3.1.77 \(\int x \sqrt {\sinh ^{-1}(a x)} \, dx\) [77]

Optimal. Leaf size=93 \[ \frac {\sqrt {\sinh ^{-1}(a x)}}{4 a^2}+\frac {1}{2} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {\sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^2}-\frac {\sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^2} \]

[Out]

-1/32*erf(2^(1/2)*arcsinh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^2-1/32*erfi(2^(1/2)*arcsinh(a*x)^(1/2))*2^(1/2)*Pi^(1
/2)/a^2+1/4*arcsinh(a*x)^(1/2)/a^2+1/2*x^2*arcsinh(a*x)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5777, 5819, 3393, 3388, 2211, 2235, 2236} \begin {gather*} -\frac {\sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^2}-\frac {\sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^2}+\frac {\sqrt {\sinh ^{-1}(a x)}}{4 a^2}+\frac {1}{2} x^2 \sqrt {\sinh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[ArcSinh[a*x]]/(4*a^2) + (x^2*Sqrt[ArcSinh[a*x]])/2 - (Sqrt[Pi/2]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(16*a^2
) - (Sqrt[Pi/2]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(16*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 3393

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 5777

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 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(1/(b*
c^(m + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Subst[Int[x^n*Sinh[-a/b + x/b]^m*Cosh[-a/b + x/b]^(2*p + 1),
x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p + 2, 0] && IGtQ[m,
0]

Rubi steps

\begin {align*} \int x \sqrt {\sinh ^{-1}(a x)} \, dx &=\frac {1}{2} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {1}{4} a \int \frac {x^2}{\sqrt {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}} \, dx\\ &=\frac {1}{2} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {\text {Subst}\left (\int \frac {\sinh ^2(x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{4 a^2}\\ &=\frac {1}{2} x^2 \sqrt {\sinh ^{-1}(a x)}+\frac {\text {Subst}\left (\int \left (\frac {1}{2 \sqrt {x}}-\frac {\cosh (2 x)}{2 \sqrt {x}}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{4 a^2}\\ &=\frac {\sqrt {\sinh ^{-1}(a x)}}{4 a^2}+\frac {1}{2} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {\text {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a^2}\\ &=\frac {\sqrt {\sinh ^{-1}(a x)}}{4 a^2}+\frac {1}{2} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {\text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^2}-\frac {\text {Subst}\left (\int \frac {e^{2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^2}\\ &=\frac {\sqrt {\sinh ^{-1}(a x)}}{4 a^2}+\frac {1}{2} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {\text {Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{8 a^2}-\frac {\text {Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{8 a^2}\\ &=\frac {\sqrt {\sinh ^{-1}(a x)}}{4 a^2}+\frac {1}{2} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {\sqrt {\frac {\pi }{2}} \text {erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^2}-\frac {\sqrt {\frac {\pi }{2}} \text {erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 52, normalized size = 0.56 \begin {gather*} \frac {\frac {\sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {3}{2},-2 \sinh ^{-1}(a x)\right )}{\sqrt {-\sinh ^{-1}(a x)}}+\Gamma \left (\frac {3}{2},2 \sinh ^{-1}(a x)\right )}{8 \sqrt {2} a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[ArcSinh[a*x]]*Gamma[3/2, -2*ArcSinh[a*x]])/Sqrt[-ArcSinh[a*x]] + Gamma[3/2, 2*ArcSinh[a*x]])/(8*Sqrt[2]
*a^2)

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Maple [A]
time = 2.48, size = 75, normalized size = 0.81

method	result	size
default	\(\frac {\sqrt {2}\, \left (8 \sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\, \sqrt {\pi }\, a^{2} x^{2}+4 \sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\, \sqrt {\pi }-\pi \erf \left (\sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\right )-\pi \erfi \left (\sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\right )\right )}{32 \sqrt {\pi }\, a^{2}}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*2^(1/2)*(8*2^(1/2)*arcsinh(a*x)^(1/2)*Pi^(1/2)*a^2*x^2+4*2^(1/2)*arcsinh(a*x)^(1/2)*Pi^(1/2)-Pi*erf(2^(1/
2)*arcsinh(a*x)^(1/2))-Pi*erfi(2^(1/2)*arcsinh(a*x)^(1/2)))/Pi^(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 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 = -0.01 \begin {gather*} \int 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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