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

Optimal. Leaf size=52 \[ \frac{2 \sqrt{a^2 x^2+1}}{a^2}+\frac{\sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2}{a^2}-\frac{2 x \sinh ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.0779984, antiderivative size = 52, normalized size of antiderivative = 1., 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 = {5717, 5653, 261} \[ \frac{2 \sqrt{a^2 x^2+1}}{a^2}+\frac{\sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2}{a^2}-\frac{2 x \sinh ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5717

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

Rule 5653

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.0328073, size = 48, normalized size = 0.92 \[ \frac{2 \sqrt{a^2 x^2+1}+\sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2-2 a x \sinh ^{-1}(a x)}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 64, normalized size = 1.2 \begin{align*}{\frac{1}{{a}^{2}} \left ( \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}{a}^{2}{x}^{2}+ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}-2\,{\it Arcsinh} \left ( ax \right ) \sqrt{{a}^{2}{x}^{2}+1}ax+2\,{a}^{2}{x}^{2}+2 \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.21249, size = 65, normalized size = 1.25 \begin{align*} \frac{\sqrt{a^{2} x^{2} + 1} \operatorname{arsinh}\left (a x\right )^{2}}{a^{2}} - \frac{2 \,{\left (a x \operatorname{arsinh}\left (a x\right ) - \sqrt{a^{2} x^{2} + 1}\right )}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 3.11211, size = 157, normalized size = 3.02 \begin{align*} -\frac{2 \, a x \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right ) - \sqrt{a^{2} x^{2} + 1} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2} - 2 \, \sqrt{a^{2} x^{2} + 1}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.986788, size = 49, normalized size = 0.94 \begin{align*} \begin{cases} - \frac{2 x \operatorname{asinh}{\left (a x \right )}}{a} + \frac{\sqrt{a^{2} x^{2} + 1} \operatorname{asinh}^{2}{\left (a x \right )}}{a^{2}} + \frac{2 \sqrt{a^{2} x^{2} + 1}}{a^{2}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*x*asinh(a*x)/a + sqrt(a**2*x**2 + 1)*asinh(a*x)**2/a**2 + 2*sqrt(a**2*x**2 + 1)/a**2, Ne(a, 0)),
(0, True))

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Giac [A]  time = 1.3837, size = 100, normalized size = 1.92 \begin{align*} \frac{\sqrt{a^{2} x^{2} + 1} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2}}{a^{2}} - \frac{2 \,{\left (x \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right ) - \frac{\sqrt{a^{2} x^{2} + 1}}{a}\right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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