3.4.83 \(\int \frac {x^2}{\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)} \, dx\) [383]

Optimal. Leaf size=27 \[ \frac {\text {Chi}\left (2 \sinh ^{-1}(a x)\right )}{2 a^3}-\frac {\log \left (\sinh ^{-1}(a x)\right )}{2 a^3} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {5819, 3393, 3382} \begin {gather*} \frac {\text {Chi}\left (2 \sinh ^{-1}(a x)\right )}{2 a^3}-\frac {\log \left (\sinh ^{-1}(a x)\right )}{2 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(Sqrt[1 + a^2*x^2]*ArcSinh[a*x]),x]

[Out]

CoshIntegral[2*ArcSinh[a*x]]/(2*a^3) - Log[ArcSinh[a*x]]/(2*a^3)

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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(1/(b*
c^(m + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Subst[Int[x^n*Sinh[-a/b + x/b]^m*Cosh[-a/b + x/b]^(2*p + 1),
x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p + 2, 0] && IGtQ[m,
0]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 22, normalized size = 0.81 \begin {gather*} \frac {\text {Chi}\left (2 \sinh ^{-1}(a x)\right )-\log \left (\sinh ^{-1}(a x)\right )}{2 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(Sqrt[1 + a^2*x^2]*ArcSinh[a*x]),x]

[Out]

(CoshIntegral[2*ArcSinh[a*x]] - Log[ArcSinh[a*x]])/(2*a^3)

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Maple [A]
time = 1.62, size = 21, normalized size = 0.78

method	result	size
default	\(-\frac {\ln \left (\arcsinh \left (a x \right )\right )-\hyperbolicCosineIntegral \left (2 \arcsinh \left (a x \right )\right )}{2 a^{3}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/arcsinh(a*x)/(a^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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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)/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^2/(sqrt(a^2*x^2 + 1)*arcsinh(a*x)), 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)/(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(x^2/(sqrt(a^2*x^2 + 1)*arcsinh(a*x)), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/asinh(a*x)/(a**2*x**2+1)**(1/2),x)

[Out]

Integral(x**2/(sqrt(a**2*x**2 + 1)*asinh(a*x)), 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)/(a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^2/(sqrt(a^2*x^2 + 1)*arcsinh(a*x)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {x^2}{\mathrm {asinh}\left (a\,x\right )\,\sqrt {a^2\,x^2+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(asinh(a*x)*(a^2*x^2 + 1)^(1/2)),x)

[Out]

int(x^2/(asinh(a*x)*(a^2*x^2 + 1)^(1/2)), x)

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