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

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

[Out]

1/3*x^3*arcsinh(a*x^2)-2/9*x*(a^2*x^4+1)^(1/2)/a+1/9*(a*x^2+1)*(cos(2*arctan(x*a^(1/2)))^2)^(1/2)/cos(2*arctan
(x*a^(1/2)))*EllipticF(sin(2*arctan(x*a^(1/2))),1/2*2^(1/2))*((a^2*x^4+1)/(a*x^2+1)^2)^(1/2)/a^(3/2)/(a^2*x^4+
1)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 12

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

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 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]

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*}

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Mathematica [C]  time = 0.13, size = 75, normalized size = 0.74 \[ \frac {1}{9} \left (-\frac {2 \left (a^2 x^5+x\right )}{a \sqrt {a^2 x^4+1}}-\frac {2 \sqrt {i a} F\left (\left .i \sinh ^{-1}\left (\sqrt {i a} x\right )\right |-1\right )}{a^2}+3 x^3 \sinh ^{-1}\left (a x^2\right )\right ) \]

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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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \operatorname {arsinh}\left (a x^{2}\right ), x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {arsinh}\left (a x^{2}\right )\,{d x} \]

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)

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maple [C]  time = 0.01, size = 89, normalized size = 0.88 \[ \frac {x^{3} \arcsinh \left (a \,x^{2}\right )}{3}-\frac {2 a \left (\frac {x \sqrt {a^{2} x^{4}+1}}{3 a^{2}}-\frac {\sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, \EllipticF \left (x \sqrt {i a}, i\right )}{3 a^{2} \sqrt {i a}\, \sqrt {a^{2} x^{4}+1}}\right )}{3} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, x^{3} \log \left (a x^{2} + \sqrt {a^{2} x^{4} + 1}\right ) - \frac {2}{9} \, x^{3} - 2 \, a \int \frac {x^{4}}{3 \, {\left (a^{3} x^{6} + a x^{2} + {\left (a^{2} x^{4} + 1\right )}^{\frac {3}{2}}\right )}}\,{d x} - \frac {i \, \sqrt {2} {\left (\log \left (\frac {i \, \sqrt {2} {\left (2 \, a x + \sqrt {2} \sqrt {a}\right )}}{2 \, \sqrt {a}} + 1\right ) - \log \left (-\frac {i \, \sqrt {2} {\left (2 \, a x + \sqrt {2} \sqrt {a}\right )}}{2 \, \sqrt {a}} + 1\right )\right )}}{12 \, a^{\frac {3}{2}}} - \frac {i \, \sqrt {2} {\left (\log \left (\frac {i \, \sqrt {2} {\left (2 \, a x - \sqrt {2} \sqrt {a}\right )}}{2 \, \sqrt {a}} + 1\right ) - \log \left (-\frac {i \, \sqrt {2} {\left (2 \, a x - \sqrt {2} \sqrt {a}\right )}}{2 \, \sqrt {a}} + 1\right )\right )}}{12 \, a^{\frac {3}{2}}} - \frac {\sqrt {2} \log \left (a x^{2} + \sqrt {2} \sqrt {a} x + 1\right )}{12 \, a^{\frac {3}{2}}} + \frac {\sqrt {2} \log \left (a x^{2} - \sqrt {2} \sqrt {a} x + 1\right )}{12 \, a^{\frac {3}{2}}} \]

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*a*x + sqrt(2)*sqrt(a))/sqrt(a) + 1) - log(-1/2*I*sqrt(2)*(2*a*x
+ sqrt(2)*sqrt(a))/sqrt(a) + 1))/a^(3/2) - 1/12*I*sqrt(2)*(log(1/2*I*sqrt(2)*(2*a*x - sqrt(2)*sqrt(a))/sqrt(a)
+ 1) - log(-1/2*I*sqrt(2)*(2*a*x - sqrt(2)*sqrt(a))/sqrt(a) + 1))/a^(3/2) - 1/12*sqrt(2)*log(a*x^2 + sqrt(2)*
sqrt(a)*x + 1)/a^(3/2) + 1/12*sqrt(2)*log(a*x^2 - sqrt(2)*sqrt(a)*x + 1)/a^(3/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {asinh}\left (a\,x^2\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*asinh(a*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {asinh}{\left (a x^{2} \right )}\, dx \]

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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