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

Optimal. Leaf size=44 \[ -\frac {\tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {a}} \]

[Out]

-1/2*arctanh(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 = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1114, 724, 206} \[ -\frac {\tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {a}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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 {\tanh ^{-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 = 44, normalized size = 1.00 \[ -\frac {\tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.86, size = 124, normalized size = 2.82 \[ \left [\frac {\log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right )}{4 \, \sqrt {a}}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right )}{2 \, a}\right ] \]

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*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)/sqrt(
a), 1/2*sqrt(-a)*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))/a]

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giac [A]  time = 0.45, size = 38, normalized size = 0.86 \[ \frac {\arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} \]

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 = 0.01, size = 39, normalized size = 0.89 \[ -\frac {\ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{2 \sqrt {a}} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 2.23, size = 44, normalized size = 1.00 \[ -\frac {\ln \left (\frac {1}{x^2}\right )}{2\,\sqrt {a}}-\frac {\ln \left (2\,a+2\,\sqrt {a}\,\sqrt {c\,x^4+b\,x^2+a}+b\,x^2\right )}{2\,\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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