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

Optimal. Leaf size=101 \[ \frac{\left (a x^2+1\right ) \sqrt{\frac{a^2 x^4+1}{\left (a x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\sqrt{a} x\right ),\frac{1}{2}\right )}{9 a^{3/2} \sqrt{a^2 x^4+1}}-\frac{2 x \sqrt{a^2 x^4+1}}{9 a}+\frac{1}{3} x^3 \sinh ^{-1}\left (a x^2\right ) \]

[Out]

(-2*x*Sqrt[1 + a^2*x^4])/(9*a) + (x^3*ArcSinh[a*x^2])/3 + ((1 + a*x^2)*Sqrt[(1 + a^2*x^4)/(1 + a*x^2)^2]*Ellip
ticF[2*ArcTan[Sqrt[a]*x], 1/2])/(9*a^(3/2)*Sqrt[1 + a^2*x^4])

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Rubi [A]  time = 0.043684, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5902, 12, 321, 220} \[ -\frac{2 x \sqrt{a^2 x^4+1}}{9 a}+\frac{\left (a x^2+1\right ) \sqrt{\frac{a^2 x^4+1}{\left (a x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt{a} x\right )|\frac{1}{2}\right )}{9 a^{3/2} \sqrt{a^2 x^4+1}}+\frac{1}{3} x^3 \sinh ^{-1}\left (a x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x*Sqrt[1 + a^2*x^4])/(9*a) + (x^3*ArcSinh[a*x^2])/3 + ((1 + a*x^2)*Sqrt[(1 + a^2*x^4)/(1 + a*x^2)^2]*Ellip
ticF[2*ArcTan[Sqrt[a]*x], 1/2])/(9*a^(3/2)*Sqrt[1 + a^2*x^4])

Rule 5902

Int[((a_.) + ArcSinh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSin
h[u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/Sqrt[1 + u^2],
x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)
^(m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

\begin{align*} \int x^2 \sinh ^{-1}\left (a x^2\right ) \, dx &=\frac{1}{3} x^3 \sinh ^{-1}\left (a x^2\right )-\frac{1}{3} \int \frac{2 a x^4}{\sqrt{1+a^2 x^4}} \, dx\\ &=\frac{1}{3} x^3 \sinh ^{-1}\left (a x^2\right )-\frac{1}{3} (2 a) \int \frac{x^4}{\sqrt{1+a^2 x^4}} \, dx\\ &=-\frac{2 x \sqrt{1+a^2 x^4}}{9 a}+\frac{1}{3} x^3 \sinh ^{-1}\left (a x^2\right )+\frac{2 \int \frac{1}{\sqrt{1+a^2 x^4}} \, dx}{9 a}\\ &=-\frac{2 x \sqrt{1+a^2 x^4}}{9 a}+\frac{1}{3} x^3 \sinh ^{-1}\left (a x^2\right )+\frac{\left (1+a x^2\right ) \sqrt{\frac{1+a^2 x^4}{\left (1+a x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt{a} x\right )|\frac{1}{2}\right )}{9 a^{3/2} \sqrt{1+a^2 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.120905, size = 75, normalized size = 0.74 \[ \frac{1}{9} \left (-\frac{2 \sqrt{i a} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{i a} x\right ),-1\right )}{a^2}-\frac{2 \left (a^2 x^5+x\right )}{a \sqrt{a^2 x^4+1}}+3 x^3 \sinh ^{-1}\left (a x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*(x + a^2*x^5))/(a*Sqrt[1 + a^2*x^4]) + 3*x^3*ArcSinh[a*x^2] - (2*Sqrt[I*a]*EllipticF[I*ArcSinh[Sqrt[I*a]*
x], -1])/a^2)/9

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Maple [C]  time = 0.016, size = 89, normalized size = 0.9 \begin{align*}{\frac{{x}^{3}{\it Arcsinh} \left ( a{x}^{2} \right ) }{3}}-{\frac{2\,a}{3} \left ({\frac{x}{3\,{a}^{2}}\sqrt{{a}^{2}{x}^{4}+1}}-{\frac{1}{3\,{a}^{2}}\sqrt{1-ia{x}^{2}}\sqrt{1+ia{x}^{2}}{\it EllipticF} \left ( x\sqrt{ia},i \right ){\frac{1}{\sqrt{ia}}}{\frac{1}{\sqrt{{a}^{2}{x}^{4}+1}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \, x^{3} \log \left (a x^{2} + \sqrt{a^{2} x^{4} + 1}\right ) - \frac{2}{9} \, x^{3} - 2 \, a \int \frac{x^{4}}{3 \,{\left (a^{3} x^{6} + a x^{2} +{\left (a^{2} x^{4} + 1\right )}^{\frac{3}{2}}\right )}}\,{d x} - \frac{i \, \sqrt{2}{\left (\log \left (\frac{i \, \sqrt{2}{\left (2 \, \sqrt{a^{2}} x + \sqrt{2}{\left (a^{2}\right )}^{\frac{1}{4}}\right )}}{2 \,{\left (a^{2}\right )}^{\frac{1}{4}}} + 1\right ) - \log \left (-\frac{i \, \sqrt{2}{\left (2 \, \sqrt{a^{2}} x + \sqrt{2}{\left (a^{2}\right )}^{\frac{1}{4}}\right )}}{2 \,{\left (a^{2}\right )}^{\frac{1}{4}}} + 1\right )\right )}}{12 \,{\left (a^{2}\right )}^{\frac{3}{4}}} - \frac{i \, \sqrt{2}{\left (\log \left (\frac{i \, \sqrt{2}{\left (2 \, \sqrt{a^{2}} x - \sqrt{2}{\left (a^{2}\right )}^{\frac{1}{4}}\right )}}{2 \,{\left (a^{2}\right )}^{\frac{1}{4}}} + 1\right ) - \log \left (-\frac{i \, \sqrt{2}{\left (2 \, \sqrt{a^{2}} x - \sqrt{2}{\left (a^{2}\right )}^{\frac{1}{4}}\right )}}{2 \,{\left (a^{2}\right )}^{\frac{1}{4}}} + 1\right )\right )}}{12 \,{\left (a^{2}\right )}^{\frac{3}{4}}} - \frac{\sqrt{2} \log \left (\sqrt{a^{2}} x^{2} + \sqrt{2}{\left (a^{2}\right )}^{\frac{1}{4}} x + 1\right )}{12 \,{\left (a^{2}\right )}^{\frac{3}{4}}} + \frac{\sqrt{2} \log \left (\sqrt{a^{2}} x^{2} - \sqrt{2}{\left (a^{2}\right )}^{\frac{1}{4}} x + 1\right )}{12 \,{\left (a^{2}\right )}^{\frac{3}{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*log(a*x^2 + sqrt(a^2*x^4 + 1)) - 2/9*x^3 - 2*a*integrate(1/3*x^4/(a^3*x^6 + a*x^2 + (a^2*x^4 + 1)^(3/2
)), x) - 1/12*I*sqrt(2)*(log(1/2*I*sqrt(2)*(2*sqrt(a^2)*x + sqrt(2)*(a^2)^(1/4))/(a^2)^(1/4) + 1) - log(-1/2*I
*sqrt(2)*(2*sqrt(a^2)*x + sqrt(2)*(a^2)^(1/4))/(a^2)^(1/4) + 1))/(a^2)^(3/4) - 1/12*I*sqrt(2)*(log(1/2*I*sqrt(
2)*(2*sqrt(a^2)*x - sqrt(2)*(a^2)^(1/4))/(a^2)^(1/4) + 1) - log(-1/2*I*sqrt(2)*(2*sqrt(a^2)*x - sqrt(2)*(a^2)^
(1/4))/(a^2)^(1/4) + 1))/(a^2)^(3/4) - 1/12*sqrt(2)*log(sqrt(a^2)*x^2 + sqrt(2)*(a^2)^(1/4)*x + 1)/(a^2)^(3/4)
+ 1/12*sqrt(2)*log(sqrt(a^2)*x^2 - sqrt(2)*(a^2)^(1/4)*x + 1)/(a^2)^(3/4)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} \operatorname{arsinh}\left (a x^{2}\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*arcsinh(a*x^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{asinh}{\left (a x^{2} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*asinh(a*x**2),x)

[Out]

Integral(x**2*asinh(a*x**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arcsinh(a*x^2), x)