3.1.38 \(\int \frac {\sinh ^{-1}(a x)^4}{x} \, dx\) [38]

Optimal. Leaf size=97 \[ -\frac {1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x)^3 \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text {PolyLog}\left (3,e^{2 \sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x) \text {PolyLog}\left (4,e^{2 \sinh ^{-1}(a x)}\right )-\frac {3}{2} \text {PolyLog}\left (5,e^{2 \sinh ^{-1}(a x)}\right ) \]

[Out]

-1/5*arcsinh(a*x)^5+arcsinh(a*x)^4*ln(1-(a*x+(a^2*x^2+1)^(1/2))^2)+2*arcsinh(a*x)^3*polylog(2,(a*x+(a^2*x^2+1)
^(1/2))^2)-3*arcsinh(a*x)^2*polylog(3,(a*x+(a^2*x^2+1)^(1/2))^2)+3*arcsinh(a*x)*polylog(4,(a*x+(a^2*x^2+1)^(1/
2))^2)-3/2*polylog(5,(a*x+(a^2*x^2+1)^(1/2))^2)

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Rubi [A]
time = 0.08, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5775, 3797, 2221, 2611, 6744, 2320, 6724} \begin {gather*} 2 \sinh ^{-1}(a x)^3 \text {Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x) \text {Li}_4\left (e^{2 \sinh ^{-1}(a x)}\right )-\frac {3}{2} \text {Li}_5\left (e^{2 \sinh ^{-1}(a x)}\right )-\frac {1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*ArcSinh[a*x]^5 + ArcSinh[a*x]^4*Log[1 - E^(2*ArcSinh[a*x])] + 2*ArcSinh[a*x]^3*PolyLog[2, E^(2*ArcSinh[a*
x])] - 3*ArcSinh[a*x]^2*PolyLog[3, E^(2*ArcSinh[a*x])] + 3*ArcSinh[a*x]*PolyLog[4, E^(2*ArcSinh[a*x])] - (3*Po
lyLog[5, E^(2*ArcSinh[a*x])])/2

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5775

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

Rule 6724

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 6744

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} \, dx &=\text {Subst}\left (\int x^4 \coth (x) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac {1}{5} \sinh ^{-1}(a x)^5-2 \text {Subst}\left (\int \frac {e^{2 x} x^4}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac {1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )-4 \text {Subst}\left (\int x^3 \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac {1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x)^3 \text {Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-6 \text {Subst}\left (\int x^2 \text {Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac {1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x)^3 \text {Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(a x)}\right )+6 \text {Subst}\left (\int x \text {Li}_3\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac {1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x)^3 \text {Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x) \text {Li}_4\left (e^{2 \sinh ^{-1}(a x)}\right )-3 \text {Subst}\left (\int \text {Li}_4\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac {1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x)^3 \text {Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x) \text {Li}_4\left (e^{2 \sinh ^{-1}(a x)}\right )-\frac {3}{2} \text {Subst}\left (\int \frac {\text {Li}_4(x)}{x} \, dx,x,e^{2 \sinh ^{-1}(a x)}\right )\\ &=-\frac {1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x)^3 \text {Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x) \text {Li}_4\left (e^{2 \sinh ^{-1}(a x)}\right )-\frac {3}{2} \text {Li}_5\left (e^{2 \sinh ^{-1}(a x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 97, normalized size = 1.00 \begin {gather*} -\frac {1}{5} \sinh ^{-1}(a x)^5+\sinh ^{-1}(a x)^4 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x)^3 \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(a x)}\right )-3 \sinh ^{-1}(a x)^2 \text {PolyLog}\left (3,e^{2 \sinh ^{-1}(a x)}\right )+3 \sinh ^{-1}(a x) \text {PolyLog}\left (4,e^{2 \sinh ^{-1}(a x)}\right )-\frac {3}{2} \text {PolyLog}\left (5,e^{2 \sinh ^{-1}(a x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*ArcSinh[a*x]^5 + ArcSinh[a*x]^4*Log[1 - E^(2*ArcSinh[a*x])] + 2*ArcSinh[a*x]^3*PolyLog[2, E^(2*ArcSinh[a*
x])] - 3*ArcSinh[a*x]^2*PolyLog[3, E^(2*ArcSinh[a*x])] + 3*ArcSinh[a*x]*PolyLog[4, E^(2*ArcSinh[a*x])] - (3*Po
lyLog[5, E^(2*ArcSinh[a*x])])/2

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Maple [A]
time = 1.31, size = 257, normalized size = 2.65

method	result	size
derivativedivides	\(-\frac {\arcsinh \left (a x \right )^{5}}{5}+\arcsinh \left (a x \right )^{4} \ln \left (1+a x +\sqrt {a^{2} x^{2}+1}\right )+4 \arcsinh \left (a x \right )^{3} \polylog \left (2, -a x -\sqrt {a^{2} x^{2}+1}\right )-12 \arcsinh \left (a x \right )^{2} \polylog \left (3, -a x -\sqrt {a^{2} x^{2}+1}\right )+24 \arcsinh \left (a x \right ) \polylog \left (4, -a x -\sqrt {a^{2} x^{2}+1}\right )-24 \polylog \left (5, -a x -\sqrt {a^{2} x^{2}+1}\right )+\arcsinh \left (a x \right )^{4} \ln \left (1-a x -\sqrt {a^{2} x^{2}+1}\right )+4 \arcsinh \left (a x \right )^{3} \polylog \left (2, a x +\sqrt {a^{2} x^{2}+1}\right )-12 \arcsinh \left (a x \right )^{2} \polylog \left (3, a x +\sqrt {a^{2} x^{2}+1}\right )+24 \arcsinh \left (a x \right ) \polylog \left (4, a x +\sqrt {a^{2} x^{2}+1}\right )-24 \polylog \left (5, a x +\sqrt {a^{2} x^{2}+1}\right )\)	\(257\)
default	\(-\frac {\arcsinh \left (a x \right )^{5}}{5}+\arcsinh \left (a x \right )^{4} \ln \left (1+a x +\sqrt {a^{2} x^{2}+1}\right )+4 \arcsinh \left (a x \right )^{3} \polylog \left (2, -a x -\sqrt {a^{2} x^{2}+1}\right )-12 \arcsinh \left (a x \right )^{2} \polylog \left (3, -a x -\sqrt {a^{2} x^{2}+1}\right )+24 \arcsinh \left (a x \right ) \polylog \left (4, -a x -\sqrt {a^{2} x^{2}+1}\right )-24 \polylog \left (5, -a x -\sqrt {a^{2} x^{2}+1}\right )+\arcsinh \left (a x \right )^{4} \ln \left (1-a x -\sqrt {a^{2} x^{2}+1}\right )+4 \arcsinh \left (a x \right )^{3} \polylog \left (2, a x +\sqrt {a^{2} x^{2}+1}\right )-12 \arcsinh \left (a x \right )^{2} \polylog \left (3, a x +\sqrt {a^{2} x^{2}+1}\right )+24 \arcsinh \left (a x \right ) \polylog \left (4, a x +\sqrt {a^{2} x^{2}+1}\right )-24 \polylog \left (5, a x +\sqrt {a^{2} x^{2}+1}\right )\)	\(257\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)^4/x,x,method=_RETURNVERBOSE)

[Out]

-1/5*arcsinh(a*x)^5+arcsinh(a*x)^4*ln(1+a*x+(a^2*x^2+1)^(1/2))+4*arcsinh(a*x)^3*polylog(2,-a*x-(a^2*x^2+1)^(1/
2))-12*arcsinh(a*x)^2*polylog(3,-a*x-(a^2*x^2+1)^(1/2))+24*arcsinh(a*x)*polylog(4,-a*x-(a^2*x^2+1)^(1/2))-24*p
olylog(5,-a*x-(a^2*x^2+1)^(1/2))+arcsinh(a*x)^4*ln(1-a*x-(a^2*x^2+1)^(1/2))+4*arcsinh(a*x)^3*polylog(2,a*x+(a^
2*x^2+1)^(1/2))-12*arcsinh(a*x)^2*polylog(3,a*x+(a^2*x^2+1)^(1/2))+24*arcsinh(a*x)*polylog(4,a*x+(a^2*x^2+1)^(
1/2))-24*polylog(5,a*x+(a^2*x^2+1)^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(a*x)^4/x, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(a*x)^4/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x)**4/x, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(a*x)^4/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asinh(a*x)^4/x, x)

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