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

Optimal. Leaf size=223 \[ 2 a^3 \sinh ^{-1}(a x)^2 \text {Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )-2 a^3 \sinh ^{-1}(a x)^2 \text {Li}_2\left (e^{\sinh ^{-1}(a x)}\right )-4 a^3 \sinh ^{-1}(a x) \text {Li}_3\left (-e^{\sinh ^{-1}(a x)}\right )+4 a^3 \sinh ^{-1}(a x) \text {Li}_3\left (e^{\sinh ^{-1}(a x)}\right )-4 a^3 \text {Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+4 a^3 \text {Li}_2\left (e^{\sinh ^{-1}(a x)}\right )+4 a^3 \text {Li}_4\left (-e^{\sinh ^{-1}(a x)}\right )-4 a^3 \text {Li}_4\left (e^{\sinh ^{-1}(a x)}\right )+\frac {4}{3} a^3 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-8 a^3 \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-\frac {2 a \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{3 x^2}-\frac {2 a^2 \sinh ^{-1}(a x)^2}{x}-\frac {\sinh ^{-1}(a x)^4}{3 x^3} \]

[Out]

-2*a^2*arcsinh(a*x)^2/x-1/3*arcsinh(a*x)^4/x^3-8*a^3*arcsinh(a*x)*arctanh(a*x+(a^2*x^2+1)^(1/2))+4/3*a^3*arcsi
nh(a*x)^3*arctanh(a*x+(a^2*x^2+1)^(1/2))-4*a^3*polylog(2,-a*x-(a^2*x^2+1)^(1/2))+2*a^3*arcsinh(a*x)^2*polylog(
2,-a*x-(a^2*x^2+1)^(1/2))+4*a^3*polylog(2,a*x+(a^2*x^2+1)^(1/2))-2*a^3*arcsinh(a*x)^2*polylog(2,a*x+(a^2*x^2+1
)^(1/2))-4*a^3*arcsinh(a*x)*polylog(3,-a*x-(a^2*x^2+1)^(1/2))+4*a^3*arcsinh(a*x)*polylog(3,a*x+(a^2*x^2+1)^(1/
2))+4*a^3*polylog(4,-a*x-(a^2*x^2+1)^(1/2))-4*a^3*polylog(4,a*x+(a^2*x^2+1)^(1/2))-2/3*a*arcsinh(a*x)^3*(a^2*x
^2+1)^(1/2)/x^2

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Rubi [A]  time = 0.39, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 19, 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, 6609, 2282, 6589, 2279, 2391} \[ 2 a^3 \sinh ^{-1}(a x)^2 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )-2 a^3 \sinh ^{-1}(a x)^2 \text {PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )-4 a^3 \sinh ^{-1}(a x) \text {PolyLog}\left (3,-e^{\sinh ^{-1}(a x)}\right )+4 a^3 \sinh ^{-1}(a x) \text {PolyLog}\left (3,e^{\sinh ^{-1}(a x)}\right )-4 a^3 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )+4 a^3 \text {PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )+4 a^3 \text {PolyLog}\left (4,-e^{\sinh ^{-1}(a x)}\right )-4 a^3 \text {PolyLog}\left (4,e^{\sinh ^{-1}(a x)}\right )-\frac {2 a \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{3 x^2}-\frac {2 a^2 \sinh ^{-1}(a x)^2}{x}+\frac {4}{3} a^3 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-8 a^3 \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-\frac {\sinh ^{-1}(a x)^4}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*ArcSinh[a*x]^2)/x - (2*a*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3)/(3*x^2) - ArcSinh[a*x]^4/(3*x^3) - 8*a^3*Ar
cSinh[a*x]*ArcTanh[E^ArcSinh[a*x]] + (4*a^3*ArcSinh[a*x]^3*ArcTanh[E^ArcSinh[a*x]])/3 - 4*a^3*PolyLog[2, -E^Ar
cSinh[a*x]] + 2*a^3*ArcSinh[a*x]^2*PolyLog[2, -E^ArcSinh[a*x]] + 4*a^3*PolyLog[2, E^ArcSinh[a*x]] - 2*a^3*ArcS
inh[a*x]^2*PolyLog[2, E^ArcSinh[a*x]] - 4*a^3*ArcSinh[a*x]*PolyLog[3, -E^ArcSinh[a*x]] + 4*a^3*ArcSinh[a*x]*Po
lyLog[3, E^ArcSinh[a*x]] + 4*a^3*PolyLog[4, -E^ArcSinh[a*x]] - 4*a^3*PolyLog[4, E^ArcSinh[a*x]]

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 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 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]

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]

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]

Rubi steps

\begin {align*} \int \frac {\sinh ^{-1}(a x)^4}{x^4} \, dx &=-\frac {\sinh ^{-1}(a x)^4}{3 x^3}+\frac {1}{3} (4 a) \int \frac {\sinh ^{-1}(a x)^3}{x^3 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {2 a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 x^2}-\frac {\sinh ^{-1}(a x)^4}{3 x^3}+\left (2 a^2\right ) \int \frac {\sinh ^{-1}(a x)^2}{x^2} \, dx-\frac {1}{3} \left (2 a^3\right ) \int \frac {\sinh ^{-1}(a x)^3}{x \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {2 a^2 \sinh ^{-1}(a x)^2}{x}-\frac {2 a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 x^2}-\frac {\sinh ^{-1}(a x)^4}{3 x^3}-\frac {1}{3} \left (2 a^3\right ) \operatorname {Subst}\left (\int x^3 \text {csch}(x) \, dx,x,\sinh ^{-1}(a x)\right )+\left (4 a^3\right ) \int \frac {\sinh ^{-1}(a x)}{x \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {2 a^2 \sinh ^{-1}(a x)^2}{x}-\frac {2 a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 x^2}-\frac {\sinh ^{-1}(a x)^4}{3 x^3}+\frac {4}{3} a^3 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )+\left (2 a^3\right ) \operatorname {Subst}\left (\int x^2 \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )-\left (2 a^3\right ) \operatorname {Subst}\left (\int x^2 \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )+\left (4 a^3\right ) \operatorname {Subst}\left (\int x \text {csch}(x) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac {2 a^2 \sinh ^{-1}(a x)^2}{x}-\frac {2 a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 x^2}-\frac {\sinh ^{-1}(a x)^4}{3 x^3}-8 a^3 \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )+\frac {4}{3} a^3 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )+2 a^3 \sinh ^{-1}(a x)^2 \text {Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )-2 a^3 \sinh ^{-1}(a x)^2 \text {Li}_2\left (e^{\sinh ^{-1}(a x)}\right )-\left (4 a^3\right ) \operatorname {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )+\left (4 a^3\right ) \operatorname {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )-\left (4 a^3\right ) \operatorname {Subst}\left (\int x \text {Li}_2\left (-e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )+\left (4 a^3\right ) \operatorname {Subst}\left (\int x \text {Li}_2\left (e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac {2 a^2 \sinh ^{-1}(a x)^2}{x}-\frac {2 a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 x^2}-\frac {\sinh ^{-1}(a x)^4}{3 x^3}-8 a^3 \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )+\frac {4}{3} a^3 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )+2 a^3 \sinh ^{-1}(a x)^2 \text {Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )-2 a^3 \sinh ^{-1}(a x)^2 \text {Li}_2\left (e^{\sinh ^{-1}(a x)}\right )-4 a^3 \sinh ^{-1}(a x) \text {Li}_3\left (-e^{\sinh ^{-1}(a x)}\right )+4 a^3 \sinh ^{-1}(a x) \text {Li}_3\left (e^{\sinh ^{-1}(a x)}\right )-\left (4 a^3\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )+\left (4 a^3\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )+\left (4 a^3\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )-\left (4 a^3\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac {2 a^2 \sinh ^{-1}(a x)^2}{x}-\frac {2 a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 x^2}-\frac {\sinh ^{-1}(a x)^4}{3 x^3}-8 a^3 \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )+\frac {4}{3} a^3 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-4 a^3 \text {Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+2 a^3 \sinh ^{-1}(a x)^2 \text {Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+4 a^3 \text {Li}_2\left (e^{\sinh ^{-1}(a x)}\right )-2 a^3 \sinh ^{-1}(a x)^2 \text {Li}_2\left (e^{\sinh ^{-1}(a x)}\right )-4 a^3 \sinh ^{-1}(a x) \text {Li}_3\left (-e^{\sinh ^{-1}(a x)}\right )+4 a^3 \sinh ^{-1}(a x) \text {Li}_3\left (e^{\sinh ^{-1}(a x)}\right )+\left (4 a^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )-\left (4 a^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )\\ &=-\frac {2 a^2 \sinh ^{-1}(a x)^2}{x}-\frac {2 a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 x^2}-\frac {\sinh ^{-1}(a x)^4}{3 x^3}-8 a^3 \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )+\frac {4}{3} a^3 \sinh ^{-1}(a x)^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-4 a^3 \text {Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+2 a^3 \sinh ^{-1}(a x)^2 \text {Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+4 a^3 \text {Li}_2\left (e^{\sinh ^{-1}(a x)}\right )-2 a^3 \sinh ^{-1}(a x)^2 \text {Li}_2\left (e^{\sinh ^{-1}(a x)}\right )-4 a^3 \sinh ^{-1}(a x) \text {Li}_3\left (-e^{\sinh ^{-1}(a x)}\right )+4 a^3 \sinh ^{-1}(a x) \text {Li}_3\left (e^{\sinh ^{-1}(a x)}\right )+4 a^3 \text {Li}_4\left (-e^{\sinh ^{-1}(a x)}\right )-4 a^3 \text {Li}_4\left (e^{\sinh ^{-1}(a x)}\right )\\ \end {align*}

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Mathematica [A]  time = 2.64, size = 355, normalized size = 1.59 \[ \frac {1}{24} a^3 \left (-\frac {8 \sinh ^4\left (\frac {1}{2} \sinh ^{-1}(a x)\right ) \sinh ^{-1}(a x)^4}{a^3 x^3}-48 \sinh ^{-1}(a x)^2 \text {Li}_2\left (e^{\sinh ^{-1}(a x)}\right )-96 \sinh ^{-1}(a x) \text {Li}_3\left (-e^{-\sinh ^{-1}(a x)}\right )+96 \sinh ^{-1}(a x) \text {Li}_3\left (e^{\sinh ^{-1}(a x)}\right )-48 \left (\sinh ^{-1}(a x)^2-2\right ) \text {Li}_2\left (-e^{-\sinh ^{-1}(a x)}\right )-96 \text {Li}_2\left (e^{-\sinh ^{-1}(a x)}\right )-96 \text {Li}_4\left (-e^{-\sinh ^{-1}(a x)}\right )-96 \text {Li}_4\left (e^{\sinh ^{-1}(a x)}\right )+4 \sinh ^{-1}(a x)^4+16 \sinh ^{-1}(a x)^3 \log \left (e^{-\sinh ^{-1}(a x)}+1\right )-16 \sinh ^{-1}(a x)^3 \log \left (1-e^{\sinh ^{-1}(a x)}\right )+96 \sinh ^{-1}(a x) \log \left (1-e^{-\sinh ^{-1}(a x)}\right )-96 \sinh ^{-1}(a x) \log \left (e^{-\sinh ^{-1}(a x)}+1\right )-2 \sinh ^{-1}(a x)^4 \tanh \left (\frac {1}{2} \sinh ^{-1}(a x)\right )+24 \sinh ^{-1}(a x)^2 \tanh \left (\frac {1}{2} \sinh ^{-1}(a x)\right )+2 \sinh ^{-1}(a x)^4 \coth \left (\frac {1}{2} \sinh ^{-1}(a x)\right )-24 \sinh ^{-1}(a x)^2 \coth \left (\frac {1}{2} \sinh ^{-1}(a x)\right )-\frac {1}{2} a x \sinh ^{-1}(a x)^4 \text {csch}^4\left (\frac {1}{2} \sinh ^{-1}(a x)\right )-4 \sinh ^{-1}(a x)^3 \text {csch}^2\left (\frac {1}{2} \sinh ^{-1}(a x)\right )-4 \sinh ^{-1}(a x)^3 \text {sech}^2\left (\frac {1}{2} \sinh ^{-1}(a x)\right )-2 \pi ^4\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^3*(-2*Pi^4 + 4*ArcSinh[a*x]^4 - 24*ArcSinh[a*x]^2*Coth[ArcSinh[a*x]/2] + 2*ArcSinh[a*x]^4*Coth[ArcSinh[a*x]
/2] - 4*ArcSinh[a*x]^3*Csch[ArcSinh[a*x]/2]^2 - (a*x*ArcSinh[a*x]^4*Csch[ArcSinh[a*x]/2]^4)/2 + 96*ArcSinh[a*x
]*Log[1 - E^(-ArcSinh[a*x])] - 96*ArcSinh[a*x]*Log[1 + E^(-ArcSinh[a*x])] + 16*ArcSinh[a*x]^3*Log[1 + E^(-ArcS
inh[a*x])] - 16*ArcSinh[a*x]^3*Log[1 - E^ArcSinh[a*x]] - 48*(-2 + ArcSinh[a*x]^2)*PolyLog[2, -E^(-ArcSinh[a*x]
)] - 96*PolyLog[2, E^(-ArcSinh[a*x])] - 48*ArcSinh[a*x]^2*PolyLog[2, E^ArcSinh[a*x]] - 96*ArcSinh[a*x]*PolyLog
[3, -E^(-ArcSinh[a*x])] + 96*ArcSinh[a*x]*PolyLog[3, E^ArcSinh[a*x]] - 96*PolyLog[4, -E^(-ArcSinh[a*x])] - 96*
PolyLog[4, E^ArcSinh[a*x]] - 4*ArcSinh[a*x]^3*Sech[ArcSinh[a*x]/2]^2 - (8*ArcSinh[a*x]^4*Sinh[ArcSinh[a*x]/2]^
4)/(a^3*x^3) + 24*ArcSinh[a*x]^2*Tanh[ArcSinh[a*x]/2] - 2*ArcSinh[a*x]^4*Tanh[ArcSinh[a*x]/2]))/24

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(a*x)^4/x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(a*x)^4/x^4, x)

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maple [A]  time = 0.48, size = 372, normalized size = 1.67 \[ -\frac {2 a \arcsinh \left (a x \right )^{3} \sqrt {a^{2} x^{2}+1}}{3 x^{2}}-\frac {2 a^{2} \arcsinh \left (a x \right )^{2}}{x}-\frac {\arcsinh \left (a x \right )^{4}}{3 x^{3}}-\frac {2 a^{3} \arcsinh \left (a x \right )^{3} \ln \left (1-a x -\sqrt {a^{2} x^{2}+1}\right )}{3}-2 a^{3} \arcsinh \left (a x \right )^{2} \polylog \left (2, a x +\sqrt {a^{2} x^{2}+1}\right )+4 a^{3} \arcsinh \left (a x \right ) \polylog \left (3, a x +\sqrt {a^{2} x^{2}+1}\right )-4 a^{3} \polylog \left (4, a x +\sqrt {a^{2} x^{2}+1}\right )+\frac {2 a^{3} \arcsinh \left (a x \right )^{3} \ln \left (a x +\sqrt {a^{2} x^{2}+1}+1\right )}{3}+2 a^{3} \arcsinh \left (a x \right )^{2} \polylog \left (2, -a x -\sqrt {a^{2} x^{2}+1}\right )-4 a^{3} \arcsinh \left (a x \right ) \polylog \left (3, -a x -\sqrt {a^{2} x^{2}+1}\right )+4 a^{3} \polylog \left (4, -a x -\sqrt {a^{2} x^{2}+1}\right )+4 a^{3} \arcsinh \left (a x \right ) \ln \left (1-a x -\sqrt {a^{2} x^{2}+1}\right )+4 a^{3} \polylog \left (2, a x +\sqrt {a^{2} x^{2}+1}\right )-4 a^{3} \arcsinh \left (a x \right ) \ln \left (a x +\sqrt {a^{2} x^{2}+1}+1\right )-4 a^{3} \polylog \left (2, -a x -\sqrt {a^{2} x^{2}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*log(a*x + sqrt(a^2*x^2 + 1))^4/x^3 + integrate(4/3*(a^3*x^2 + sqrt(a^2*x^2 + 1)*a^2*x + a)*log(a*x + sqrt
(a^2*x^2 + 1))^3/(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 = -0.00 \[ \int \frac {{\mathrm {asinh}\left (a\,x\right )}^4}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asinh(a*x)^4/x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x)**4/x**4, x)

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