3.60 \(\int \frac{x^4}{\sinh ^{-1}(a x)^3} \, dx\)

Optimal. Leaf size=97 \[ \frac{\text{Chi}\left (\sinh ^{-1}(a x)\right )}{16 a^5}-\frac{27 \text{Chi}\left (3 \sinh ^{-1}(a x)\right )}{32 a^5}+\frac{25 \text{Chi}\left (5 \sinh ^{-1}(a x)\right )}{32 a^5}-\frac{x^4 \sqrt{a^2 x^2+1}}{2 a \sinh ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac{5 x^5}{2 \sinh ^{-1}(a x)} \]

[Out]

-(x^4*Sqrt[1 + a^2*x^2])/(2*a*ArcSinh[a*x]^2) - (2*x^3)/(a^2*ArcSinh[a*x]) - (5*x^5)/(2*ArcSinh[a*x]) + CoshIn
tegral[ArcSinh[a*x]]/(16*a^5) - (27*CoshIntegral[3*ArcSinh[a*x]])/(32*a^5) + (25*CoshIntegral[5*ArcSinh[a*x]])
/(32*a^5)

________________________________________________________________________________________

Rubi [A]  time = 0.354339, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5667, 5774, 5669, 5448, 3301} \[ \frac{\text{Chi}\left (\sinh ^{-1}(a x)\right )}{16 a^5}-\frac{27 \text{Chi}\left (3 \sinh ^{-1}(a x)\right )}{32 a^5}+\frac{25 \text{Chi}\left (5 \sinh ^{-1}(a x)\right )}{32 a^5}-\frac{x^4 \sqrt{a^2 x^2+1}}{2 a \sinh ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac{5 x^5}{2 \sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Int[x^4/ArcSinh[a*x]^3,x]

[Out]

-(x^4*Sqrt[1 + a^2*x^2])/(2*a*ArcSinh[a*x]^2) - (2*x^3)/(a^2*ArcSinh[a*x]) - (5*x^5)/(2*ArcSinh[a*x]) + CoshIn
tegral[ArcSinh[a*x]]/(16*a^5) - (27*CoshIntegral[3*ArcSinh[a*x]])/(32*a^5) + (25*CoshIntegral[5*ArcSinh[a*x]])
/(32*a^5)

Rule 5667

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[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcSinh[c*x])^(n +
1))/Sqrt[1 + c^2*x^2], x], x] - Dist[m/(b*c*(n + 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] && LtQ[n, -2]

Rule 5774

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x
)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && LtQ[n, -
1] && GtQ[d, 0]

Rule 5669

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Sinh[x]^m*Cosh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

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^4}{\sinh ^{-1}(a x)^3} \, dx &=-\frac{x^4 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}+\frac{2 \int \frac{x^3}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2} \, dx}{a}+\frac{1}{2} (5 a) \int \frac{x^5}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2} \, dx\\ &=-\frac{x^4 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac{5 x^5}{2 \sinh ^{-1}(a x)}+\frac{25}{2} \int \frac{x^4}{\sinh ^{-1}(a x)} \, dx+\frac{6 \int \frac{x^2}{\sinh ^{-1}(a x)} \, dx}{a^2}\\ &=-\frac{x^4 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac{5 x^5}{2 \sinh ^{-1}(a x)}+\frac{6 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^2(x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a^5}+\frac{25 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^4(x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^5}\\ &=-\frac{x^4 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac{5 x^5}{2 \sinh ^{-1}(a x)}+\frac{6 \operatorname{Subst}\left (\int \left (-\frac{\cosh (x)}{4 x}+\frac{\cosh (3 x)}{4 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a^5}+\frac{25 \operatorname{Subst}\left (\int \left (\frac{\cosh (x)}{8 x}-\frac{3 \cosh (3 x)}{16 x}+\frac{\cosh (5 x)}{16 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^5}\\ &=-\frac{x^4 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac{5 x^5}{2 \sinh ^{-1}(a x)}+\frac{25 \operatorname{Subst}\left (\int \frac{\cosh (5 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{32 a^5}-\frac{3 \operatorname{Subst}\left (\int \frac{\cosh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^5}+\frac{3 \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^5}+\frac{25 \operatorname{Subst}\left (\int \frac{\cosh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^5}-\frac{75 \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{32 a^5}\\ &=-\frac{x^4 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac{5 x^5}{2 \sinh ^{-1}(a x)}+\frac{\text{Chi}\left (\sinh ^{-1}(a x)\right )}{16 a^5}-\frac{27 \text{Chi}\left (3 \sinh ^{-1}(a x)\right )}{32 a^5}+\frac{25 \text{Chi}\left (5 \sinh ^{-1}(a x)\right )}{32 a^5}\\ \end{align*}

Mathematica [A]  time = 0.140226, size = 102, normalized size = 1.05 \[ -\frac{16 a^4 x^4 \sqrt{a^2 x^2+1}+80 a^5 x^5 \sinh ^{-1}(a x)+64 a^3 x^3 \sinh ^{-1}(a x)-2 \sinh ^{-1}(a x)^2 \text{Chi}\left (\sinh ^{-1}(a x)\right )+27 \sinh ^{-1}(a x)^2 \text{Chi}\left (3 \sinh ^{-1}(a x)\right )-25 \sinh ^{-1}(a x)^2 \text{Chi}\left (5 \sinh ^{-1}(a x)\right )}{32 a^5 \sinh ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcSinh[a*x]^3,x]

[Out]

-(16*a^4*x^4*Sqrt[1 + a^2*x^2] + 64*a^3*x^3*ArcSinh[a*x] + 80*a^5*x^5*ArcSinh[a*x] - 2*ArcSinh[a*x]^2*CoshInte
gral[ArcSinh[a*x]] + 27*ArcSinh[a*x]^2*CoshIntegral[3*ArcSinh[a*x]] - 25*ArcSinh[a*x]^2*CoshIntegral[5*ArcSinh
[a*x]])/(32*a^5*ArcSinh[a*x]^2)

________________________________________________________________________________________

Maple [A]  time = 0.039, size = 120, normalized size = 1.2 \begin{align*}{\frac{1}{{a}^{5}} \left ( -{\frac{1}{16\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{ax}{16\,{\it Arcsinh} \left ( ax \right ) }}+{\frac{{\it Chi} \left ({\it Arcsinh} \left ( ax \right ) \right ) }{16}}+{\frac{3\,\cosh \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{32\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}}+{\frac{9\,\sinh \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{32\,{\it Arcsinh} \left ( ax \right ) }}-{\frac{27\,{\it Chi} \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{32}}-{\frac{\cosh \left ( 5\,{\it Arcsinh} \left ( ax \right ) \right ) }{32\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}}-{\frac{5\,\sinh \left ( 5\,{\it Arcsinh} \left ( ax \right ) \right ) }{32\,{\it Arcsinh} \left ( ax \right ) }}+{\frac{25\,{\it Chi} \left ( 5\,{\it Arcsinh} \left ( ax \right ) \right ) }{32}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arcsinh(a*x)^3,x)

[Out]

1/a^5*(-1/16/arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)-1/16*a*x/arcsinh(a*x)+1/16*Chi(arcsinh(a*x))+3/32/arcsinh(a*x)^2
*cosh(3*arcsinh(a*x))+9/32/arcsinh(a*x)*sinh(3*arcsinh(a*x))-27/32*Chi(3*arcsinh(a*x))-1/32/arcsinh(a*x)^2*cos
h(5*arcsinh(a*x))-5/32/arcsinh(a*x)*sinh(5*arcsinh(a*x))+25/32*Chi(5*arcsinh(a*x)))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arcsinh(a*x)^3,x, algorithm="maxima")

[Out]

-1/2*(a^8*x^11 + 3*a^6*x^9 + 3*a^4*x^7 + a^2*x^5 + (a^5*x^8 + a^3*x^6)*(a^2*x^2 + 1)^(3/2) + (3*a^6*x^9 + 5*a^
4*x^7 + 2*a^2*x^5)*(a^2*x^2 + 1) + (5*a^8*x^11 + 15*a^6*x^9 + 15*a^4*x^7 + 5*a^2*x^5 + (5*a^5*x^8 + 8*a^3*x^6
+ 3*a*x^4)*(a^2*x^2 + 1)^(3/2) + (15*a^6*x^9 + 31*a^4*x^7 + 20*a^2*x^5 + 4*x^3)*(a^2*x^2 + 1) + (15*a^7*x^10 +
38*a^5*x^8 + 32*a^3*x^6 + 9*a*x^4)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 + 1)) + (3*a^7*x^10 + 7*a^5*x^8
+ 5*a^3*x^6 + a*x^4)*sqrt(a^2*x^2 + 1))/((a^8*x^6 + 3*a^6*x^4 + (a^2*x^2 + 1)^(3/2)*a^5*x^3 + 3*a^4*x^2 + 3*(a
^6*x^4 + a^4*x^2)*(a^2*x^2 + 1) + a^2 + 3*(a^7*x^5 + 2*a^5*x^3 + a^3*x)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*
x^2 + 1))^2) + integrate(1/2*(25*a^10*x^12 + 100*a^8*x^10 + 150*a^6*x^8 + 100*a^4*x^6 + 25*a^2*x^4 + (25*a^6*x
^8 + 24*a^4*x^6 + 3*a^2*x^4)*(a^2*x^2 + 1)^2 + (100*a^7*x^9 + 172*a^5*x^7 + 87*a^3*x^5 + 12*a*x^3)*(a^2*x^2 +
1)^(3/2) + 3*(50*a^8*x^10 + 124*a^6*x^8 + 105*a^4*x^6 + 35*a^2*x^4 + 4*x^2)*(a^2*x^2 + 1) + (100*a^9*x^11 + 32
4*a^7*x^9 + 381*a^5*x^7 + 193*a^3*x^5 + 36*a*x^3)*sqrt(a^2*x^2 + 1))/((a^10*x^8 + 4*a^8*x^6 + (a^2*x^2 + 1)^2*
a^6*x^4 + 6*a^6*x^4 + 4*a^4*x^2 + 4*(a^7*x^5 + a^5*x^3)*(a^2*x^2 + 1)^(3/2) + 6*(a^8*x^6 + 2*a^6*x^4 + a^4*x^2
)*(a^2*x^2 + 1) + a^2 + 4*(a^9*x^7 + 3*a^7*x^5 + 3*a^5*x^3 + a^3*x)*sqrt(a^2*x^2 + 1))*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^{4}}{\operatorname{arsinh}\left (a x\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arcsinh(a*x)^3,x, algorithm="fricas")

[Out]

integral(x^4/arcsinh(a*x)^3, x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\operatorname{asinh}^{3}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/asinh(a*x)**3,x)

[Out]

Integral(x**4/asinh(a*x)**3, x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\operatorname{arsinh}\left (a x\right )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arcsinh(a*x)^3,x, algorithm="giac")

[Out]

integrate(x^4/arcsinh(a*x)^3, x)