3.2.8 \(\int \frac {x}{\sinh ^{-1}(a x)^{5/2}} \, dx\) [108]

Optimal. Leaf size=118 \[ -\frac {2 x \sqrt {1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\sinh ^{-1}(a x)}}-\frac {8 x^2}{3 \sqrt {\sinh ^{-1}(a x)}}-\frac {2 \sqrt {2 \pi } \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^2}+\frac {2 \sqrt {2 \pi } \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^2} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 118, 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 = {5779, 5818, 5780, 5556, 12, 3389, 2211, 2235, 2236, 5783} \begin {gather*} -\frac {2 \sqrt {2 \pi } \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^2}+\frac {2 \sqrt {2 \pi } \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^2}-\frac {2 x \sqrt {a^2 x^2+1}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\sinh ^{-1}(a x)}}-\frac {8 x^2}{3 \sqrt {\sinh ^{-1}(a x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/ArcSinh[a*x]^(5/2),x]

[Out]

(-2*x*Sqrt[1 + a^2*x^2])/(3*a*ArcSinh[a*x]^(3/2)) - 4/(3*a^2*Sqrt[ArcSinh[a*x]]) - (8*x^2)/(3*Sqrt[ArcSinh[a*x
]]) - (2*Sqrt[2*Pi]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(3*a^2) + (2*Sqrt[2*Pi]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]])
/(3*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 5779

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[c*((m + 1)/(b*(n + 1))), Int[x^(m + 1)*((a + b*ArcSinh[c*x])^(n +
1)/Sqrt[1 + c^2*x^2]), x], x] - Dist[m/(b*c*(n + 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] && LtQ[n, -2]

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 5818

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

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 98, normalized size = 0.83 \begin {gather*} -\frac {2 \sinh ^{-1}(a x) \left (e^{-2 \sinh ^{-1}(a x)}+e^{2 \sinh ^{-1}(a x)}-\sqrt {2} \sqrt {-\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-2 \sinh ^{-1}(a x)\right )-\sqrt {2} \sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},2 \sinh ^{-1}(a x)\right )\right )+\sinh \left (2 \sinh ^{-1}(a x)\right )}{3 a^2 \sinh ^{-1}(a x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/ArcSinh[a*x]^(5/2),x]

[Out]

-1/3*(2*ArcSinh[a*x]*(E^(-2*ArcSinh[a*x]) + E^(2*ArcSinh[a*x]) - Sqrt[2]*Sqrt[-ArcSinh[a*x]]*Gamma[1/2, -2*Arc
Sinh[a*x]] - Sqrt[2]*Sqrt[ArcSinh[a*x]]*Gamma[1/2, 2*ArcSinh[a*x]]) + Sinh[2*ArcSinh[a*x]])/(a^2*ArcSinh[a*x]^
(3/2))

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Maple [A]
time = 2.45, size = 119, normalized size = 1.01

method	result	size
default	\(-\frac {\sqrt {2}\, \left (4 \arcsinh \left (a x \right )^{\frac {3}{2}} \sqrt {\pi }\, \sqrt {2}\, a^{2} x^{2}+\sqrt {\arcsinh \left (a x \right )}\, \sqrt {\pi }\, \sqrt {a^{2} x^{2}+1}\, \sqrt {2}\, a x +2 \arcsinh \left (a x \right )^{\frac {3}{2}} \sqrt {\pi }\, \sqrt {2}+2 \arcsinh \left (a x \right )^{2} \pi \erf \left (\sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\right )-2 \arcsinh \left (a x \right )^{2} \pi \erfi \left (\sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\right )\right )}{3 \sqrt {\pi }\, a^{2} \arcsinh \left (a x \right )^{2}}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/arcsinh(a*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*2^(1/2)*(4*arcsinh(a*x)^(3/2)*Pi^(1/2)*2^(1/2)*a^2*x^2+arcsinh(a*x)^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)*2^(1
/2)*a*x+2*arcsinh(a*x)^(3/2)*Pi^(1/2)*2^(1/2)+2*arcsinh(a*x)^2*Pi*erf(2^(1/2)*arcsinh(a*x)^(1/2))-2*arcsinh(a*
x)^2*Pi*erfi(2^(1/2)*arcsinh(a*x)^(1/2)))/Pi^(1/2)/a^2/arcsinh(a*x)^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)^(5/2),x, algorithm="maxima")

[Out]

integrate(x/arcsinh(a*x)^(5/2), 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)^(5/2),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 {5}{2}}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/asinh(a*x)**(5/2),x)

[Out]

Integral(x/asinh(a*x)**(5/2), 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)^(5/2),x, algorithm="giac")

[Out]

integrate(x/arcsinh(a*x)^(5/2), 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 )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/asinh(a*x)^(5/2),x)

[Out]

int(x/asinh(a*x)^(5/2), x)

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