3.1.74 \(\int x^4 \sqrt {\sinh ^{-1}(a x)} \, dx\) [74]

Optimal. Leaf size=182 \[ \frac {1}{5} x^5 \sqrt {\sinh ^{-1}(a x)}+\frac {\sqrt {\pi } \text {Erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{32 a^5}-\frac {\sqrt {\frac {\pi }{3}} \text {Erf}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{64 a^5}+\frac {\sqrt {\frac {\pi }{5}} \text {Erf}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{320 a^5}-\frac {\sqrt {\pi } \text {Erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{32 a^5}+\frac {\sqrt {\frac {\pi }{3}} \text {Erfi}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{64 a^5}-\frac {\sqrt {\frac {\pi }{5}} \text {Erfi}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{320 a^5} \]

[Out]

1/1600*erf(5^(1/2)*arcsinh(a*x)^(1/2))*5^(1/2)*Pi^(1/2)/a^5-1/1600*erfi(5^(1/2)*arcsinh(a*x)^(1/2))*5^(1/2)*Pi
^(1/2)/a^5-1/192*erf(3^(1/2)*arcsinh(a*x)^(1/2))*3^(1/2)*Pi^(1/2)/a^5+1/192*erfi(3^(1/2)*arcsinh(a*x)^(1/2))*3
^(1/2)*Pi^(1/2)/a^5+1/32*erf(arcsinh(a*x)^(1/2))*Pi^(1/2)/a^5-1/32*erfi(arcsinh(a*x)^(1/2))*Pi^(1/2)/a^5+1/5*x
^5*arcsinh(a*x)^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5777, 5819, 3393, 3389, 2211, 2235, 2236} \begin {gather*} \frac {\sqrt {\pi } \text {Erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{32 a^5}-\frac {\sqrt {\frac {\pi }{3}} \text {Erf}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{64 a^5}+\frac {\sqrt {\frac {\pi }{5}} \text {Erf}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{320 a^5}-\frac {\sqrt {\pi } \text {Erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{32 a^5}+\frac {\sqrt {\frac {\pi }{3}} \text {Erfi}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{64 a^5}-\frac {\sqrt {\frac {\pi }{5}} \text {Erfi}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{320 a^5}+\frac {1}{5} x^5 \sqrt {\sinh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[ArcSinh[a*x]],x]

[Out]

(x^5*Sqrt[ArcSinh[a*x]])/5 + (Sqrt[Pi]*Erf[Sqrt[ArcSinh[a*x]]])/(32*a^5) - (Sqrt[Pi/3]*Erf[Sqrt[3]*Sqrt[ArcSin
h[a*x]]])/(64*a^5) + (Sqrt[Pi/5]*Erf[Sqrt[5]*Sqrt[ArcSinh[a*x]]])/(320*a^5) - (Sqrt[Pi]*Erfi[Sqrt[ArcSinh[a*x]
]])/(32*a^5) + (Sqrt[Pi/3]*Erfi[Sqrt[3]*Sqrt[ArcSinh[a*x]]])/(64*a^5) - (Sqrt[Pi/5]*Erfi[Sqrt[5]*Sqrt[ArcSinh[
a*x]]])/(320*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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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 5819

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

Rubi steps

\begin {align*} \int x^4 \sqrt {\sinh ^{-1}(a x)} \, dx &=\frac {1}{5} x^5 \sqrt {\sinh ^{-1}(a x)}-\frac {1}{10} a \int \frac {x^5}{\sqrt {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}} \, dx\\ &=\frac {1}{5} x^5 \sqrt {\sinh ^{-1}(a x)}-\frac {\text {Subst}\left (\int \frac {\sinh ^5(x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{10 a^5}\\ &=\frac {1}{5} x^5 \sqrt {\sinh ^{-1}(a x)}+\frac {i \text {Subst}\left (\int \left (\frac {5 i \sinh (x)}{8 \sqrt {x}}-\frac {5 i \sinh (3 x)}{16 \sqrt {x}}+\frac {i \sinh (5 x)}{16 \sqrt {x}}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{10 a^5}\\ &=\frac {1}{5} x^5 \sqrt {\sinh ^{-1}(a x)}-\frac {\text {Subst}\left (\int \frac {\sinh (5 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{160 a^5}+\frac {\text {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{32 a^5}-\frac {\text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^5}\\ &=\frac {1}{5} x^5 \sqrt {\sinh ^{-1}(a x)}+\frac {\text {Subst}\left (\int \frac {e^{-5 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{320 a^5}-\frac {\text {Subst}\left (\int \frac {e^{5 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{320 a^5}-\frac {\text {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a^5}+\frac {\text {Subst}\left (\int \frac {e^{3 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a^5}+\frac {\text {Subst}\left (\int \frac {e^{-x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{32 a^5}-\frac {\text {Subst}\left (\int \frac {e^x}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{32 a^5}\\ &=\frac {1}{5} x^5 \sqrt {\sinh ^{-1}(a x)}+\frac {\text {Subst}\left (\int e^{-5 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{160 a^5}-\frac {\text {Subst}\left (\int e^{5 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{160 a^5}-\frac {\text {Subst}\left (\int e^{-3 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{32 a^5}+\frac {\text {Subst}\left (\int e^{3 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{32 a^5}+\frac {\text {Subst}\left (\int e^{-x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5}-\frac {\text {Subst}\left (\int e^{x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{16 a^5}\\ &=\frac {1}{5} x^5 \sqrt {\sinh ^{-1}(a x)}+\frac {\sqrt {\pi } \text {erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{32 a^5}-\frac {\sqrt {\frac {\pi }{3}} \text {erf}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{64 a^5}+\frac {\sqrt {\frac {\pi }{5}} \text {erf}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{320 a^5}-\frac {\sqrt {\pi } \text {erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{32 a^5}+\frac {\sqrt {\frac {\pi }{3}} \text {erfi}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{64 a^5}-\frac {\sqrt {\frac {\pi }{5}} \text {erfi}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{320 a^5}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 161, normalized size = 0.88 \begin {gather*} \frac {\frac {\sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {3}{2},-5 \sinh ^{-1}(a x)\right )}{160 \sqrt {5} \sqrt {-\sinh ^{-1}(a x)}}-\frac {\sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {3}{2},-3 \sinh ^{-1}(a x)\right )}{32 \sqrt {3} \sqrt {-\sinh ^{-1}(a x)}}+\frac {\sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {3}{2},-\sinh ^{-1}(a x)\right )}{16 \sqrt {-\sinh ^{-1}(a x)}}-\frac {1}{16} \Gamma \left (\frac {3}{2},\sinh ^{-1}(a x)\right )+\frac {\Gamma \left (\frac {3}{2},3 \sinh ^{-1}(a x)\right )}{32 \sqrt {3}}-\frac {\Gamma \left (\frac {3}{2},5 \sinh ^{-1}(a x)\right )}{160 \sqrt {5}}}{a^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*Sqrt[ArcSinh[a*x]],x]

[Out]

((Sqrt[ArcSinh[a*x]]*Gamma[3/2, -5*ArcSinh[a*x]])/(160*Sqrt[5]*Sqrt[-ArcSinh[a*x]]) - (Sqrt[ArcSinh[a*x]]*Gamm
a[3/2, -3*ArcSinh[a*x]])/(32*Sqrt[3]*Sqrt[-ArcSinh[a*x]]) + (Sqrt[ArcSinh[a*x]]*Gamma[3/2, -ArcSinh[a*x]])/(16
*Sqrt[-ArcSinh[a*x]]) - Gamma[3/2, ArcSinh[a*x]]/16 + Gamma[3/2, 3*ArcSinh[a*x]]/(32*Sqrt[3]) - Gamma[3/2, 5*A
rcSinh[a*x]]/(160*Sqrt[5]))/a^5

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Maple [F]
time = 6.61, size = 0, normalized size = 0.00 \[\int x^{4} \sqrt {\arcsinh \left (a x \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arcsinh(a*x)^(1/2),x)

[Out]

int(x^4*arcsinh(a*x)^(1/2),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)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4*sqrt(arcsinh(a*x)), 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)^(1/2),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^{4} \sqrt {\operatorname {asinh}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*asinh(a*x)**(1/2),x)

[Out]

Integral(x**4*sqrt(asinh(a*x)), 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)^(1/2),x, algorithm="giac")

[Out]

integrate(x^4*sqrt(arcsinh(a*x)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^4\,\sqrt {\mathrm {asinh}\left (a\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*asinh(a*x)^(1/2),x)

[Out]

int(x^4*asinh(a*x)^(1/2), x)

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