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

Optimal. Leaf size=152 \[ -\frac{15 \sqrt{\frac{\pi }{2}} \text{Erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{256 a^2}-\frac{15 \sqrt{\frac{\pi }{2}} \text{Erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{256 a^2}-\frac{5 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac{\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac{15 \sqrt{\sinh ^{-1}(a x)}}{64 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^{5/2}+\frac{15}{32} x^2 \sqrt{\sinh ^{-1}(a x)} \]

[Out]

(15*Sqrt[ArcSinh[a*x]])/(64*a^2) + (15*x^2*Sqrt[ArcSinh[a*x]])/32 - (5*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^(3/2))
/(8*a) + ArcSinh[a*x]^(5/2)/(4*a^2) + (x^2*ArcSinh[a*x]^(5/2))/2 - (15*Sqrt[Pi/2]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x
]]])/(256*a^2) - (15*Sqrt[Pi/2]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(256*a^2)

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Rubi [A]  time = 0.336808, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.9, Rules used = {5663, 5758, 5675, 5779, 3312, 3307, 2180, 2204, 2205} \[ -\frac{15 \sqrt{\frac{\pi }{2}} \text{Erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{256 a^2}-\frac{15 \sqrt{\frac{\pi }{2}} \text{Erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{256 a^2}-\frac{5 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac{\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac{15 \sqrt{\sinh ^{-1}(a x)}}{64 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^{5/2}+\frac{15}{32} x^2 \sqrt{\sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Sqrt[ArcSinh[a*x]])/(64*a^2) + (15*x^2*Sqrt[ArcSinh[a*x]])/32 - (5*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^(3/2))
/(8*a) + ArcSinh[a*x]^(5/2)/(4*a^2) + (x^2*ArcSinh[a*x]^(5/2))/2 - (15*Sqrt[Pi/2]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x
]]])/(256*a^2) - (15*Sqrt[Pi/2]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(256*a^2)

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 \sinh ^{-1}(a x)^{5/2} \, dx &=\frac{1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac{1}{4} (5 a) \int \frac{x^2 \sinh ^{-1}(a x)^{3/2}}{\sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{5 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^{5/2}+\frac{15}{16} \int x \sqrt{\sinh ^{-1}(a x)} \, dx+\frac{5 \int \frac{\sinh ^{-1}(a x)^{3/2}}{\sqrt{1+a^2 x^2}} \, dx}{8 a}\\ &=\frac{15}{32} x^2 \sqrt{\sinh ^{-1}(a x)}-\frac{5 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac{\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac{1}{64} (15 a) \int \frac{x^2}{\sqrt{1+a^2 x^2} \sqrt{\sinh ^{-1}(a x)}} \, dx\\ &=\frac{15}{32} x^2 \sqrt{\sinh ^{-1}(a x)}-\frac{5 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac{\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac{15 \operatorname{Subst}\left (\int \frac{\sinh ^2(x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a^2}\\ &=\frac{15}{32} x^2 \sqrt{\sinh ^{-1}(a x)}-\frac{5 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac{\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^{5/2}+\frac{15 \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 )}{64 a^2}\\ &=\frac{15 \sqrt{\sinh ^{-1}(a x)}}{64 a^2}+\frac{15}{32} x^2 \sqrt{\sinh ^{-1}(a x)}-\frac{5 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac{\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac{15 \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{128 a^2}\\ &=\frac{15 \sqrt{\sinh ^{-1}(a x)}}{64 a^2}+\frac{15}{32} x^2 \sqrt{\sinh ^{-1}(a x)}-\frac{5 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac{\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac{15 \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{256 a^2}-\frac{15 \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{256 a^2}\\ &=\frac{15 \sqrt{\sinh ^{-1}(a x)}}{64 a^2}+\frac{15}{32} x^2 \sqrt{\sinh ^{-1}(a x)}-\frac{5 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac{\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac{15 \operatorname{Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{128 a^2}-\frac{15 \operatorname{Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{128 a^2}\\ &=\frac{15 \sqrt{\sinh ^{-1}(a x)}}{64 a^2}+\frac{15}{32} x^2 \sqrt{\sinh ^{-1}(a x)}-\frac{5 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac{\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac{15 \sqrt{\frac{\pi }{2}} \text{erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{256 a^2}-\frac{15 \sqrt{\frac{\pi }{2}} \text{erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{256 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0321814, size = 52, normalized size = 0.34 \[ \frac{\frac{\sqrt{\sinh ^{-1}(a x)} \text{Gamma}\left (\frac{7}{2},-2 \sinh ^{-1}(a x)\right )}{\sqrt{-\sinh ^{-1}(a x)}}+\text{Gamma}\left (\frac{7}{2},2 \sinh ^{-1}(a x)\right )}{32 \sqrt{2} a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((Sqrt[ArcSinh[a*x]]*Gamma[7/2, -2*ArcSinh[a*x]])/Sqrt[-ArcSinh[a*x]] + Gamma[7/2, 2*ArcSinh[a*x]])/(32*Sqrt[2
]*a^2)

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Maple [A]  time = 0.095, size = 136, normalized size = 0.9 \begin{align*} -{\frac{\sqrt{2}}{512\,\sqrt{\pi }{a}^{2}} \left ( -128\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{5/2}\sqrt{2}\sqrt{\pi }{x}^{2}{a}^{2}+160\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3/2}\sqrt{2}\sqrt{\pi }\sqrt{{a}^{2}{x}^{2}+1}xa-120\,\sqrt{2}\sqrt{{\it Arcsinh} \left ( ax \right ) }\sqrt{\pi }{x}^{2}{a}^{2}-64\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{5/2}\sqrt{2}\sqrt{\pi }-60\,\sqrt{2}\sqrt{{\it Arcsinh} \left ( ax \right ) }\sqrt{\pi }+15\,\pi \,{\it Erf} \left ( \sqrt{2}\sqrt{{\it Arcsinh} \left ( ax \right ) } \right ) +15\,\pi \,{\it erfi} \left ( \sqrt{2}\sqrt{{\it Arcsinh} \left ( ax \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/512*2^(1/2)*(-128*arcsinh(a*x)^(5/2)*2^(1/2)*Pi^(1/2)*x^2*a^2+160*arcsinh(a*x)^(3/2)*2^(1/2)*Pi^(1/2)*(a^2*
x^2+1)^(1/2)*x*a-120*2^(1/2)*arcsinh(a*x)^(1/2)*Pi^(1/2)*x^2*a^2-64*arcsinh(a*x)^(5/2)*2^(1/2)*Pi^(1/2)-60*2^(
1/2)*arcsinh(a*x)^(1/2)*Pi^(1/2)+15*Pi*erf(2^(1/2)*arcsinh(a*x)^(1/2))+15*Pi*erfi(2^(1/2)*arcsinh(a*x)^(1/2)))
/Pi^(1/2)/a^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int 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(-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*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 \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)