left (\int \sinh ^2(x) \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=\frac{3 \sqrt{\frac{1-a x}{1+a x}} (1+a x)}{8 x^2}-\frac{3 (1-a x) (1+a x) \text{sech}^{-1}(a x)}{4 x^2}+\frac{3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{4 x^2}-\frac{1}{4} a^2 \text{sech}^{-1}(a x)^3-\frac{(1-a x) (1+a x) \text{sech}^{-1}(a x)^3}{2 x^2}-\frac{1}{8} \left (3 a^2\right ) \operatorname{Subst}\left (\int 1 \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=\frac{3 \sqrt{\frac{1-a x}{1+a x}} (1+a x)}{8 x^2}-\frac{3}{8} a^2 \text{sech}^{-1}(a x)-\frac{3 (1-a x) (1+a x) \text{sech}^{-1}(a x)}{4 x^2}+\frac{3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{4 x^2}-\frac{1}{4} a^2 \text{sech}^{-1}(a x)^3-\frac{(1-a x) (1+a x) \text{sech}^{-1}(a x)^3}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.133262, size = 147, normalized size = 1.07 \[ \frac{-3 a^2 x^2 \log (x)+3 a^2 x^2 \log \left (a x \sqrt{\frac{1-a x}{a x+1}}+\sqrt{\frac{1-a x}{a x+1}}+1\right )+2 \left (a^2 x^2-2\right ) \text{sech}^{-1}(a x)^3+3 \sqrt{\frac{1-a x}{a x+1}} (a x+1)+6 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2-6 \text{sech}^{-1}(a x)}{8 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x) - 6*ArcSech[a*x] + 6*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x]^2
+ 2*(-2 + a^2*x^2)*ArcSech[a*x]^3 - 3*a^2*x^2*Log[x] + 3*a^2*x^2*Log[1 + Sqrt[(1 - a*x)/(1 + a*x)] + a*x*Sqrt
[(1 - a*x)/(1 + a*x)]])/(8*x^2)

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Maple [A]  time = 0.217, size = 126, normalized size = 0.9 \begin{align*}{a}^{2} \left ( -{\frac{ \left ({\rm arcsech} \left (ax\right ) \right ) ^{3}}{2\,{a}^{2}{x}^{2}}}+{\frac{3\, \left ({\rm arcsech} \left (ax\right ) \right ) ^{2}}{4\,ax}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}}}+{\frac{ \left ({\rm arcsech} \left (ax\right ) \right ) ^{3}}{4}}-{\frac{3\,{\rm arcsech} \left (ax\right )}{4\,{a}^{2}{x}^{2}}}+{\frac{3}{8\,ax}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}}}+{\frac{3\,{\rm arcsech} \left (ax\right )}{8}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(-1/2*arcsech(a*x)^3/a^2/x^2+3/4*arcsech(a*x)^2/a/x*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)+1/4*arcsech(a
*x)^3-3/4/a^2/x^2*arcsech(a*x)+3/8/a/x*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)+3/8*arcsech(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 )^{3}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(a*x)^3/x^3, x)

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Fricas [A]  time = 2.03459, size = 382, normalized size = 2.79 \begin{align*} \frac{6 \, 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} + 2 \,{\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 )^{3} + 3 \, a x \sqrt{-\frac{a^{2} x^{2} - 1}{a^{2} x^{2}}} + 3 \,{\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 )}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(6*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 + 2*(a^2*x^2 -
2)*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x))^3 + 3*a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 3*(a^2*x^2
- 2)*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x)))/x^2

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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^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(a*x)**3/x**3, 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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(a*x)^3/x^3, x)