3.300 \(\int x^2 \sinh ^{-1}(\frac {a}{x}) \, dx\)

Optimal. Leaf size=56 \[ \frac {1}{6} a x^2 \sqrt {\frac {a^2}{x^2}+1}-\frac {1}{6} a^3 \tanh ^{-1}\left (\sqrt {\frac {a^2}{x^2}+1}\right )+\frac {1}{3} x^3 \text {csch}^{-1}\left (\frac {x}{a}\right ) \]

[Out]

1/3*x^3*arccsch(x/a)-1/6*a^3*arctanh((1+a^2/x^2)^(1/2))+1/6*a*x^2*(1+a^2/x^2)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5892, 6284, 266, 51, 63, 208} \[ \frac {1}{6} a x^2 \sqrt {\frac {a^2}{x^2}+1}-\frac {1}{6} a^3 \tanh ^{-1}\left (\sqrt {\frac {a^2}{x^2}+1}\right )+\frac {1}{3} x^3 \text {csch}^{-1}\left (\frac {x}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcSinh[a/x],x]

[Out]

(a*Sqrt[1 + a^2/x^2]*x^2)/6 + (x^3*ArcCsch[x/a])/3 - (a^3*ArcTanh[Sqrt[1 + a^2/x^2]])/6

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5892

Int[ArcSinh[(c_.)/((a_.) + (b_.)*(x_)^(n_.))]^(m_.)*(u_.), x_Symbol] :> Int[u*ArcCsch[a/c + (b*x^n)/c]^m, x] /
; FreeQ[{a, b, c, n, m}, x]

Rule 6284

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

Rubi steps

\begin {align*} \int x^2 \sinh ^{-1}\left (\frac {a}{x}\right ) \, dx &=\int x^2 \text {csch}^{-1}\left (\frac {x}{a}\right ) \, dx\\ &=\frac {1}{3} x^3 \text {csch}^{-1}\left (\frac {x}{a}\right )+\frac {1}{3} a \int \frac {x}{\sqrt {1+\frac {a^2}{x^2}}} \, dx\\ &=\frac {1}{3} x^3 \text {csch}^{-1}\left (\frac {x}{a}\right )-\frac {1}{6} a \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1+a^2 x}} \, dx,x,\frac {1}{x^2}\right )\\ &=\frac {1}{6} a \sqrt {1+\frac {a^2}{x^2}} x^2+\frac {1}{3} x^3 \text {csch}^{-1}\left (\frac {x}{a}\right )+\frac {1}{12} a^3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+a^2 x}} \, dx,x,\frac {1}{x^2}\right )\\ &=\frac {1}{6} a \sqrt {1+\frac {a^2}{x^2}} x^2+\frac {1}{3} x^3 \text {csch}^{-1}\left (\frac {x}{a}\right )+\frac {1}{6} a \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{a^2}+\frac {x^2}{a^2}} \, dx,x,\sqrt {1+\frac {a^2}{x^2}}\right )\\ &=\frac {1}{6} a \sqrt {1+\frac {a^2}{x^2}} x^2+\frac {1}{3} x^3 \text {csch}^{-1}\left (\frac {x}{a}\right )-\frac {1}{6} a^3 \tanh ^{-1}\left (\sqrt {1+\frac {a^2}{x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 57, normalized size = 1.02 \[ \frac {1}{6} \left (a x^2 \sqrt {\frac {a^2}{x^2}+1}+a^3 \left (-\log \left (x \left (\sqrt {\frac {a^2}{x^2}+1}+1\right )\right )\right )+2 x^3 \sinh ^{-1}\left (\frac {a}{x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcSinh[a/x],x]

[Out]

(a*Sqrt[1 + a^2/x^2]*x^2 + 2*x^3*ArcSinh[a/x] - a^3*Log[(1 + Sqrt[1 + a^2/x^2])*x])/6

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fricas [B]  time = 0.68, size = 122, normalized size = 2.18 \[ \frac {1}{6} \, a^{3} \log \left (x \sqrt {\frac {a^{2} + x^{2}}{x^{2}}} - x\right ) + \frac {1}{6} \, a x^{2} \sqrt {\frac {a^{2} + x^{2}}{x^{2}}} + \frac {1}{3} \, {\left (x^{3} - 1\right )} \log \left (\frac {x \sqrt {\frac {a^{2} + x^{2}}{x^{2}}} + a}{x}\right ) + \frac {1}{3} \, \log \left (x \sqrt {\frac {a^{2} + x^{2}}{x^{2}}} + a - x\right ) - \frac {1}{3} \, \log \left (x \sqrt {\frac {a^{2} + x^{2}}{x^{2}}} - a - x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a^3*log(x*sqrt((a^2 + x^2)/x^2) - x) + 1/6*a*x^2*sqrt((a^2 + x^2)/x^2) + 1/3*(x^3 - 1)*log((x*sqrt((a^2 +
x^2)/x^2) + a)/x) + 1/3*log(x*sqrt((a^2 + x^2)/x^2) + a - x) - 1/3*log(x*sqrt((a^2 + x^2)/x^2) - a - x)

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giac [A]  time = 0.55, size = 74, normalized size = 1.32 \[ -\frac {1}{6} \, a^{3} \log \left ({\left | a \right |}\right ) \mathrm {sgn}\relax (x) + \frac {1}{3} \, x^{3} \log \left (\sqrt {\frac {a^{2}}{x^{2}} + 1} + \frac {a}{x}\right ) + \frac {a^{3} \log \left (-x + \sqrt {a^{2} + x^{2}}\right )}{6 \, \mathrm {sgn}\relax (x)} + \frac {\sqrt {a^{2} + x^{2}} a x}{6 \, \mathrm {sgn}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a^3*log(abs(a))*sgn(x) + 1/3*x^3*log(sqrt(a^2/x^2 + 1) + a/x) + 1/6*a^3*log(-x + sqrt(a^2 + x^2))/sgn(x)
+ 1/6*sqrt(a^2 + x^2)*a*x/sgn(x)

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maple [A]  time = 0.03, size = 54, normalized size = 0.96 \[ -a^{3} \left (-\frac {x^{3} \arcsinh \left (\frac {a}{x}\right )}{3 a^{3}}-\frac {x^{2} \sqrt {1+\frac {a^{2}}{x^{2}}}}{6 a^{2}}+\frac {\arctanh \left (\frac {1}{\sqrt {1+\frac {a^{2}}{x^{2}}}}\right )}{6}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arcsinh(a/x),x)

[Out]

-a^3*(-1/3/a^3*x^3*arcsinh(a/x)-1/6/a^2*x^2*(1+a^2/x^2)^(1/2)+1/6*arctanh(1/(1+a^2/x^2)^(1/2)))

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maxima [A]  time = 0.46, size = 69, normalized size = 1.23 \[ \frac {1}{3} \, x^{3} \operatorname {arsinh}\left (\frac {a}{x}\right ) - \frac {1}{12} \, {\left (a^{2} \log \left (\sqrt {\frac {a^{2}}{x^{2}} + 1} + 1\right ) - a^{2} \log \left (\sqrt {\frac {a^{2}}{x^{2}} + 1} - 1\right ) - 2 \, x^{2} \sqrt {\frac {a^{2}}{x^{2}} + 1}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arcsinh(a/x) - 1/12*(a^2*log(sqrt(a^2/x^2 + 1) + 1) - a^2*log(sqrt(a^2/x^2 + 1) - 1) - 2*x^2*sqrt(a^2/
x^2 + 1))*a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x^2\,\mathrm {asinh}\left (\frac {a}{x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*asinh(a/x),x)

[Out]

int(x^2*asinh(a/x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*asinh(a/x),x)

[Out]

Integral(x**2*asinh(a/x), x)

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