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

Optimal. Leaf size=88 \[ -\frac{3}{2} \text{sech}^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a x)}\right )+\frac{3}{2} \text{sech}^{-1}(a x) \text{PolyLog}\left (3,-e^{2 \text{sech}^{-1}(a x)}\right )-\frac{3}{4} \text{PolyLog}\left (4,-e^{2 \text{sech}^{-1}(a x)}\right )+\frac{1}{4} \text{sech}^{-1}(a x)^4-\text{sech}^{-1}(a x)^3 \log \left (e^{2 \text{sech}^{-1}(a x)}+1\right ) \]

[Out]

ArcSech[a*x]^4/4 - ArcSech[a*x]^3*Log[1 + E^(2*ArcSech[a*x])] - (3*ArcSech[a*x]^2*PolyLog[2, -E^(2*ArcSech[a*x
])])/2 + (3*ArcSech[a*x]*PolyLog[3, -E^(2*ArcSech[a*x])])/2 - (3*PolyLog[4, -E^(2*ArcSech[a*x])])/4

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Rubi [A]  time = 0.107518, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {6285, 3718, 2190, 2531, 6609, 2282, 6589} \[ -\frac{3}{2} \text{sech}^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a x)}\right )+\frac{3}{2} \text{sech}^{-1}(a x) \text{PolyLog}\left (3,-e^{2 \text{sech}^{-1}(a x)}\right )-\frac{3}{4} \text{PolyLog}\left (4,-e^{2 \text{sech}^{-1}(a x)}\right )+\frac{1}{4} \text{sech}^{-1}(a x)^4-\text{sech}^{-1}(a x)^3 \log \left (e^{2 \text{sech}^{-1}(a x)}+1\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcSech[a*x]^4/4 - ArcSech[a*x]^3*Log[1 + E^(2*ArcSech[a*x])] - (3*ArcSech[a*x]^2*PolyLog[2, -E^(2*ArcSech[a*x
])])/2 + (3*ArcSech[a*x]*PolyLog[3, -E^(2*ArcSech[a*x])])/2 - (3*PolyLog[4, -E^(2*ArcSech[a*x])])/4

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 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{\text{sech}^{-1}(a x)^3}{x} \, dx &=-\operatorname{Subst}\left (\int x^3 \tanh (x) \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=\frac{1}{4} \text{sech}^{-1}(a x)^4-2 \operatorname{Subst}\left (\int \frac{e^{2 x} x^3}{1+e^{2 x}} \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=\frac{1}{4} \text{sech}^{-1}(a x)^4-\text{sech}^{-1}(a x)^3 \log \left (1+e^{2 \text{sech}^{-1}(a x)}\right )+3 \operatorname{Subst}\left (\int x^2 \log \left (1+e^{2 x}\right ) \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=\frac{1}{4} \text{sech}^{-1}(a x)^4-\text{sech}^{-1}(a x)^3 \log \left (1+e^{2 \text{sech}^{-1}(a x)}\right )-\frac{3}{2} \text{sech}^{-1}(a x)^2 \text{Li}_2\left (-e^{2 \text{sech}^{-1}(a x)}\right )+3 \operatorname{Subst}\left (\int x \text{Li}_2\left (-e^{2 x}\right ) \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=\frac{1}{4} \text{sech}^{-1}(a x)^4-\text{sech}^{-1}(a x)^3 \log \left (1+e^{2 \text{sech}^{-1}(a x)}\right )-\frac{3}{2} \text{sech}^{-1}(a x)^2 \text{Li}_2\left (-e^{2 \text{sech}^{-1}(a x)}\right )+\frac{3}{2} \text{sech}^{-1}(a x) \text{Li}_3\left (-e^{2 \text{sech}^{-1}(a x)}\right )-\frac{3}{2} \operatorname{Subst}\left (\int \text{Li}_3\left (-e^{2 x}\right ) \, dx,x,\text{sech}^{-1}(a x)\right )\\ &=\frac{1}{4} \text{sech}^{-1}(a x)^4-\text{sech}^{-1}(a x)^3 \log \left (1+e^{2 \text{sech}^{-1}(a x)}\right )-\frac{3}{2} \text{sech}^{-1}(a x)^2 \text{Li}_2\left (-e^{2 \text{sech}^{-1}(a x)}\right )+\frac{3}{2} \text{sech}^{-1}(a x) \text{Li}_3\left (-e^{2 \text{sech}^{-1}(a x)}\right )-\frac{3}{4} \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 \text{sech}^{-1}(a x)}\right )\\ &=\frac{1}{4} \text{sech}^{-1}(a x)^4-\text{sech}^{-1}(a x)^3 \log \left (1+e^{2 \text{sech}^{-1}(a x)}\right )-\frac{3}{2} \text{sech}^{-1}(a x)^2 \text{Li}_2\left (-e^{2 \text{sech}^{-1}(a x)}\right )+\frac{3}{2} \text{sech}^{-1}(a x) \text{Li}_3\left (-e^{2 \text{sech}^{-1}(a x)}\right )-\frac{3}{4} \text{Li}_4\left (-e^{2 \text{sech}^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.0493463, size = 84, normalized size = 0.95 \[ \frac{1}{4} \left (6 \text{sech}^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}(a x)}\right )+6 \text{sech}^{-1}(a x) \text{PolyLog}\left (3,-e^{-2 \text{sech}^{-1}(a x)}\right )+3 \text{PolyLog}\left (4,-e^{-2 \text{sech}^{-1}(a x)}\right )-\text{sech}^{-1}(a x)^4-4 \text{sech}^{-1}(a x)^3 \log \left (e^{-2 \text{sech}^{-1}(a x)}+1\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-ArcSech[a*x]^4 - 4*ArcSech[a*x]^3*Log[1 + E^(-2*ArcSech[a*x])] + 6*ArcSech[a*x]^2*PolyLog[2, -E^(-2*ArcSech[
a*x])] + 6*ArcSech[a*x]*PolyLog[3, -E^(-2*ArcSech[a*x])] + 3*PolyLog[4, -E^(-2*ArcSech[a*x])])/4

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Maple [A]  time = 0.244, size = 181, normalized size = 2.1 \begin{align*}{\frac{ \left ({\rm arcsech} \left (ax\right ) \right ) ^{4}}{4}}- \left ({\rm arcsech} \left (ax\right ) \right ) ^{3}\ln \left ( 1+ \left ({\frac{1}{ax}}+\sqrt{{\frac{1}{ax}}-1}\sqrt{1+{\frac{1}{ax}}} \right ) ^{2} \right ) -{\frac{3\, \left ({\rm arcsech} \left (ax\right ) \right ) ^{2}}{2}{\it polylog} \left ( 2,- \left ({\frac{1}{ax}}+\sqrt{{\frac{1}{ax}}-1}\sqrt{1+{\frac{1}{ax}}} \right ) ^{2} \right ) }+{\frac{3\,{\rm arcsech} \left (ax\right )}{2}{\it polylog} \left ( 3,- \left ({\frac{1}{ax}}+\sqrt{{\frac{1}{ax}}-1}\sqrt{1+{\frac{1}{ax}}} \right ) ^{2} \right ) }-{\frac{3}{4}{\it polylog} \left ( 4,- \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(arcsech(a*x)^3/x,x)

[Out]

1/4*arcsech(a*x)^4-arcsech(a*x)^3*ln(1+(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))^2)-3/2*arcsech(a*x)^2*polylog(2
,-(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))^2)+3/2*arcsech(a*x)*polylog(3,-(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2
))^2)-3/4*polylog(4,-(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))^2)

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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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(a*x)^3/x, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arsech}\left (a x\right )^{3}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsech(a*x)^3/x, x)

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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}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(a*x)^3/x, x)