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

Optimal. Leaf size=67 \[ -\frac{x^3 \sqrt{a^2 x^2+1}}{16 a}+\frac{3 x \sqrt{a^2 x^2+1}}{32 a^3}-\frac{3 \sinh ^{-1}(a x)}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0264488, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {5661, 321, 215} \[ -\frac{x^3 \sqrt{a^2 x^2+1}}{16 a}+\frac{3 x \sqrt{a^2 x^2+1}}{32 a^3}-\frac{3 \sinh ^{-1}(a x)}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.016575, size = 49, normalized size = 0.73 \[ \frac{a x \sqrt{a^2 x^2+1} \left (3-2 a^2 x^2\right )+\left (8 a^4 x^4-3\right ) \sinh ^{-1}(a x)}{32 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 58, normalized size = 0.9 \begin{align*}{\frac{1}{{a}^{4}} \left ({\frac{{a}^{4}{x}^{4}{\it Arcsinh} \left ( ax \right ) }{4}}-{\frac{{a}^{3}{x}^{3}}{16}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{3\,ax}{32}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{3\,{\it Arcsinh} \left ( ax \right ) }{32}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.17873, size = 96, normalized size = 1.43 \begin{align*} \frac{1}{4} \, x^{4} \operatorname{arsinh}\left (a x\right ) - \frac{1}{32} \,{\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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arcsinh(a*x) - 1/32*(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

________________________________________________________________________________________

Fricas [A]  time = 1.86331, size = 131, normalized size = 1.96 \begin{align*} \frac{{\left (8 \, a^{4} x^{4} - 3\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right ) -{\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \sqrt{a^{2} x^{2} + 1}}{32 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 1.1598, size = 61, normalized size = 0.91 \begin{align*} \begin{cases} \frac{x^{4} \operatorname{asinh}{\left (a x \right )}}{4} - \frac{x^{3} \sqrt{a^{2} x^{2} + 1}}{16 a} + \frac{3 x \sqrt{a^{2} x^{2} + 1}}{32 a^{3}} - \frac{3 \operatorname{asinh}{\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),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.37658, size = 108, normalized size = 1.61 \begin{align*} \frac{1}{4} \, x^{4} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right ) - \frac{1}{32} \,{\left (\sqrt{a^{2} x^{2} + 1} x{\left (\frac{2 \, x^{2}}{a^{2}} - \frac{3}{a^{4}}\right )} - \frac{3 \, \log \left (-x{\left | a \right |} + \sqrt{a^{2} x^{2} + 1}\right )}{a^{4}{\left | a \right |}}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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