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

Optimal. Leaf size=53 \[ -\frac{\log (x)}{a^2}-\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)}{a^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a x)^2 \]

[Out]

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

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Rubi [A]  time = 0.0567211, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6285, 5418, 4184, 3475} \[ -\frac{\log (x)}{a^2}-\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)}{a^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a x)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x])/a^2) + (x^2*ArcSech[a*x]^2)/2 - Log[x]/a^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 5418

Int[(x_)^(m_.)*Sech[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tanh[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> -Simp[(
x^(m - n + 1)*Sech[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sech[a + b*x^n]^p, x],
x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3475

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

Rubi steps

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

Mathematica [A]  time = 0.0595508, size = 53, normalized size = 1. \[ -\frac{\log (x)}{a^2}-\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)}{a^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a x)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.258, size = 101, normalized size = 1.9 \begin{align*} -{\frac{{\rm arcsech} \left (ax\right )}{{a}^{2}}}+{\frac{{x}^{2} \left ({\rm arcsech} \left (ax\right ) \right ) ^{2}}{2}}-{\frac{{\rm arcsech} \left (ax\right )x}{a}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}}}+{\frac{1}{{a}^{2}}\ln \left ( 1+ \left ({\frac{1}{ax}}+\sqrt{{\frac{1}{ax}}-1}\sqrt{1+{\frac{1}{ax}}} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.04775, size = 54, normalized size = 1.02 \begin{align*} \frac{1}{2} \, x^{2} \operatorname{arsech}\left (a x\right )^{2} - \frac{x \sqrt{\frac{1}{a^{2} x^{2}} - 1} \operatorname{arsech}\left (a x\right )}{a} - \frac{\log \left (x\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.63633, size = 236, normalized size = 4.45 \begin{align*} \frac{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 \, 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 (x\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.11023, size = 42, normalized size = 0.79 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{asech}^{2}{\left (a x \right )}}{2} - \frac{\sqrt{- a^{2} x^{2} + 1} \operatorname{asech}{\left (a x \right )}}{a^{2}} - \frac{\log{\left (x \right )}}{a^{2}} & \text{for}\: a \neq 0 \\\infty x^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2*asech(a*x)**2/2 - sqrt(-a**2*x**2 + 1)*asech(a*x)/a**2 - log(x)/a**2, Ne(a, 0)), (oo*x**2, Tru
e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsech(a*x)^2, x)