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

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

[Out]

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

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Rubi [A]  time = 0.0343985, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2029, 206} \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a x^3+b x^4}}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a*x^3 + b*x^4],x]

[Out]

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

Mathematica [A]  time = 0.0166366, size = 59, normalized size = 1.84 \[ \frac{2 \sqrt{a} x^{3/2} \sqrt{\frac{b x}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{b} \sqrt{x^3 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[a*x^3 + b*x^4],x]

[Out]

(2*Sqrt[a]*x^(3/2)*Sqrt[1 + (b*x)/a]*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(Sqrt[b]*Sqrt[x^3*(a + b*x)])

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Maple [B]  time = 0.003, size = 56, normalized size = 1.8 \begin{align*}{x\sqrt{x \left ( bx+a \right ) }\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{b{x}^{2}+ax}\sqrt{b}+2\,bx+a \right ){\frac{1}{\sqrt{b}}}} \right ){\frac{1}{\sqrt{b{x}^{4}+a{x}^{3}}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(b*x^4+a*x^3)^(1/2)*x*(x*(b*x+a))^(1/2)*ln(1/2*(2*(b*x^2+a*x)^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))/b^(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^{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/sqrt(b*x^4 + a*x^3), x)

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Fricas [A]  time = 0.820835, size = 173, normalized size = 5.41 \begin{align*} \left [\frac{\log \left (\frac{2 \, b x^{2} + a x + 2 \, \sqrt{b x^{4} + a x^{3}} \sqrt{b}}{x}\right )}{\sqrt{b}}, -\frac{2 \, \sqrt{-b} \arctan \left (\frac{\sqrt{b x^{4} + a x^{3}} \sqrt{-b}}{b x^{2}}\right )}{b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[log((2*b*x^2 + a*x + 2*sqrt(b*x^4 + a*x^3)*sqrt(b))/x)/sqrt(b), -2*sqrt(-b)*arctan(sqrt(b*x^4 + a*x^3)*sqrt(-
b)/(b*x^2))/b]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/sqrt(x**3*(a + b*x)), x)

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Giac [A]  time = 1.39866, size = 31, normalized size = 0.97 \begin{align*} -\frac{2 \, \arctan \left (\frac{\sqrt{b + \frac{a}{x}}}{\sqrt{-b}}\right )}{\sqrt{-b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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