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

Optimal. Leaf size=75 \[ \frac{\sqrt{a} \left (a x^2+1\right ) \sqrt{\frac{a^2 x^4+1}{\left (a x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\sqrt{a} x\right ),\frac{1}{2}\right )}{\sqrt{a^2 x^4+1}}-\frac{\sinh ^{-1}\left (a x^2\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.0225441, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {5902, 12, 220} \[ \frac{\sqrt{a} \left (a x^2+1\right ) \sqrt{\frac{a^2 x^4+1}{\left (a x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt{a} x\right )|\frac{1}{2}\right )}{\sqrt{a^2 x^4+1}}-\frac{\sinh ^{-1}\left (a x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 0.0369385, size = 42, normalized size = 0.56 \[ -\frac{\sinh ^{-1}\left (a x^2\right )+2 \sqrt{i a} x \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{i a} x\right ),-1\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.006, size = 66, normalized size = 0.9 \begin{align*} -{\frac{{\it Arcsinh} \left ( a{x}^{2} \right ) }{x}}+2\,{\frac{a\sqrt{1-ia{x}^{2}}\sqrt{1+ia{x}^{2}}{\it EllipticF} \left ( x\sqrt{ia},i \right ) }{\sqrt{ia}\sqrt{{a}^{2}{x}^{4}+1}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsinh(a*x^2)/x+2*a/(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}{4} \, a^{2}{\left (\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 )}}{{\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 )}}{{\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 )}{{\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 )}{{\left (a^{2}\right )}^{\frac{3}{4}}}\right )} + 2 \, a \int \frac{1}{a^{3} x^{6} + a x^{2} +{\left (a^{2} x^{4} + 1\right )}^{\frac{3}{2}}}\,{d x} - \frac{\log \left (a x^{2} + \sqrt{a^{2} x^{4} + 1}\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(a*x^2)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x**2)/x**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(a*x^2)/x^2, x)