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

Optimal. Leaf size=94 \[ -\frac{5 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^{3/2}}{2 a}+\frac{15 \sqrt{\pi } \text{Erf}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{16 a}-\frac{15 \sqrt{\pi } \text{Erfi}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{16 a}+x \sinh ^{-1}(a x)^{5/2}+\frac{15}{4} x \sqrt{\sinh ^{-1}(a x)} \]

[Out]

(15*x*Sqrt[ArcSinh[a*x]])/4 - (5*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^(3/2))/(2*a) + x*ArcSinh[a*x]^(5/2) + (15*Sqrt
[Pi]*Erf[Sqrt[ArcSinh[a*x]]])/(16*a) - (15*Sqrt[Pi]*Erfi[Sqrt[ArcSinh[a*x]]])/(16*a)

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Rubi [A]  time = 0.184399, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.875, Rules used = {5653, 5717, 5779, 3308, 2180, 2204, 2205} \[ -\frac{5 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^{3/2}}{2 a}+\frac{15 \sqrt{\pi } \text{Erf}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{16 a}-\frac{15 \sqrt{\pi } \text{Erfi}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{16 a}+x \sinh ^{-1}(a x)^{5/2}+\frac{15}{4} x \sqrt{\sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*x*Sqrt[ArcSinh[a*x]])/4 - (5*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^(3/2))/(2*a) + x*ArcSinh[a*x]^(5/2) + (15*Sqrt
[Pi]*Erf[Sqrt[ArcSinh[a*x]]])/(16*a) - (15*Sqrt[Pi]*Erfi[Sqrt[ArcSinh[a*x]]])/(16*a)

Rule 5653

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

Rule 5717

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

Rule 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

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]

Rubi steps

\begin{align*} \int \sinh ^{-1}(a x)^{5/2} \, dx &=x \sinh ^{-1}(a x)^{5/2}-\frac{1}{2} (5 a) \int \frac{x \sinh ^{-1}(a x)^{3/2}}{\sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{5 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{2 a}+x \sinh ^{-1}(a x)^{5/2}+\frac{15}{4} \int \sqrt{\sinh ^{-1}(a x)} \, dx\\ &=\frac{15}{4} x \sqrt{\sinh ^{-1}(a x)}-\frac{5 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{2 a}+x \sinh ^{-1}(a x)^{5/2}-\frac{1}{8} (15 a) \int \frac{x}{\sqrt{1+a^2 x^2} \sqrt{\sinh ^{-1}(a x)}} \, dx\\ &=\frac{15}{4} x \sqrt{\sinh ^{-1}(a x)}-\frac{5 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{2 a}+x \sinh ^{-1}(a x)^{5/2}-\frac{15 \operatorname{Subst}\left (\int \frac{\sinh (x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a}\\ &=\frac{15}{4} x \sqrt{\sinh ^{-1}(a x)}-\frac{5 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{2 a}+x \sinh ^{-1}(a x)^{5/2}+\frac{15 \operatorname{Subst}\left (\int \frac{e^{-x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a}-\frac{15 \operatorname{Subst}\left (\int \frac{e^x}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a}\\ &=\frac{15}{4} x \sqrt{\sinh ^{-1}(a x)}-\frac{5 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{2 a}+x \sinh ^{-1}(a x)^{5/2}+\frac{15 \operatorname{Subst}\left (\int e^{-x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{8 a}-\frac{15 \operatorname{Subst}\left (\int e^{x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{8 a}\\ &=\frac{15}{4} x \sqrt{\sinh ^{-1}(a x)}-\frac{5 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{2 a}+x \sinh ^{-1}(a x)^{5/2}+\frac{15 \sqrt{\pi } \text{erf}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{16 a}-\frac{15 \sqrt{\pi } \text{erfi}\left (\sqrt{\sinh ^{-1}(a x)}\right )}{16 a}\\ \end{align*}

Mathematica [A]  time = 0.0476373, size = 45, normalized size = 0.48 \[ -\frac{\frac{\sqrt{-\sinh ^{-1}(a x)} \text{Gamma}\left (\frac{7}{2},-\sinh ^{-1}(a x)\right )}{\sqrt{\sinh ^{-1}(a x)}}+\text{Gamma}\left (\frac{7}{2},\sinh ^{-1}(a x)\right )}{2 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((Sqrt[-ArcSinh[a*x]]*Gamma[7/2, -ArcSinh[a*x]])/Sqrt[ArcSinh[a*x]] + Gamma[7/2, ArcSinh[a*x]])/(2*a)

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Maple [A]  time = 0.072, size = 78, normalized size = 0.8 \begin{align*}{\frac{1}{16\,\sqrt{\pi }a} \left ( 16\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{5/2}\sqrt{\pi }xa-40\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3/2}\sqrt{\pi }\sqrt{{a}^{2}{x}^{2}+1}+60\,\sqrt{{\it Arcsinh} \left ( ax \right ) }\sqrt{\pi }xa+15\,\pi \,{\it Erf} \left ( \sqrt{{\it Arcsinh} \left ( ax \right ) } \right ) -15\,\pi \,{\it erfi} \left ( \sqrt{{\it Arcsinh} \left ( ax \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*arcsinh(a*x)^(5/2)*Pi^(1/2)*x*a-40*arcsinh(a*x)^(3/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)+60*arcsinh(a*x)^(1/2
)*Pi^(1/2)*x*a+15*Pi*erf(arcsinh(a*x)^(1/2))-15*Pi*erfi(arcsinh(a*x)^(1/2)))/Pi^(1/2)/a

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

[Out]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

integrate(arcsinh(a*x)^(5/2), x)