3.1.64 \(\int \frac {1}{\sinh ^{-1}(a x)^3} \, dx\) [64]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 = {5773, 5818, 5774, 3382} \begin {gather*} -\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3382

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]

Rule 5773

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_), 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 5818

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

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 {\text {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 = 0.02, size = 47, normalized size = 0.94 \begin {gather*} -\frac {\sqrt {1+a^2 x^2}+a x \sinh ^{-1}(a x)-\sinh ^{-1}(a x)^2 \text {Chi}\left (\sinh ^{-1}(a x)\right )}{2 a \sinh ^{-1}(a x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 1.20, size = 42, normalized size = 0.84

method	result	size
derivativedivides	\(\frac {-\frac {\sqrt {a^{2} x^{2}+1}}{2 \arcsinh \left (a x \right )^{2}}-\frac {a x}{2 \arcsinh \left (a x \right )}+\frac {\hyperbolicCosineIntegral \left (\arcsinh \left (a x \right )\right )}{2}}{a}\)	\(42\)
default	\(\frac {-\frac {\sqrt {a^{2} x^{2}+1}}{2 \arcsinh \left (a x \right )^{2}}-\frac {a x}{2 \arcsinh \left (a x \right )}+\frac {\hyperbolicCosineIntegral \left (\arcsinh \left (a x \right )\right )}{2}}{a}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\operatorname {asinh}^{3}{\left (a x \right )}}\, dx \end {gather*}

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)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\mathrm {asinh}\left (a\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/asinh(a*x)^3, x)

________________________________________________________________________________________