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

Optimal. Leaf size=102 \[ \frac{3 \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a x)}\right )}{2 a^2}-\frac{3 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2}{2 a^2}-\frac{3 \text{sech}^{-1}(a x)^2}{2 a^2}+\frac{3 \text{sech}^{-1}(a x) \log \left (e^{2 \text{sech}^{-1}(a x)}+1\right )}{a^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a x)^3 \]

[Out]

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

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Rubi [A]  time = 0.121762, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.875, Rules used = {6285, 5418, 4184, 3718, 2190, 2279, 2391} \[ \frac{3 \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a x)}\right )}{2 a^2}-\frac{3 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)^2}{2 a^2}-\frac{3 \text{sech}^{-1}(a x)^2}{2 a^2}+\frac{3 \text{sech}^{-1}(a x) \log \left (e^{2 \text{sech}^{-1}(a x)}+1\right )}{a^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a x)^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcSech[a*x]^2)/(2*a^2) - (3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x]^2)/(2*a^2) + (x^2*ArcSech[a*
x]^3)/2 + (3*ArcSech[a*x]*Log[1 + E^(2*ArcSech[a*x])])/a^2 + (3*PolyLog[2, -E^(2*ArcSech[a*x])])/(2*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 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 2279

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int x \text{sech}^{-1}(a x)^3 \, dx &=-\frac{\operatorname{Subst}\left (\int x^3 \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)^3-\frac{3 \operatorname{Subst}\left (\int x^2 \text{sech}^2(x) \, dx,x,\text{sech}^{-1}(a x)\right )}{2 a^2}\\ &=-\frac{3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{2 a^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a x)^3+\frac{3 \operatorname{Subst}\left (\int x \tanh (x) \, dx,x,\text{sech}^{-1}(a x)\right )}{a^2}\\ &=-\frac{3 \text{sech}^{-1}(a x)^2}{2 a^2}-\frac{3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{2 a^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a x)^3+\frac{6 \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1+e^{2 x}} \, dx,x,\text{sech}^{-1}(a x)\right )}{a^2}\\ &=-\frac{3 \text{sech}^{-1}(a x)^2}{2 a^2}-\frac{3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{2 a^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a x)^3+\frac{3 \text{sech}^{-1}(a x) \log \left (1+e^{2 \text{sech}^{-1}(a x)}\right )}{a^2}-\frac{3 \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\text{sech}^{-1}(a x)\right )}{a^2}\\ &=-\frac{3 \text{sech}^{-1}(a x)^2}{2 a^2}-\frac{3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{2 a^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a x)^3+\frac{3 \text{sech}^{-1}(a x) \log \left (1+e^{2 \text{sech}^{-1}(a x)}\right )}{a^2}-\frac{3 \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \text{sech}^{-1}(a x)}\right )}{2 a^2}\\ &=-\frac{3 \text{sech}^{-1}(a x)^2}{2 a^2}-\frac{3 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)^2}{2 a^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a x)^3+\frac{3 \text{sech}^{-1}(a x) \log \left (1+e^{2 \text{sech}^{-1}(a x)}\right )}{a^2}+\frac{3 \text{Li}_2\left (-e^{2 \text{sech}^{-1}(a x)}\right )}{2 a^2}\\ \end{align*}

Mathematica [A]  time = 0.360192, size = 101, normalized size = 0.99 \[ \frac{\text{sech}^{-1}(a x) \left (a^2 x^2 \text{sech}^{-1}(a x)^2-3 \left (a x \sqrt{\frac{1-a x}{a x+1}}+\sqrt{\frac{1-a x}{a x+1}}-1\right ) \text{sech}^{-1}(a x)+6 \log \left (e^{-2 \text{sech}^{-1}(a x)}+1\right )\right )-3 \text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}(a x)}\right )}{2 a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.267, size = 152, normalized size = 1.5 \begin{align*}{\frac{{x}^{2} \left ({\rm arcsech} \left (ax\right ) \right ) ^{3}}{2}}-{\frac{3\,x \left ({\rm arcsech} \left (ax\right ) \right ) ^{2}}{2\,a}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}}}-{\frac{3\, \left ({\rm arcsech} \left (ax\right ) \right ) ^{2}}{2\,{a}^{2}}}+3\,{\frac{{\rm arcsech} \left (ax\right )}{{a}^{2}}\ln \left ( 1+ \left ({\frac{1}{ax}}+\sqrt{{\frac{1}{ax}}-1}\sqrt{1+{\frac{1}{ax}}} \right ) ^{2} \right ) }+{\frac{3}{2\,{a}^{2}}{\it polylog} \left ( 2,- \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)^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{arsech}\left (a x\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsech(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*arcsech(a*x)^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{asech}^{3}{\left (a x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*asech(a*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{arsech}\left (a x\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsech(a*x)^3, x)