3.1.4 \(\int x \text {sech}^{-1}(a x)^2 \, dx\) [4]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6420, 5526, 4269, 3556} \begin {gather*} -\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 \end {gather*}

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 3556

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

Rule 4269

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 5526

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 6420

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 x \text {sech}^{-1}(a x)^2 \, dx &=-\frac {\text {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 {\text {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 {\text {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*}

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Mathematica [A]
time = 0.04, size = 53, normalized size = 1.00 \begin {gather*} -\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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(99\) vs. \(2(49)=98\).
time = 0.32, size = 100, normalized size = 1.89

method	result	size
derivativedivides	\(\frac {-2 \,\mathrm {arcsech}\left (a x \right )+\frac {\mathrm {arcsech}\left (a x \right ) \left (\mathrm {arcsech}\left (a x \right ) a^{2} x^{2}-2 \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, a x +2\right )}{2}+\ln \left (1+\left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )^{2}\right )}{a^{2}}\)	\(100\)
default	\(\frac {-2 \,\mathrm {arcsech}\left (a x \right )+\frac {\mathrm {arcsech}\left (a x \right ) \left (\mathrm {arcsech}\left (a x \right ) a^{2} x^{2}-2 \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, a x +2\right )}{2}+\ln \left (1+\left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )^{2}\right )}{a^{2}}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arcsech(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 40, normalized size = 0.75 \begin {gather*} \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (49) = 98\).
time = 0.34, size = 106, normalized size = 2.00 \begin {gather*} \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 {gather*}

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 = 0.26, size = 42, normalized size = 0.79 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x\,{\mathrm {acosh}\left (\frac {1}{a\,x}\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*acosh(1/(a*x))^2, x)

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