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

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

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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 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 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 \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*}

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Mathematica [C]  time = 0.04, size = 42, normalized size = 0.56 \[ -\frac {\sinh ^{-1}\left (a x^2\right )+2 \sqrt {i a} x F\left (\left .i \sinh ^{-1}\left (\sqrt {i a} x\right )\right |-1\right )}{x} \]

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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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arsinh}\left (a x^{2}\right )}{x^{2}}, x\right ) \]

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

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)

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maple [C]  time = 0.01, size = 66, normalized size = 0.88 \[ -\frac {\arcsinh \left (a \,x^{2}\right )}{x}+\frac {2 a \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, \EllipticF \left (x \sqrt {i a}, i\right )}{\sqrt {i a}\, \sqrt {a^{2} x^{4}+1}} \]

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

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*a*x + sqrt(2)*sqrt(a))/sqrt(a) + 1) - log(-1/2*I*sqrt(2)*(2*a*x + sq
rt(2)*sqrt(a))/sqrt(a) + 1))/a^(3/2) + I*sqrt(2)*(log(1/2*I*sqrt(2)*(2*a*x - sqrt(2)*sqrt(a))/sqrt(a) + 1) - l
og(-1/2*I*sqrt(2)*(2*a*x - sqrt(2)*sqrt(a))/sqrt(a) + 1))/a^(3/2) + sqrt(2)*log(a*x^2 + sqrt(2)*sqrt(a)*x + 1)
/a^(3/2) - sqrt(2)*log(a*x^2 - sqrt(2)*sqrt(a)*x + 1)/a^(3/2)) + 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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {asinh}\left (a\,x^2\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asinh(a*x^2)/x^2, x)

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

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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