\(\int \frac {x}{\sqrt {a+b x^2+c x^4}} \, dx\) [958]

Optimal result

Integrand size = 18, antiderivative size = 43 \[ \int \frac {x}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {\text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1121, 635, 212} \[ \int \frac {x}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {\text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {c}} \]

[In]

Int[x/Sqrt[a + b*x^2 + c*x^4],x]

[Out]

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

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 635

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

Rule 1121

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \frac {x}{\sqrt {a+b x^2+c x^4}} \, dx=-\frac {\log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )}{2 \sqrt {c}} \]

[In]

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

[Out]

-1/2*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]]/Sqrt[c]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81

method	result	size
default	\(\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 \sqrt {c}}\)	\(35\)
elliptic	\(\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 \sqrt {c}}\)	\(35\)
pseudoelliptic	\(\frac {-\ln \left (2\right )+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )}{2 \sqrt {c}}\)	\(43\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.74 \[ \int \frac {x}{\sqrt {a+b x^2+c x^4}} \, dx=\left [\frac {\log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right )}{4 \, \sqrt {c}}, -\frac {\sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right )}{2 \, c}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {x}{\sqrt {a+b x^2+c x^4}} \, dx=\int \frac {x}{\sqrt {a + b x^{2} + c x^{4}}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\sqrt {a+b x^2+c x^4}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(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 deta

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88 \[ \int \frac {x}{\sqrt {a+b x^2+c x^4}} \, dx=-\frac {\log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{2 \, \sqrt {c}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.37 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.79 \[ \int \frac {x}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )}{2\,\sqrt {c}} \]

[In]

int(x/(a + b*x^2 + c*x^4)^(1/2),x)

[Out]

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