3.179 \(\int \frac {x^2 \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx\)

Optimal. Leaf size=49 \[ -\frac {\sinh ^{-1}(a x)^2}{4 a^3}+\frac {x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{2 a^2}-\frac {x^2}{4 a} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5758, 5675, 30} \[ \frac {x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{2 a^2}-\frac {\sinh ^{-1}(a x)^2}{4 a^3}-\frac {x^2}{4 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]

Rule 5758

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(e*m), x] + (-Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)
^(m - 2)*(a + b*ArcSinh[c*x])^n)/Sqrt[d + e*x^2], x], x] - Dist[(b*f*n*Sqrt[1 + c^2*x^2])/(c*m*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}, x] && EqQ[e, c^2*d] &&
GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {x^2 \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx &=\frac {x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{2 a^2}-\frac {\int \frac {\sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{2 a^2}-\frac {\int x \, dx}{2 a}\\ &=-\frac {x^2}{4 a}+\frac {x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{2 a^2}-\frac {\sinh ^{-1}(a x)^2}{4 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 42, normalized size = 0.86 \[ -\frac {a^2 x^2-2 a x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)+\sinh ^{-1}(a x)^2}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 62, normalized size = 1.27 \[ -\frac {a^{2} x^{2} - 2 \, \sqrt {a^{2} x^{2} + 1} a x \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) + \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2}}{4 \, a^{3}} \]

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]

-1/4*(a^2*x^2 - 2*sqrt(a^2*x^2 + 1)*a*x*log(a*x + sqrt(a^2*x^2 + 1)) + log(a*x + sqrt(a^2*x^2 + 1))^2)/a^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \operatorname {arsinh}\left (a x\right )}{\sqrt {a^{2} x^{2} + 1}}\,{d x} \]

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*arcsinh(a*x)/sqrt(a^2*x^2 + 1), x)

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maple [A]  time = 0.00, size = 40, normalized size = 0.82 \[ -\frac {-2 \arcsinh \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a x +a^{2} x^{2}+\arcsinh \left (a x \right )^{2}+1}{4 a^{3}} \]

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)

[Out]

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

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maxima [A]  time = 0.54, size = 55, normalized size = 1.12 \[ -\frac {1}{4} \, a {\left (\frac {x^{2}}{a^{2}} - \frac {\operatorname {arsinh}\left (a x\right )^{2}}{a^{4}}\right )} + \frac {1}{2} \, {\left (\frac {\sqrt {a^{2} x^{2} + 1} x}{a^{2}} - \frac {\operatorname {arsinh}\left (a x\right )}{a^{3}}\right )} \operatorname {arsinh}\left (a x\right ) \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^2\,\mathrm {asinh}\left (a\,x\right )}{\sqrt {a^2\,x^2+1}} \,d x \]

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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sympy [A]  time = 0.75, size = 42, normalized size = 0.86 \[ \begin {cases} - \frac {x^{2}}{4 a} + \frac {x \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{2 a^{2}} - \frac {\operatorname {asinh}^{2}{\left (a x \right )}}{4 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

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]

Piecewise((-x**2/(4*a) + x*sqrt(a**2*x**2 + 1)*asinh(a*x)/(2*a**2) - asinh(a*x)**2/(4*a**3), Ne(a, 0)), (0, Tr
ue))

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