3.52 \(\int \frac{x^5}{\sinh ^{-1}(a x)^2} \, dx\)

Optimal. Leaf size=70 \[ \frac{5 \text{Chi}\left (2 \sinh ^{-1}(a x)\right )}{16 a^6}-\frac{\text{Chi}\left (4 \sinh ^{-1}(a x)\right )}{2 a^6}+\frac{3 \text{Chi}\left (6 \sinh ^{-1}(a x)\right )}{16 a^6}-\frac{x^5 \sqrt{a^2 x^2+1}}{a \sinh ^{-1}(a x)} \]

[Out]

-((x^5*Sqrt[1 + a^2*x^2])/(a*ArcSinh[a*x])) + (5*CoshIntegral[2*ArcSinh[a*x]])/(16*a^6) - CoshIntegral[4*ArcSi
nh[a*x]]/(2*a^6) + (3*CoshIntegral[6*ArcSinh[a*x]])/(16*a^6)

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Rubi [A]  time = 0.0664206, antiderivative size = 70, 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, 3301} \[ \frac{5 \text{Chi}\left (2 \sinh ^{-1}(a x)\right )}{16 a^6}-\frac{\text{Chi}\left (4 \sinh ^{-1}(a x)\right )}{2 a^6}+\frac{3 \text{Chi}\left (6 \sinh ^{-1}(a x)\right )}{16 a^6}-\frac{x^5 \sqrt{a^2 x^2+1}}{a \sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((x^5*Sqrt[1 + a^2*x^2])/(a*ArcSinh[a*x])) + (5*CoshIntegral[2*ArcSinh[a*x]])/(16*a^6) - CoshIntegral[4*ArcSi
nh[a*x]]/(2*a^6) + (3*CoshIntegral[6*ArcSinh[a*x]])/(16*a^6)

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 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \frac{x^5}{\sinh ^{-1}(a x)^2} \, dx &=-\frac{x^5 \sqrt{1+a^2 x^2}}{a \sinh ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \left (\frac{5 \cosh (2 x)}{16 x}-\frac{\cosh (4 x)}{2 x}+\frac{3 \cosh (6 x)}{16 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a^6}\\ &=-\frac{x^5 \sqrt{1+a^2 x^2}}{a \sinh ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \frac{\cosh (6 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^6}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^6}-\frac{\operatorname{Subst}\left (\int \frac{\cosh (4 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^6}\\ &=-\frac{x^5 \sqrt{1+a^2 x^2}}{a \sinh ^{-1}(a x)}+\frac{5 \text{Chi}\left (2 \sinh ^{-1}(a x)\right )}{16 a^6}-\frac{\text{Chi}\left (4 \sinh ^{-1}(a x)\right )}{2 a^6}+\frac{3 \text{Chi}\left (6 \sinh ^{-1}(a x)\right )}{16 a^6}\\ \end{align*}

Mathematica [A]  time = 0.0408332, size = 78, normalized size = 1.11 \[ -\frac{-10 \sinh ^{-1}(a x) \text{Chi}\left (2 \sinh ^{-1}(a x)\right )+16 \sinh ^{-1}(a x) \text{Chi}\left (4 \sinh ^{-1}(a x)\right )-6 \sinh ^{-1}(a x) \text{Chi}\left (6 \sinh ^{-1}(a x)\right )+5 \sinh \left (2 \sinh ^{-1}(a x)\right )-4 \sinh \left (4 \sinh ^{-1}(a x)\right )+\sinh \left (6 \sinh ^{-1}(a x)\right )}{32 a^6 \sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-10*ArcSinh[a*x]*CoshIntegral[2*ArcSinh[a*x]] + 16*ArcSinh[a*x]*CoshIntegral[4*ArcSinh[a*x]] - 6*ArcSinh[a*x
]*CoshIntegral[6*ArcSinh[a*x]] + 5*Sinh[2*ArcSinh[a*x]] - 4*Sinh[4*ArcSinh[a*x]] + Sinh[6*ArcSinh[a*x]])/(32*a
^6*ArcSinh[a*x])

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Maple [A]  time = 0.038, size = 78, normalized size = 1.1 \begin{align*}{\frac{1}{{a}^{6}} \left ( -{\frac{5\,\sinh \left ( 2\,{\it Arcsinh} \left ( ax \right ) \right ) }{32\,{\it Arcsinh} \left ( ax \right ) }}+{\frac{5\,{\it Chi} \left ( 2\,{\it Arcsinh} \left ( ax \right ) \right ) }{16}}+{\frac{\sinh \left ( 4\,{\it Arcsinh} \left ( ax \right ) \right ) }{8\,{\it Arcsinh} \left ( ax \right ) }}-{\frac{{\it Chi} \left ( 4\,{\it Arcsinh} \left ( ax \right ) \right ) }{2}}-{\frac{\sinh \left ( 6\,{\it Arcsinh} \left ( ax \right ) \right ) }{32\,{\it Arcsinh} \left ( ax \right ) }}+{\frac{3\,{\it Chi} \left ( 6\,{\it Arcsinh} \left ( ax \right ) \right ) }{16}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^6*(-5/32/arcsinh(a*x)*sinh(2*arcsinh(a*x))+5/16*Chi(2*arcsinh(a*x))+1/8/arcsinh(a*x)*sinh(4*arcsinh(a*x))-
1/2*Chi(4*arcsinh(a*x))-1/32/arcsinh(a*x)*sinh(6*arcsinh(a*x))+3/16*Chi(6*arcsinh(a*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{a^{3} x^{8} + a x^{6} +{\left (a^{2} x^{7} + x^{5}\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{6 \, a^{5} x^{9} + 12 \, a^{3} x^{7} + 6 \, a x^{5} + 2 \,{\left (3 \, a^{3} x^{7} + 2 \, a x^{5}\right )}{\left (a^{2} x^{2} + 1\right )} +{\left (12 \, a^{4} x^{8} + 16 \, a^{2} x^{6} + 5 \, x^{4}\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^5/arcsinh(a*x)^2,x, algorithm="maxima")

[Out]

-(a^3*x^8 + a*x^6 + (a^2*x^7 + x^5)*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((6*a^5*x^9 + 12*a^3*x^7 + 6*a*x^5 + 2*(3*a^3*x^7 + 2*a*x^5)*(a^2*x^2 + 1) + (12*a^
4*x^8 + 16*a^2*x^6 + 5*x^4)*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^{5}}{\operatorname{arsinh}\left (a x\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

integral(x^5/arcsinh(a*x)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\operatorname{asinh}^{2}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**5/asinh(a*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\operatorname{arsinh}\left (a x\right )^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/arcsinh(a*x)^2,x, algorithm="giac")

[Out]

integrate(x^5/arcsinh(a*x)^2, x)