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

Optimal. Leaf size=50 \[ -\frac{\sqrt{a^2 x^2+1}}{2 a \sinh ^{-1}(a x)^2}+\frac{\text{Chi}\left (\sinh ^{-1}(a x)\right )}{2 a}-\frac{x}{2 \sinh ^{-1}(a x)} \]

[Out]

-Sqrt[1 + a^2*x^2]/(2*a*ArcSinh[a*x]^2) - x/(2*ArcSinh[a*x]) + CoshIntegral[ArcSinh[a*x]]/(2*a)

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Rubi [A]  time = 0.0826796, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {5655, 5774, 5657, 3301} \[ -\frac{\sqrt{a^2 x^2+1}}{2 a \sinh ^{-1}(a x)^2}+\frac{\text{Chi}\left (\sinh ^{-1}(a x)\right )}{2 a}-\frac{x}{2 \sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]^(-3),x]

[Out]

-Sqrt[1 + a^2*x^2]/(2*a*ArcSinh[a*x]^2) - x/(2*ArcSinh[a*x]) + CoshIntegral[ArcSinh[a*x]]/(2*a)

Rule 5655

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^(n + 1
))/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcSinh[c*x])^(n + 1))/Sqrt[1 + c^2*x^2], x], x] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]

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 5657

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

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

Mathematica [A]  time = 0.0184735, size = 47, normalized size = 0.94 \[ -\frac{\sqrt{a^2 x^2+1}+\sinh ^{-1}(a x)^2 \left (-\text{Chi}\left (\sinh ^{-1}(a x)\right )\right )+a x \sinh ^{-1}(a x)}{2 a \sinh ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a*x]^(-3),x]

[Out]

-(Sqrt[1 + a^2*x^2] + a*x*ArcSinh[a*x] - ArcSinh[a*x]^2*CoshIntegral[ArcSinh[a*x]])/(2*a*ArcSinh[a*x]^2)

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Maple [A]  time = 0.022, size = 42, normalized size = 0.8 \begin{align*}{\frac{1}{a} \left ( -{\frac{1}{2\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{ax}{2\,{\it Arcsinh} \left ( ax \right ) }}+{\frac{{\it Chi} \left ({\it Arcsinh} \left ( ax \right ) \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(-1/2/arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)-1/2*a*x/arcsinh(a*x)+1/2*Chi(arcsinh(a*x)))

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(a*x)^(-3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x)**(-3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(a*x)^(-3), x)