3.30 \(\int \frac {\sinh ^{-1}(a x)^3}{x^4} \, dx\)

Optimal. Leaf size=151 \[ a^3 \sinh ^{-1}(a x) \text {Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )-a^3 \sinh ^{-1}(a x) \text {Li}_2\left (e^{\sinh ^{-1}(a x)}\right )-a^3 \text {Li}_3\left (-e^{\sinh ^{-1}(a x)}\right )+a^3 \text {Li}_3\left (e^{\sinh ^{-1}(a x)}\right )+a^3 \sinh ^{-1}(a x)^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-\frac {a \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{2 x^2}-\frac {a^2 \sinh ^{-1}(a x)}{x}-a^3 \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right )-\frac {\sinh ^{-1}(a x)^3}{3 x^3} \]

[Out]

-a^2*arcsinh(a*x)/x-1/3*arcsinh(a*x)^3/x^3+a^3*arcsinh(a*x)^2*arctanh(a*x+(a^2*x^2+1)^(1/2))-a^3*arctanh((a^2*
x^2+1)^(1/2))+a^3*arcsinh(a*x)*polylog(2,-a*x-(a^2*x^2+1)^(1/2))-a^3*arcsinh(a*x)*polylog(2,a*x+(a^2*x^2+1)^(1
/2))-a^3*polylog(3,-a*x-(a^2*x^2+1)^(1/2))+a^3*polylog(3,a*x+(a^2*x^2+1)^(1/2))-1/2*a*arcsinh(a*x)^2*(a^2*x^2+
1)^(1/2)/x^2

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Rubi [A]  time = 0.28, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5661, 5747, 5760, 4182, 2531, 2282, 6589, 266, 63, 208} \[ a^3 \sinh ^{-1}(a x) \text {PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )-a^3 \sinh ^{-1}(a x) \text {PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )-a^3 \text {PolyLog}\left (3,-e^{\sinh ^{-1}(a x)}\right )+a^3 \text {PolyLog}\left (3,e^{\sinh ^{-1}(a x)}\right )-\frac {a \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{2 x^2}-a^3 \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right )-\frac {a^2 \sinh ^{-1}(a x)}{x}+a^3 \sinh ^{-1}(a x)^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-\frac {\sinh ^{-1}(a x)^3}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]^3/x^4,x]

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 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 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 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5747

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] + (-Dist[(c^2*(m + 2*p + 3))/(f^2
*(m + 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^
2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

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

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Mathematica [A]  time = 2.25, size = 268, normalized size = 1.77 \[ \frac {1}{48} a^3 \left (-\frac {16 \sinh ^{-1}(a x)^3 \sinh ^4\left (\frac {1}{2} \sinh ^{-1}(a x)\right )}{a^3 x^3}-48 \sinh ^{-1}(a x) \text {Li}_2\left (-e^{-\sinh ^{-1}(a x)}\right )+48 \sinh ^{-1}(a x) \text {Li}_2\left (e^{-\sinh ^{-1}(a x)}\right )-48 \text {Li}_3\left (-e^{-\sinh ^{-1}(a x)}\right )+48 \text {Li}_3\left (e^{-\sinh ^{-1}(a x)}\right )-24 \sinh ^{-1}(a x)^2 \log \left (1-e^{-\sinh ^{-1}(a x)}\right )+24 \sinh ^{-1}(a x)^2 \log \left (e^{-\sinh ^{-1}(a x)}+1\right )-4 \sinh ^{-1}(a x)^3 \tanh \left (\frac {1}{2} \sinh ^{-1}(a x)\right )+24 \sinh ^{-1}(a x) \tanh \left (\frac {1}{2} \sinh ^{-1}(a x)\right )+4 \sinh ^{-1}(a x)^3 \coth \left (\frac {1}{2} \sinh ^{-1}(a x)\right )-24 \sinh ^{-1}(a x) \coth \left (\frac {1}{2} \sinh ^{-1}(a x)\right )-a x \sinh ^{-1}(a x)^3 \text {csch}^4\left (\frac {1}{2} \sinh ^{-1}(a x)\right )-6 \sinh ^{-1}(a x)^2 \text {csch}^2\left (\frac {1}{2} \sinh ^{-1}(a x)\right )-6 \sinh ^{-1}(a x)^2 \text {sech}^2\left (\frac {1}{2} \sinh ^{-1}(a x)\right )+48 \log \left (\tanh \left (\frac {1}{2} \sinh ^{-1}(a x)\right )\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSinh[a*x]^3/x^4,x]

[Out]

(a^3*(-24*ArcSinh[a*x]*Coth[ArcSinh[a*x]/2] + 4*ArcSinh[a*x]^3*Coth[ArcSinh[a*x]/2] - 6*ArcSinh[a*x]^2*Csch[Ar
cSinh[a*x]/2]^2 - a*x*ArcSinh[a*x]^3*Csch[ArcSinh[a*x]/2]^4 - 24*ArcSinh[a*x]^2*Log[1 - E^(-ArcSinh[a*x])] + 2
4*ArcSinh[a*x]^2*Log[1 + E^(-ArcSinh[a*x])] + 48*Log[Tanh[ArcSinh[a*x]/2]] - 48*ArcSinh[a*x]*PolyLog[2, -E^(-A
rcSinh[a*x])] + 48*ArcSinh[a*x]*PolyLog[2, E^(-ArcSinh[a*x])] - 48*PolyLog[3, -E^(-ArcSinh[a*x])] + 48*PolyLog
[3, E^(-ArcSinh[a*x])] - 6*ArcSinh[a*x]^2*Sech[ArcSinh[a*x]/2]^2 - (16*ArcSinh[a*x]^3*Sinh[ArcSinh[a*x]/2]^4)/
(a^3*x^3) + 24*ArcSinh[a*x]*Tanh[ArcSinh[a*x]/2] - 4*ArcSinh[a*x]^3*Tanh[ArcSinh[a*x]/2]))/48

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arsinh}\left (a x\right )^{3}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/x^4,x, algorithm="fricas")

[Out]

integral(arcsinh(a*x)^3/x^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsinh}\left (a x\right )^{3}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/x^4,x, algorithm="giac")

[Out]

integrate(arcsinh(a*x)^3/x^4, x)

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maple [A]  time = 0.54, size = 228, normalized size = 1.51 \[ -\frac {a \arcsinh \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}}{2 x^{2}}-\frac {a^{2} \arcsinh \left (a x \right )}{x}-\frac {\arcsinh \left (a x \right )^{3}}{3 x^{3}}-\frac {a^{3} \arcsinh \left (a x \right )^{2} \ln \left (1-a x -\sqrt {a^{2} x^{2}+1}\right )}{2}-a^{3} \arcsinh \left (a x \right ) \polylog \left (2, a x +\sqrt {a^{2} x^{2}+1}\right )+a^{3} \polylog \left (3, a x +\sqrt {a^{2} x^{2}+1}\right )+\frac {a^{3} \arcsinh \left (a x \right )^{2} \ln \left (a x +\sqrt {a^{2} x^{2}+1}+1\right )}{2}+a^{3} \arcsinh \left (a x \right ) \polylog \left (2, -a x -\sqrt {a^{2} x^{2}+1}\right )-a^{3} \polylog \left (3, -a x -\sqrt {a^{2} x^{2}+1}\right )-2 a^{3} \arctanh \left (a x +\sqrt {a^{2} x^{2}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)^3/x^4,x)

[Out]

-1/2*a*arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)/x^2-a^2*arcsinh(a*x)/x-1/3*arcsinh(a*x)^3/x^3-1/2*a^3*arcsinh(a*x)^2*l
n(1-a*x-(a^2*x^2+1)^(1/2))-a^3*arcsinh(a*x)*polylog(2,a*x+(a^2*x^2+1)^(1/2))+a^3*polylog(3,a*x+(a^2*x^2+1)^(1/
2))+1/2*a^3*arcsinh(a*x)^2*ln(a*x+(a^2*x^2+1)^(1/2)+1)+a^3*arcsinh(a*x)*polylog(2,-a*x-(a^2*x^2+1)^(1/2))-a^3*
polylog(3,-a*x-(a^2*x^2+1)^(1/2))-2*a^3*arctanh(a*x+(a^2*x^2+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {\log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{3}}{3 \, x^{3}} + \int \frac {{\left (a^{3} x^{2} + \sqrt {a^{2} x^{2} + 1} a^{2} x + a\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2}}{a^{3} x^{6} + a x^{4} + {\left (a^{2} x^{5} + x^{3}\right )} \sqrt {a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/x^4,x, algorithm="maxima")

[Out]

-1/3*log(a*x + sqrt(a^2*x^2 + 1))^3/x^3 + integrate((a^3*x^2 + sqrt(a^2*x^2 + 1)*a^2*x + a)*log(a*x + sqrt(a^2
*x^2 + 1))^2/(a^3*x^6 + a*x^4 + (a^2*x^5 + x^3)*sqrt(a^2*x^2 + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {asinh}\left (a\,x\right )}^3}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a*x)^3/x^4,x)

[Out]

int(asinh(a*x)^3/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}^{3}{\left (a x \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x)**3/x**4,x)

[Out]

Integral(asinh(a*x)**3/x**4, x)

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