3.53 \(\int \frac{x^4}{\sinh ^{-1}(a x)^2} \, dx\)

Optimal. Leaf size=68 \[ \frac{\text{Shi}\left (\sinh ^{-1}(a x)\right )}{8 a^5}-\frac{9 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{16 a^5}+\frac{5 \text{Shi}\left (5 \sinh ^{-1}(a x)\right )}{16 a^5}-\frac{x^4 \sqrt{a^2 x^2+1}}{a \sinh ^{-1}(a x)} \]

[Out]

-((x^4*Sqrt[1 + a^2*x^2])/(a*ArcSinh[a*x])) + SinhIntegral[ArcSinh[a*x]]/(8*a^5) - (9*SinhIntegral[3*ArcSinh[a
*x]])/(16*a^5) + (5*SinhIntegral[5*ArcSinh[a*x]])/(16*a^5)

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Rubi [A]  time = 0.0658968, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5665, 3298} \[ \frac{\text{Shi}\left (\sinh ^{-1}(a x)\right )}{8 a^5}-\frac{9 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{16 a^5}+\frac{5 \text{Shi}\left (5 \sinh ^{-1}(a x)\right )}{16 a^5}-\frac{x^4 \sqrt{a^2 x^2+1}}{a \sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((x^4*Sqrt[1 + a^2*x^2])/(a*ArcSinh[a*x])) + SinhIntegral[ArcSinh[a*x]]/(8*a^5) - (9*SinhIntegral[3*ArcSinh[a
*x]])/(16*a^5) + (5*SinhIntegral[5*ArcSinh[a*x]])/(16*a^5)

Rule 5665

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*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -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^4}{\sinh ^{-1}(a x)^2} \, dx &=-\frac{x^4 \sqrt{1+a^2 x^2}}{a \sinh ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \left (\frac{\sinh (x)}{8 x}-\frac{9 \sinh (3 x)}{16 x}+\frac{5 \sinh (5 x)}{16 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a^5}\\ &=-\frac{x^4 \sqrt{1+a^2 x^2}}{a \sinh ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a^5}+\frac{5 \operatorname{Subst}\left (\int \frac{\sinh (5 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^5}-\frac{9 \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^5}\\ &=-\frac{x^4 \sqrt{1+a^2 x^2}}{a \sinh ^{-1}(a x)}+\frac{\text{Shi}\left (\sinh ^{-1}(a x)\right )}{8 a^5}-\frac{9 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{16 a^5}+\frac{5 \text{Shi}\left (5 \sinh ^{-1}(a x)\right )}{16 a^5}\\ \end{align*}

Mathematica [A]  time = 0.202217, size = 60, normalized size = 0.88 \[ \frac{-\frac{16 a^4 x^4 \sqrt{a^2 x^2+1}}{\sinh ^{-1}(a x)}+2 \text{Shi}\left (\sinh ^{-1}(a x)\right )-9 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )+5 \text{Shi}\left (5 \sinh ^{-1}(a x)\right )}{16 a^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-16*a^4*x^4*Sqrt[1 + a^2*x^2])/ArcSinh[a*x] + 2*SinhIntegral[ArcSinh[a*x]] - 9*SinhIntegral[3*ArcSinh[a*x]]
+ 5*SinhIntegral[5*ArcSinh[a*x]])/(16*a^5)

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Maple [A]  time = 0.028, size = 80, normalized size = 1.2 \begin{align*}{\frac{1}{{a}^{5}} \left ( -{\frac{1}{8\,{\it Arcsinh} \left ( ax \right ) }\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{{\it Shi} \left ({\it Arcsinh} \left ( ax \right ) \right ) }{8}}+{\frac{3\,\cosh \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{16\,{\it Arcsinh} \left ( ax \right ) }}-{\frac{9\,{\it Shi} \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{16}}-{\frac{\cosh \left ( 5\,{\it Arcsinh} \left ( ax \right ) \right ) }{16\,{\it Arcsinh} \left ( ax \right ) }}+{\frac{5\,{\it Shi} \left ( 5\,{\it Arcsinh} \left ( ax \right ) \right ) }{16}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^5*(-1/8/arcsinh(a*x)*(a^2*x^2+1)^(1/2)+1/8*Shi(arcsinh(a*x))+3/16/arcsinh(a*x)*cosh(3*arcsinh(a*x))-9/16*S
hi(3*arcsinh(a*x))-1/16/arcsinh(a*x)*cosh(5*arcsinh(a*x))+5/16*Shi(5*arcsinh(a*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{a^{3} x^{7} + a x^{5} +{\left (a^{2} x^{6} + x^{4}\right )} \sqrt{a^{2} x^{2} + 1}}{{\left (a^{3} x^{2} + \sqrt{a^{2} x^{2} + 1} a^{2} x + a\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )} + \int \frac{5 \, a^{5} x^{8} + 10 \, a^{3} x^{6} + 5 \, a x^{4} +{\left (5 \, a^{3} x^{6} + 3 \, a x^{4}\right )}{\left (a^{2} x^{2} + 1\right )} +{\left (10 \, a^{4} x^{7} + 13 \, a^{2} x^{5} + 4 \, x^{3}\right )} \sqrt{a^{2} x^{2} + 1}}{{\left (a^{5} x^{4} +{\left (a^{2} x^{2} + 1\right )} a^{3} x^{2} + 2 \, a^{3} x^{2} + 2 \,{\left (a^{4} x^{3} + a^{2} x\right )} \sqrt{a^{2} x^{2} + 1} + a\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^3*x^7 + a*x^5 + (a^2*x^6 + x^4)*sqrt(a^2*x^2 + 1))/((a^3*x^2 + sqrt(a^2*x^2 + 1)*a^2*x + a)*log(a*x + sqrt
(a^2*x^2 + 1))) + integrate((5*a^5*x^8 + 10*a^3*x^6 + 5*a*x^4 + (5*a^3*x^6 + 3*a*x^4)*(a^2*x^2 + 1) + (10*a^4*
x^7 + 13*a^2*x^5 + 4*x^3)*sqrt(a^2*x^2 + 1))/((a^5*x^4 + (a^2*x^2 + 1)*a^3*x^2 + 2*a^3*x^2 + 2*(a^4*x^3 + a^2*
x)*sqrt(a^2*x^2 + 1) + a)*log(a*x + sqrt(a^2*x^2 + 1))), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{4}}{\operatorname{arsinh}\left (a x\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

integral(x^4/arcsinh(a*x)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\operatorname{asinh}^{2}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4/asinh(a*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\operatorname{arsinh}\left (a x\right )^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/arcsinh(a*x)^2, x)