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

Optimal. Leaf size=88 \[ -\frac {\sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{x}+3 a \sinh ^{-1}(a x) \text {Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-\frac {3}{2} a \text {Li}_3\left (e^{2 \sinh ^{-1}(a x)}\right )-a \sinh ^{-1}(a x)^3+3 a \sinh ^{-1}(a x)^2 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right ) \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5723, 5659, 3716, 2190, 2531, 2282, 6589} \[ 3 a \sinh ^{-1}(a x) \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(a x)}\right )-\frac {3}{2} a \text {PolyLog}\left (3,e^{2 \sinh ^{-1}(a x)}\right )-\frac {\sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{x}-a \sinh ^{-1}(a x)^3+3 a \sinh ^{-1}(a x)^2 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 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 3716

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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5659

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

Rule 5723

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[(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*Arc
Sinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

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

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Mathematica [C]  time = 0.21, size = 97, normalized size = 1.10 \[ \frac {1}{8} a \left (-\frac {8 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{a x}+24 \sinh ^{-1}(a x) \text {Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-12 \text {Li}_3\left (e^{2 \sinh ^{-1}(a x)}\right )-8 \sinh ^{-1}(a x)^3+24 \sinh ^{-1}(a x)^2 \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+i \pi ^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(I*Pi^3 - 8*ArcSinh[a*x]^3 - (8*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3)/(a*x) + 24*ArcSinh[a*x]^2*Log[1 - E^(2*Ar
cSinh[a*x])] + 24*ArcSinh[a*x]*PolyLog[2, E^(2*ArcSinh[a*x])] - 12*PolyLog[3, E^(2*ArcSinh[a*x])]))/8

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} x^{2} + 1} \operatorname {arsinh}\left (a x\right )^{3}}{a^{2} x^{4} + x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/x^2/(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^4 + x^2), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.18, size = 187, normalized size = 2.12 \[ \frac {\left (a x -\sqrt {a^{2} x^{2}+1}\right ) \arcsinh \left (a x \right )^{3}}{x}-2 a \arcsinh \left (a x \right )^{3}+3 a \arcsinh \left (a x \right )^{2} \ln \left (1+a x +\sqrt {a^{2} x^{2}+1}\right )+6 a \arcsinh \left (a x \right ) \polylog \left (2, -a x -\sqrt {a^{2} x^{2}+1}\right )-6 a \polylog \left (3, -a x -\sqrt {a^{2} x^{2}+1}\right )+3 a \arcsinh \left (a x \right )^{2} \ln \left (1-a x -\sqrt {a^{2} x^{2}+1}\right )+6 a \arcsinh \left (a x \right ) \polylog \left (2, a x +\sqrt {a^{2} x^{2}+1}\right )-6 a \polylog \left (3, a x +\sqrt {a^{2} x^{2}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {asinh}\left (a\,x\right )}^3}{x^2\,\sqrt {a^2\,x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}^{3}{\left (a x \right )}}{x^{2} \sqrt {a^{2} x^{2} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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