3.1.78 \(\int \frac {\sinh ^{-1}(a+b x)^3}{x} \, dx\) [78]

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

[Out]

-1/4*arcsinh(b*x+a)^4+arcsinh(b*x+a)^3*ln(1-(b*x+a+(1+(b*x+a)^2)^(1/2))/(a-(a^2+1)^(1/2)))+arcsinh(b*x+a)^3*ln
(1-(b*x+a+(1+(b*x+a)^2)^(1/2))/(a+(a^2+1)^(1/2)))+3*arcsinh(b*x+a)^2*polylog(2,(b*x+a+(1+(b*x+a)^2)^(1/2))/(a-
(a^2+1)^(1/2)))+3*arcsinh(b*x+a)^2*polylog(2,(b*x+a+(1+(b*x+a)^2)^(1/2))/(a+(a^2+1)^(1/2)))-6*arcsinh(b*x+a)*p
olylog(3,(b*x+a+(1+(b*x+a)^2)^(1/2))/(a-(a^2+1)^(1/2)))-6*arcsinh(b*x+a)*polylog(3,(b*x+a+(1+(b*x+a)^2)^(1/2))
/(a+(a^2+1)^(1/2)))+6*polylog(4,(b*x+a+(1+(b*x+a)^2)^(1/2))/(a-(a^2+1)^(1/2)))+6*polylog(4,(b*x+a+(1+(b*x+a)^2
)^(1/2))/(a+(a^2+1)^(1/2)))

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Rubi [A]
time = 0.27, antiderivative size = 275, normalized size of antiderivative = 1.00, 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 = {5859, 5827, 5680, 2221, 2611, 6744, 2320, 6724} \begin {gather*} 3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )-6 \sinh ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )-6 \sinh ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )+6 \text {Li}_4\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+6 \text {Li}_4\left (\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)}}{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 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]^3/x,x]

[Out]

-1/4*ArcSinh[a + b*x]^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 -
Sqrt[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]
*PolyLog[3, E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])] - 6*ArcSinh[a + b*x]*PolyLog[3, E^ArcSinh[a + b*x]/(a + Sq
rt[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 + Sq
rt[1 + a^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 5680

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 5827

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 5859

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 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+b x)^3}{x} \, dx &=\frac {\text {Subst}\left (\int \frac {\sinh ^{-1}(x)^3}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b}\\ &=\frac {\text {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 {\text {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 {\text {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 \text {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 \text {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 \text {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 \text {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 \text {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 \text {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 \text {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 \text {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*}

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Mathematica [A]
time = 0.02, size = 346, normalized size = 1.26 \begin {gather*} -\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)}}{\left (-\frac {a}{b}-\frac {\sqrt {1+a^2}}{b}\right ) b}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1+\frac {e^{\sinh ^{-1}(a+b x)}}{\left (-\frac {a}{b}+\frac {\sqrt {1+a^2}}{b}\right ) b}\right )+3 \sinh ^{-1}(a+b x)^2 \text {PolyLog}\left (2,-\frac {e^{\sinh ^{-1}(a+b x)}}{\left (-\frac {a}{b}-\frac {\sqrt {1+a^2}}{b}\right ) b}\right )+3 \sinh ^{-1}(a+b x)^2 \text {PolyLog}\left (2,-\frac {e^{\sinh ^{-1}(a+b x)}}{\left (-\frac {a}{b}+\frac {\sqrt {1+a^2}}{b}\right ) b}\right )-6 \sinh ^{-1}(a+b x) \text {PolyLog}\left (3,-\frac {e^{\sinh ^{-1}(a+b x)}}{\left (-\frac {a}{b}-\frac {\sqrt {1+a^2}}{b}\right ) b}\right )-6 \sinh ^{-1}(a+b x) \text {PolyLog}\left (3,-\frac {e^{\sinh ^{-1}(a+b x)}}{\left (-\frac {a}{b}+\frac {\sqrt {1+a^2}}{b}\right ) b}\right )+6 \text {PolyLog}\left (4,\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+6 \text {PolyLog}\left (4,\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]^3/x,x]

[Out]

-1/4*ArcSinh[a + b*x]^4 + ArcSinh[a + b*x]^3*Log[1 + E^ArcSinh[a + b*x]/((-(a/b) - Sqrt[1 + a^2]/b)*b)] + ArcS
inh[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^A
rcSinh[a + b*x]/(a - Sqrt[1 + a^2])] + 6*PolyLog[4, E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])]

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Maple [F]
time = 1.67, size = 0, normalized size = 0.00 \[\int \frac {\arcsinh \left (b x +a \right )^{3}}{x}\, dx\]

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)

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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(b*x+a)^3/x,x, algorithm="maxima")

[Out]

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

[Out]

integral(arcsinh(b*x + a)^3/x, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asinh}^{3}{\left (a + b x \right )}}{x}\, dx \end {gather*}

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)

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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(b*x+a)^3/x,x, algorithm="giac")

[Out]

integrate(arcsinh(b*x + a)^3/x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {asinh}\left (a+b\,x\right )}^3}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a + b*x)^3/x,x)

[Out]

int(asinh(a + b*x)^3/x, x)

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