3.51 \(\int \frac{x}{\sqrt{a x^2+b x^3+c x^4}} \, dx\)

Optimal. Leaf size=71 \[ \frac{x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{\sqrt{c} \sqrt{a x^2+b x^3+c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0366343, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {1914, 621, 206} \[ \frac{x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{\sqrt{c} \sqrt{a x^2+b x^3+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a*x^2 + b*x^3 + c*x^4],x]

[Out]

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

Rule 1914

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[(x^(q/2)*Sqrt[a
+ b*x^(n - q) + c*x^(2*(n - q))])/Sqrt[a*x^q + b*x^n + c*x^(2*n - q)], Int[x^(m - q/2)/Sqrt[a + b*x^(n - q) +
c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] &&
EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q,
1]))

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

Mathematica [A]  time = 0.0412742, size = 66, normalized size = 0.93 \[ \frac{x \sqrt{a+b x+c x^2} \log \left (2 \sqrt{c} \sqrt{a+b x+c x^2}+b+2 c x\right )}{\sqrt{c} \sqrt{x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[a*x^2 + b*x^3 + c*x^4],x]

[Out]

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

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Maple [A]  time = 0.006, size = 65, normalized size = 0.9 \begin{align*}{x\sqrt{c{x}^{2}+bx+a}\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{c{x}^{2}+bx+a}\sqrt{c}+2\,cx+b \right ){\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c{x}^{4}+b{x}^{3}+a{x}^{2}}}}{\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(c*x^4+b*x^3+a*x^2)^(1/2),x)

[Out]

1/(c*x^4+b*x^3+a*x^2)^(1/2)*x*(c*x^2+b*x+a)^(1/2)*ln(1/2*(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c*x+b)/c^(1/2))/c^(1
/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{c x^{4} + b x^{3} + a x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(c*x^4+b*x^3+a*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(x/sqrt(c*x^4 + b*x^3 + a*x^2), x)

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Fricas [A]  time = 1.54597, size = 298, normalized size = 4.2 \begin{align*} \left [\frac{\log \left (-\frac{8 \, c^{2} x^{3} + 8 \, b c x^{2} + 4 \, \sqrt{c x^{4} + b x^{3} + a x^{2}}{\left (2 \, c x + b\right )} \sqrt{c} +{\left (b^{2} + 4 \, a c\right )} x}{x}\right )}{2 \, \sqrt{c}}, -\frac{\sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{3} + a x^{2}}{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right )}{c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(c*x^4+b*x^3+a*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/2*log(-(8*c^2*x^3 + 8*b*c*x^2 + 4*sqrt(c*x^4 + b*x^3 + a*x^2)*(2*c*x + b)*sqrt(c) + (b^2 + 4*a*c)*x)/x)/sqr
t(c), -sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x^3 + a*x^2)*(2*c*x + b)*sqrt(-c)/(c^2*x^3 + b*c*x^2 + a*c*x))/c]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{x^{2} \left (a + b x + c x^{2}\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(c*x**4+b*x**3+a*x**2)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.15887, size = 50, normalized size = 0.7 \begin{align*} -\frac{2 \, \arctan \left (\frac{\sqrt{c + \frac{b}{x} + \frac{a}{x^{2}}} - \frac{\sqrt{a}}{x}}{\sqrt{-c}}\right )}{\sqrt{-c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(c*x^4+b*x^3+a*x^2)^(1/2),x, algorithm="giac")

[Out]

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