3.1.79 \(\int \frac {\sinh ^{-1}(a+b x)^3}{x^2} \, dx\) [79]

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

[Out]

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

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Rubi [A]
time = 0.40, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5859, 5828, 5843, 3403, 2296, 2221, 2611, 2320, 6724} \begin {gather*} -\frac {6 b \sinh ^{-1}(a+b x) \text {Li}_2\left (\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 {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}+\frac {6 b \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}-\frac {6 b \text {Li}_3\left (\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)}}{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} \end {gather*}

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 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 2296

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 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 3403

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 5828

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 5843

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 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]

Rubi steps

\begin {align*} \int \frac {\sinh ^{-1}(a+b x)^3}{x^2} \, dx &=\frac {\text {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 \text {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 \text {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 \text {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 \text {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 \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 0.09, size = 259, normalized size = 0.97 \begin {gather*} -\frac {\sqrt {1+a^2} \sinh ^{-1}(a+b x)^3-3 b x \sinh ^{-1}(a+b x)^2 \log \left (\frac {a+\sqrt {1+a^2}-e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )+3 b x \sinh ^{-1}(a+b x)^2 \log \left (\frac {-a+\sqrt {1+a^2}+e^{\sinh ^{-1}(a+b x)}}{-a+\sqrt {1+a^2}}\right )+6 b x \sinh ^{-1}(a+b x) \text {PolyLog}\left (2,\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )-6 b x \sinh ^{-1}(a+b x) \text {PolyLog}\left (2,\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )-6 b x \text {PolyLog}\left (3,\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+6 b x \text {PolyLog}\left (3,\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )}{\sqrt {1+a^2} x} \end {gather*}

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 = 4.09, size = 0, normalized size = 0.00 \[\int \frac {\arcsinh \left (b x +a \right )^{3}}{x^{2}}\, dx\]

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]
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^2,x, algorithm="maxima")

[Out]

-log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^3/x + integrate(3*(b^3*x^2 + 2*a*b^2*x + a^2*b + sqrt(b^2*x^
2 + 2*a*b*x + a^2 + 1)*(b^2*x + a*b) + b)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2/(b^3*x^4 + 3*a*b^
2*x^3 + (3*a^2*b + b)*x^2 + (a^3 + a)*x + (b^2*x^3 + 2*a*b*x^2 + (a^2 + 1)*x)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1
)), 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^2,x, algorithm="fricas")

[Out]

integral(arcsinh(b*x + a)^3/x^2, 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^{2}}\, dx \end {gather*}

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 = 0.00, 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^2,x, algorithm="giac")

[Out]

integrate(arcsinh(b*x + a)^3/x^2, 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^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a + b*x)^3/x^2,x)

[Out]

int(asinh(a + b*x)^3/x^2, x)

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