3.1.95 \(\int \frac {x}{\sqrt {\sinh ^{-1}(a x)}} \, dx\) [95]

Optimal. Leaf size=63 \[ -\frac {\sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a^2}+\frac {\sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a^2} \]

[Out]

-1/8*erf(2^(1/2)*arcsinh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^2+1/8*erfi(2^(1/2)*arcsinh(a*x)^(1/2))*2^(1/2)*Pi^(1/2
)/a^2

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Rubi [A]
time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5780, 5556, 12, 3389, 2211, 2235, 2236} \begin {gather*} \frac {\sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a^2}-\frac {\sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[ArcSinh[a*x]],x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2211

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] &&  !TrueQ[$UseGamma]

Rule 2235

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 2236

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 3389

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 5556

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 5780

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/(b*c^(m + 1)), Subst[Int[x^n*Sinh
[-a/b + x/b]^m*Cosh[-a/b + x/b], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 52, normalized size = 0.83 \begin {gather*} \frac {\frac {\sqrt {-\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-2 \sinh ^{-1}(a x)\right )}{\sqrt {\sinh ^{-1}(a x)}}+\Gamma \left (\frac {1}{2},2 \sinh ^{-1}(a x)\right )}{4 \sqrt {2} a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[ArcSinh[a*x]],x]

[Out]

((Sqrt[-ArcSinh[a*x]]*Gamma[1/2, -2*ArcSinh[a*x]])/Sqrt[ArcSinh[a*x]] + Gamma[1/2, 2*ArcSinh[a*x]])/(4*Sqrt[2]
*a^2)

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Maple [A]
time = 2.06, size = 37, normalized size = 0.59

method	result	size
default	\(-\frac {\sqrt {\pi }\, \sqrt {2}\, \left (\erf \left (\sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\right )-\erfi \left (\sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\right )\right )}{8 a^{2}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/arcsinh(a*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/8*Pi^(1/2)*2^(1/2)*(erf(2^(1/2)*arcsinh(a*x)^(1/2))-erfi(2^(1/2)*arcsinh(a*x)^(1/2)))/a^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/sqrt(arcsinh(a*x)), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {\operatorname {asinh}{\left (a x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/sqrt(asinh(a*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/sqrt(arcsinh(a*x)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x}{\sqrt {\mathrm {asinh}\left (a\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x/asinh(a*x)^(1/2), x)

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