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

Optimal. Leaf size=43 \[ \frac {\tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {c}} \]

[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)

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Rubi [A]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1107, 621, 206} \[ \frac {\tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {c}} \]

Antiderivative was successfully verified.

[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 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 621

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 1107

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

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Mathematica [A]  time = 0.01, size = 43, normalized size = 1.00 \[ \frac {\tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Integrate[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])

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fricas [A]  time = 0.82, size = 118, normalized size = 2.74 \[ \left [\frac {\log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right )}{4 \, \sqrt {c}}, -\frac {\sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right )}{2 \, c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [A]  time = 0.21, size = 40, normalized size = 0.93 \[ -\frac {\log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right )}{2 \, \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 35, normalized size = 0.81 \[ \frac {\ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(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 zero or nonzero?

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mupad [B]  time = 4.69, size = 34, normalized size = 0.79 \[ \frac {\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )}{2\,\sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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