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

Optimal. Leaf size=49 \[ -\frac{\tanh ^{-1}\left (\frac{x \left (2 a+b x^2\right )}{2 \sqrt{a} \sqrt{a x^2+b x^4+c x^6}}\right )}{2 \sqrt{a}} \]

[Out]

-ArcTanh[(x*(2*a + b*x^2))/(2*Sqrt[a]*Sqrt[a*x^2 + b*x^4 + c*x^6])]/(2*Sqrt[a])

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Rubi [A]  time = 0.0151489, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {1996, 1904, 206} \[ -\frac{\tanh ^{-1}\left (\frac{x \left (2 a+b x^2\right )}{2 \sqrt{a} \sqrt{a x^2+b x^4+c x^6}}\right )}{2 \sqrt{a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[(x*(2*a + b*x^2))/(2*Sqrt[a]*Sqrt[a*x^2 + b*x^4 + c*x^6])]/(2*Sqrt[a])

Rule 1996

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedTrinomialQ[u, x] &&  !Gen
eralizedTrinomialMatchQ[u, x]

Rule 1904

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[-2/(n - 2), Subst[Int[1/(4*a
- x^2), x], x, (x*(2*a + b*x^(n - 2)))/Sqrt[a*x^2 + b*x^n + c*x^r]], x] /; FreeQ[{a, b, c, n, r}, x] && EqQ[r
, 2*n - 2] && PosQ[n - 2] && NeQ[b^2 - 4*a*c, 0]

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])

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{x^2 \left (a+b x^2+c x^4\right )}} \, dx &=\int \frac{1}{\sqrt{a x^2+b x^4+c x^6}} \, dx\\ &=-\operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{x \left (2 a+b x^2\right )}{\sqrt{a x^2+b x^4+c x^6}}\right )\\ &=-\frac{\tanh ^{-1}\left (\frac{x \left (2 a+b x^2\right )}{2 \sqrt{a} \sqrt{a x^2+b x^4+c x^6}}\right )}{2 \sqrt{a}}\\ \end{align*}

Mathematica [A]  time = 0.0163743, size = 81, normalized size = 1.65 \[ -\frac{x \sqrt{a+b x^2+c x^4} \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{2 \sqrt{a} \sqrt{x^2 \left (a+b x^2+c x^4\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 72, normalized size = 1.5 \begin{align*} -{\frac{x}{2}\sqrt{c{x}^{4}+b{x}^{2}+a}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{{x}^{2} \left ( c{x}^{4}+b{x}^{2}+a \right ) }}}{\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(x^2*(c*x^4+b*x^2+a))^(1/2)*x*(c*x^4+b*x^2+a)^(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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{{\left (c x^{4} + b x^{2} + a\right )} x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.32455, size = 311, normalized size = 6.35 \begin{align*} \left [\frac{\log \left (-\frac{{\left (b^{2} + 4 \, a c\right )} x^{5} + 8 \, a b x^{3} + 8 \, a^{2} x - 4 \, \sqrt{c x^{6} + b x^{4} + a x^{2}}{\left (b x^{2} + 2 \, a\right )} \sqrt{a}}{x^{5}}\right )}{4 \, \sqrt{a}}, \frac{\sqrt{-a} \arctan \left (\frac{\sqrt{c x^{6} + b x^{4} + a x^{2}}{\left (b x^{2} + 2 \, a\right )} \sqrt{-a}}{2 \,{\left (a c x^{5} + a b x^{3} + a^{2} x\right )}}\right )}{2 \, a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*log(-((b^2 + 4*a*c)*x^5 + 8*a*b*x^3 + 8*a^2*x - 4*sqrt(c*x^6 + b*x^4 + a*x^2)*(b*x^2 + 2*a)*sqrt(a))/x^5)
/sqrt(a), 1/2*sqrt(-a)*arctan(1/2*sqrt(c*x^6 + b*x^4 + a*x^2)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^5 + a*b*x^3 + a^2*
x))/a]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{{\left (c x^{4} + b x^{2} + a\right )} x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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