3.1.83 \(\int x \sinh ^{-1}(a x)^{3/2} \, dx\) [83]

Optimal. Leaf size=122 \[ -\frac {3 x \sqrt {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}}{8 a}+\frac {\sinh ^{-1}(a x)^{3/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{3/2}-\frac {3 \sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{64 a^2}+\frac {3 \sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{64 a^2} \]

[Out]

1/4*arcsinh(a*x)^(3/2)/a^2+1/2*x^2*arcsinh(a*x)^(3/2)-3/128*erf(2^(1/2)*arcsinh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a
^2+3/128*erfi(2^(1/2)*arcsinh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^2-3/8*x*(a^2*x^2+1)^(1/2)*arcsinh(a*x)^(1/2)/a

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Rubi [A]
time = 0.14, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5777, 5812, 5783, 5780, 5556, 12, 3389, 2211, 2235, 2236} \begin {gather*} -\frac {3 \sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{64 a^2}+\frac {3 \sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{64 a^2}-\frac {3 x \sqrt {a^2 x^2+1} \sqrt {\sinh ^{-1}(a x)}}{8 a}+\frac {\sinh ^{-1}(a x)^{3/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x*Sqrt[1 + a^2*x^2]*Sqrt[ArcSinh[a*x]])/(8*a) + ArcSinh[a*x]^(3/2)/(4*a^2) + (x^2*ArcSinh[a*x]^(3/2))/2 -
(3*Sqrt[Pi/2]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(64*a^2) + (3*Sqrt[Pi/2]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(64*
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 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 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]

Rule 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -1]

Rule 5812

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

Rubi steps

\begin {align*} \int x \sinh ^{-1}(a x)^{3/2} \, dx &=\frac {1}{2} x^2 \sinh ^{-1}(a x)^{3/2}-\frac {1}{4} (3 a) \int \frac {x^2 \sqrt {\sinh ^{-1}(a x)}}{\sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {3 x \sqrt {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}}{8 a}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{3/2}+\frac {3}{16} \int \frac {x}{\sqrt {\sinh ^{-1}(a x)}} \, dx+\frac {3 \int \frac {\sqrt {\sinh ^{-1}(a x)}}{\sqrt {1+a^2 x^2}} \, dx}{8 a}\\ &=-\frac {3 x \sqrt {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}}{8 a}+\frac {\sinh ^{-1}(a x)^{3/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{3/2}+\frac {3 \text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^2}\\ &=-\frac {3 x \sqrt {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}}{8 a}+\frac {\sinh ^{-1}(a x)^{3/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{3/2}+\frac {3 \text {Subst}\left (\int \frac {\sinh (2 x)}{2 \sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^2}\\ &=-\frac {3 x \sqrt {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}}{8 a}+\frac {\sinh ^{-1}(a x)^{3/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{3/2}+\frac {3 \text {Subst}\left (\int \frac {\sinh (2 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{32 a^2}\\ &=-\frac {3 x \sqrt {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}}{8 a}+\frac {\sinh ^{-1}(a x)^{3/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{3/2}-\frac {3 \text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a^2}+\frac {3 \text {Subst}\left (\int \frac {e^{2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a^2}\\ &=-\frac {3 x \sqrt {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}}{8 a}+\frac {\sinh ^{-1}(a x)^{3/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{3/2}-\frac {3 \text {Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{32 a^2}+\frac {3 \text {Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{32 a^2}\\ &=-\frac {3 x \sqrt {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}}{8 a}+\frac {\sinh ^{-1}(a x)^{3/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{3/2}-\frac {3 \sqrt {\frac {\pi }{2}} \text {erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{64 a^2}+\frac {3 \sqrt {\frac {\pi }{2}} \text {erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{64 a^2}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 52, normalized size = 0.43 \begin {gather*} \frac {\frac {\sqrt {-\sinh ^{-1}(a x)} \Gamma \left (\frac {5}{2},-2 \sinh ^{-1}(a x)\right )}{\sqrt {\sinh ^{-1}(a x)}}+\Gamma \left (\frac {5}{2},2 \sinh ^{-1}(a x)\right )}{16 \sqrt {2} a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[-ArcSinh[a*x]]*Gamma[5/2, -2*ArcSinh[a*x]])/Sqrt[ArcSinh[a*x]] + Gamma[5/2, 2*ArcSinh[a*x]])/(16*Sqrt[2
]*a^2)

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Maple [A]
time = 2.47, size = 102, normalized size = 0.84

method	result	size
default	\(-\frac {\sqrt {2}\, \left (-32 \arcsinh \left (a x \right )^{\frac {3}{2}} \sqrt {\pi }\, \sqrt {2}\, a^{2} x^{2}+24 \sqrt {\arcsinh \left (a x \right )}\, \sqrt {\pi }\, \sqrt {a^{2} x^{2}+1}\, \sqrt {2}\, a x -16 \arcsinh \left (a x \right )^{\frac {3}{2}} \sqrt {\pi }\, \sqrt {2}+3 \pi \erf \left (\sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\right )-3 \pi \erfi \left (\sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\right )\right )}{128 \sqrt {\pi }\, a^{2}}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*2^(1/2)*(-32*arcsinh(a*x)^(3/2)*Pi^(1/2)*2^(1/2)*a^2*x^2+24*arcsinh(a*x)^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1/
2)*2^(1/2)*a*x-16*arcsinh(a*x)^(3/2)*Pi^(1/2)*2^(1/2)+3*Pi*erf(2^(1/2)*arcsinh(a*x)^(1/2))-3*Pi*erfi(2^(1/2)*a
rcsinh(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)^(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 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 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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