3.15 \(\int x \sinh ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=59 \[ -\frac{x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{2 a}+\frac{\sinh ^{-1}(a x)^2}{4 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^2+\frac{x^2}{4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0918002, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5661, 5758, 5675, 30} \[ -\frac{x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{2 a}+\frac{\sinh ^{-1}(a x)^2}{4 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^2+\frac{x^2}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -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]

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 30

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

Rubi steps

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

Mathematica [A]  time = 0.0282382, size = 53, normalized size = 0.9 \[ \frac{a^2 x^2-2 a x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)+\left (2 a^2 x^2+1\right ) \sinh ^{-1}(a x)^2}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.026, size = 59, normalized size = 1. \begin{align*}{\frac{1}{{a}^{2}} \left ({\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2} \left ({a}^{2}{x}^{2}+1 \right ) }{2}}-{\frac{{\it Arcsinh} \left ( ax \right ) ax}{2}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}{4}}+{\frac{{a}^{2}{x}^{2}}{4}}+{\frac{1}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.12721, size = 135, normalized size = 2.29 \begin{align*} \frac{1}{2} \, x^{2} \operatorname{arsinh}\left (a x\right )^{2} + \frac{1}{4} \, a^{2}{\left (\frac{x^{2}}{a^{2}} - \frac{\log \left (\frac{a^{2} x}{\sqrt{a^{2}}} + \sqrt{a^{2} x^{2} + 1}\right )^{2}}{a^{4}}\right )} - \frac{1}{2} \, a{\left (\frac{\sqrt{a^{2} x^{2} + 1} x}{a^{2}} - \frac{\operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}} a^{2}}\right )} \operatorname{arsinh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.00562, size = 166, normalized size = 2.81 \begin{align*} \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 ) +{\left (2 \, a^{2} x^{2} + 1\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2}}{4 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Sympy [A]  time = 0.541131, size = 51, normalized size = 0.86 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{asinh}^{2}{\left (a x \right )}}{2} + \frac{x^{2}}{4} - \frac{x \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}{\left (a x \right )}}{2 a} + \frac{\operatorname{asinh}^{2}{\left (a x \right )}}{4 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,x)

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{arsinh}\left (a x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsinh(a*x)^2, x)