3.290 \(\int \frac {\sinh ^{-1}(a x^2)}{x^4} \, dx\)

Optimal. Leaf size=197 \[ -\frac {2 a \sqrt {a^2 x^4+1}}{3 x}+\frac {2 a^2 x \sqrt {a^2 x^4+1}}{3 \left (a x^2+1\right )}+\frac {a^{3/2} \left (a x^2+1\right ) \sqrt {\frac {a^2 x^4+1}{\left (a x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{3 \sqrt {a^2 x^4+1}}-\frac {2 a^{3/2} \left (a x^2+1\right ) \sqrt {\frac {a^2 x^4+1}{\left (a x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{3 \sqrt {a^2 x^4+1}}-\frac {\sinh ^{-1}\left (a x^2\right )}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5902, 12, 325, 305, 220, 1196} \[ \frac {2 a^2 x \sqrt {a^2 x^4+1}}{3 \left (a x^2+1\right )}-\frac {2 a \sqrt {a^2 x^4+1}}{3 x}+\frac {a^{3/2} \left (a x^2+1\right ) \sqrt {\frac {a^2 x^4+1}{\left (a x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{3 \sqrt {a^2 x^4+1}}-\frac {2 a^{3/2} \left (a x^2+1\right ) \sqrt {\frac {a^2 x^4+1}{\left (a x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{3 \sqrt {a^2 x^4+1}}-\frac {\sinh ^{-1}\left (a x^2\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x^2]/x^4,x]

[Out]

(-2*a*Sqrt[1 + a^2*x^4])/(3*x) + (2*a^2*x*Sqrt[1 + a^2*x^4])/(3*(1 + a*x^2)) - ArcSinh[a*x^2]/(3*x^3) - (2*a^(
3/2)*(1 + a*x^2)*Sqrt[(1 + a^2*x^4)/(1 + a*x^2)^2]*EllipticE[2*ArcTan[Sqrt[a]*x], 1/2])/(3*Sqrt[1 + a^2*x^4])
+ (a^(3/2)*(1 + a*x^2)*Sqrt[(1 + a^2*x^4)/(1 + a*x^2)^2]*EllipticF[2*ArcTan[Sqrt[a]*x], 1/2])/(3*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 305

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1196

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)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/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^4} \, dx &=-\frac {\sinh ^{-1}\left (a x^2\right )}{3 x^3}+\frac {1}{3} \int \frac {2 a}{x^2 \sqrt {1+a^2 x^4}} \, dx\\ &=-\frac {\sinh ^{-1}\left (a x^2\right )}{3 x^3}+\frac {1}{3} (2 a) \int \frac {1}{x^2 \sqrt {1+a^2 x^4}} \, dx\\ &=-\frac {2 a \sqrt {1+a^2 x^4}}{3 x}-\frac {\sinh ^{-1}\left (a x^2\right )}{3 x^3}+\frac {1}{3} \left (2 a^3\right ) \int \frac {x^2}{\sqrt {1+a^2 x^4}} \, dx\\ &=-\frac {2 a \sqrt {1+a^2 x^4}}{3 x}-\frac {\sinh ^{-1}\left (a x^2\right )}{3 x^3}+\frac {1}{3} \left (2 a^2\right ) \int \frac {1}{\sqrt {1+a^2 x^4}} \, dx-\frac {1}{3} \left (2 a^2\right ) \int \frac {1-a x^2}{\sqrt {1+a^2 x^4}} \, dx\\ &=-\frac {2 a \sqrt {1+a^2 x^4}}{3 x}+\frac {2 a^2 x \sqrt {1+a^2 x^4}}{3 \left (1+a x^2\right )}-\frac {\sinh ^{-1}\left (a x^2\right )}{3 x^3}-\frac {2 a^{3/2} \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 )}{3 \sqrt {1+a^2 x^4}}+\frac {a^{3/2} \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 )}{3 \sqrt {1+a^2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.17, size = 88, normalized size = 0.45 \[ \frac {1}{3} \left (-\frac {2 a \sqrt {a^2 x^4+1}}{x}+\frac {2 a^2 \left (E\left (\left .i \sinh ^{-1}\left (\sqrt {i a} x\right )\right |-1\right )-F\left (\left .i \sinh ^{-1}\left (\sqrt {i a} x\right )\right |-1\right )\right )}{\sqrt {i a}}-\frac {\sinh ^{-1}\left (a x^2\right )}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a*x^2]/x^4,x]

[Out]

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

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arsinh}\left (a x^{2}\right )}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x^2)/x^4,x, algorithm="fricas")

[Out]

integral(arcsinh(a*x^2)/x^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsinh}\left (a x^{2}\right )}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x^2)/x^4,x, algorithm="giac")

[Out]

integrate(arcsinh(a*x^2)/x^4, x)

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maple [C]  time = 0.01, size = 101, normalized size = 0.51 \[ -\frac {\arcsinh \left (a \,x^{2}\right )}{3 x^{3}}+\frac {2 a \left (-\frac {\sqrt {a^{2} x^{4}+1}}{x}+\frac {i a \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, \left (\EllipticF \left (x \sqrt {i a}, i\right )-\EllipticE \left (x \sqrt {i a}, i\right )\right )}{\sqrt {i a}\, \sqrt {a^{2} x^{4}+1}}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x^2)/x^4,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{12} i \, \sqrt {2} a^{\frac {3}{2}} {\left (\log \left (\frac {i \, \sqrt {2} {\left (2 \, a x + \sqrt {2} \sqrt {a}\right )}}{2 \, \sqrt {a}} + 1\right ) - \log \left (-\frac {i \, \sqrt {2} {\left (2 \, a x + \sqrt {2} \sqrt {a}\right )}}{2 \, \sqrt {a}} + 1\right )\right )} - \frac {1}{12} i \, \sqrt {2} a^{\frac {3}{2}} {\left (\log \left (\frac {i \, \sqrt {2} {\left (2 \, a x - \sqrt {2} \sqrt {a}\right )}}{2 \, \sqrt {a}} + 1\right ) - \log \left (-\frac {i \, \sqrt {2} {\left (2 \, a x - \sqrt {2} \sqrt {a}\right )}}{2 \, \sqrt {a}} + 1\right )\right )} + \frac {1}{12} \, \sqrt {2} a^{\frac {3}{2}} \log \left (a x^{2} + \sqrt {2} \sqrt {a} x + 1\right ) - \frac {1}{12} \, \sqrt {2} a^{\frac {3}{2}} \log \left (a x^{2} - \sqrt {2} \sqrt {a} x + 1\right ) + 2 \, a \int \frac {1}{3 \, {\left (a^{3} x^{8} + a x^{4} + {\left (a^{2} x^{6} + x^{2}\right )} \sqrt {a^{2} x^{4} + 1}\right )}}\,{d x} - \frac {\log \left (a x^{2} + \sqrt {a^{2} x^{4} + 1}\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x^2)/x^4,x, algorithm="maxima")

[Out]

-1/12*I*sqrt(2)*a^(3/2)*(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)) - 1/12*I*sqrt(2)*a^(3/2)*(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)) + 1/12*sqrt(2)*a^(3/2)*log(a*x^2 + sqrt(2)*
sqrt(a)*x + 1) - 1/12*sqrt(2)*a^(3/2)*log(a*x^2 - sqrt(2)*sqrt(a)*x + 1) + 2*a*integrate(1/3/(a^3*x^8 + a*x^4
+ (a^2*x^6 + x^2)*sqrt(a^2*x^4 + 1)), x) - 1/3*log(a*x^2 + sqrt(a^2*x^4 + 1))/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {asinh}\left (a\,x^2\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a*x^2)/x^4,x)

[Out]

int(asinh(a*x^2)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}{\left (a x^{2} \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x**2)/x**4,x)

[Out]

Integral(asinh(a*x**2)/x**4, x)

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