3.1.99 \(\int \frac {x^4}{\sinh ^{-1}(a x)^{3/2}} \, dx\) [99]

Optimal. Leaf size=188 \[ -\frac {2 x^4 \sqrt {1+a^2 x^2}}{a \sqrt {\sinh ^{-1}(a x)}}-\frac {\sqrt {\pi } \text {Erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{8 a^5}+\frac {3 \sqrt {3 \pi } \text {Erf}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5}-\frac {\sqrt {5 \pi } \text {Erf}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5}+\frac {\sqrt {\pi } \text {Erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{8 a^5}-\frac {3 \sqrt {3 \pi } \text {Erfi}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5}+\frac {\sqrt {5 \pi } \text {Erfi}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5} \]

[Out]

-1/8*erf(arcsinh(a*x)^(1/2))*Pi^(1/2)/a^5+1/8*erfi(arcsinh(a*x)^(1/2))*Pi^(1/2)/a^5+3/16*erf(3^(1/2)*arcsinh(a
*x)^(1/2))*3^(1/2)*Pi^(1/2)/a^5-3/16*erfi(3^(1/2)*arcsinh(a*x)^(1/2))*3^(1/2)*Pi^(1/2)/a^5-1/16*erf(5^(1/2)*ar
csinh(a*x)^(1/2))*5^(1/2)*Pi^(1/2)/a^5+1/16*erfi(5^(1/2)*arcsinh(a*x)^(1/2))*5^(1/2)*Pi^(1/2)/a^5-2*x^4*(a^2*x
^2+1)^(1/2)/a/arcsinh(a*x)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5778, 3389, 2211, 2235, 2236} \begin {gather*} -\frac {\sqrt {\pi } \text {Erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{8 a^5}+\frac {3 \sqrt {3 \pi } \text {Erf}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5}-\frac {\sqrt {5 \pi } \text {Erf}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5}+\frac {\sqrt {\pi } \text {Erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{8 a^5}-\frac {3 \sqrt {3 \pi } \text {Erfi}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5}+\frac {\sqrt {5 \pi } \text {Erfi}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5}-\frac {2 x^4 \sqrt {a^2 x^2+1}}{a \sqrt {\sinh ^{-1}(a x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/ArcSinh[a*x]^(3/2),x]

[Out]

(-2*x^4*Sqrt[1 + a^2*x^2])/(a*Sqrt[ArcSinh[a*x]]) - (Sqrt[Pi]*Erf[Sqrt[ArcSinh[a*x]]])/(8*a^5) + (3*Sqrt[3*Pi]
*Erf[Sqrt[3]*Sqrt[ArcSinh[a*x]]])/(16*a^5) - (Sqrt[5*Pi]*Erf[Sqrt[5]*Sqrt[ArcSinh[a*x]]])/(16*a^5) + (Sqrt[Pi]
*Erfi[Sqrt[ArcSinh[a*x]]])/(8*a^5) - (3*Sqrt[3*Pi]*Erfi[Sqrt[3]*Sqrt[ArcSinh[a*x]]])/(16*a^5) + (Sqrt[5*Pi]*Er
fi[Sqrt[5]*Sqrt[ArcSinh[a*x]]])/(16*a^5)

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 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^4}{\sinh ^{-1}(a x)^{3/2}} \, dx &=-\frac {2 x^4 \sqrt {1+a^2 x^2}}{a \sqrt {\sinh ^{-1}(a x)}}+\frac {2 \text {Subst}\left (\int \left (\frac {\sinh (x)}{8 \sqrt {x}}-\frac {9 \sinh (3 x)}{16 \sqrt {x}}+\frac {5 \sinh (5 x)}{16 \sqrt {x}}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a^5}\\ &=-\frac {2 x^4 \sqrt {1+a^2 x^2}}{a \sqrt {\sinh ^{-1}(a x)}}+\frac {\text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{4 a^5}+\frac {5 \text {Subst}\left (\int \frac {\sinh (5 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a^5}-\frac {9 \text {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a^5}\\ &=-\frac {2 x^4 \sqrt {1+a^2 x^2}}{a \sqrt {\sinh ^{-1}(a x)}}-\frac {\text {Subst}\left (\int \frac {e^{-x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a^5}+\frac {\text {Subst}\left (\int \frac {e^x}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a^5}-\frac {5 \text {Subst}\left (\int \frac {e^{-5 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^5}+\frac {5 \text {Subst}\left (\int \frac {e^{5 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^5}+\frac {9 \text {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^5}-\frac {9 \text {Subst}\left (\int \frac {e^{3 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^5}\\ &=-\frac {2 x^4 \sqrt {1+a^2 x^2}}{a \sqrt {\sinh ^{-1}(a x)}}-\frac {\text {Subst}\left (\int e^{-x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{4 a^5}+\frac {\text {Subst}\left (\int e^{x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{4 a^5}-\frac {5 \text {Subst}\left (\int e^{-5 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{8 a^5}+\frac {5 \text {Subst}\left (\int e^{5 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{8 a^5}+\frac {9 \text {Subst}\left (\int e^{-3 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{8 a^5}-\frac {9 \text {Subst}\left (\int e^{3 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{8 a^5}\\ &=-\frac {2 x^4 \sqrt {1+a^2 x^2}}{a \sqrt {\sinh ^{-1}(a x)}}-\frac {\sqrt {\pi } \text {erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{8 a^5}+\frac {3 \sqrt {3 \pi } \text {erf}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5}-\frac {\sqrt {5 \pi } \text {erf}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5}+\frac {\sqrt {\pi } \text {erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{8 a^5}-\frac {3 \sqrt {3 \pi } \text {erfi}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5}+\frac {\sqrt {5 \pi } \text {erfi}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 216, normalized size = 1.15 \begin {gather*} \frac {-e^{-5 \sinh ^{-1}(a x)}+3 e^{-3 \sinh ^{-1}(a x)}-2 e^{-\sinh ^{-1}(a x)}-2 e^{\sinh ^{-1}(a x)}+3 e^{3 \sinh ^{-1}(a x)}-e^{5 \sinh ^{-1}(a x)}+\sqrt {5} \sqrt {-\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-5 \sinh ^{-1}(a x)\right )-3 \sqrt {3} \sqrt {-\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-3 \sinh ^{-1}(a x)\right )+2 \sqrt {-\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-\sinh ^{-1}(a x)\right )+2 \sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},\sinh ^{-1}(a x)\right )-3 \sqrt {3} \sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},3 \sinh ^{-1}(a x)\right )+\sqrt {5} \sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},5 \sinh ^{-1}(a x)\right )}{16 a^5 \sqrt {\sinh ^{-1}(a x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcSinh[a*x]^(3/2),x]

[Out]

(-E^(-5*ArcSinh[a*x]) + 3/E^(3*ArcSinh[a*x]) - 2/E^ArcSinh[a*x] - 2*E^ArcSinh[a*x] + 3*E^(3*ArcSinh[a*x]) - E^
(5*ArcSinh[a*x]) + Sqrt[5]*Sqrt[-ArcSinh[a*x]]*Gamma[1/2, -5*ArcSinh[a*x]] - 3*Sqrt[3]*Sqrt[-ArcSinh[a*x]]*Gam
ma[1/2, -3*ArcSinh[a*x]] + 2*Sqrt[-ArcSinh[a*x]]*Gamma[1/2, -ArcSinh[a*x]] + 2*Sqrt[ArcSinh[a*x]]*Gamma[1/2, A
rcSinh[a*x]] - 3*Sqrt[3]*Sqrt[ArcSinh[a*x]]*Gamma[1/2, 3*ArcSinh[a*x]] + Sqrt[5]*Sqrt[ArcSinh[a*x]]*Gamma[1/2,
5*ArcSinh[a*x]])/(16*a^5*Sqrt[ArcSinh[a*x]])

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Maple [F]
time = 6.02, size = 0, normalized size = 0.00 \[\int \frac {x^{4}}{\arcsinh \left (a x \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arcsinh(a*x)^(3/2),x)

[Out]

int(x^4/arcsinh(a*x)^(3/2),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^4/arcsinh(a*x)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^4/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^4/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^{4}}{\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**4/asinh(a*x)**(3/2),x)

[Out]

Integral(x**4/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^4/arcsinh(a*x)^(3/2),x, algorithm="giac")

[Out]

integrate(x^4/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^4}{{\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^4/asinh(a*x)^(3/2),x)

[Out]

int(x^4/asinh(a*x)^(3/2), x)

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