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

Optimal. Leaf size=34 \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x^2+b x^3}}\right )}{\sqrt{b}} \]

[Out]

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

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Rubi [A]  time = 0.0424737, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2029, 206} \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x^2+b x^3}}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/Sqrt[a*x^2 + b*x^3],x]

[Out]

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

Mathematica [A]  time = 0.0161086, size = 55, normalized size = 1.62 \[ \frac{2 \sqrt{a} x \sqrt{\frac{b x}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{b} \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.005, size = 58, normalized size = 1.7 \begin{align*}{\sqrt{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}^{3}+a{x}^{2}}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.840399, size = 189, normalized size = 5.56 \begin{align*} \left [\frac{\log \left (\frac{2 \, b x^{2} + a x + 2 \, \sqrt{b x^{3} + a x^{2}} \sqrt{b} \sqrt{x}}{x}\right )}{\sqrt{b}}, -\frac{2 \, \sqrt{-b} \arctan \left (\frac{\sqrt{b x^{3} + a x^{2}} \sqrt{-b}}{b x^{\frac{3}{2}}}\right )}{b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.39691, size = 31, normalized size = 0.91 \begin{align*} -\frac{2 \, \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{\sqrt{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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