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

Optimal. Leaf size=111 \[ \frac{6 i \text{sech}^{-1}(a x) \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a x)}\right )}{a}-\frac{6 i \text{sech}^{-1}(a x) \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a x)}\right )}{a}-\frac{6 i \text{PolyLog}\left (3,-i e^{\text{sech}^{-1}(a x)}\right )}{a}+\frac{6 i \text{PolyLog}\left (3,i e^{\text{sech}^{-1}(a x)}\right )}{a}+x \text{sech}^{-1}(a x)^3-\frac{6 \text{sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a x)}\right )}{a} \]

[Out]

x*ArcSech[a*x]^3 - (6*ArcSech[a*x]^2*ArcTan[E^ArcSech[a*x]])/a + ((6*I)*ArcSech[a*x]*PolyLog[2, (-I)*E^ArcSech
[a*x]])/a - ((6*I)*ArcSech[a*x]*PolyLog[2, I*E^ArcSech[a*x]])/a - ((6*I)*PolyLog[3, (-I)*E^ArcSech[a*x]])/a +
((6*I)*PolyLog[3, I*E^ArcSech[a*x]])/a

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Rubi [A]  time = 0.0884604, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {6279, 5418, 4180, 2531, 2282, 6589} \[ \frac{6 i \text{sech}^{-1}(a x) \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a x)}\right )}{a}-\frac{6 i \text{sech}^{-1}(a x) \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a x)}\right )}{a}-\frac{6 i \text{PolyLog}\left (3,-i e^{\text{sech}^{-1}(a x)}\right )}{a}+\frac{6 i \text{PolyLog}\left (3,i e^{\text{sech}^{-1}(a x)}\right )}{a}+x \text{sech}^{-1}(a x)^3-\frac{6 \text{sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a x)}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*ArcSech[a*x]^3 - (6*ArcSech[a*x]^2*ArcTan[E^ArcSech[a*x]])/a + ((6*I)*ArcSech[a*x]*PolyLog[2, (-I)*E^ArcSech
[a*x]])/a - ((6*I)*ArcSech[a*x]*PolyLog[2, I*E^ArcSech[a*x]])/a - ((6*I)*PolyLog[3, (-I)*E^ArcSech[a*x]])/a +
((6*I)*PolyLog[3, I*E^ArcSech[a*x]])/a

Rule 6279

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Dist[c^(-1), Subst[Int[(a + b*x)^n*Sech[x]*Tanh[x]
, x], x, ArcSech[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[n, 0]

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 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && 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 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 \text{sech}^{-1}(a x)^3 \, dx &=-\frac{\operatorname{Subst}\left (\int x^3 \text{sech}(x) \tanh (x) \, dx,x,\text{sech}^{-1}(a x)\right )}{a}\\ &=x \text{sech}^{-1}(a x)^3-\frac{3 \operatorname{Subst}\left (\int x^2 \text{sech}(x) \, dx,x,\text{sech}^{-1}(a x)\right )}{a}\\ &=x \text{sech}^{-1}(a x)^3-\frac{6 \text{sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a x)}\right )}{a}+\frac{(6 i) \operatorname{Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\text{sech}^{-1}(a x)\right )}{a}-\frac{(6 i) \operatorname{Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\text{sech}^{-1}(a x)\right )}{a}\\ &=x \text{sech}^{-1}(a x)^3-\frac{6 \text{sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a x)}\right )}{a}+\frac{6 i \text{sech}^{-1}(a x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a x)}\right )}{a}-\frac{6 i \text{sech}^{-1}(a x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a x)}\right )}{a}-\frac{(6 i) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\text{sech}^{-1}(a x)\right )}{a}+\frac{(6 i) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\text{sech}^{-1}(a x)\right )}{a}\\ &=x \text{sech}^{-1}(a x)^3-\frac{6 \text{sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a x)}\right )}{a}+\frac{6 i \text{sech}^{-1}(a x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a x)}\right )}{a}-\frac{6 i \text{sech}^{-1}(a x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a x)}\right )}{a}-\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a x)}\right )}{a}+\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a x)}\right )}{a}\\ &=x \text{sech}^{-1}(a x)^3-\frac{6 \text{sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a x)}\right )}{a}+\frac{6 i \text{sech}^{-1}(a x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a x)}\right )}{a}-\frac{6 i \text{sech}^{-1}(a x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a x)}\right )}{a}-\frac{6 i \text{Li}_3\left (-i e^{\text{sech}^{-1}(a x)}\right )}{a}+\frac{6 i \text{Li}_3\left (i e^{\text{sech}^{-1}(a x)}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.116257, size = 128, normalized size = 1.15 \[ x \text{sech}^{-1}(a x)^3-\frac{3 i \left (-2 \text{sech}^{-1}(a x) \left (\text{PolyLog}\left (2,-i e^{-\text{sech}^{-1}(a x)}\right )-\text{PolyLog}\left (2,i e^{-\text{sech}^{-1}(a x)}\right )\right )-2 \left (\text{PolyLog}\left (3,-i e^{-\text{sech}^{-1}(a x)}\right )-\text{PolyLog}\left (3,i e^{-\text{sech}^{-1}(a x)}\right )\right )+\text{sech}^{-1}(a x)^2 \left (-\left (\log \left (1-i e^{-\text{sech}^{-1}(a x)}\right )-\log \left (1+i e^{-\text{sech}^{-1}(a x)}\right )\right )\right )\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.304, size = 0, normalized size = 0. \begin{align*} \int \left ({\rm arcsech} \left (ax\right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} x \log \left (\sqrt{a x + 1} \sqrt{-a x + 1} + 1\right )^{3} - \int \frac{a^{2} x^{2} \log \left (a\right )^{3} +{\left (a^{2} x^{2} - 1\right )} \log \left (x\right )^{3} + 3 \,{\left (a^{2} x^{2} \log \left (a\right ) +{\left (a^{2} x^{2}{\left (\log \left (a\right ) + 1\right )} +{\left (a^{2} x^{2} - 1\right )} \log \left (x\right ) - \log \left (a\right )\right )} \sqrt{a x + 1} \sqrt{-a x + 1} +{\left (a^{2} x^{2} - 1\right )} \log \left (x\right ) - \log \left (a\right )\right )} \log \left (\sqrt{a x + 1} \sqrt{-a x + 1} + 1\right )^{2} - \log \left (a\right )^{3} + 3 \,{\left (a^{2} x^{2} \log \left (a\right ) - \log \left (a\right )\right )} \log \left (x\right )^{2} +{\left (a^{2} x^{2} \log \left (a\right )^{3} +{\left (a^{2} x^{2} - 1\right )} \log \left (x\right )^{3} - \log \left (a\right )^{3} + 3 \,{\left (a^{2} x^{2} \log \left (a\right ) - \log \left (a\right )\right )} \log \left (x\right )^{2} + 3 \,{\left (a^{2} x^{2} \log \left (a\right )^{2} - \log \left (a\right )^{2}\right )} \log \left (x\right )\right )} \sqrt{a x + 1} \sqrt{-a x + 1} - 3 \,{\left (a^{2} x^{2} \log \left (a\right )^{2} +{\left (a^{2} x^{2} - 1\right )} \log \left (x\right )^{2} +{\left (a^{2} x^{2} \log \left (a\right )^{2} +{\left (a^{2} x^{2} - 1\right )} \log \left (x\right )^{2} - \log \left (a\right )^{2} + 2 \,{\left (a^{2} x^{2} \log \left (a\right ) - \log \left (a\right )\right )} \log \left (x\right )\right )} \sqrt{a x + 1} \sqrt{-a x + 1} - \log \left (a\right )^{2} + 2 \,{\left (a^{2} x^{2} \log \left (a\right ) - \log \left (a\right )\right )} \log \left (x\right )\right )} \log \left (\sqrt{a x + 1} \sqrt{-a x + 1} + 1\right ) + 3 \,{\left (a^{2} x^{2} \log \left (a\right )^{2} - \log \left (a\right )^{2}\right )} \log \left (x\right )}{a^{2} x^{2} +{\left (a^{2} x^{2} - 1\right )} \sqrt{a x + 1} \sqrt{-a x + 1} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(sqrt(a*x + 1)*sqrt(-a*x + 1) + 1)^3 - integrate((a^2*x^2*log(a)^3 + (a^2*x^2 - 1)*log(x)^3 + 3*(a^2*x^2*
log(a) + (a^2*x^2*(log(a) + 1) + (a^2*x^2 - 1)*log(x) - log(a))*sqrt(a*x + 1)*sqrt(-a*x + 1) + (a^2*x^2 - 1)*l
og(x) - log(a))*log(sqrt(a*x + 1)*sqrt(-a*x + 1) + 1)^2 - log(a)^3 + 3*(a^2*x^2*log(a) - log(a))*log(x)^2 + (a
^2*x^2*log(a)^3 + (a^2*x^2 - 1)*log(x)^3 - log(a)^3 + 3*(a^2*x^2*log(a) - log(a))*log(x)^2 + 3*(a^2*x^2*log(a)
^2 - log(a)^2)*log(x))*sqrt(a*x + 1)*sqrt(-a*x + 1) - 3*(a^2*x^2*log(a)^2 + (a^2*x^2 - 1)*log(x)^2 + (a^2*x^2*
log(a)^2 + (a^2*x^2 - 1)*log(x)^2 - log(a)^2 + 2*(a^2*x^2*log(a) - log(a))*log(x))*sqrt(a*x + 1)*sqrt(-a*x + 1
) - log(a)^2 + 2*(a^2*x^2*log(a) - log(a))*log(x))*log(sqrt(a*x + 1)*sqrt(-a*x + 1) + 1) + 3*(a^2*x^2*log(a)^2
- log(a)^2)*log(x))/(a^2*x^2 + (a^2*x^2 - 1)*sqrt(a*x + 1)*sqrt(-a*x + 1) - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsech(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(a*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(a*x)^3, x)