3.2.6 \(\int \frac {x^3}{\sinh ^{-1}(a x)^{5/2}} \, dx\) [106]

Optimal. Leaf size=167 \[ -\frac {2 x^3 \sqrt {1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^2}{a^2 \sqrt {\sinh ^{-1}(a x)}}-\frac {16 x^4}{3 \sqrt {\sinh ^{-1}(a x)}}-\frac {2 \sqrt {\pi } \text {Erf}\left (2 \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}+\frac {\sqrt {2 \pi } \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}+\frac {2 \sqrt {\pi } \text {Erfi}\left (2 \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}-\frac {\sqrt {2 \pi } \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4} \]

[Out]

-2/3*erf(2*arcsinh(a*x)^(1/2))*Pi^(1/2)/a^4+2/3*erfi(2*arcsinh(a*x)^(1/2))*Pi^(1/2)/a^4+1/3*erf(2^(1/2)*arcsin
h(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^4-1/3*erfi(2^(1/2)*arcsinh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^4-2/3*x^3*(a^2*x^2+
1)^(1/2)/a/arcsinh(a*x)^(3/2)-4*x^2/a^2/arcsinh(a*x)^(1/2)-16/3*x^4/arcsinh(a*x)^(1/2)

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Rubi [A]
time = 0.31, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5779, 5818, 5780, 5556, 3389, 2211, 2235, 2236, 12} \begin {gather*} -\frac {2 \sqrt {\pi } \text {Erf}\left (2 \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}+\frac {\sqrt {2 \pi } \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}+\frac {2 \sqrt {\pi } \text {Erfi}\left (2 \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}-\frac {\sqrt {2 \pi } \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}-\frac {4 x^2}{a^2 \sqrt {\sinh ^{-1}(a x)}}-\frac {2 x^3 \sqrt {a^2 x^2+1}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {16 x^4}{3 \sqrt {\sinh ^{-1}(a x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/ArcSinh[a*x]^(5/2),x]

[Out]

(-2*x^3*Sqrt[1 + a^2*x^2])/(3*a*ArcSinh[a*x]^(3/2)) - (4*x^2)/(a^2*Sqrt[ArcSinh[a*x]]) - (16*x^4)/(3*Sqrt[ArcS
inh[a*x]]) - (2*Sqrt[Pi]*Erf[2*Sqrt[ArcSinh[a*x]]])/(3*a^4) + (Sqrt[2*Pi]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(3*
a^4) + (2*Sqrt[Pi]*Erfi[2*Sqrt[ArcSinh[a*x]]])/(3*a^4) - (Sqrt[2*Pi]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(3*a^4)

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 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^3}{\sinh ^{-1}(a x)^{5/2}} \, dx &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}+\frac {2 \int \frac {x^2}{\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}} \, dx}{a}+\frac {1}{3} (8 a) \int \frac {x^4}{\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}} \, dx\\ &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^2}{a^2 \sqrt {\sinh ^{-1}(a x)}}-\frac {16 x^4}{3 \sqrt {\sinh ^{-1}(a x)}}+\frac {64}{3} \int \frac {x^3}{\sqrt {\sinh ^{-1}(a x)}} \, dx+\frac {8 \int \frac {x}{\sqrt {\sinh ^{-1}(a x)}} \, dx}{a^2}\\ &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^2}{a^2 \sqrt {\sinh ^{-1}(a x)}}-\frac {16 x^4}{3 \sqrt {\sinh ^{-1}(a x)}}+\frac {8 \text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a^4}+\frac {64 \text {Subst}\left (\int \frac {\cosh (x) \sinh ^3(x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^4}\\ &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^2}{a^2 \sqrt {\sinh ^{-1}(a x)}}-\frac {16 x^4}{3 \sqrt {\sinh ^{-1}(a x)}}+\frac {8 \text {Subst}\left (\int \frac {\sinh (2 x)}{2 \sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a^4}+\frac {64 \text {Subst}\left (\int \left (-\frac {\sinh (2 x)}{4 \sqrt {x}}+\frac {\sinh (4 x)}{8 \sqrt {x}}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^4}\\ &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^2}{a^2 \sqrt {\sinh ^{-1}(a x)}}-\frac {16 x^4}{3 \sqrt {\sinh ^{-1}(a x)}}+\frac {8 \text {Subst}\left (\int \frac {\sinh (4 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^4}+\frac {4 \text {Subst}\left (\int \frac {\sinh (2 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a^4}-\frac {16 \text {Subst}\left (\int \frac {\sinh (2 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^4}\\ &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^2}{a^2 \sqrt {\sinh ^{-1}(a x)}}-\frac {16 x^4}{3 \sqrt {\sinh ^{-1}(a x)}}-\frac {4 \text {Subst}\left (\int \frac {e^{-4 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^4}+\frac {4 \text {Subst}\left (\int \frac {e^{4 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^4}-\frac {2 \text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a^4}+\frac {2 \text {Subst}\left (\int \frac {e^{2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a^4}+\frac {8 \text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^4}-\frac {8 \text {Subst}\left (\int \frac {e^{2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^4}\\ &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^2}{a^2 \sqrt {\sinh ^{-1}(a x)}}-\frac {16 x^4}{3 \sqrt {\sinh ^{-1}(a x)}}-\frac {8 \text {Subst}\left (\int e^{-4 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}+\frac {8 \text {Subst}\left (\int e^{4 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}-\frac {4 \text {Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{a^4}+\frac {4 \text {Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{a^4}+\frac {16 \text {Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}-\frac {16 \text {Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}\\ &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^2}{a^2 \sqrt {\sinh ^{-1}(a x)}}-\frac {16 x^4}{3 \sqrt {\sinh ^{-1}(a x)}}-\frac {2 \sqrt {\pi } \text {erf}\left (2 \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}+\frac {\sqrt {2 \pi } \text {erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}+\frac {2 \sqrt {\pi } \text {erfi}\left (2 \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}-\frac {\sqrt {2 \pi } \text {erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a^4}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 174, normalized size = 1.04 \begin {gather*} \frac {4 \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 )-4 \sinh ^{-1}(a x) \left (e^{-4 \sinh ^{-1}(a x)}+e^{4 \sinh ^{-1}(a x)}-2 \sqrt {-\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-4 \sinh ^{-1}(a x)\right )-2 \sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},4 \sinh ^{-1}(a x)\right )\right )+2 \sinh \left (2 \sinh ^{-1}(a x)\right )-\sinh \left (4 \sinh ^{-1}(a x)\right )}{12 a^4 \sinh ^{-1}(a x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/ArcSinh[a*x]^(5/2),x]

[Out]

(4*ArcSinh[a*x]*(E^(-2*ArcSinh[a*x]) + E^(2*ArcSinh[a*x]) - Sqrt[2]*Sqrt[-ArcSinh[a*x]]*Gamma[1/2, -2*ArcSinh[
a*x]] - Sqrt[2]*Sqrt[ArcSinh[a*x]]*Gamma[1/2, 2*ArcSinh[a*x]]) - 4*ArcSinh[a*x]*(E^(-4*ArcSinh[a*x]) + E^(4*Ar
cSinh[a*x]) - 2*Sqrt[-ArcSinh[a*x]]*Gamma[1/2, -4*ArcSinh[a*x]] - 2*Sqrt[ArcSinh[a*x]]*Gamma[1/2, 4*ArcSinh[a*
x]]) + 2*Sinh[2*ArcSinh[a*x]] - Sinh[4*ArcSinh[a*x]])/(12*a^4*ArcSinh[a*x]^(3/2))

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Maple [F]
time = 4.17, size = 0, normalized size = 0.00 \[\int \frac {x^{3}}{\arcsinh \left (a x \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arcsinh(a*x)^(5/2),x)

[Out]

int(x^3/arcsinh(a*x)^(5/2),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^3/arcsinh(a*x)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^3/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^3/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^{3}}{\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**3/asinh(a*x)**(5/2),x)

[Out]

Integral(x**3/asinh(a*x)**(5/2), x)

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Giac [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^3/arcsinh(a*x)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3}{{\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^3/asinh(a*x)^(5/2),x)

[Out]

int(x^3/asinh(a*x)^(5/2), x)

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