3.347 \(\int \frac{\sinh ^{-1}(a x)^3}{x \sqrt{1+a^2 x^2}} \, dx\)

Optimal. Leaf size=102 \[ -3 \sinh ^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )+6 \sinh ^{-1}(a x) \text{PolyLog}\left (3,-e^{\sinh ^{-1}(a x)}\right )-6 \sinh ^{-1}(a x) \text{PolyLog}\left (3,e^{\sinh ^{-1}(a x)}\right )-6 \text{PolyLog}\left (4,-e^{\sinh ^{-1}(a x)}\right )+6 \text{PolyLog}\left (4,e^{\sinh ^{-1}(a x)}\right )-2 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right ) \]

[Out]

-2*ArcSinh[a*x]^3*ArcTanh[E^ArcSinh[a*x]] - 3*ArcSinh[a*x]^2*PolyLog[2, -E^ArcSinh[a*x]] + 3*ArcSinh[a*x]^2*Po
lyLog[2, E^ArcSinh[a*x]] + 6*ArcSinh[a*x]*PolyLog[3, -E^ArcSinh[a*x]] - 6*ArcSinh[a*x]*PolyLog[3, E^ArcSinh[a*
x]] - 6*PolyLog[4, -E^ArcSinh[a*x]] + 6*PolyLog[4, E^ArcSinh[a*x]]

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Rubi [A]  time = 0.162071, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {5760, 4182, 2531, 6609, 2282, 6589} \[ -3 \sinh ^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )+6 \sinh ^{-1}(a x) \text{PolyLog}\left (3,-e^{\sinh ^{-1}(a x)}\right )-6 \sinh ^{-1}(a x) \text{PolyLog}\left (3,e^{\sinh ^{-1}(a x)}\right )-6 \text{PolyLog}\left (4,-e^{\sinh ^{-1}(a x)}\right )+6 \text{PolyLog}\left (4,e^{\sinh ^{-1}(a x)}\right )-2 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]^3/(x*Sqrt[1 + a^2*x^2]),x]

[Out]

-2*ArcSinh[a*x]^3*ArcTanh[E^ArcSinh[a*x]] - 3*ArcSinh[a*x]^2*PolyLog[2, -E^ArcSinh[a*x]] + 3*ArcSinh[a*x]^2*Po
lyLog[2, E^ArcSinh[a*x]] + 6*ArcSinh[a*x]*PolyLog[3, -E^ArcSinh[a*x]] - 6*ArcSinh[a*x]*PolyLog[3, E^ArcSinh[a*
x]] - 6*PolyLog[4, -E^ArcSinh[a*x]] + 6*PolyLog[4, E^ArcSinh[a*x]]

Rule 5760

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c^(m
+ 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*Sinh[x]^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[
e, c^2*d] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, 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{\sinh ^{-1}(a x)^3}{x \sqrt{1+a^2 x^2}} \, dx &=\operatorname{Subst}\left (\int x^3 \text{csch}(x) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-2 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-3 \operatorname{Subst}\left (\int x^2 \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )+3 \operatorname{Subst}\left (\int x^2 \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-2 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text{Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x)^2 \text{Li}_2\left (e^{\sinh ^{-1}(a x)}\right )+6 \operatorname{Subst}\left (\int x \text{Li}_2\left (-e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )-6 \operatorname{Subst}\left (\int x \text{Li}_2\left (e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-2 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text{Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x)^2 \text{Li}_2\left (e^{\sinh ^{-1}(a x)}\right )+6 \sinh ^{-1}(a x) \text{Li}_3\left (-e^{\sinh ^{-1}(a x)}\right )-6 \sinh ^{-1}(a x) \text{Li}_3\left (e^{\sinh ^{-1}(a x)}\right )-6 \operatorname{Subst}\left (\int \text{Li}_3\left (-e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )+6 \operatorname{Subst}\left (\int \text{Li}_3\left (e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-2 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text{Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x)^2 \text{Li}_2\left (e^{\sinh ^{-1}(a x)}\right )+6 \sinh ^{-1}(a x) \text{Li}_3\left (-e^{\sinh ^{-1}(a x)}\right )-6 \sinh ^{-1}(a x) \text{Li}_3\left (e^{\sinh ^{-1}(a x)}\right )-6 \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )+6 \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )\\ &=-2 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text{Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x)^2 \text{Li}_2\left (e^{\sinh ^{-1}(a x)}\right )+6 \sinh ^{-1}(a x) \text{Li}_3\left (-e^{\sinh ^{-1}(a x)}\right )-6 \sinh ^{-1}(a x) \text{Li}_3\left (e^{\sinh ^{-1}(a x)}\right )-6 \text{Li}_4\left (-e^{\sinh ^{-1}(a x)}\right )+6 \text{Li}_4\left (e^{\sinh ^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.128932, size = 146, normalized size = 1.43 \[ \frac{1}{8} \left (24 \sinh ^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(a x)}\right )+24 \sinh ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )+48 \sinh ^{-1}(a x) \text{PolyLog}\left (3,-e^{-\sinh ^{-1}(a x)}\right )-48 \sinh ^{-1}(a x) \text{PolyLog}\left (3,e^{\sinh ^{-1}(a x)}\right )+48 \text{PolyLog}\left (4,-e^{-\sinh ^{-1}(a x)}\right )+48 \text{PolyLog}\left (4,e^{\sinh ^{-1}(a x)}\right )-2 \sinh ^{-1}(a x)^4-8 \sinh ^{-1}(a x)^3 \log \left (e^{-\sinh ^{-1}(a x)}+1\right )+8 \sinh ^{-1}(a x)^3 \log \left (1-e^{\sinh ^{-1}(a x)}\right )+\pi ^4\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSinh[a*x]^3/(x*Sqrt[1 + a^2*x^2]),x]

[Out]

(Pi^4 - 2*ArcSinh[a*x]^4 - 8*ArcSinh[a*x]^3*Log[1 + E^(-ArcSinh[a*x])] + 8*ArcSinh[a*x]^3*Log[1 - E^ArcSinh[a*
x]] + 24*ArcSinh[a*x]^2*PolyLog[2, -E^(-ArcSinh[a*x])] + 24*ArcSinh[a*x]^2*PolyLog[2, E^ArcSinh[a*x]] + 48*Arc
Sinh[a*x]*PolyLog[3, -E^(-ArcSinh[a*x])] - 48*ArcSinh[a*x]*PolyLog[3, E^ArcSinh[a*x]] + 48*PolyLog[4, -E^(-Arc
Sinh[a*x])] + 48*PolyLog[4, E^ArcSinh[a*x]])/8

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Maple [A]  time = 0.056, size = 197, normalized size = 1.9 \begin{align*} - \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}\ln \left ( 1+ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) -3\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}{\it polylog} \left ( 2,-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) +6\,{\it Arcsinh} \left ( ax \right ){\it polylog} \left ( 3,-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) -6\,{\it polylog} \left ( 4,-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) + \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}\ln \left ( 1-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) +3\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}{\it polylog} \left ( 2,ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) -6\,{\it Arcsinh} \left ( ax \right ){\it polylog} \left ( 3,ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) +6\,{\it polylog} \left ( 4,ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)^3/x/(a^2*x^2+1)^(1/2),x)

[Out]

-arcsinh(a*x)^3*ln(1+a*x+(a^2*x^2+1)^(1/2))-3*arcsinh(a*x)^2*polylog(2,-a*x-(a^2*x^2+1)^(1/2))+6*arcsinh(a*x)*
polylog(3,-a*x-(a^2*x^2+1)^(1/2))-6*polylog(4,-a*x-(a^2*x^2+1)^(1/2))+arcsinh(a*x)^3*ln(1-a*x-(a^2*x^2+1)^(1/2
))+3*arcsinh(a*x)^2*polylog(2,a*x+(a^2*x^2+1)^(1/2))-6*arcsinh(a*x)*polylog(3,a*x+(a^2*x^2+1)^(1/2))+6*polylog
(4,a*x+(a^2*x^2+1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/x/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(arcsinh(a*x)^3/(sqrt(a^2*x^2 + 1)*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a^{2} x^{2} + 1} \operatorname{arsinh}\left (a x\right )^{3}}{a^{2} x^{3} + x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/x/(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*x^2 + 1)*arcsinh(a*x)^3/(a^2*x^3 + x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asinh}^{3}{\left (a x \right )}}{x \sqrt{a^{2} x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x)**3/x/(a**2*x**2+1)**(1/2),x)

[Out]

Integral(asinh(a*x)**3/(x*sqrt(a**2*x**2 + 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/x/(a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(arcsinh(a*x)^3/(sqrt(a^2*x^2 + 1)*x), x)