3.51 \(\int \frac{x^6}{\sinh ^{-1}(a x)^2} \, dx\)

Optimal. Leaf size=82 \[ -\frac{5 \text{Shi}\left (\sinh ^{-1}(a x)\right )}{64 a^7}+\frac{27 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{64 a^7}-\frac{25 \text{Shi}\left (5 \sinh ^{-1}(a x)\right )}{64 a^7}+\frac{7 \text{Shi}\left (7 \sinh ^{-1}(a x)\right )}{64 a^7}-\frac{x^6 \sqrt{a^2 x^2+1}}{a \sinh ^{-1}(a x)} \]

[Out]

-((x^6*Sqrt[1 + a^2*x^2])/(a*ArcSinh[a*x])) - (5*SinhIntegral[ArcSinh[a*x]])/(64*a^7) + (27*SinhIntegral[3*Arc
Sinh[a*x]])/(64*a^7) - (25*SinhIntegral[5*ArcSinh[a*x]])/(64*a^7) + (7*SinhIntegral[7*ArcSinh[a*x]])/(64*a^7)

________________________________________________________________________________________

Rubi [A]  time = 0.0811772, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5665, 3298} \[ -\frac{5 \text{Shi}\left (\sinh ^{-1}(a x)\right )}{64 a^7}+\frac{27 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{64 a^7}-\frac{25 \text{Shi}\left (5 \sinh ^{-1}(a x)\right )}{64 a^7}+\frac{7 \text{Shi}\left (7 \sinh ^{-1}(a x)\right )}{64 a^7}-\frac{x^6 \sqrt{a^2 x^2+1}}{a \sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Int[x^6/ArcSinh[a*x]^2,x]

[Out]

-((x^6*Sqrt[1 + a^2*x^2])/(a*ArcSinh[a*x])) - (5*SinhIntegral[ArcSinh[a*x]])/(64*a^7) + (27*SinhIntegral[3*Arc
Sinh[a*x]])/(64*a^7) - (25*SinhIntegral[5*ArcSinh[a*x]])/(64*a^7) + (7*SinhIntegral[7*ArcSinh[a*x]])/(64*a^7)

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^6}{\sinh ^{-1}(a x)^2} \, dx &=-\frac{x^6 \sqrt{1+a^2 x^2}}{a \sinh ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \left (-\frac{5 \sinh (x)}{64 x}+\frac{27 \sinh (3 x)}{64 x}-\frac{25 \sinh (5 x)}{64 x}+\frac{7 \sinh (7 x)}{64 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a^7}\\ &=-\frac{x^6 \sqrt{1+a^2 x^2}}{a \sinh ^{-1}(a x)}-\frac{5 \operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a^7}+\frac{7 \operatorname{Subst}\left (\int \frac{\sinh (7 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a^7}-\frac{25 \operatorname{Subst}\left (\int \frac{\sinh (5 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a^7}+\frac{27 \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a^7}\\ &=-\frac{x^6 \sqrt{1+a^2 x^2}}{a \sinh ^{-1}(a x)}-\frac{5 \text{Shi}\left (\sinh ^{-1}(a x)\right )}{64 a^7}+\frac{27 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{64 a^7}-\frac{25 \text{Shi}\left (5 \sinh ^{-1}(a x)\right )}{64 a^7}+\frac{7 \text{Shi}\left (7 \sinh ^{-1}(a x)\right )}{64 a^7}\\ \end{align*}

Mathematica [A]  time = 0.279956, size = 85, normalized size = 1.04 \[ -\frac{64 a^6 x^6 \sqrt{a^2 x^2+1}+5 \sinh ^{-1}(a x) \text{Shi}\left (\sinh ^{-1}(a x)\right )-27 \sinh ^{-1}(a x) \text{Shi}\left (3 \sinh ^{-1}(a x)\right )+25 \sinh ^{-1}(a x) \text{Shi}\left (5 \sinh ^{-1}(a x)\right )-7 \sinh ^{-1}(a x) \text{Shi}\left (7 \sinh ^{-1}(a x)\right )}{64 a^7 \sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/ArcSinh[a*x]^2,x]

[Out]

-(64*a^6*x^6*Sqrt[1 + a^2*x^2] + 5*ArcSinh[a*x]*SinhIntegral[ArcSinh[a*x]] - 27*ArcSinh[a*x]*SinhIntegral[3*Ar
cSinh[a*x]] + 25*ArcSinh[a*x]*SinhIntegral[5*ArcSinh[a*x]] - 7*ArcSinh[a*x]*SinhIntegral[7*ArcSinh[a*x]])/(64*
a^7*ArcSinh[a*x])

________________________________________________________________________________________

Maple [A]  time = 0.043, size = 104, normalized size = 1.3 \begin{align*}{\frac{1}{{a}^{7}} \left ({\frac{5}{64\,{\it Arcsinh} \left ( ax \right ) }\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{5\,{\it Shi} \left ({\it Arcsinh} \left ( ax \right ) \right ) }{64}}-{\frac{9\,\cosh \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{64\,{\it Arcsinh} \left ( ax \right ) }}+{\frac{27\,{\it Shi} \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{64}}+{\frac{5\,\cosh \left ( 5\,{\it Arcsinh} \left ( ax \right ) \right ) }{64\,{\it Arcsinh} \left ( ax \right ) }}-{\frac{25\,{\it Shi} \left ( 5\,{\it Arcsinh} \left ( ax \right ) \right ) }{64}}-{\frac{\cosh \left ( 7\,{\it Arcsinh} \left ( ax \right ) \right ) }{64\,{\it Arcsinh} \left ( ax \right ) }}+{\frac{7\,{\it Shi} \left ( 7\,{\it Arcsinh} \left ( ax \right ) \right ) }{64}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/arcsinh(a*x)^2,x)

[Out]

1/a^7*(5/64/arcsinh(a*x)*(a^2*x^2+1)^(1/2)-5/64*Shi(arcsinh(a*x))-9/64/arcsinh(a*x)*cosh(3*arcsinh(a*x))+27/64
*Shi(3*arcsinh(a*x))+5/64/arcsinh(a*x)*cosh(5*arcsinh(a*x))-25/64*Shi(5*arcsinh(a*x))-1/64/arcsinh(a*x)*cosh(7
*arcsinh(a*x))+7/64*Shi(7*arcsinh(a*x)))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{a^{3} x^{9} + a x^{7} +{\left (a^{2} x^{8} + x^{6}\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{7 \, a^{5} x^{10} + 14 \, a^{3} x^{8} + 7 \, a x^{6} +{\left (7 \, a^{3} x^{8} + 5 \, a x^{6}\right )}{\left (a^{2} x^{2} + 1\right )} +{\left (14 \, a^{4} x^{9} + 19 \, a^{2} x^{7} + 6 \, x^{5}\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^6/arcsinh(a*x)^2,x, algorithm="maxima")

[Out]

-(a^3*x^9 + a*x^7 + (a^2*x^8 + x^6)*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((7*a^5*x^10 + 14*a^3*x^8 + 7*a*x^6 + (7*a^3*x^8 + 5*a*x^6)*(a^2*x^2 + 1) + (14*a^4
*x^9 + 19*a^2*x^7 + 6*x^5)*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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{6}}{\operatorname{arsinh}\left (a x\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

integral(x^6/arcsinh(a*x)^2, x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{\operatorname{asinh}^{2}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/asinh(a*x)**2,x)

[Out]

Integral(x**6/asinh(a*x)**2, x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{\operatorname{arsinh}\left (a x\right )^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/arcsinh(a*x)^2,x, algorithm="giac")

[Out]

integrate(x^6/arcsinh(a*x)^2, x)