3.1 \(\int x^4 \sinh ^{-1}(a x) \, dx\)

Optimal. Leaf size=72 \[ -\frac{\left (a^2 x^2+1\right )^{5/2}}{25 a^5}+\frac{2 \left (a^2 x^2+1\right )^{3/2}}{15 a^5}-\frac{\sqrt{a^2 x^2+1}}{5 a^5}+\frac{1}{5} x^5 \sinh ^{-1}(a x) \]

[Out]

-Sqrt[1 + a^2*x^2]/(5*a^5) + (2*(1 + a^2*x^2)^(3/2))/(15*a^5) - (1 + a^2*x^2)^(5/2)/(25*a^5) + (x^5*ArcSinh[a*
x])/5

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Rubi [A]  time = 0.0441345, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {5661, 266, 43} \[ -\frac{\left (a^2 x^2+1\right )^{5/2}}{25 a^5}+\frac{2 \left (a^2 x^2+1\right )^{3/2}}{15 a^5}-\frac{\sqrt{a^2 x^2+1}}{5 a^5}+\frac{1}{5} x^5 \sinh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcSinh[a*x],x]

[Out]

-Sqrt[1 + a^2*x^2]/(5*a^5) + (2*(1 + a^2*x^2)^(3/2))/(15*a^5) - (1 + a^2*x^2)^(5/2)/(25*a^5) + (x^5*ArcSinh[a*
x])/5

Rule 5661

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

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^4 \sinh ^{-1}(a x) \, dx &=\frac{1}{5} x^5 \sinh ^{-1}(a x)-\frac{1}{5} a \int \frac{x^5}{\sqrt{1+a^2 x^2}} \, dx\\ &=\frac{1}{5} x^5 \sinh ^{-1}(a x)-\frac{1}{10} a \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+a^2 x}} \, dx,x,x^2\right )\\ &=\frac{1}{5} x^5 \sinh ^{-1}(a x)-\frac{1}{10} a \operatorname{Subst}\left (\int \left (\frac{1}{a^4 \sqrt{1+a^2 x}}-\frac{2 \sqrt{1+a^2 x}}{a^4}+\frac{\left (1+a^2 x\right )^{3/2}}{a^4}\right ) \, dx,x,x^2\right )\\ &=-\frac{\sqrt{1+a^2 x^2}}{5 a^5}+\frac{2 \left (1+a^2 x^2\right )^{3/2}}{15 a^5}-\frac{\left (1+a^2 x^2\right )^{5/2}}{25 a^5}+\frac{1}{5} x^5 \sinh ^{-1}(a x)\\ \end{align*}

Mathematica [A]  time = 0.0347631, size = 50, normalized size = 0.69 \[ \frac{1}{5} x^5 \sinh ^{-1}(a x)-\frac{\sqrt{a^2 x^2+1} \left (3 a^4 x^4-4 a^2 x^2+8\right )}{75 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*ArcSinh[a*x],x]

[Out]

-(Sqrt[1 + a^2*x^2]*(8 - 4*a^2*x^2 + 3*a^4*x^4))/(75*a^5) + (x^5*ArcSinh[a*x])/5

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Maple [A]  time = 0.031, size = 69, normalized size = 1. \begin{align*}{\frac{1}{{a}^{5}} \left ({\frac{{a}^{5}{x}^{5}{\it Arcsinh} \left ( ax \right ) }{5}}-{\frac{{a}^{4}{x}^{4}}{25}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{4\,{a}^{2}{x}^{2}}{75}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{8}{75}\sqrt{{a}^{2}{x}^{2}+1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arcsinh(a*x),x)

[Out]

1/a^5*(1/5*a^5*x^5*arcsinh(a*x)-1/25*a^4*x^4*(a^2*x^2+1)^(1/2)+4/75*a^2*x^2*(a^2*x^2+1)^(1/2)-8/75*(a^2*x^2+1)
^(1/2))

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Maxima [A]  time = 1.13708, size = 92, normalized size = 1.28 \begin{align*} \frac{1}{5} \, x^{5} \operatorname{arsinh}\left (a x\right ) - \frac{1}{75} \,{\left (\frac{3 \, \sqrt{a^{2} x^{2} + 1} x^{4}}{a^{2}} - \frac{4 \, \sqrt{a^{2} x^{2} + 1} x^{2}}{a^{4}} + \frac{8 \, \sqrt{a^{2} x^{2} + 1}}{a^{6}}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x),x, algorithm="maxima")

[Out]

1/5*x^5*arcsinh(a*x) - 1/75*(3*sqrt(a^2*x^2 + 1)*x^4/a^2 - 4*sqrt(a^2*x^2 + 1)*x^2/a^4 + 8*sqrt(a^2*x^2 + 1)/a
^6)*a

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Fricas [A]  time = 1.83797, size = 135, normalized size = 1.88 \begin{align*} \frac{15 \, a^{5} x^{5} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right ) -{\left (3 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 8\right )} \sqrt{a^{2} x^{2} + 1}}{75 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x),x, algorithm="fricas")

[Out]

1/75*(15*a^5*x^5*log(a*x + sqrt(a^2*x^2 + 1)) - (3*a^4*x^4 - 4*a^2*x^2 + 8)*sqrt(a^2*x^2 + 1))/a^5

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Sympy [A]  time = 2.16633, size = 70, normalized size = 0.97 \begin{align*} \begin{cases} \frac{x^{5} \operatorname{asinh}{\left (a x \right )}}{5} - \frac{x^{4} \sqrt{a^{2} x^{2} + 1}}{25 a} + \frac{4 x^{2} \sqrt{a^{2} x^{2} + 1}}{75 a^{3}} - \frac{8 \sqrt{a^{2} x^{2} + 1}}{75 a^{5}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*asinh(a*x),x)

[Out]

Piecewise((x**5*asinh(a*x)/5 - x**4*sqrt(a**2*x**2 + 1)/(25*a) + 4*x**2*sqrt(a**2*x**2 + 1)/(75*a**3) - 8*sqrt
(a**2*x**2 + 1)/(75*a**5), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.30164, size = 90, normalized size = 1.25 \begin{align*} \frac{1}{5} \, x^{5} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right ) - \frac{3 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{5}{2}} - 10 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}} + 15 \, \sqrt{a^{2} x^{2} + 1}}{75 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x),x, algorithm="giac")

[Out]

1/5*x^5*log(a*x + sqrt(a^2*x^2 + 1)) - 1/75*(3*(a^2*x^2 + 1)^(5/2) - 10*(a^2*x^2 + 1)^(3/2) + 15*sqrt(a^2*x^2
+ 1))/a^5