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

Optimal. Leaf size=60 \[ \frac{\sqrt{a x^2+b x^3}}{b \sqrt{x}}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x^2+b x^3}}\right )}{b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0833389, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2024, 2029, 206} \[ \frac{\sqrt{a x^2+b x^3}}{b \sqrt{x}}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x^2+b x^3}}\right )}{b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/Sqrt[a*x^2 + b*x^3],x]

[Out]

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

Rule 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

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

Mathematica [A]  time = 0.0392275, size = 73, normalized size = 1.22 \[ \frac{\sqrt{b} x^{3/2} (a+b x)-a^{3/2} x \sqrt{\frac{b x}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{3/2} \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 79, normalized size = 1.3 \begin{align*}{\frac{1}{2}\sqrt{x} \left ( 2\,{b}^{5/2}{x}^{2}+2\,{b}^{3/2}xa-a\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 ) b \right ){\frac{1}{\sqrt{b{x}^{3}+a{x}^{2}}}}{b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^(1/2)*(2*b^(5/2)*x^2+2*b^(3/2)*x*a-a*(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)/(b*x^3+a*x^2)^(1/2)/b^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.885123, size = 315, normalized size = 5.25 \begin{align*} \left [\frac{a \sqrt{b} x \log \left (\frac{2 \, b x^{2} + a x - 2 \, \sqrt{b x^{3} + a x^{2}} \sqrt{b} \sqrt{x}}{x}\right ) + 2 \, \sqrt{b x^{3} + a x^{2}} b \sqrt{x}}{2 \, b^{2} x}, \frac{a \sqrt{-b} x \arctan \left (\frac{\sqrt{b x^{3} + a x^{2}} \sqrt{-b}}{b x^{\frac{3}{2}}}\right ) + \sqrt{b x^{3} + a x^{2}} b \sqrt{x}}{b^{2} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{3}{2}}}{\sqrt{x^{2} \left (a + b x\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.35451, size = 51, normalized size = 0.85 \begin{align*} \frac{a \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{b^{\frac{3}{2}}} + \frac{\sqrt{b x + a} \sqrt{x}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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