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

Optimal. Leaf size=32 \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a x+b x^4}}\right )}{3 \sqrt{b}} \]

[Out]

(2*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a*x + b*x^4]])/(3*Sqrt[b])

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Rubi [A]  time = 0.0322918, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2029, 206} \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a x+b x^4}}\right )}{3 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a*x + b*x^4],x]

[Out]

(2*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a*x + b*x^4]])/(3*Sqrt[b])

Rule 2029

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

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 x+b x^4}} \, dx &=\frac{2}{3} \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^2}{\sqrt{a x+b x^4}}\right )\\ &=\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a x+b x^4}}\right )}{3 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0131221, size = 61, normalized size = 1.91 \[ \frac{2 \sqrt{x} \sqrt{a+b x^3} \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a+b x^3}}\right )}{3 \sqrt{b} \sqrt{x \left (a+b x^3\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[a*x + b*x^4],x]

[Out]

(2*Sqrt[x]*Sqrt[a + b*x^3]*ArcTanh[(Sqrt[b]*x^(3/2))/Sqrt[a + b*x^3]])/(3*Sqrt[b]*Sqrt[x*(a + b*x^3)])

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Maple [C]  time = 0.015, size = 979, normalized size = 30.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*((-3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3)
)*x/(-1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(x-1/b*(-b^2*a)^(1/3)))^(1/2)*(x-1/b*(-b^2*a)^(1/3)
)^2*(1/b*(-b^2*a)^(1/3)*(x+1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(-1/2/b*(-b^2*a)^(1/3)-1/2*I*3
^(1/2)/b*(-b^2*a)^(1/3))/(x-1/b*(-b^2*a)^(1/3)))^(1/2)*(1/b*(-b^2*a)^(1/3)*(x+1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/
2)/b*(-b^2*a)^(1/3))/(-1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(x-1/b*(-b^2*a)^(1/3)))^(1/2)/(-3/
2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*b/(-b^2*a)^(1/3)/(b*x*(x-1/b*(-b^2*a)^(1/3))*(x+1/2/b*(-b^2
*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*(x+1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3)))^(1/2)*(1/b*
(-b^2*a)^(1/3)*EllipticF(((-3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*x/(-1/2/b*(-b^2*a)^(1/3)+1/2*
I*3^(1/2)/b*(-b^2*a)^(1/3))/(x-1/b*(-b^2*a)^(1/3)))^(1/2),((3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3
))*(1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))
/(3/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3)))^(1/2))-1/b*(-b^2*a)^(1/3)*EllipticPi(((-3/2/b*(-b^2*a)
^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*x/(-1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(x-1/b*(-b^2*a
)^(1/3)))^(1/2),(-1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(-3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*
(-b^2*a)^(1/3)),((3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*(1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(
-b^2*a)^(1/3))/(1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(3/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b
^2*a)^(1/3)))^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{b x^{4} + a x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/sqrt(b*x^4 + a*x), x)

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Fricas [A]  time = 1.89581, size = 225, normalized size = 7.03 \begin{align*} \left [\frac{\log \left (-8 \, b^{2} x^{6} - 8 \, a b x^{3} - a^{2} - 4 \,{\left (2 \, b x^{4} + a x\right )} \sqrt{b x^{4} + a x} \sqrt{b}\right )}{6 \, \sqrt{b}}, -\frac{\sqrt{-b} \arctan \left (\frac{2 \, \sqrt{b x^{4} + a x} \sqrt{-b} x}{2 \, b x^{3} + a}\right )}{3 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{x \left (a + b x^{3}\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/sqrt(x*(a + b*x**3)), x)

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Giac [A]  time = 1.30927, size = 31, normalized size = 0.97 \begin{align*} -\frac{2 \, \arctan \left (\frac{\sqrt{b + \frac{a}{x^{3}}}}{\sqrt{-b}}\right )}{3 \, \sqrt{-b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*arctan(sqrt(b + a/x^3)/sqrt(-b))/sqrt(-b)