3.1.55 \(\int \frac {x^2}{\sinh ^{-1}(a x)^2} \, dx\) [55]

Optimal. Leaf size=54 \[ -\frac {x^2 \sqrt {1+a^2 x^2}}{a \sinh ^{-1}(a x)}-\frac {\text {Shi}\left (\sinh ^{-1}(a x)\right )}{4 a^3}+\frac {3 \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{4 a^3} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 54, 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 = {5778, 3379} \begin {gather*} -\frac {\text {Shi}\left (\sinh ^{-1}(a x)\right )}{4 a^3}+\frac {3 \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{4 a^3}-\frac {x^2 \sqrt {a^2 x^2+1}}{a \sinh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3379

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]

Rule 5778

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^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), Si
nh[-a/b + x/b]^(m - 1)*(m + (m + 1)*Sinh[-a/b + x/b]^2), x], x], x, a + b*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^2}{\sinh ^{-1}(a x)^2} \, dx &=-\frac {x^2 \sqrt {1+a^2 x^2}}{a \sinh ^{-1}(a x)}+\frac {\text {Subst}\left (\int \left (-\frac {\sinh (x)}{4 x}+\frac {3 \sinh (3 x)}{4 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a^3}\\ &=-\frac {x^2 \sqrt {1+a^2 x^2}}{a \sinh ^{-1}(a x)}-\frac {\text {Subst}\left (\int \frac {\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{4 a^3}+\frac {3 \text {Subst}\left (\int \frac {\sinh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{4 a^3}\\ &=-\frac {x^2 \sqrt {1+a^2 x^2}}{a \sinh ^{-1}(a x)}-\frac {\text {Shi}\left (\sinh ^{-1}(a x)\right )}{4 a^3}+\frac {3 \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{4 a^3}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 49, normalized size = 0.91 \begin {gather*} -\frac {\frac {4 a^2 x^2 \sqrt {1+a^2 x^2}}{\sinh ^{-1}(a x)}+\text {Shi}\left (\sinh ^{-1}(a x)\right )-3 \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{4 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.33, size = 56, normalized size = 1.04

method result size
derivativedivides \(\frac {\frac {\sqrt {a^{2} x^{2}+1}}{4 \arcsinh \left (a x \right )}-\frac {\hyperbolicSineIntegral \left (\arcsinh \left (a x \right )\right )}{4}-\frac {\cosh \left (3 \arcsinh \left (a x \right )\right )}{4 \arcsinh \left (a x \right )}+\frac {3 \hyperbolicSineIntegral \left (3 \arcsinh \left (a x \right )\right )}{4}}{a^{3}}\) \(56\)
default \(\frac {\frac {\sqrt {a^{2} x^{2}+1}}{4 \arcsinh \left (a x \right )}-\frac {\hyperbolicSineIntegral \left (\arcsinh \left (a x \right )\right )}{4}-\frac {\cosh \left (3 \arcsinh \left (a x \right )\right )}{4 \arcsinh \left (a x \right )}+\frac {3 \hyperbolicSineIntegral \left (3 \arcsinh \left (a x \right )\right )}{4}}{a^{3}}\) \(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/arcsinh(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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

[Out]

-(a^3*x^5 + a*x^3 + (a^2*x^4 + x^2)*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((3*a^5*x^6 + 6*a^3*x^4 + 3*a*x^2 + (3*a^3*x^4 + a*x^2)*(a^2*x^2 + 1) + (6*a^4*x^5
+ 7*a^2*x^3 + 2*x)*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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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(x^2/arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

integral(x^2/arcsinh(a*x)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/asinh(a*x)**2, x)

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

[Out]

integrate(x^2/arcsinh(a*x)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2}{{\mathrm {asinh}\left (a\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2/asinh(a*x)^2, x)

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