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

Optimal. Leaf size=161 \[ -\frac{\sqrt{\pi } \text{Erf}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{6 a^3}+\frac{\sqrt{3 \pi } \text{Erf}\left (\sqrt{3} \sqrt{\sinh ^{-1}(a x)}\right )}{2 a^3}-\frac{\sqrt{\pi } \text{Erfi}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{6 a^3}+\frac{\sqrt{3 \pi } \text{Erfi}\left (\sqrt{3} \sqrt{\sinh ^{-1}(a x)}\right )}{2 a^3}-\frac{2 x^2 \sqrt{a^2 x^2+1}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{4 x^3}{\sqrt{\sinh ^{-1}(a x)}} \]

[Out]

(-2*x^2*Sqrt[1 + a^2*x^2])/(3*a*ArcSinh[a*x]^(3/2)) - (8*x)/(3*a^2*Sqrt[ArcSinh[a*x]]) - (4*x^3)/Sqrt[ArcSinh[
a*x]] - (Sqrt[Pi]*Erf[Sqrt[ArcSinh[a*x]]])/(6*a^3) + (Sqrt[3*Pi]*Erf[Sqrt[3]*Sqrt[ArcSinh[a*x]]])/(2*a^3) - (S
qrt[Pi]*Erfi[Sqrt[ArcSinh[a*x]]])/(6*a^3) + (Sqrt[3*Pi]*Erfi[Sqrt[3]*Sqrt[ArcSinh[a*x]]])/(2*a^3)

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Rubi [A]  time = 0.372393, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {5667, 5774, 5669, 5448, 3307, 2180, 2204, 2205, 5657} \[ -\frac{\sqrt{\pi } \text{Erf}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{6 a^3}+\frac{\sqrt{3 \pi } \text{Erf}\left (\sqrt{3} \sqrt{\sinh ^{-1}(a x)}\right )}{2 a^3}-\frac{\sqrt{\pi } \text{Erfi}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{6 a^3}+\frac{\sqrt{3 \pi } \text{Erfi}\left (\sqrt{3} \sqrt{\sinh ^{-1}(a x)}\right )}{2 a^3}-\frac{2 x^2 \sqrt{a^2 x^2+1}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{4 x^3}{\sqrt{\sinh ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^2*Sqrt[1 + a^2*x^2])/(3*a*ArcSinh[a*x]^(3/2)) - (8*x)/(3*a^2*Sqrt[ArcSinh[a*x]]) - (4*x^3)/Sqrt[ArcSinh[
a*x]] - (Sqrt[Pi]*Erf[Sqrt[ArcSinh[a*x]]])/(6*a^3) + (Sqrt[3*Pi]*Erf[Sqrt[3]*Sqrt[ArcSinh[a*x]]])/(2*a^3) - (S
qrt[Pi]*Erfi[Sqrt[ArcSinh[a*x]]])/(6*a^3) + (Sqrt[3*Pi]*Erfi[Sqrt[3]*Sqrt[ArcSinh[a*x]]])/(2*a^3)

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 3307

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

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 5657

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

Rubi steps

\begin{align*} \int \frac{x^2}{\sinh ^{-1}(a x)^{5/2}} \, dx &=-\frac{2 x^2 \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}+\frac{4 \int \frac{x}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}} \, dx}{3 a}+(2 a) \int \frac{x^3}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}} \, dx\\ &=-\frac{2 x^2 \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{4 x^3}{\sqrt{\sinh ^{-1}(a x)}}+12 \int \frac{x^2}{\sqrt{\sinh ^{-1}(a x)}} \, dx+\frac{8 \int \frac{1}{\sqrt{\sinh ^{-1}(a x)}} \, dx}{3 a^2}\\ &=-\frac{2 x^2 \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{4 x^3}{\sqrt{\sinh ^{-1}(a x)}}+\frac{8 \operatorname{Subst}\left (\int \frac{\cosh (x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^3}+\frac{12 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^2(x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a^3}\\ &=-\frac{2 x^2 \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{4 x^3}{\sqrt{\sinh ^{-1}(a x)}}+\frac{4 \operatorname{Subst}\left (\int \frac{e^{-x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^3}+\frac{4 \operatorname{Subst}\left (\int \frac{e^x}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^3}+\frac{12 \operatorname{Subst}\left (\int \left (-\frac{\cosh (x)}{4 \sqrt{x}}+\frac{\cosh (3 x)}{4 \sqrt{x}}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a^3}\\ &=-\frac{2 x^2 \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{4 x^3}{\sqrt{\sinh ^{-1}(a x)}}+\frac{8 \operatorname{Subst}\left (\int e^{-x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{3 a^3}+\frac{8 \operatorname{Subst}\left (\int e^{x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{3 a^3}-\frac{3 \operatorname{Subst}\left (\int \frac{\cosh (x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a^3}+\frac{3 \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a^3}\\ &=-\frac{2 x^2 \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{4 x^3}{\sqrt{\sinh ^{-1}(a x)}}+\frac{4 \sqrt{\pi } \text{erf}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{3 a^3}+\frac{4 \sqrt{\pi } \text{erfi}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{3 a^3}+\frac{3 \operatorname{Subst}\left (\int \frac{e^{-3 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^3}-\frac{3 \operatorname{Subst}\left (\int \frac{e^{-x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^3}-\frac{3 \operatorname{Subst}\left (\int \frac{e^x}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^3}+\frac{3 \operatorname{Subst}\left (\int \frac{e^{3 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^3}\\ &=-\frac{2 x^2 \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{4 x^3}{\sqrt{\sinh ^{-1}(a x)}}+\frac{4 \sqrt{\pi } \text{erf}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{3 a^3}+\frac{4 \sqrt{\pi } \text{erfi}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{3 a^3}+\frac{3 \operatorname{Subst}\left (\int e^{-3 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{a^3}-\frac{3 \operatorname{Subst}\left (\int e^{-x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{a^3}-\frac{3 \operatorname{Subst}\left (\int e^{x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{a^3}+\frac{3 \operatorname{Subst}\left (\int e^{3 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{a^3}\\ &=-\frac{2 x^2 \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 \sqrt{\sinh ^{-1}(a x)}}-\frac{4 x^3}{\sqrt{\sinh ^{-1}(a x)}}-\frac{\sqrt{\pi } \text{erf}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{6 a^3}+\frac{\sqrt{3 \pi } \text{erf}\left (\sqrt{3} \sqrt{\sinh ^{-1}(a x)}\right )}{2 a^3}-\frac{\sqrt{\pi } \text{erfi}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{6 a^3}+\frac{\sqrt{3 \pi } \text{erfi}\left (\sqrt{3} \sqrt{\sinh ^{-1}(a x)}\right )}{2 a^3}\\ \end{align*}

Mathematica [A]  time = 0.138446, size = 225, normalized size = 1.4 \[ \frac{-\frac{6 \sqrt{3} \left (-\sinh ^{-1}(a x)\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-3 \sinh ^{-1}(a x)\right )+e^{3 \sinh ^{-1}(a x)} \left (6 \sinh ^{-1}(a x)+1\right )}{12 \sinh ^{-1}(a x)^{3/2}}+\frac{2 \left (-\sinh ^{-1}(a x)\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\sinh ^{-1}(a x)\right )+e^{\sinh ^{-1}(a x)} \left (2 \sinh ^{-1}(a x)+1\right )}{12 \sinh ^{-1}(a x)^{3/2}}+\frac{e^{-\sinh ^{-1}(a x)} \left (2 e^{\sinh ^{-1}(a x)} \sinh ^{-1}(a x)^{3/2} \text{Gamma}\left (\frac{1}{2},\sinh ^{-1}(a x)\right )-2 \sinh ^{-1}(a x)+1\right )}{12 \sinh ^{-1}(a x)^{3/2}}-\frac{e^{-3 \sinh ^{-1}(a x)} \left (6 \sqrt{3} e^{3 \sinh ^{-1}(a x)} \sinh ^{-1}(a x)^{3/2} \text{Gamma}\left (\frac{1}{2},3 \sinh ^{-1}(a x)\right )-6 \sinh ^{-1}(a x)+1\right )}{12 \sinh ^{-1}(a x)^{3/2}}}{a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(E^(3*ArcSinh[a*x])*(1 + 6*ArcSinh[a*x]) + 6*Sqrt[3]*(-ArcSinh[a*x])^(3/2)*Gamma[1/2, -3*ArcSinh[a*x]])/(12*
ArcSinh[a*x]^(3/2)) + (E^ArcSinh[a*x]*(1 + 2*ArcSinh[a*x]) + 2*(-ArcSinh[a*x])^(3/2)*Gamma[1/2, -ArcSinh[a*x]]
)/(12*ArcSinh[a*x]^(3/2)) + (1 - 2*ArcSinh[a*x] + 2*E^ArcSinh[a*x]*ArcSinh[a*x]^(3/2)*Gamma[1/2, ArcSinh[a*x]]
)/(12*E^ArcSinh[a*x]*ArcSinh[a*x]^(3/2)) - (1 - 6*ArcSinh[a*x] + 6*Sqrt[3]*E^(3*ArcSinh[a*x])*ArcSinh[a*x]^(3/
2)*Gamma[1/2, 3*ArcSinh[a*x]])/(12*E^(3*ArcSinh[a*x])*ArcSinh[a*x]^(3/2)))/a^3

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Maple [F]  time = 0.103, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2} \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\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^2/arcsinh(a*x)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^2/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^2/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^{2}}{\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**2/asinh(a*x)**(5/2),x)

[Out]

Integral(x**2/asinh(a*x)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\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^2/arcsinh(a*x)^(5/2),x, algorithm="giac")

[Out]

integrate(x^2/arcsinh(a*x)^(5/2), x)