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

Optimal. Leaf size=130 \[ \frac {\sqrt {\pi } \text {erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{4 a^3}-\frac {\sqrt {3 \pi } \text {erf}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a^3}-\frac {\sqrt {\pi } \text {erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{4 a^3}+\frac {\sqrt {3 \pi } \text {erfi}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a^3}-\frac {2 x^2 \sqrt {a^2 x^2+1}}{a \sqrt {\sinh ^{-1}(a x)}} \]

[Out]

1/4*erf(arcsinh(a*x)^(1/2))*Pi^(1/2)/a^3-1/4*erfi(arcsinh(a*x)^(1/2))*Pi^(1/2)/a^3-1/4*erf(3^(1/2)*arcsinh(a*x
)^(1/2))*3^(1/2)*Pi^(1/2)/a^3+1/4*erfi(3^(1/2)*arcsinh(a*x)^(1/2))*3^(1/2)*Pi^(1/2)/a^3-2*x^2*(a^2*x^2+1)^(1/2
)/a/arcsinh(a*x)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5665, 3308, 2180, 2204, 2205} \[ \frac {\sqrt {\pi } \text {Erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{4 a^3}-\frac {\sqrt {3 \pi } \text {Erf}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a^3}-\frac {\sqrt {\pi } \text {Erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{4 a^3}+\frac {\sqrt {3 \pi } \text {Erfi}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a^3}-\frac {2 x^2 \sqrt {a^2 x^2+1}}{a \sqrt {\sinh ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^2*Sqrt[1 + a^2*x^2])/(a*Sqrt[ArcSinh[a*x]]) + (Sqrt[Pi]*Erf[Sqrt[ArcSinh[a*x]]])/(4*a^3) - (Sqrt[3*Pi]*E
rf[Sqrt[3]*Sqrt[ArcSinh[a*x]]])/(4*a^3) - (Sqrt[Pi]*Erfi[Sqrt[ArcSinh[a*x]]])/(4*a^3) + (Sqrt[3*Pi]*Erfi[Sqrt[
3]*Sqrt[ArcSinh[a*x]]])/(4*a^3)

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 5665

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*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, 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^2}{\sinh ^{-1}(a x)^{3/2}} \, dx &=-\frac {2 x^2 \sqrt {1+a^2 x^2}}{a \sqrt {\sinh ^{-1}(a x)}}+\frac {2 \operatorname {Subst}\left (\int \left (-\frac {\sinh (x)}{4 \sqrt {x}}+\frac {3 \sinh (3 x)}{4 \sqrt {x}}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a^3}\\ &=-\frac {2 x^2 \sqrt {1+a^2 x^2}}{a \sqrt {\sinh ^{-1}(a x)}}-\frac {\operatorname {Subst}\left (\int \frac {\sinh (x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^3}+\frac {3 \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^3}\\ &=-\frac {2 x^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 )}{4 a^3}-\frac {\operatorname {Subst}\left (\int \frac {e^x}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{4 a^3}-\frac {3 \operatorname {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{4 a^3}+\frac {3 \operatorname {Subst}\left (\int \frac {e^{3 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{4 a^3}\\ &=-\frac {2 x^2 \sqrt {1+a^2 x^2}}{a \sqrt {\sinh ^{-1}(a x)}}+\frac {\operatorname {Subst}\left (\int e^{-x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{2 a^3}-\frac {\operatorname {Subst}\left (\int e^{x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{2 a^3}-\frac {3 \operatorname {Subst}\left (\int e^{-3 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{2 a^3}+\frac {3 \operatorname {Subst}\left (\int e^{3 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{2 a^3}\\ &=-\frac {2 x^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 )}{4 a^3}-\frac {\sqrt {3 \pi } \text {erf}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a^3}-\frac {\sqrt {\pi } \text {erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{4 a^3}+\frac {\sqrt {3 \pi } \text {erfi}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 140, normalized size = 1.08 \[ \frac {-e^{-3 \sinh ^{-1}(a x)}+e^{-\sinh ^{-1}(a x)}+e^{\sinh ^{-1}(a x)}-e^{3 \sinh ^{-1}(a x)}+\sqrt {3} \sqrt {-\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-3 \sinh ^{-1}(a x)\right )-\sqrt {-\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-\sinh ^{-1}(a x)\right )-\sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},\sinh ^{-1}(a x)\right )+\sqrt {3} \sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},3 \sinh ^{-1}(a x)\right )}{4 a^3 \sqrt {\sinh ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-E^(-3*ArcSinh[a*x]) + E^(-ArcSinh[a*x]) + E^ArcSinh[a*x] - E^(3*ArcSinh[a*x]) + Sqrt[3]*Sqrt[-ArcSinh[a*x]]*
Gamma[1/2, -3*ArcSinh[a*x]] - Sqrt[-ArcSinh[a*x]]*Gamma[1/2, -ArcSinh[a*x]] - Sqrt[ArcSinh[a*x]]*Gamma[1/2, Ar
cSinh[a*x]] + Sqrt[3]*Sqrt[ArcSinh[a*x]]*Gamma[1/2, 3*ArcSinh[a*x]])/(4*a^3*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(x^2/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 \[ \int \frac {x^{2}}{\operatorname {arsinh}\left (a x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/arcsinh(a*x)^(3/2), x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\arcsinh \left (a x \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/arcsinh(a*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2/asinh(a*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/asinh(a*x)**(3/2), x)

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