3.112 \(\int \frac{x^4 \sinh ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx\)

Optimal. Leaf size=86 \[ \frac{3 x^2}{16 a^3}+\frac{x^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{4 a^2}-\frac{3 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{8 a^4}+\frac{3 \sinh ^{-1}(a x)^2}{16 a^5}-\frac{x^4}{16 a} \]

[Out]

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

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Rubi [A]  time = 0.154876, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {5758, 5675, 30} \[ \frac{3 x^2}{16 a^3}+\frac{x^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{4 a^2}-\frac{3 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{8 a^4}+\frac{3 \sinh ^{-1}(a x)^2}{16 a^5}-\frac{x^4}{16 a} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*ArcSinh[a*x])/Sqrt[1 + a^2*x^2],x]

[Out]

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

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

Mathematica [A]  time = 0.046194, size = 63, normalized size = 0.73 \[ \frac{-a^4 x^4+3 a^2 x^2+2 a x \sqrt{a^2 x^2+1} \left (2 a^2 x^2-3\right ) \sinh ^{-1}(a x)+3 \sinh ^{-1}(a x)^2}{16 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*ArcSinh[a*x])/Sqrt[1 + a^2*x^2],x]

[Out]

(3*a^2*x^2 - a^4*x^4 + 2*a*x*Sqrt[1 + a^2*x^2]*(-3 + 2*a^2*x^2)*ArcSinh[a*x] + 3*ArcSinh[a*x]^2)/(16*a^5)

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Maple [A]  time = 0.038, size = 74, normalized size = 0.9 \begin{align*}{\frac{1}{16\,{a}^{5}} \left ( 4\,{\it Arcsinh} \left ( ax \right ) \sqrt{{a}^{2}{x}^{2}+1}{a}^{3}{x}^{3}-{x}^{4}{a}^{4}-6\,{\it Arcsinh} \left ( ax \right ) \sqrt{{a}^{2}{x}^{2}+1}ax+3\,{a}^{2}{x}^{2}+3\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}+4 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arcsinh(a*x)/(a^2*x^2+1)^(1/2),x)

[Out]

1/16*(4*arcsinh(a*x)*(a^2*x^2+1)^(1/2)*a^3*x^3-x^4*a^4-6*arcsinh(a*x)*(a^2*x^2+1)^(1/2)*a*x+3*a^2*x^2+3*arcsin
h(a*x)^2+4)/a^5

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Maxima [A]  time = 1.12058, size = 138, normalized size = 1.6 \begin{align*} -\frac{1}{16} \,{\left (\frac{x^{4}}{a^{2}} - \frac{3 \, x^{2}}{a^{4}} + \frac{3 \, \operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )^{2}}{a^{6}}\right )} a + \frac{1}{8} \,{\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 )} \operatorname{arsinh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x)/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-1/16*(x^4/a^2 - 3*x^2/a^4 + 3*arcsinh(a^2*x/sqrt(a^2))^2/a^6)*a + 1/8*(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))*arcsinh(a*x)

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Fricas [A]  time = 2.5808, size = 188, normalized size = 2.19 \begin{align*} -\frac{a^{4} x^{4} - 3 \, a^{2} x^{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 ) - 3 \, \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2}}{16 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x)/(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

-1/16*(a^4*x^4 - 3*a^2*x^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)) - 3*log(a*x
+ sqrt(a^2*x^2 + 1))^2)/a^5

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Sympy [A]  time = 2.97125, size = 82, normalized size = 0.95 \begin{align*} \begin{cases} - \frac{x^{4}}{16 a} + \frac{x^{3} \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}{\left (a x \right )}}{4 a^{2}} + \frac{3 x^{2}}{16 a^{3}} - \frac{3 x \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}{\left (a x \right )}}{8 a^{4}} + \frac{3 \operatorname{asinh}^{2}{\left (a x \right )}}{16 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)/(a**2*x**2+1)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x)/(a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^4*arcsinh(a*x)/sqrt(a^2*x^2 + 1), x)