3.13 \(\int x^3 \sinh ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=96 \[ -\frac{3 x^2}{32 a^2}-\frac{x^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{8 a}+\frac{3 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{16 a^3}-\frac{3 \sinh ^{-1}(a x)^2}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^2+\frac{x^4}{32} \]

[Out]

(-3*x^2)/(32*a^2) + x^4/32 + (3*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(16*a^3) - (x^3*Sqrt[1 + a^2*x^2]*ArcSinh[a*
x])/(8*a) - (3*ArcSinh[a*x]^2)/(32*a^4) + (x^4*ArcSinh[a*x]^2)/4

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Rubi [A]  time = 0.165311, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5661, 5758, 5675, 30} \[ -\frac{3 x^2}{32 a^2}-\frac{x^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{8 a}+\frac{3 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{16 a^3}-\frac{3 \sinh ^{-1}(a x)^2}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^2+\frac{x^4}{32} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSinh[a*x]^2,x]

[Out]

(-3*x^2)/(32*a^2) + x^4/32 + (3*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(16*a^3) - (x^3*Sqrt[1 + a^2*x^2]*ArcSinh[a*
x])/(8*a) - (3*ArcSinh[a*x]^2)/(32*a^4) + (x^4*ArcSinh[a*x]^2)/4

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^3 \sinh ^{-1}(a x)^2 \, dx &=\frac{1}{4} x^4 \sinh ^{-1}(a x)^2-\frac{1}{2} a \int \frac{x^4 \sinh ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{8 a}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^2+\frac{\int x^3 \, dx}{8}+\frac{3 \int \frac{x^2 \sinh ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{8 a}\\ &=\frac{x^4}{32}+\frac{3 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{16 a^3}-\frac{x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{8 a}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^2-\frac{3 \int \frac{\sinh ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{16 a^3}-\frac{3 \int x \, dx}{16 a^2}\\ &=-\frac{3 x^2}{32 a^2}+\frac{x^4}{32}+\frac{3 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{16 a^3}-\frac{x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{8 a}-\frac{3 \sinh ^{-1}(a x)^2}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^2\\ \end{align*}

Mathematica [A]  time = 0.0444278, size = 72, normalized size = 0.75 \[ \frac{a^2 x^2 \left (a^2 x^2-3\right )-2 a x \sqrt{a^2 x^2+1} \left (2 a^2 x^2-3\right ) \sinh ^{-1}(a x)+\left (8 a^4 x^4-3\right ) \sinh ^{-1}(a x)^2}{32 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSinh[a*x]^2,x]

[Out]

(a^2*x^2*(-3 + a^2*x^2) - 2*a*x*Sqrt[1 + a^2*x^2]*(-3 + 2*a^2*x^2)*ArcSinh[a*x] + (-3 + 8*a^4*x^4)*ArcSinh[a*x
]^2)/(32*a^4)

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Maple [A]  time = 0.035, size = 118, normalized size = 1.2 \begin{align*}{\frac{1}{{a}^{4}} \left ({\frac{{a}^{2}{x}^{2} \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2} \left ({a}^{2}{x}^{2}+1 \right ) }{4}}-{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2} \left ({a}^{2}{x}^{2}+1 \right ) }{4}}-{\frac{{\it Arcsinh} \left ( ax \right ) ax}{8} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{\it Arcsinh} \left ( ax \right ) ax}{16}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{5\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}{32}}+{\frac{{a}^{2}{x}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }{32}}-{\frac{{a}^{2}{x}^{2}}{8}}-{\frac{1}{8}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsinh(a*x)^2,x)

[Out]

1/a^4*(1/4*a^2*x^2*arcsinh(a*x)^2*(a^2*x^2+1)-1/4*arcsinh(a*x)^2*(a^2*x^2+1)-1/8*arcsinh(a*x)*a*x*(a^2*x^2+1)^
(3/2)+5/16*arcsinh(a*x)*(a^2*x^2+1)^(1/2)*a*x+5/32*arcsinh(a*x)^2+1/32*a^2*x^2*(a^2*x^2+1)-1/8*a^2*x^2-1/8)

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Maxima [A]  time = 1.269, size = 173, normalized size = 1.8 \begin{align*} \frac{1}{4} \, x^{4} \operatorname{arsinh}\left (a x\right )^{2} + \frac{1}{32} \,{\left (\frac{x^{4}}{a^{2}} - \frac{3 \, x^{2}}{a^{4}} + \frac{3 \, \log \left (\frac{a^{2} x}{\sqrt{a^{2}}} + \sqrt{a^{2} x^{2} + 1}\right )^{2}}{a^{6}}\right )} a^{2} - \frac{1}{16} \,{\left (\frac{2 \, \sqrt{a^{2} x^{2} + 1} x^{3}}{a^{2}} - \frac{3 \, \sqrt{a^{2} x^{2} + 1} x}{a^{4}} + \frac{3 \, \operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}} a^{4}}\right )} a \operatorname{arsinh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsinh(a*x)^2,x, algorithm="maxima")

[Out]

1/4*x^4*arcsinh(a*x)^2 + 1/32*(x^4/a^2 - 3*x^2/a^4 + 3*log(a^2*x/sqrt(a^2) + sqrt(a^2*x^2 + 1))^2/a^6)*a^2 - 1
/16*(2*sqrt(a^2*x^2 + 1)*x^3/a^2 - 3*sqrt(a^2*x^2 + 1)*x/a^4 + 3*arcsinh(a^2*x/sqrt(a^2))/(sqrt(a^2)*a^4))*a*a
rcsinh(a*x)

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Fricas [A]  time = 1.82037, size = 205, normalized size = 2.14 \begin{align*} \frac{a^{4} x^{4} - 3 \, a^{2} x^{2} +{\left (8 \, a^{4} x^{4} - 3\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2} - 2 \,{\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \sqrt{a^{2} x^{2} + 1} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )}{32 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

1/32*(a^4*x^4 - 3*a^2*x^2 + (8*a^4*x^4 - 3)*log(a*x + sqrt(a^2*x^2 + 1))^2 - 2*(2*a^3*x^3 - 3*a*x)*sqrt(a^2*x^
2 + 1)*log(a*x + sqrt(a^2*x^2 + 1)))/a^4

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Sympy [A]  time = 2.28006, size = 90, normalized size = 0.94 \begin{align*} \begin{cases} \frac{x^{4} \operatorname{asinh}^{2}{\left (a x \right )}}{4} + \frac{x^{4}}{32} - \frac{x^{3} \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}{\left (a x \right )}}{8 a} - \frac{3 x^{2}}{32 a^{2}} + \frac{3 x \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}{\left (a x \right )}}{16 a^{3}} - \frac{3 \operatorname{asinh}^{2}{\left (a x \right )}}{32 a^{4}} & \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**3*asinh(a*x)**2,x)

[Out]

Piecewise((x**4*asinh(a*x)**2/4 + x**4/32 - x**3*sqrt(a**2*x**2 + 1)*asinh(a*x)/(8*a) - 3*x**2/(32*a**2) + 3*x
*sqrt(a**2*x**2 + 1)*asinh(a*x)/(16*a**3) - 3*asinh(a*x)**2/(32*a**4), Ne(a, 0)), (0, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{arsinh}\left (a x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsinh(a*x)^2,x, algorithm="giac")

[Out]

integrate(x^3*arcsinh(a*x)^2, x)