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

Optimal. Leaf size=30 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 \sqrt{b}} \]

[Out]

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

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Rubi [A]  time = 0.0147117, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {275, 217, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + b*x^4],x]

[Out]

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

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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{x}{\sqrt{a+b x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^2}{\sqrt{a+b x^4}}\right )\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0041508, size = 30, normalized size = 1. \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[a + b*x^4],x]

[Out]

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

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Maple [A]  time = 0.006, size = 24, normalized size = 0.8 \begin{align*}{\frac{1}{2}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ){\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52388, size = 163, normalized size = 5.43 \begin{align*} \left [\frac{\log \left (-2 \, b x^{4} - 2 \, \sqrt{b x^{4} + a} \sqrt{b} x^{2} - a\right )}{4 \, \sqrt{b}}, -\frac{\sqrt{-b} \arctan \left (\frac{\sqrt{-b} x^{2}}{\sqrt{b x^{4} + a}}\right )}{2 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.39566, size = 20, normalized size = 0.67 \begin{align*} \frac{\operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2 \sqrt{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

asinh(sqrt(b)*x**2/sqrt(a))/(2*sqrt(b))

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Giac [A]  time = 1.12209, size = 34, normalized size = 1.13 \begin{align*} -\frac{\log \left ({\left | -\sqrt{b} x^{2} + \sqrt{b x^{4} + a} \right |}\right )}{2 \, \sqrt{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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