3.108 \(\int \frac{x}{\sinh ^{-1}(a x)^{5/2}} \, dx\)

Optimal. Leaf size=118 \[ -\frac{2 \sqrt{2 \pi } \text{Erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{3 a^2}+\frac{2 \sqrt{2 \pi } \text{Erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{3 a^2}-\frac{2 x \sqrt{a^2 x^2+1}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{4}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{8 x^2}{3 \sqrt{\sinh ^{-1}(a x)}} \]

[Out]

(-2*x*Sqrt[1 + a^2*x^2])/(3*a*ArcSinh[a*x]^(3/2)) - 4/(3*a^2*Sqrt[ArcSinh[a*x]]) - (8*x^2)/(3*Sqrt[ArcSinh[a*x
]]) - (2*Sqrt[2*Pi]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(3*a^2) + (2*Sqrt[2*Pi]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]])
/(3*a^2)

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Rubi [A]  time = 0.222622, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {5667, 5774, 5669, 5448, 12, 3308, 2180, 2204, 2205, 5675} \[ -\frac{2 \sqrt{2 \pi } \text{Erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{3 a^2}+\frac{2 \sqrt{2 \pi } \text{Erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{3 a^2}-\frac{2 x \sqrt{a^2 x^2+1}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{4}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{8 x^2}{3 \sqrt{\sinh ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcSinh[a*x]^(5/2),x]

[Out]

(-2*x*Sqrt[1 + a^2*x^2])/(3*a*ArcSinh[a*x]^(3/2)) - 4/(3*a^2*Sqrt[ArcSinh[a*x]]) - (8*x^2)/(3*Sqrt[ArcSinh[a*x
]]) - (2*Sqrt[2*Pi]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(3*a^2) + (2*Sqrt[2*Pi]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]])
/(3*a^2)

Rule 5667

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

Rule 5774

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x
)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && LtQ[n, -
1] && GtQ[d, 0]

Rule 5669

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Sinh[x]^m*Cosh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3308

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 2180

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - (c*
f)/d) + (f*g*x^2)/d), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]

Rubi steps

\begin{align*} \int \frac{x}{\sinh ^{-1}(a x)^{5/2}} \, dx &=-\frac{2 x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}+\frac{2 \int \frac{1}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}} \, dx}{3 a}+\frac{1}{3} (4 a) \int \frac{x^2}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}} \, dx\\ &=-\frac{2 x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{4}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{8 x^2}{3 \sqrt{\sinh ^{-1}(a x)}}+\frac{16}{3} \int \frac{x}{\sqrt{\sinh ^{-1}(a x)}} \, dx\\ &=-\frac{2 x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{4}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{8 x^2}{3 \sqrt{\sinh ^{-1}(a x)}}+\frac{16 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^2}\\ &=-\frac{2 x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{4}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{8 x^2}{3 \sqrt{\sinh ^{-1}(a x)}}+\frac{16 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 \sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^2}\\ &=-\frac{2 x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{4}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{8 x^2}{3 \sqrt{\sinh ^{-1}(a x)}}+\frac{8 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^2}\\ &=-\frac{2 x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{4}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{8 x^2}{3 \sqrt{\sinh ^{-1}(a x)}}-\frac{4 \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^2}+\frac{4 \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^2}\\ &=-\frac{2 x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{4}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{8 x^2}{3 \sqrt{\sinh ^{-1}(a x)}}-\frac{8 \operatorname{Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{3 a^2}+\frac{8 \operatorname{Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{3 a^2}\\ &=-\frac{2 x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{4}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{8 x^2}{3 \sqrt{\sinh ^{-1}(a x)}}-\frac{2 \sqrt{2 \pi } \text{erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{3 a^2}+\frac{2 \sqrt{2 \pi } \text{erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{3 a^2}\\ \end{align*}

Mathematica [A]  time = 0.163837, size = 98, normalized size = 0.83 \[ -\frac{2 \sinh ^{-1}(a x) \left (-\sqrt{2} \sqrt{-\sinh ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},-2 \sinh ^{-1}(a x)\right )-\sqrt{2} \sqrt{\sinh ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},2 \sinh ^{-1}(a x)\right )+e^{-2 \sinh ^{-1}(a x)}+e^{2 \sinh ^{-1}(a x)}\right )+\sinh \left (2 \sinh ^{-1}(a x)\right )}{3 a^2 \sinh ^{-1}(a x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x/ArcSinh[a*x]^(5/2),x]

[Out]

-(2*ArcSinh[a*x]*(E^(-2*ArcSinh[a*x]) + E^(2*ArcSinh[a*x]) - Sqrt[2]*Sqrt[-ArcSinh[a*x]]*Gamma[1/2, -2*ArcSinh
[a*x]] - Sqrt[2]*Sqrt[ArcSinh[a*x]]*Gamma[1/2, 2*ArcSinh[a*x]]) + Sinh[2*ArcSinh[a*x]])/(3*a^2*ArcSinh[a*x]^(3
/2))

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Maple [A]  time = 0.086, size = 119, normalized size = 1. \begin{align*} -{\frac{\sqrt{2}}{3\,\sqrt{\pi }{a}^{2} \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}} \left ( 4\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3/2}\sqrt{2}\sqrt{\pi }{x}^{2}{a}^{2}+\sqrt{2}\sqrt{{\it Arcsinh} \left ( ax \right ) }\sqrt{\pi }\sqrt{{a}^{2}{x}^{2}+1}xa+2\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}\pi \,{\it Erf} \left ( \sqrt{2}\sqrt{{\it Arcsinh} \left ( ax \right ) } \right ) -2\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}\pi \,{\it erfi} \left ( \sqrt{2}\sqrt{{\it Arcsinh} \left ( ax \right ) } \right ) +2\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3/2}\sqrt{2}\sqrt{\pi } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/arcsinh(a*x)^(5/2),x)

[Out]

-1/3*2^(1/2)*(4*arcsinh(a*x)^(3/2)*2^(1/2)*Pi^(1/2)*x^2*a^2+2^(1/2)*arcsinh(a*x)^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1
/2)*x*a+2*arcsinh(a*x)^2*Pi*erf(2^(1/2)*arcsinh(a*x)^(1/2))-2*arcsinh(a*x)^2*Pi*erfi(2^(1/2)*arcsinh(a*x)^(1/2
))+2*arcsinh(a*x)^(3/2)*2^(1/2)*Pi^(1/2))/Pi^(1/2)/a^2/arcsinh(a*x)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\operatorname{arsinh}\left (a x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arcsinh(a*x)^(5/2),x, algorithm="maxima")

[Out]

integrate(x/arcsinh(a*x)^(5/2), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arcsinh(a*x)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\operatorname{asinh}^{\frac{5}{2}}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/asinh(a*x)**(5/2),x)

[Out]

Integral(x/asinh(a*x)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\operatorname{arsinh}\left (a x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arcsinh(a*x)^(5/2),x, algorithm="giac")

[Out]

integrate(x/arcsinh(a*x)^(5/2), x)