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

Optimal. Leaf size=110 \[ -\frac{x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^3}{a}-\frac{3 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{2 a}+\frac{\sinh ^{-1}(a x)^4}{4 a^2}+\frac{3 \sinh ^{-1}(a x)^2}{4 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^4+\frac{3}{2} x^2 \sinh ^{-1}(a x)^2+\frac{3 x^2}{4} \]

[Out]

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

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Rubi [A]  time = 0.239413, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5661, 5758, 5675, 30} \[ -\frac{x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^3}{a}-\frac{3 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{2 a}+\frac{\sinh ^{-1}(a x)^4}{4 a^2}+\frac{3 \sinh ^{-1}(a x)^2}{4 a^2}+\frac{1}{2} x^2 \sinh ^{-1}(a x)^4+\frac{3}{2} x^2 \sinh ^{-1}(a x)^2+\frac{3 x^2}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0454885, size = 94, normalized size = 0.85 \[ \frac{3 a^2 x^2+\left (2 a^2 x^2+1\right ) \sinh ^{-1}(a x)^4-4 a x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^3+\left (6 a^2 x^2+3\right ) \sinh ^{-1}(a x)^2-6 a x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.026, size = 105, normalized size = 1. \begin{align*}{\frac{1}{{a}^{2}} \left ({\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{4} \left ({a}^{2}{x}^{2}+1 \right ) }{2}}- \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}ax\sqrt{{a}^{2}{x}^{2}+1}-{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{4}}{4}}+{\frac{3\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2} \left ({a}^{2}{x}^{2}+1 \right ) }{2}}-{\frac{3\,{\it Arcsinh} \left ( ax \right ) ax}{2}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{3\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}{4}}+{\frac{3\,{a}^{2}{x}^{2}}{4}}+{\frac{3}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )^{4} - \int \frac{2 \,{\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 )^{3}}{a^{3} x^{3} + a x +{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.05094, size = 316, normalized size = 2.87 \begin{align*} -\frac{4 \, \sqrt{a^{2} x^{2} + 1} a x \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{3} -{\left (2 \, a^{2} x^{2} + 1\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{4} - 3 \, a^{2} x^{2} + 6 \, \sqrt{a^{2} x^{2} + 1} a x \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right ) - 3 \,{\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)^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.14062, size = 104, normalized size = 0.95 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{asinh}^{4}{\left (a x \right )}}{2} + \frac{3 x^{2} \operatorname{asinh}^{2}{\left (a x \right )}}{2} + \frac{3 x^{2}}{4} - \frac{x \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}^{3}{\left (a x \right )}}{a} - \frac{3 x \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}{\left (a x \right )}}{2 a} + \frac{\operatorname{asinh}^{4}{\left (a x \right )}}{4 a^{2}} + \frac{3 \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)**4,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{arsinh}\left (a x\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsinh(a*x)^4, x)