3.12 \(\int x^4 \sinh ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=117 \[ -\frac{8 x^3}{225 a^2}-\frac{2 x^4 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{25 a}+\frac{8 x^2 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{75 a^3}-\frac{16 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{75 a^5}+\frac{16 x}{75 a^4}+\frac{1}{5} x^5 \sinh ^{-1}(a x)^2+\frac{2 x^5}{125} \]

[Out]

(16*x)/(75*a^4) - (8*x^3)/(225*a^2) + (2*x^5)/125 - (16*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(75*a^5) + (8*x^2*Sqrt
[1 + a^2*x^2]*ArcSinh[a*x])/(75*a^3) - (2*x^4*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(25*a) + (x^5*ArcSinh[a*x]^2)/5

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Rubi [A]  time = 0.188201, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5661, 5758, 5717, 8, 30} \[ -\frac{8 x^3}{225 a^2}-\frac{2 x^4 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{25 a}+\frac{8 x^2 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{75 a^3}-\frac{16 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{75 a^5}+\frac{16 x}{75 a^4}+\frac{1}{5} x^5 \sinh ^{-1}(a x)^2+\frac{2 x^5}{125} \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcSinh[a*x]^2,x]

[Out]

(16*x)/(75*a^4) - (8*x^3)/(225*a^2) + (2*x^5)/125 - (16*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(75*a^5) + (8*x^2*Sqrt
[1 + a^2*x^2]*ArcSinh[a*x])/(75*a^3) - (2*x^4*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(25*a) + (x^5*ArcSinh[a*x]^2)/5

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 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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^4 \sinh ^{-1}(a x)^2 \, dx &=\frac{1}{5} x^5 \sinh ^{-1}(a x)^2-\frac{1}{5} (2 a) \int \frac{x^5 \sinh ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{2 x^4 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{25 a}+\frac{1}{5} x^5 \sinh ^{-1}(a x)^2+\frac{2 \int x^4 \, dx}{25}+\frac{8 \int \frac{x^3 \sinh ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{25 a}\\ &=\frac{2 x^5}{125}+\frac{8 x^2 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{75 a^3}-\frac{2 x^4 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{25 a}+\frac{1}{5} x^5 \sinh ^{-1}(a x)^2-\frac{16 \int \frac{x \sinh ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{75 a^3}-\frac{8 \int x^2 \, dx}{75 a^2}\\ &=-\frac{8 x^3}{225 a^2}+\frac{2 x^5}{125}-\frac{16 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{75 a^5}+\frac{8 x^2 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{75 a^3}-\frac{2 x^4 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{25 a}+\frac{1}{5} x^5 \sinh ^{-1}(a x)^2+\frac{16 \int 1 \, dx}{75 a^4}\\ &=\frac{16 x}{75 a^4}-\frac{8 x^3}{225 a^2}+\frac{2 x^5}{125}-\frac{16 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{75 a^5}+\frac{8 x^2 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{75 a^3}-\frac{2 x^4 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{25 a}+\frac{1}{5} x^5 \sinh ^{-1}(a x)^2\\ \end{align*}

Mathematica [A]  time = 0.0642595, size = 75, normalized size = 0.64 \[ \frac{-\frac{40 x^3}{a^2}-\frac{30 \sqrt{a^2 x^2+1} \left (3 a^4 x^4-4 a^2 x^2+8\right ) \sinh ^{-1}(a x)}{a^5}+\frac{240 x}{a^4}+225 x^5 \sinh ^{-1}(a x)^2+18 x^5}{1125} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*ArcSinh[a*x]^2,x]

[Out]

((240*x)/a^4 - (40*x^3)/a^2 + 18*x^5 - (30*Sqrt[1 + a^2*x^2]*(8 - 4*a^2*x^2 + 3*a^4*x^4)*ArcSinh[a*x])/a^5 + 2
25*x^5*ArcSinh[a*x]^2)/1125

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Maple [A]  time = 0.131, size = 153, normalized size = 1.3 \begin{align*}{\frac{1}{{a}^{5}} \left ({\frac{{a}^{3}{x}^{3} \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2} \left ({a}^{2}{x}^{2}+1 \right ) }{5}}-{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}ax \left ({a}^{2}{x}^{2}+1 \right ) }{5}}+{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}ax}{5}}-{\frac{2\,{\it Arcsinh} \left ( ax \right ){a}^{2}{x}^{2}}{25} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{14\,{\it Arcsinh} \left ( ax \right ){a}^{2}{x}^{2}}{75}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{16\,{\it Arcsinh} \left ( ax \right ) }{75}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{2\,ax \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}{125}}+{\frac{298\,ax}{1125}}-{\frac{76\,ax \left ({a}^{2}{x}^{2}+1 \right ) }{1125}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arcsinh(a*x)^2,x)

[Out]

1/a^5*(1/5*a^3*x^3*arcsinh(a*x)^2*(a^2*x^2+1)-1/5*arcsinh(a*x)^2*a*x*(a^2*x^2+1)+1/5*arcsinh(a*x)^2*a*x-2/25*a
rcsinh(a*x)*a^2*x^2*(a^2*x^2+1)^(3/2)+14/75*arcsinh(a*x)*a^2*x^2*(a^2*x^2+1)^(1/2)-16/75*arcsinh(a*x)*(a^2*x^2
+1)^(1/2)+2/125*a*x*(a^2*x^2+1)^2+298/1125*a*x-76/1125*a*x*(a^2*x^2+1))

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Maxima [A]  time = 1.15904, size = 134, normalized size = 1.15 \begin{align*} \frac{1}{5} \, x^{5} \operatorname{arsinh}\left (a x\right )^{2} - \frac{2}{75} \,{\left (\frac{3 \, \sqrt{a^{2} x^{2} + 1} x^{4}}{a^{2}} - \frac{4 \, \sqrt{a^{2} x^{2} + 1} x^{2}}{a^{4}} + \frac{8 \, \sqrt{a^{2} x^{2} + 1}}{a^{6}}\right )} a \operatorname{arsinh}\left (a x\right ) + \frac{2 \,{\left (9 \, a^{4} x^{5} - 20 \, a^{2} x^{3} + 120 \, x\right )}}{1125 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x)^2,x, algorithm="maxima")

[Out]

1/5*x^5*arcsinh(a*x)^2 - 2/75*(3*sqrt(a^2*x^2 + 1)*x^4/a^2 - 4*sqrt(a^2*x^2 + 1)*x^2/a^4 + 8*sqrt(a^2*x^2 + 1)
/a^6)*a*arcsinh(a*x) + 2/1125*(9*a^4*x^5 - 20*a^2*x^3 + 120*x)/a^4

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Fricas [A]  time = 1.7856, size = 234, normalized size = 2. \begin{align*} \frac{225 \, a^{5} x^{5} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2} + 18 \, a^{5} x^{5} - 40 \, a^{3} x^{3} - 30 \,{\left (3 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 8\right )} \sqrt{a^{2} x^{2} + 1} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right ) + 240 \, a x}{1125 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

1/1125*(225*a^5*x^5*log(a*x + sqrt(a^2*x^2 + 1))^2 + 18*a^5*x^5 - 40*a^3*x^3 - 30*(3*a^4*x^4 - 4*a^2*x^2 + 8)*
sqrt(a^2*x^2 + 1)*log(a*x + sqrt(a^2*x^2 + 1)) + 240*a*x)/a^5

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Sympy [A]  time = 4.13875, size = 114, normalized size = 0.97 \begin{align*} \begin{cases} \frac{x^{5} \operatorname{asinh}^{2}{\left (a x \right )}}{5} + \frac{2 x^{5}}{125} - \frac{2 x^{4} \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}{\left (a x \right )}}{25 a} - \frac{8 x^{3}}{225 a^{2}} + \frac{8 x^{2} \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}{\left (a x \right )}}{75 a^{3}} + \frac{16 x}{75 a^{4}} - \frac{16 \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}{\left (a x \right )}}{75 a^{5}} & \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**4*asinh(a*x)**2,x)

[Out]

Piecewise((x**5*asinh(a*x)**2/5 + 2*x**5/125 - 2*x**4*sqrt(a**2*x**2 + 1)*asinh(a*x)/(25*a) - 8*x**3/(225*a**2
) + 8*x**2*sqrt(a**2*x**2 + 1)*asinh(a*x)/(75*a**3) + 16*x/(75*a**4) - 16*sqrt(a**2*x**2 + 1)*asinh(a*x)/(75*a
**5), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.40166, size = 153, normalized size = 1.31 \begin{align*} \frac{1}{5} \, x^{5} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2} + \frac{2}{1125} \, a{\left (\frac{9 \, a^{4} x^{5} - 20 \, a^{2} x^{3} + 120 \, x}{a^{5}} - \frac{15 \,{\left (3 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{5}{2}} - 10 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}} + 15 \, \sqrt{a^{2} x^{2} + 1}\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )}{a^{6}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x)^2,x, algorithm="giac")

[Out]

1/5*x^5*log(a*x + sqrt(a^2*x^2 + 1))^2 + 2/1125*a*((9*a^4*x^5 - 20*a^2*x^3 + 120*x)/a^5 - 15*(3*(a^2*x^2 + 1)^
(5/2) - 10*(a^2*x^2 + 1)^(3/2) + 15*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 + 1))/a^6)