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

Optimal. Leaf size=275 \[ 3 \sinh ^{-1}(a+b x)^2 \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )+3 \sinh ^{-1}(a+b x)^2 \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )-6 \sinh ^{-1}(a+b x) \text{PolyLog}\left (3,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )-6 \sinh ^{-1}(a+b x) \text{PolyLog}\left (3,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )+6 \text{PolyLog}\left (4,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )+6 \text{PolyLog}\left (4,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )-\frac{1}{4} \sinh ^{-1}(a+b x)^4 \]

[Out]

-ArcSinh[a + b*x]^4/4 + ArcSinh[a + b*x]^3*Log[1 - E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])] + ArcSinh[a + b*x]^
3*Log[1 - E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])] + 3*ArcSinh[a + b*x]^2*PolyLog[2, E^ArcSinh[a + b*x]/(a - Sq
rt[1 + a^2])] + 3*ArcSinh[a + b*x]^2*PolyLog[2, E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])] - 6*ArcSinh[a + b*x]*P
olyLog[3, E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])] - 6*ArcSinh[a + b*x]*PolyLog[3, E^ArcSinh[a + b*x]/(a + Sqrt
[1 + a^2])] + 6*PolyLog[4, E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])] + 6*PolyLog[4, E^ArcSinh[a + b*x]/(a + Sqrt
[1 + a^2])]

________________________________________________________________________________________

Rubi [A]  time = 0.398384, antiderivative size = 275, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {5865, 5799, 5561, 2190, 2531, 6609, 2282, 6589} \[ 3 \sinh ^{-1}(a+b x)^2 \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )+3 \sinh ^{-1}(a+b x)^2 \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )-6 \sinh ^{-1}(a+b x) \text{PolyLog}\left (3,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )-6 \sinh ^{-1}(a+b x) \text{PolyLog}\left (3,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )+6 \text{PolyLog}\left (4,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )+6 \text{PolyLog}\left (4,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )-\frac{1}{4} \sinh ^{-1}(a+b x)^4 \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]^3/x,x]

[Out]

-ArcSinh[a + b*x]^4/4 + ArcSinh[a + b*x]^3*Log[1 - E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])] + ArcSinh[a + b*x]^
3*Log[1 - E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])] + 3*ArcSinh[a + b*x]^2*PolyLog[2, E^ArcSinh[a + b*x]/(a - Sq
rt[1 + a^2])] + 3*ArcSinh[a + b*x]^2*PolyLog[2, E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])] - 6*ArcSinh[a + b*x]*P
olyLog[3, E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])] - 6*ArcSinh[a + b*x]*PolyLog[3, E^ArcSinh[a + b*x]/(a + Sqrt
[1 + a^2])] + 6*PolyLog[4, E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])] + 6*PolyLog[4, E^ArcSinh[a + b*x]/(a + 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 5799

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

Rule 5561

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(c + d*x))/(a - Rt[a^2 + b^2, 2] + b*E^(c +
d*x)), x] + Int[((e + f*x)^m*E^(c + d*x))/(a + Rt[a^2 + b^2, 2] + b*E^(c + d*x)), x]) /; FreeQ[{a, b, c, d, e,
f}, x] && IGtQ[m, 0] && NeQ[a^2 + b^2, 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 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]

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

Mathematica [A]  time = 0.0328524, size = 346, normalized size = 1.26 \[ 3 \sinh ^{-1}(a+b x)^2 \text{PolyLog}\left (2,-\frac{e^{\sinh ^{-1}(a+b x)}}{b \left (-\frac{\sqrt{a^2+1}}{b}-\frac{a}{b}\right )}\right )+3 \sinh ^{-1}(a+b x)^2 \text{PolyLog}\left (2,-\frac{e^{\sinh ^{-1}(a+b x)}}{b \left (\frac{\sqrt{a^2+1}}{b}-\frac{a}{b}\right )}\right )-6 \sinh ^{-1}(a+b x) \text{PolyLog}\left (3,-\frac{e^{\sinh ^{-1}(a+b x)}}{b \left (-\frac{\sqrt{a^2+1}}{b}-\frac{a}{b}\right )}\right )-6 \sinh ^{-1}(a+b x) \text{PolyLog}\left (3,-\frac{e^{\sinh ^{-1}(a+b x)}}{b \left (\frac{\sqrt{a^2+1}}{b}-\frac{a}{b}\right )}\right )+6 \text{PolyLog}\left (4,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )+6 \text{PolyLog}\left (4,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )+\sinh ^{-1}(a+b x)^3 \log \left (\frac{e^{\sinh ^{-1}(a+b x)}}{b \left (-\frac{\sqrt{a^2+1}}{b}-\frac{a}{b}\right )}+1\right )+\sinh ^{-1}(a+b x)^3 \log \left (\frac{e^{\sinh ^{-1}(a+b x)}}{b \left (\frac{\sqrt{a^2+1}}{b}-\frac{a}{b}\right )}+1\right )-\frac{1}{4} \sinh ^{-1}(a+b x)^4 \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]^3/x,x]

[Out]

-ArcSinh[a + b*x]^4/4 + ArcSinh[a + b*x]^3*Log[1 + E^ArcSinh[a + b*x]/((-(a/b) - Sqrt[1 + a^2]/b)*b)] + ArcSin
h[a + b*x]^3*Log[1 + E^ArcSinh[a + b*x]/((-(a/b) + Sqrt[1 + a^2]/b)*b)] + 3*ArcSinh[a + b*x]^2*PolyLog[2, -(E^
ArcSinh[a + b*x]/((-(a/b) - Sqrt[1 + a^2]/b)*b))] + 3*ArcSinh[a + b*x]^2*PolyLog[2, -(E^ArcSinh[a + b*x]/((-(a
/b) + Sqrt[1 + a^2]/b)*b))] - 6*ArcSinh[a + b*x]*PolyLog[3, -(E^ArcSinh[a + b*x]/((-(a/b) - Sqrt[1 + a^2]/b)*b
))] - 6*ArcSinh[a + b*x]*PolyLog[3, -(E^ArcSinh[a + b*x]/((-(a/b) + Sqrt[1 + a^2]/b)*b))] + 6*PolyLog[4, E^Arc
Sinh[a + b*x]/(a - Sqrt[1 + a^2])] + 6*PolyLog[4, E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])]

________________________________________________________________________________________

Maple [F]  time = 0.106, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{3}}{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(b*x + a)^3/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a + b*x)**3/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(b*x + a)^3/x, x)