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

Optimal. Leaf size=268 \[ -\frac{6 b \sinh ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )}{\sqrt{a^2+1}}+\frac{6 b \sinh ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )}{\sqrt{a^2+1}}+\frac{6 b \text{PolyLog}\left (3,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )}{\sqrt{a^2+1}}-\frac{6 b \text{PolyLog}\left (3,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )}{\sqrt{a^2+1}}-\frac{3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )}{\sqrt{a^2+1}}+\frac{3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )}{\sqrt{a^2+1}}-\frac{\sinh ^{-1}(a+b x)^3}{x} \]

[Out]

-(ArcSinh[a + b*x]^3/x) - (3*b*ArcSinh[a + b*x]^2*Log[1 - E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])])/Sqrt[1 + a^
2] + (3*b*ArcSinh[a + b*x]^2*Log[1 - E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])])/Sqrt[1 + a^2] - (6*b*ArcSinh[a +
b*x]*PolyLog[2, E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])])/Sqrt[1 + a^2] + (6*b*ArcSinh[a + b*x]*PolyLog[2, E^A
rcSinh[a + b*x]/(a + Sqrt[1 + a^2])])/Sqrt[1 + a^2] + (6*b*PolyLog[3, E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])])
/Sqrt[1 + a^2] - (6*b*PolyLog[3, E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])])/Sqrt[1 + a^2]

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Rubi [A]  time = 0.580218, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {5865, 5801, 5831, 3322, 2264, 2190, 2531, 2282, 6589} \[ -\frac{6 b \sinh ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )}{\sqrt{a^2+1}}+\frac{6 b \sinh ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )}{\sqrt{a^2+1}}+\frac{6 b \text{PolyLog}\left (3,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )}{\sqrt{a^2+1}}-\frac{6 b \text{PolyLog}\left (3,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )}{\sqrt{a^2+1}}-\frac{3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )}{\sqrt{a^2+1}}+\frac{3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )}{\sqrt{a^2+1}}-\frac{\sinh ^{-1}(a+b x)^3}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]^3/x^2,x]

[Out]

-(ArcSinh[a + b*x]^3/x) - (3*b*ArcSinh[a + b*x]^2*Log[1 - E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])])/Sqrt[1 + a^
2] + (3*b*ArcSinh[a + b*x]^2*Log[1 - E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])])/Sqrt[1 + a^2] - (6*b*ArcSinh[a +
b*x]*PolyLog[2, E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])])/Sqrt[1 + a^2] + (6*b*ArcSinh[a + b*x]*PolyLog[2, E^A
rcSinh[a + b*x]/(a + Sqrt[1 + a^2])])/Sqrt[1 + a^2] + (6*b*PolyLog[3, E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])])
/Sqrt[1 + a^2] - (6*b*PolyLog[3, E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])])/Sqrt[1 + a^2]

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rule 5801

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

Rule 5831

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

Rule 3322

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol] :> Dist[2,
Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(-(I*b) + 2*a*E^(-(I*e) + f*fz*x) + I*b*E^(2*(-(I*e) + f*fz*x))), x], x]
/; FreeQ[{a, b, c, d, e, f, fz}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

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

Mathematica [A]  time = 0.112068, size = 259, normalized size = 0.97 \[ -\frac{6 b x \sinh ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )-6 b x \sinh ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )-6 b x \text{PolyLog}\left (3,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )+6 b x \text{PolyLog}\left (3,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )+\sqrt{a^2+1} \sinh ^{-1}(a+b x)^3-3 b x \sinh ^{-1}(a+b x)^2 \log \left (\frac{\sqrt{a^2+1}-e^{\sinh ^{-1}(a+b x)}+a}{\sqrt{a^2+1}+a}\right )+3 b x \sinh ^{-1}(a+b x)^2 \log \left (\frac{\sqrt{a^2+1}+e^{\sinh ^{-1}(a+b x)}-a}{\sqrt{a^2+1}-a}\right )}{\sqrt{a^2+1} x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]^3/x^2,x]

[Out]

-((Sqrt[1 + a^2]*ArcSinh[a + b*x]^3 - 3*b*x*ArcSinh[a + b*x]^2*Log[(a + Sqrt[1 + a^2] - E^ArcSinh[a + b*x])/(a
+ Sqrt[1 + a^2])] + 3*b*x*ArcSinh[a + b*x]^2*Log[(-a + Sqrt[1 + a^2] + E^ArcSinh[a + b*x])/(-a + Sqrt[1 + a^2
])] + 6*b*x*ArcSinh[a + b*x]*PolyLog[2, E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])] - 6*b*x*ArcSinh[a + b*x]*PolyL
og[2, E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])] - 6*b*x*PolyLog[3, E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])] + 6*b
*x*PolyLog[3, E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])])/(Sqrt[1 + a^2]*x))

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Maple [F]  time = 0.177, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{3}}{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(b*x + a)^3/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a + b*x)**3/x**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(b*x + a)^3/x^2, x)