3.381 \(\int \frac{x^3}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)} \, dx\)

Optimal. Leaf size=27 \[ \frac{\text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{4 a^4}-\frac{3 \text{Shi}\left (\sinh ^{-1}(a x)\right )}{4 a^4} \]

[Out]

(-3*SinhIntegral[ArcSinh[a*x]])/(4*a^4) + SinhIntegral[3*ArcSinh[a*x]]/(4*a^4)

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Rubi [A]  time = 0.155289, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {5779, 3312, 3298} \[ \frac{\text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{4 a^4}-\frac{3 \text{Shi}\left (\sinh ^{-1}(a x)\right )}{4 a^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(Sqrt[1 + a^2*x^2]*ArcSinh[a*x]),x]

[Out]

(-3*SinhIntegral[ArcSinh[a*x]])/(4*a^4) + SinhIntegral[3*ArcSinh[a*x]]/(4*a^4)

Rule 5779

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

Rule 3312

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 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \frac{x^3}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sinh ^3(x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a^4}\\ &=\frac{i \operatorname{Subst}\left (\int \left (\frac{3 i \sinh (x)}{4 x}-\frac{i \sinh (3 x)}{4 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a^4}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sinh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{4 a^4}-\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{4 a^4}\\ &=-\frac{3 \text{Shi}\left (\sinh ^{-1}(a x)\right )}{4 a^4}+\frac{\text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{4 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0774442, size = 22, normalized size = 0.81 \[ \frac{\text{Shi}\left (3 \sinh ^{-1}(a x)\right )-3 \text{Shi}\left (\sinh ^{-1}(a x)\right )}{4 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(Sqrt[1 + a^2*x^2]*ArcSinh[a*x]),x]

[Out]

(-3*SinhIntegral[ArcSinh[a*x]] + SinhIntegral[3*ArcSinh[a*x]])/(4*a^4)

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Maple [A]  time = 0.043, size = 23, normalized size = 0.9 \begin{align*} -{\frac{3\,{\it Shi} \left ({\it Arcsinh} \left ( ax \right ) \right ) -{\it Shi} \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{4\,{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arcsinh(a*x)/(a^2*x^2+1)^(1/2),x)

[Out]

-1/4*(3*Shi(arcsinh(a*x))-Shi(3*arcsinh(a*x)))/a^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\sqrt{a^{2} x^{2} + 1} \operatorname{arsinh}\left (a x\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arcsinh(a*x)/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^3/(sqrt(a^2*x^2 + 1)*arcsinh(a*x)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{3}}{\sqrt{a^{2} x^{2} + 1} \operatorname{arsinh}\left (a x\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arcsinh(a*x)/(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(x^3/(sqrt(a^2*x^2 + 1)*arcsinh(a*x)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\sqrt{a^{2} x^{2} + 1} \operatorname{asinh}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/asinh(a*x)/(a**2*x**2+1)**(1/2),x)

[Out]

Integral(x**3/(sqrt(a**2*x**2 + 1)*asinh(a*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\sqrt{a^{2} x^{2} + 1} \operatorname{arsinh}\left (a x\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arcsinh(a*x)/(a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^3/(sqrt(a^2*x^2 + 1)*arcsinh(a*x)), x)