3.54 \(\int \frac {x^3}{\sinh ^{-1}(a x)^2} \, dx\)

Optimal. Leaf size=56 \[ -\frac {\text {Chi}\left (2 \sinh ^{-1}(a x)\right )}{2 a^4}+\frac {\text {Chi}\left (4 \sinh ^{-1}(a x)\right )}{2 a^4}-\frac {x^3 \sqrt {a^2 x^2+1}}{a \sinh ^{-1}(a x)} \]

[Out]

-1/2*Chi(2*arcsinh(a*x))/a^4+1/2*Chi(4*arcsinh(a*x))/a^4-x^3*(a^2*x^2+1)^(1/2)/a/arcsinh(a*x)

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Rubi [A]  time = 0.05, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5665, 3301} \[ -\frac {\text {Chi}\left (2 \sinh ^{-1}(a x)\right )}{2 a^4}+\frac {\text {Chi}\left (4 \sinh ^{-1}(a x)\right )}{2 a^4}-\frac {x^3 \sqrt {a^2 x^2+1}}{a \sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Int[x^3/ArcSinh[a*x]^2,x]

[Out]

-((x^3*Sqrt[1 + a^2*x^2])/(a*ArcSinh[a*x])) - CoshIntegral[2*ArcSinh[a*x]]/(2*a^4) + CoshIntegral[4*ArcSinh[a*
x]]/(2*a^4)

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 56, normalized size = 1.00 \[ -\frac {4 \sinh ^{-1}(a x) \text {Chi}\left (2 \sinh ^{-1}(a x)\right )-4 \sinh ^{-1}(a x) \text {Chi}\left (4 \sinh ^{-1}(a x)\right )-2 \sinh \left (2 \sinh ^{-1}(a x)\right )+\sinh \left (4 \sinh ^{-1}(a x)\right )}{8 a^4 \sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/ArcSinh[a*x]^2,x]

[Out]

-1/8*(4*ArcSinh[a*x]*CoshIntegral[2*ArcSinh[a*x]] - 4*ArcSinh[a*x]*CoshIntegral[4*ArcSinh[a*x]] - 2*Sinh[2*Arc
Sinh[a*x]] + Sinh[4*ArcSinh[a*x]])/(a^4*ArcSinh[a*x])

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fricas [F]  time = 0.39, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{3}}{\operatorname {arsinh}\left (a x\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

integral(x^3/arcsinh(a*x)^2, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arcsinh(a*x)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.16, size = 54, normalized size = 0.96 \[ \frac {\frac {\sinh \left (2 \arcsinh \left (a x \right )\right )}{4 \arcsinh \left (a x \right )}-\frac {\Chi \left (2 \arcsinh \left (a x \right )\right )}{2}-\frac {\sinh \left (4 \arcsinh \left (a x \right )\right )}{8 \arcsinh \left (a x \right )}+\frac {\Chi \left (4 \arcsinh \left (a x \right )\right )}{2}}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arcsinh(a*x)^2,x)

[Out]

1/a^4*(1/4/arcsinh(a*x)*sinh(2*arcsinh(a*x))-1/2*Chi(2*arcsinh(a*x))-1/8/arcsinh(a*x)*sinh(4*arcsinh(a*x))+1/2
*Chi(4*arcsinh(a*x)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {a^{3} x^{6} + a x^{4} + {\left (a^{2} x^{5} + x^{3}\right )} \sqrt {a^{2} x^{2} + 1}}{{\left (a^{3} x^{2} + \sqrt {a^{2} x^{2} + 1} a^{2} x + a\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )} + \int \frac {4 \, a^{5} x^{7} + 8 \, a^{3} x^{5} + 4 \, a x^{3} + 2 \, {\left (2 \, a^{3} x^{5} + a x^{3}\right )} {\left (a^{2} x^{2} + 1\right )} + {\left (8 \, a^{4} x^{6} + 10 \, a^{2} x^{4} + 3 \, x^{2}\right )} \sqrt {a^{2} x^{2} + 1}}{{\left (a^{5} x^{4} + {\left (a^{2} x^{2} + 1\right )} a^{3} x^{2} + 2 \, a^{3} x^{2} + 2 \, {\left (a^{4} x^{3} + a^{2} x\right )} \sqrt {a^{2} x^{2} + 1} + a\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arcsinh(a*x)^2,x, algorithm="maxima")

[Out]

-(a^3*x^6 + a*x^4 + (a^2*x^5 + x^3)*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((4*a^5*x^7 + 8*a^3*x^5 + 4*a*x^3 + 2*(2*a^3*x^5 + a*x^3)*(a^2*x^2 + 1) + (8*a^4*x^
6 + 10*a^2*x^4 + 3*x^2)*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^3}{{\mathrm {asinh}\left (a\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/asinh(a*x)^2,x)

[Out]

int(x^3/asinh(a*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/asinh(a*x)**2,x)

[Out]

Integral(x**3/asinh(a*x)**2, x)

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