3.2.2 \(\int \frac {x}{\sinh ^{-1}(a x)^{3/2}} \, dx\) [102]

Optimal. Leaf size=84 \[ -\frac {2 x \sqrt {1+a^2 x^2}}{a \sqrt {\sinh ^{-1}(a x)}}+\frac {\sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{a^2}+\frac {\sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{a^2} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5778, 3388, 2211, 2235, 2236} \begin {gather*} \frac {\sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{a^2}+\frac {\sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{a^2}-\frac {2 x \sqrt {a^2 x^2+1}}{a \sqrt {\sinh ^{-1}(a x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3388

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 5778

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x^m*Sqrt[1 + c^2*x^2]*((a + b*ArcSi
nh[c*x])^(n + 1)/(b*c*(n + 1))), x] - Dist[1/(b^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), Si
nh[-a/b + x/b]^(m - 1)*(m + (m + 1)*Sinh[-a/b + x/b]^2), x], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}
, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 78, normalized size = 0.93 \begin {gather*} \frac {\sqrt {-\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-2 \sinh ^{-1}(a x)\right )}{\sqrt {2} a^2 \sqrt {\sinh ^{-1}(a x)}}-\frac {\Gamma \left (\frac {1}{2},2 \sinh ^{-1}(a x)\right )}{\sqrt {2} a^2}-\frac {\sinh \left (2 \sinh ^{-1}(a x)\right )}{a^2 \sqrt {\sinh ^{-1}(a x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 2.59, size = 82, normalized size = 0.98

method	result	size
default	\(-\frac {\sqrt {2}\, \left (2 \sqrt {\arcsinh \left (a x \right )}\, \sqrt {\pi }\, \sqrt {a^{2} x^{2}+1}\, \sqrt {2}\, a x -\arcsinh \left (a x \right ) \pi \erf \left (\sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\right )-\arcsinh \left (a x \right ) \pi \erfi \left (\sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\right )\right )}{2 \sqrt {\pi }\, a^{2} \arcsinh \left (a x \right )}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/arcsinh(a*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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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)^(3/2),x, algorithm="maxima")

[Out]

integrate(x/arcsinh(a*x)^(3/2), 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)^(3/2),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}{\operatorname {asinh}^{\frac {3}{2}}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/asinh(a*x)**(3/2), 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)^(3/2),x, algorithm="giac")

[Out]

integrate(x/arcsinh(a*x)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{{\mathrm {asinh}\left (a\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x/asinh(a*x)^(3/2), x)

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