3.3.86 \(\int \sinh ^{-1}(a x^2) \, dx\) [286]

Optimal. Leaf size=162 \[ -\frac {2 x \sqrt {1+a^2 x^4}}{1+a x^2}+x \sinh ^{-1}\left (a x^2\right )+\frac {2 \left (1+a x^2\right ) \sqrt {\frac {1+a^2 x^4}{\left (1+a x^2\right )^2}} E\left (2 \text {ArcTan}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{\sqrt {a} \sqrt {1+a^2 x^4}}-\frac {\left (1+a x^2\right ) \sqrt {\frac {1+a^2 x^4}{\left (1+a x^2\right )^2}} F\left (2 \text {ArcTan}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{\sqrt {a} \sqrt {1+a^2 x^4}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {5874, 12, 311, 226, 1210} \begin {gather*} -\frac {\left (a x^2+1\right ) \sqrt {\frac {a^2 x^4+1}{\left (a x^2+1\right )^2}} F\left (2 \text {ArcTan}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{\sqrt {a} \sqrt {a^2 x^4+1}}+\frac {2 \left (a x^2+1\right ) \sqrt {\frac {a^2 x^4+1}{\left (a x^2+1\right )^2}} E\left (2 \text {ArcTan}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{\sqrt {a} \sqrt {a^2 x^4+1}}-\frac {2 x \sqrt {a^2 x^4+1}}{a x^2+1}+x \sinh ^{-1}\left (a x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x*Sqrt[1 + a^2*x^4])/(1 + a*x^2) + x*ArcSinh[a*x^2] + (2*(1 + a*x^2)*Sqrt[(1 + a^2*x^4)/(1 + a*x^2)^2]*Ell
ipticE[2*ArcTan[Sqrt[a]*x], 1/2])/(Sqrt[a]*Sqrt[1 + a^2*x^4]) - ((1 + a*x^2)*Sqrt[(1 + a^2*x^4)/(1 + a*x^2)^2]
*EllipticF[2*ArcTan[Sqrt[a]*x], 1/2])/(Sqrt[a]*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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

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

Rule 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 5874

Int[ArcSinh[u_], x_Symbol] :> Simp[x*ArcSinh[u], x] - Int[SimplifyIntegrand[x*(D[u, x]/Sqrt[1 + u^2]), x], x]
/; InverseFunctionFreeQ[u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.01, size = 35, normalized size = 0.22 \begin {gather*} x \sinh ^{-1}\left (a x^2\right )-\frac {2}{3} a x^3 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-a^2 x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*ArcSinh[a*x^2] - (2*a*x^3*Hypergeometric2F1[1/2, 3/4, 7/4, -(a^2*x^4)])/3

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Maple [C] Result contains complex when optimal does not.
time = 0.15, size = 77, normalized size = 0.48

method	result	size
default	\(x \arcsinh \left (a \,x^{2}\right )-\frac {2 i \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, \left (\EllipticF \left (x \sqrt {i a}, i\right )-\EllipticE \left (x \sqrt {i a}, i\right )\right )}{\sqrt {i a}\, \sqrt {a^{2} x^{4}+1}}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x^2),x,method=_RETURNVERBOSE)

[Out]

x*arcsinh(a*x^2)-2*I/(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)-EllipticE(x*(I*a)^(1/2),I))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*integrate(x^2/(a^3*x^6 + a*x^2 + (a^2*x^4 + 1)^(3/2)), x) + x*log(a*x^2 + sqrt(a^2*x^4 + 1)) - 2*x - 1/4*
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))/sqrt(a) - 1/4*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))/sqrt(a) + 1/4*sqrt(2)*log(a*x^2 + sqrt(2)*sqrt(a)*x + 1)/s
qrt(a) - 1/4*sqrt(2)*log(a*x^2 - sqrt(2)*sqrt(a)*x + 1)/sqrt(a)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x**2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(a*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asinh(a*x^2), x)

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