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

Optimal. Leaf size=64 \[ -\frac {2 \sqrt {a^2 x^2+1}}{a \sqrt {\sinh ^{-1}(a x)}}-\frac {\sqrt {\pi } \text {erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{a}+\frac {\sqrt {\pi } \text {erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5655, 5779, 3308, 2180, 2204, 2205} \[ -\frac {2 \sqrt {a^2 x^2+1}}{a \sqrt {\sinh ^{-1}(a x)}}-\frac {\sqrt {\pi } \text {Erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{a}+\frac {\sqrt {\pi } \text {Erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 + a^2*x^2])/(a*Sqrt[ArcSinh[a*x]]) - (Sqrt[Pi]*Erf[Sqrt[ArcSinh[a*x]]])/a + (Sqrt[Pi]*Erfi[Sqrt[Arc
Sinh[a*x]]])/a

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 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 5655

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 69, normalized size = 1.08 \[ \frac {-e^{-\sinh ^{-1}(a x)}-e^{\sinh ^{-1}(a x)}+\sqrt {-\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-\sinh ^{-1}(a x)\right )+\sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},\sinh ^{-1}(a x)\right )}{a \sqrt {\sinh ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arcsinh(a*x)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\operatorname {arsinh}\left (a x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arcsinh(a*x)^(3/2),x, algorithm="giac")

[Out]

integrate(arcsinh(a*x)^(-3/2), x)

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maple [A]  time = 0.36, size = 65, normalized size = 1.02 \[ -\frac {\arcsinh \left (a x \right ) \pi \erf \left (\sqrt {\arcsinh \left (a x \right )}\right )-\arcsinh \left (a x \right ) \pi \erfi \left (\sqrt {\arcsinh \left (a x \right )}\right )+2 \sqrt {\arcsinh \left (a x \right )}\, \sqrt {\pi }\, \sqrt {a^{2} x^{2}+1}}{\sqrt {\pi }\, a \arcsinh \left (a x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arcsinh(a*x)^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\operatorname {arsinh}\left (a x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arcsinh(a*x)^(3/2),x, algorithm="maxima")

[Out]

integrate(arcsinh(a*x)^(-3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\mathrm {asinh}\left (a\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/asinh(a*x)^(3/2),x)

[Out]

int(1/asinh(a*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\operatorname {asinh}^{\frac {3}{2}}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/asinh(a*x)**(3/2),x)

[Out]

Integral(asinh(a*x)**(-3/2), x)

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