3.1764 \(\int (a+\frac{b}{x})^{3/2} \sqrt{x} \, dx\)

Optimal. Leaf size=70 \[ -2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )+\frac{2}{3} x^{3/2} \left (a+\frac{b}{x}\right )^{3/2}+2 b \sqrt{x} \sqrt{a+\frac{b}{x}} \]

[Out]

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

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Rubi [A]  time = 0.0362546, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {337, 277, 217, 206} \[ -2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )+\frac{2}{3} x^{3/2} \left (a+\frac{b}{x}\right )^{3/2}+2 b \sqrt{x} \sqrt{a+\frac{b}{x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(3/2)*Sqrt[x],x]

[Out]

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

Rule 337

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, -Dist[k/c, Subst[
Int[(a + b/(c^n*x^(k*n)))^p/x^(k*(m + 1) + 1), x], x, 1/(c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && ILtQ[n,
0] && FractionQ[m]

Rule 277

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 \left (a+\frac{b}{x}\right )^{3/2} \sqrt{x} \, dx &=-\left (2 \operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^{3/2}}{x^4} \, dx,x,\frac{1}{\sqrt{x}}\right )\right )\\ &=\frac{2}{3} \left (a+\frac{b}{x}\right )^{3/2} x^{3/2}-(2 b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{x^2} \, dx,x,\frac{1}{\sqrt{x}}\right )\\ &=2 b \sqrt{a+\frac{b}{x}} \sqrt{x}+\frac{2}{3} \left (a+\frac{b}{x}\right )^{3/2} x^{3/2}-\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{\sqrt{x}}\right )\\ &=2 b \sqrt{a+\frac{b}{x}} \sqrt{x}+\frac{2}{3} \left (a+\frac{b}{x}\right )^{3/2} x^{3/2}-\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{1}{\sqrt{a+\frac{b}{x}} \sqrt{x}}\right )\\ &=2 b \sqrt{a+\frac{b}{x}} \sqrt{x}+\frac{2}{3} \left (a+\frac{b}{x}\right )^{3/2} x^{3/2}-2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x}} \sqrt{x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.012255, size = 54, normalized size = 0.77 \[ \frac{2 a x^{3/2} \sqrt{a+\frac{b}{x}} \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};-\frac{b}{a x}\right )}{3 \sqrt{\frac{b}{a x}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(3/2)*Sqrt[x],x]

[Out]

(2*a*Sqrt[a + b/x]*x^(3/2)*Hypergeometric2F1[-3/2, -3/2, -1/2, -(b/(a*x))])/(3*Sqrt[1 + b/(a*x)])

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Maple [A]  time = 0.01, size = 63, normalized size = 0.9 \begin{align*} -{\frac{2}{3}\sqrt{{\frac{ax+b}{x}}}\sqrt{x} \left ( 3\,{b}^{3/2}{\it Artanh} \left ({\frac{\sqrt{ax+b}}{\sqrt{b}}} \right ) -xa\sqrt{ax+b}-4\,b\sqrt{ax+b} \right ){\frac{1}{\sqrt{ax+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*((a*x+b)/x)^(1/2)*x^(1/2)*(3*b^(3/2)*arctanh((a*x+b)^(1/2)/b^(1/2))-x*a*(a*x+b)^(1/2)-4*b*(a*x+b)^(1/2))/
(a*x+b)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53829, size = 293, normalized size = 4.19 \begin{align*} \left [b^{\frac{3}{2}} \log \left (\frac{a x - 2 \, \sqrt{b} \sqrt{x} \sqrt{\frac{a x + b}{x}} + 2 \, b}{x}\right ) + \frac{2}{3} \,{\left (a x + 4 \, b\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}, 2 \, \sqrt{-b} b \arctan \left (\frac{\sqrt{-b} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{b}\right ) + \frac{2}{3} \,{\left (a x + 4 \, b\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 16.3857, size = 71, normalized size = 1.01 \begin{align*} \frac{2 a \sqrt{b} x \sqrt{\frac{a x}{b} + 1}}{3} + \frac{8 b^{\frac{3}{2}} \sqrt{\frac{a x}{b} + 1}}{3} + b^{\frac{3}{2}} \log{\left (\frac{a x}{b} \right )} - 2 b^{\frac{3}{2}} \log{\left (\sqrt{\frac{a x}{b} + 1} + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*sqrt(b)*x*sqrt(a*x/b + 1)/3 + 8*b**(3/2)*sqrt(a*x/b + 1)/3 + b**(3/2)*log(a*x/b) - 2*b**(3/2)*log(sqrt(a*x
/b + 1) + 1)

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Giac [A]  time = 1.21822, size = 59, normalized size = 0.84 \begin{align*} \frac{2 \, b^{2} \arctan \left (\frac{\sqrt{a x + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} + \frac{2}{3} \,{\left (a x + b\right )}^{\frac{3}{2}} + 2 \, \sqrt{a x + b} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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