3.87 \(\int x^3 \sinh ^{-1}(a x)^{5/2} \, dx\)

Optimal. Leaf size=247 \[ -\frac{15 \sqrt{\pi } \text{Erf}\left (2 \sqrt{\sinh ^{-1}(a x)}\right )}{16384 a^4}+\frac{15 \sqrt{\frac{\pi }{2}} \text{Erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{512 a^4}-\frac{15 \sqrt{\pi } \text{Erfi}\left (2 \sqrt{\sinh ^{-1}(a x)}\right )}{16384 a^4}+\frac{15 \sqrt{\frac{\pi }{2}} \text{Erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{512 a^4}-\frac{5 x^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^{3/2}}{32 a}-\frac{45 x^2 \sqrt{\sinh ^{-1}(a x)}}{256 a^2}+\frac{15 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^{3/2}}{64 a^3}-\frac{3 \sinh ^{-1}(a x)^{5/2}}{32 a^4}-\frac{225 \sqrt{\sinh ^{-1}(a x)}}{2048 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^{5/2}+\frac{15}{256} x^4 \sqrt{\sinh ^{-1}(a x)} \]

[Out]

(-225*Sqrt[ArcSinh[a*x]])/(2048*a^4) - (45*x^2*Sqrt[ArcSinh[a*x]])/(256*a^2) + (15*x^4*Sqrt[ArcSinh[a*x]])/256
+ (15*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^(3/2))/(64*a^3) - (5*x^3*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^(3/2))/(32*a)
- (3*ArcSinh[a*x]^(5/2))/(32*a^4) + (x^4*ArcSinh[a*x]^(5/2))/4 - (15*Sqrt[Pi]*Erf[2*Sqrt[ArcSinh[a*x]]])/(1638
4*a^4) + (15*Sqrt[Pi/2]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(512*a^4) - (15*Sqrt[Pi]*Erfi[2*Sqrt[ArcSinh[a*x]]])/
(16384*a^4) + (15*Sqrt[Pi/2]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(512*a^4)

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Rubi [A]  time = 0.71077, antiderivative size = 247, normalized size of antiderivative = 1., number of steps used = 27, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {5663, 5758, 5675, 5779, 3312, 3307, 2180, 2204, 2205} \[ -\frac{15 \sqrt{\pi } \text{Erf}\left (2 \sqrt{\sinh ^{-1}(a x)}\right )}{16384 a^4}+\frac{15 \sqrt{\frac{\pi }{2}} \text{Erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{512 a^4}-\frac{15 \sqrt{\pi } \text{Erfi}\left (2 \sqrt{\sinh ^{-1}(a x)}\right )}{16384 a^4}+\frac{15 \sqrt{\frac{\pi }{2}} \text{Erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{512 a^4}-\frac{5 x^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^{3/2}}{32 a}-\frac{45 x^2 \sqrt{\sinh ^{-1}(a x)}}{256 a^2}+\frac{15 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^{3/2}}{64 a^3}-\frac{3 \sinh ^{-1}(a x)^{5/2}}{32 a^4}-\frac{225 \sqrt{\sinh ^{-1}(a x)}}{2048 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^{5/2}+\frac{15}{256} x^4 \sqrt{\sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSinh[a*x]^(5/2),x]

[Out]

(-225*Sqrt[ArcSinh[a*x]])/(2048*a^4) - (45*x^2*Sqrt[ArcSinh[a*x]])/(256*a^2) + (15*x^4*Sqrt[ArcSinh[a*x]])/256
+ (15*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^(3/2))/(64*a^3) - (5*x^3*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^(3/2))/(32*a)
- (3*ArcSinh[a*x]^(5/2))/(32*a^4) + (x^4*ArcSinh[a*x]^(5/2))/4 - (15*Sqrt[Pi]*Erf[2*Sqrt[ArcSinh[a*x]]])/(1638
4*a^4) + (15*Sqrt[Pi/2]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(512*a^4) - (15*Sqrt[Pi]*Erfi[2*Sqrt[ArcSinh[a*x]]])/
(16384*a^4) + (15*Sqrt[Pi/2]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(512*a^4)

Rule 5663

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^(m + 1)*(a + b*ArcSinh[c*x])^n)/
(m + 1), x] - Dist[(b*c*n)/(m + 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] && GtQ[n, 0]

Rule 5758

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

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
]

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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]

Rubi steps

\begin{align*} \int x^3 \sinh ^{-1}(a x)^{5/2} \, dx &=\frac{1}{4} x^4 \sinh ^{-1}(a x)^{5/2}-\frac{1}{8} (5 a) \int \frac{x^4 \sinh ^{-1}(a x)^{3/2}}{\sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{5 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{32 a}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^{5/2}+\frac{15}{64} \int x^3 \sqrt{\sinh ^{-1}(a x)} \, dx+\frac{15 \int \frac{x^2 \sinh ^{-1}(a x)^{3/2}}{\sqrt{1+a^2 x^2}} \, dx}{32 a}\\ &=\frac{15}{256} x^4 \sqrt{\sinh ^{-1}(a x)}+\frac{15 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{64 a^3}-\frac{5 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{32 a}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^{5/2}-\frac{15 \int \frac{\sinh ^{-1}(a x)^{3/2}}{\sqrt{1+a^2 x^2}} \, dx}{64 a^3}-\frac{45 \int x \sqrt{\sinh ^{-1}(a x)} \, dx}{128 a^2}-\frac{1}{512} (15 a) \int \frac{x^4}{\sqrt{1+a^2 x^2} \sqrt{\sinh ^{-1}(a x)}} \, dx\\ &=-\frac{45 x^2 \sqrt{\sinh ^{-1}(a x)}}{256 a^2}+\frac{15}{256} x^4 \sqrt{\sinh ^{-1}(a x)}+\frac{15 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{64 a^3}-\frac{5 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{32 a}-\frac{3 \sinh ^{-1}(a x)^{5/2}}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^{5/2}-\frac{15 \operatorname{Subst}\left (\int \frac{\sinh ^4(x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{512 a^4}+\frac{45 \int \frac{x^2}{\sqrt{1+a^2 x^2} \sqrt{\sinh ^{-1}(a x)}} \, dx}{512 a}\\ &=-\frac{45 x^2 \sqrt{\sinh ^{-1}(a x)}}{256 a^2}+\frac{15}{256} x^4 \sqrt{\sinh ^{-1}(a x)}+\frac{15 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{64 a^3}-\frac{5 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{32 a}-\frac{3 \sinh ^{-1}(a x)^{5/2}}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^{5/2}-\frac{15 \operatorname{Subst}\left (\int \left (\frac{3}{8 \sqrt{x}}-\frac{\cosh (2 x)}{2 \sqrt{x}}+\frac{\cosh (4 x)}{8 \sqrt{x}}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{512 a^4}+\frac{45 \operatorname{Subst}\left (\int \frac{\sinh ^2(x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{512 a^4}\\ &=-\frac{45 \sqrt{\sinh ^{-1}(a x)}}{2048 a^4}-\frac{45 x^2 \sqrt{\sinh ^{-1}(a x)}}{256 a^2}+\frac{15}{256} x^4 \sqrt{\sinh ^{-1}(a x)}+\frac{15 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{64 a^3}-\frac{5 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{32 a}-\frac{3 \sinh ^{-1}(a x)^{5/2}}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^{5/2}-\frac{15 \operatorname{Subst}\left (\int \frac{\cosh (4 x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{4096 a^4}+\frac{15 \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{1024 a^4}-\frac{45 \operatorname{Subst}\left (\int \left (\frac{1}{2 \sqrt{x}}-\frac{\cosh (2 x)}{2 \sqrt{x}}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{512 a^4}\\ &=-\frac{225 \sqrt{\sinh ^{-1}(a x)}}{2048 a^4}-\frac{45 x^2 \sqrt{\sinh ^{-1}(a x)}}{256 a^2}+\frac{15}{256} x^4 \sqrt{\sinh ^{-1}(a x)}+\frac{15 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{64 a^3}-\frac{5 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{32 a}-\frac{3 \sinh ^{-1}(a x)^{5/2}}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^{5/2}-\frac{15 \operatorname{Subst}\left (\int \frac{e^{-4 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{8192 a^4}-\frac{15 \operatorname{Subst}\left (\int \frac{e^{4 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{8192 a^4}+\frac{15 \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{2048 a^4}+\frac{15 \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{2048 a^4}+\frac{45 \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{1024 a^4}\\ &=-\frac{225 \sqrt{\sinh ^{-1}(a x)}}{2048 a^4}-\frac{45 x^2 \sqrt{\sinh ^{-1}(a x)}}{256 a^2}+\frac{15}{256} x^4 \sqrt{\sinh ^{-1}(a x)}+\frac{15 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{64 a^3}-\frac{5 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{32 a}-\frac{3 \sinh ^{-1}(a x)^{5/2}}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^{5/2}-\frac{15 \operatorname{Subst}\left (\int e^{-4 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{4096 a^4}-\frac{15 \operatorname{Subst}\left (\int e^{4 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{4096 a^4}+\frac{15 \operatorname{Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{1024 a^4}+\frac{15 \operatorname{Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{1024 a^4}+\frac{45 \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{2048 a^4}+\frac{45 \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{2048 a^4}\\ &=-\frac{225 \sqrt{\sinh ^{-1}(a x)}}{2048 a^4}-\frac{45 x^2 \sqrt{\sinh ^{-1}(a x)}}{256 a^2}+\frac{15}{256} x^4 \sqrt{\sinh ^{-1}(a x)}+\frac{15 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{64 a^3}-\frac{5 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{32 a}-\frac{3 \sinh ^{-1}(a x)^{5/2}}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^{5/2}-\frac{15 \sqrt{\pi } \text{erf}\left (2 \sqrt{\sinh ^{-1}(a x)}\right )}{16384 a^4}+\frac{15 \sqrt{\frac{\pi }{2}} \text{erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{2048 a^4}-\frac{15 \sqrt{\pi } \text{erfi}\left (2 \sqrt{\sinh ^{-1}(a x)}\right )}{16384 a^4}+\frac{15 \sqrt{\frac{\pi }{2}} \text{erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{2048 a^4}+\frac{45 \operatorname{Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{1024 a^4}+\frac{45 \operatorname{Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{1024 a^4}\\ &=-\frac{225 \sqrt{\sinh ^{-1}(a x)}}{2048 a^4}-\frac{45 x^2 \sqrt{\sinh ^{-1}(a x)}}{256 a^2}+\frac{15}{256} x^4 \sqrt{\sinh ^{-1}(a x)}+\frac{15 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{64 a^3}-\frac{5 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{32 a}-\frac{3 \sinh ^{-1}(a x)^{5/2}}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^{5/2}-\frac{15 \sqrt{\pi } \text{erf}\left (2 \sqrt{\sinh ^{-1}(a x)}\right )}{16384 a^4}+\frac{15 \sqrt{\frac{\pi }{2}} \text{erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{512 a^4}-\frac{15 \sqrt{\pi } \text{erfi}\left (2 \sqrt{\sinh ^{-1}(a x)}\right )}{16384 a^4}+\frac{15 \sqrt{\frac{\pi }{2}} \text{erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{512 a^4}\\ \end{align*}

Mathematica [A]  time = 0.03659, size = 101, normalized size = 0.41 \[ \frac{\sqrt{\sinh ^{-1}(a x)} \text{Gamma}\left (\frac{7}{2},-4 \sinh ^{-1}(a x)\right )-16 \sqrt{2} \sqrt{\sinh ^{-1}(a x)} \text{Gamma}\left (\frac{7}{2},-2 \sinh ^{-1}(a x)\right )+\sqrt{-\sinh ^{-1}(a x)} \left (\text{Gamma}\left (\frac{7}{2},4 \sinh ^{-1}(a x)\right )-16 \sqrt{2} \text{Gamma}\left (\frac{7}{2},2 \sinh ^{-1}(a x)\right )\right )}{2048 a^4 \sqrt{-\sinh ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3*ArcSinh[a*x]^(5/2),x]

[Out]

(Sqrt[ArcSinh[a*x]]*Gamma[7/2, -4*ArcSinh[a*x]] - 16*Sqrt[2]*Sqrt[ArcSinh[a*x]]*Gamma[7/2, -2*ArcSinh[a*x]] +
Sqrt[-ArcSinh[a*x]]*(-16*Sqrt[2]*Gamma[7/2, 2*ArcSinh[a*x]] + Gamma[7/2, 4*ArcSinh[a*x]]))/(2048*a^4*Sqrt[-Arc
Sinh[a*x]])

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Maple [F]  time = 0.099, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \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^3*arcsinh(a*x)^(5/2),x)

[Out]

int(x^3*arcsinh(a*x)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \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^3*arcsinh(a*x)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^3*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^3*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(x**3*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 x^{3} \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^3*arcsinh(a*x)^(5/2),x, algorithm="giac")

[Out]

integrate(x^3*arcsinh(a*x)^(5/2), x)