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

Optimal. Leaf size=97 \[ -\frac{3 x \sqrt{a^2 x^2+1}}{8 a}-\frac{3 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2}{4 a}+\frac{\sinh ^{-1}(a x)^3}{4 a^2}+\frac{3 \sinh ^{-1}(a x)}{8 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^3+\frac{3}{4} x^2 \sinh ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.152615, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {5661, 5758, 5675, 321, 215} \[ -\frac{3 x \sqrt{a^2 x^2+1}}{8 a}-\frac{3 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2}{4 a}+\frac{\sinh ^{-1}(a x)^3}{4 a^2}+\frac{3 \sinh ^{-1}(a x)}{8 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^3+\frac{3}{4} x^2 \sinh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x*Sqrt[1 + a^2*x^2])/(8*a) + (3*ArcSinh[a*x])/(8*a^2) + (3*x^2*ArcSinh[a*x])/4 - (3*x*Sqrt[1 + a^2*x^2]*Ar
cSinh[a*x]^2)/(4*a) + ArcSinh[a*x]^3/(4*a^2) + (x^2*ArcSinh[a*x]^3)/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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

Mathematica [A]  time = 0.0435968, size = 80, normalized size = 0.82 \[ \frac{-3 a x \sqrt{a^2 x^2+1}+\left (4 a^2 x^2+2\right ) \sinh ^{-1}(a x)^3-6 a x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2+\left (6 a^2 x^2+3\right ) \sinh ^{-1}(a x)}{8 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*x*Sqrt[1 + a^2*x^2] + (3 + 6*a^2*x^2)*ArcSinh[a*x] - 6*a*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^2 + (2 + 4*a^2
*x^2)*ArcSinh[a*x]^3)/(8*a^2)

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Maple [A]  time = 0.025, size = 88, normalized size = 0.9 \begin{align*}{\frac{1}{{a}^{2}} \left ({\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3} \left ({a}^{2}{x}^{2}+1 \right ) }{2}}-{\frac{3\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}ax}{4}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}}{4}}+{\frac{ \left ( 3\,{a}^{2}{x}^{2}+3 \right ){\it Arcsinh} \left ( ax \right ) }{4}}-{\frac{3\,ax}{8}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{3\,{\it Arcsinh} \left ( ax \right ) }{8}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, x^{2} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{3} - \int \frac{3 \,{\left (a^{3} x^{4} + \sqrt{a^{2} x^{2} + 1} a^{2} x^{3} + a x^{2}\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2}}{2 \,{\left (a^{3} x^{3} + a x +{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.08086, size = 261, normalized size = 2.69 \begin{align*} -\frac{6 \, \sqrt{a^{2} x^{2} + 1} a x \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2} - 2 \,{\left (2 \, a^{2} x^{2} + 1\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{3} + 3 \, \sqrt{a^{2} x^{2} + 1} a x - 3 \,{\left (2 \, a^{2} x^{2} + 1\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )}{8 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(6*sqrt(a^2*x^2 + 1)*a*x*log(a*x + sqrt(a^2*x^2 + 1))^2 - 2*(2*a^2*x^2 + 1)*log(a*x + sqrt(a^2*x^2 + 1))^
3 + 3*sqrt(a^2*x^2 + 1)*a*x - 3*(2*a^2*x^2 + 1)*log(a*x + sqrt(a^2*x^2 + 1)))/a^2

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Sympy [A]  time = 1.11925, size = 92, normalized size = 0.95 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{asinh}^{3}{\left (a x \right )}}{2} + \frac{3 x^{2} \operatorname{asinh}{\left (a x \right )}}{4} - \frac{3 x \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}^{2}{\left (a x \right )}}{4 a} - \frac{3 x \sqrt{a^{2} x^{2} + 1}}{8 a} + \frac{\operatorname{asinh}^{3}{\left (a x \right )}}{4 a^{2}} + \frac{3 \operatorname{asinh}{\left (a x \right )}}{8 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)**3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsinh(a*x)^3, x)