\(\int \frac {1}{\sqrt {x^2 (a+b x^2+c x^4)}} \, dx\) [134]

Optimal result

Integrand size = 20, antiderivative size = 49 \[ \int \frac {1}{\sqrt {x^2 \left (a+b x^2+c x^4\right )}} \, dx=-\frac {\text {arctanh}\left (\frac {x \left (2 a+b x^2\right )}{2 \sqrt {a} \sqrt {a x^2+b x^4+c x^6}}\right )}{2 \sqrt {a}} \]

[Out]

-1/2*arctanh(1/2*x*(b*x^2+2*a)/a^(1/2)/(c*x^6+b*x^4+a*x^2)^(1/2))/a^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2021, 1918, 212} \[ \int \frac {1}{\sqrt {x^2 \left (a+b x^2+c x^4\right )}} \, dx=-\frac {\text {arctanh}\left (\frac {x \left (2 a+b x^2\right )}{2 \sqrt {a} \sqrt {a x^2+b x^4+c x^6}}\right )}{2 \sqrt {a}} \]

[In]

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

[Out]

-1/2*ArcTanh[(x*(2*a + b*x^2))/(2*Sqrt[a]*Sqrt[a*x^2 + b*x^4 + c*x^6])]/Sqrt[a]

Rule 212

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

Rule 1918

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 2021

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\sqrt {a x^2+b x^4+c x^6}} \, dx \\ & = -\text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.59 \[ \int \frac {1}{\sqrt {x^2 \left (a+b x^2+c x^4\right )}} \, dx=\frac {x \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {x^2 \left (a+b x^2+c x^4\right )}} \]

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

Maple [A] (verified)

Time = 1.71 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.47

method	result	size
default	\(-\frac {x \sqrt {c \,x^{4}+b \,x^{2}+a}\, \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 \sqrt {x^{2} \left (c \,x^{4}+b \,x^{2}+a \right )}\, \sqrt {a}}\)	\(72\)

[In]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 2.76 \[ \int \frac {1}{\sqrt {x^2 \left (a+b x^2+c x^4\right )}} \, dx=\left [\frac {\log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{5} + 8 \, a b x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{6} + b x^{4} + a x^{2}} {\left (b x^{2} + 2 \, a\right )} \sqrt {a}}{x^{5}}\right )}{4 \, \sqrt {a}}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {c x^{6} + b x^{4} + a x^{2}} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{5} + a b x^{3} + a^{2} x\right )}}\right )}{2 \, a}\right ] \]

[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]

Sympy [F]

\[ \int \frac {1}{\sqrt {x^2 \left (a+b x^2+c x^4\right )}} \, dx=\int \frac {1}{\sqrt {x^{2} \left (a + b x^{2} + c x^{4}\right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\sqrt {x^2 \left (a+b x^2+c x^4\right )}} \, dx=\int { \frac {1}{\sqrt {{\left (c x^{4} + b x^{2} + a\right )} x^{2}}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.27 \[ \int \frac {1}{\sqrt {x^2 \left (a+b x^2+c x^4\right )}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a}} + \frac {\arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {x^2 \left (a+b x^2+c x^4\right )}} \, dx=\int \frac {1}{\sqrt {x^2\,\left (c\,x^4+b\,x^2+a\right )}} \,d x \]

[In]

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

[Out]

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