3.972 \(\int \frac {1}{x \sqrt {-a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=47 \[ -\frac {\tan ^{-1}\left (\frac {2 a-b x^2}{2 \sqrt {a} \sqrt {-a+b x^2+c x^4}}\right )}{2 \sqrt {a}} \]

[Out]

-1/2*arctan(1/2*(-b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^2-a)^(1/2))/a^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1114, 724, 204} \[ -\frac {\tan ^{-1}\left (\frac {2 a-b x^2}{2 \sqrt {a} \sqrt {-a+b x^2+c x^4}}\right )}{2 \sqrt {a}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[-a + b*x^2 + c*x^4]),x]

[Out]

-ArcTan[(2*a - b*x^2)/(2*Sqrt[a]*Sqrt[-a + b*x^2 + c*x^4])]/(2*Sqrt[a])

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {1}{x \sqrt {-a+b x^2+c x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {-a+b x+c x^2}} \, dx,x,x^2\right )\\ &=-\operatorname {Subst}\left (\int \frac {1}{-4 a-x^2} \, dx,x,\frac {-2 a+b x^2}{\sqrt {-a+b x^2+c x^4}}\right )\\ &=\frac {\tan ^{-1}\left (\frac {-2 a+b x^2}{2 \sqrt {a} \sqrt {-a+b x^2+c x^4}}\right )}{2 \sqrt {a}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 46, normalized size = 0.98 \[ \frac {\tan ^{-1}\left (\frac {b x^2-2 a}{2 \sqrt {a} \sqrt {-a+b x^2+c x^4}}\right )}{2 \sqrt {a}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[-a + b*x^2 + c*x^4]),x]

[Out]

ArcTan[(-2*a + b*x^2)/(2*Sqrt[a]*Sqrt[-a + b*x^2 + c*x^4])]/(2*Sqrt[a])

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fricas [A]  time = 0.82, size = 129, normalized size = 2.74 \[ \left [-\frac {\sqrt {-a} \log \left (\frac {{\left (b^{2} - 4 \, a c\right )} x^{4} - 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} - a} {\left (b x^{2} - 2 \, a\right )} \sqrt {-a} + 8 \, a^{2}}{x^{4}}\right )}{4 \, a}, \frac {\arctan \left (\frac {\sqrt {c x^{4} + b x^{2} - a} {\left (b x^{2} - 2 \, a\right )} \sqrt {a}}{2 \, {\left (a c x^{4} + a b x^{2} - a^{2}\right )}}\right )}{2 \, \sqrt {a}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^4+b*x^2-a)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*sqrt(-a)*log(((b^2 - 4*a*c)*x^4 - 8*a*b*x^2 - 4*sqrt(c*x^4 + b*x^2 - a)*(b*x^2 - 2*a)*sqrt(-a) + 8*a^2)/
x^4)/a, 1/2*arctan(1/2*sqrt(c*x^4 + b*x^2 - a)*(b*x^2 - 2*a)*sqrt(a)/(a*c*x^4 + a*b*x^2 - a^2))/sqrt(a)]

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giac [A]  time = 0.21, size = 36, normalized size = 0.77 \[ \frac {\arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} - a}}{\sqrt {a}}\right )}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^4+b*x^2-a)^(1/2),x, algorithm="giac")

[Out]

arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 - a))/sqrt(a))/sqrt(a)

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maple [A]  time = 0.02, size = 45, normalized size = 0.96 \[ -\frac {\ln \left (\frac {b \,x^{2}-2 a +2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{x^{2}}\right )}{2 \sqrt {-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(c*x^4+b*x^2-a)^(1/2),x)

[Out]

-1/2/(-a)^(1/2)*ln((-2*a+b*x^2+2*(-a)^(1/2)*(c*x^4+b*x^2-a)^(1/2))/x^2)

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maxima [A]  time = 2.33, size = 36, normalized size = 0.77 \[ -\frac {\arcsin \left (-\frac {b}{\sqrt {b^{2} + 4 \, a c}} + \frac {2 \, a}{\sqrt {b^{2} + 4 \, a c} x^{2}}\right )}{2 \, \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^4+b*x^2-a)^(1/2),x, algorithm="maxima")

[Out]

-1/2*arcsin(-b/sqrt(b^2 + 4*a*c) + 2*a/(sqrt(b^2 + 4*a*c)*x^2))/sqrt(a)

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mupad [B]  time = 4.52, size = 52, normalized size = 1.11 \[ -\frac {\ln \left (\frac {1}{x^2}\right )}{2\,\sqrt {-a}}-\frac {\ln \left (2\,\sqrt {-a}\,\sqrt {c\,x^4+b\,x^2-a}-2\,a+b\,x^2\right )}{2\,\sqrt {-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(b*x^2 - a + c*x^4)^(1/2)),x)

[Out]

- log(1/x^2)/(2*(-a)^(1/2)) - log(2*(-a)^(1/2)*(b*x^2 - a + c*x^4)^(1/2) - 2*a + b*x^2)/(2*(-a)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \sqrt {- a + b x^{2} + c x^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x**4+b*x**2-a)**(1/2),x)

[Out]

Integral(1/(x*sqrt(-a + b*x**2 + c*x**4)), x)

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