3.1773 \(\int (a+\frac{b}{x})^{5/2} \sqrt{x} \, dx\)

Optimal. Leaf size=94 \[ -\frac{5 b^2 \sqrt{a+\frac{b}{x}}}{\sqrt{x}}-5 a b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )+\frac{2}{3} x^{3/2} \left (a+\frac{b}{x}\right )^{5/2}+\frac{10}{3} b \sqrt{x} \left (a+\frac{b}{x}\right )^{3/2} \]

[Out]

(-5*b^2*Sqrt[a + b/x])/Sqrt[x] + (10*b*(a + b/x)^(3/2)*Sqrt[x])/3 + (2*(a + b/x)^(5/2)*x^(3/2))/3 - 5*a*b^(3/2
)*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])]

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Rubi [A]  time = 0.0476009, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {337, 277, 195, 217, 206} \[ -\frac{5 b^2 \sqrt{a+\frac{b}{x}}}{\sqrt{x}}-5 a b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )+\frac{2}{3} x^{3/2} \left (a+\frac{b}{x}\right )^{5/2}+\frac{10}{3} b \sqrt{x} \left (a+\frac{b}{x}\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)*Sqrt[x],x]

[Out]

(-5*b^2*Sqrt[a + b/x])/Sqrt[x] + (10*b*(a + b/x)^(3/2)*Sqrt[x])/3 + (2*(a + b/x)^(5/2)*x^(3/2))/3 - 5*a*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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 )^{5/2} \sqrt{x} \, dx &=-\left (2 \operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^{5/2}}{x^4} \, dx,x,\frac{1}{\sqrt{x}}\right )\right )\\ &=\frac{2}{3} \left (a+\frac{b}{x}\right )^{5/2} x^{3/2}-\frac{1}{3} (10 b) \operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^{3/2}}{x^2} \, dx,x,\frac{1}{\sqrt{x}}\right )\\ &=\frac{10}{3} b \left (a+\frac{b}{x}\right )^{3/2} \sqrt{x}+\frac{2}{3} \left (a+\frac{b}{x}\right )^{5/2} x^{3/2}-\left (10 b^2\right ) \operatorname{Subst}\left (\int \sqrt{a+b x^2} \, dx,x,\frac{1}{\sqrt{x}}\right )\\ &=-\frac{5 b^2 \sqrt{a+\frac{b}{x}}}{\sqrt{x}}+\frac{10}{3} b \left (a+\frac{b}{x}\right )^{3/2} \sqrt{x}+\frac{2}{3} \left (a+\frac{b}{x}\right )^{5/2} x^{3/2}-\left (5 a b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{\sqrt{x}}\right )\\ &=-\frac{5 b^2 \sqrt{a+\frac{b}{x}}}{\sqrt{x}}+\frac{10}{3} b \left (a+\frac{b}{x}\right )^{3/2} \sqrt{x}+\frac{2}{3} \left (a+\frac{b}{x}\right )^{5/2} x^{3/2}-\left (5 a 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 )\\ &=-\frac{5 b^2 \sqrt{a+\frac{b}{x}}}{\sqrt{x}}+\frac{10}{3} b \left (a+\frac{b}{x}\right )^{3/2} \sqrt{x}+\frac{2}{3} \left (a+\frac{b}{x}\right )^{5/2} x^{3/2}-5 a b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x}} \sqrt{x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0129692, size = 56, normalized size = 0.6 \[ \frac{2 a^2 x^{3/2} \sqrt{a+\frac{b}{x}} \, _2F_1\left (-\frac{5}{2},-\frac{3}{2};-\frac{1}{2};-\frac{b}{a x}\right )}{3 \sqrt{\frac{b}{a x}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(5/2)*Sqrt[x],x]

[Out]

(2*a^2*Sqrt[a + b/x]*x^(3/2)*Hypergeometric2F1[-5/2, -3/2, -1/2, -(b/(a*x))])/(3*Sqrt[1 + b/(a*x)])

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Maple [A]  time = 0.015, size = 91, normalized size = 1. \begin{align*} -{\frac{1}{3}\sqrt{{\frac{ax+b}{x}}} \left ( -2\,{x}^{2}{a}^{2}\sqrt{b}\sqrt{ax+b}+15\,{\it Artanh} \left ({\frac{\sqrt{ax+b}}{\sqrt{b}}} \right ) xa{b}^{2}-14\,xa{b}^{3/2}\sqrt{ax+b}+3\,{b}^{5/2}\sqrt{ax+b} \right ){\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{ax+b}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^(5/2)*x^(1/2),x)

[Out]

-1/3*((a*x+b)/x)^(1/2)/x^(1/2)*(-2*x^2*a^2*b^(1/2)*(a*x+b)^(1/2)+15*arctanh((a*x+b)^(1/2)/b^(1/2))*x*a*b^2-14*
x*a*b^(3/2)*(a*x+b)^(1/2)+3*b^(5/2)*(a*x+b)^(1/2))/(a*x+b)^(1/2)/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)^(5/2)*x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.49046, size = 374, normalized size = 3.98 \begin{align*} \left [\frac{15 \, a b^{\frac{3}{2}} x \log \left (\frac{a x - 2 \, \sqrt{b} \sqrt{x} \sqrt{\frac{a x + b}{x}} + 2 \, b}{x}\right ) + 2 \,{\left (2 \, a^{2} x^{2} + 14 \, a b x - 3 \, b^{2}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{6 \, x}, \frac{15 \, a \sqrt{-b} b x \arctan \left (\frac{\sqrt{-b} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{b}\right ) +{\left (2 \, a^{2} x^{2} + 14 \, a b x - 3 \, b^{2}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{3 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)*x^(1/2),x, algorithm="fricas")

[Out]

[1/6*(15*a*b^(3/2)*x*log((a*x - 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) + 2*(2*a^2*x^2 + 14*a*b*x - 3*b^
2)*sqrt(x)*sqrt((a*x + b)/x))/x, 1/3*(15*a*sqrt(-b)*b*x*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x + b)/x)/b) + (2*a^2*
x^2 + 14*a*b*x - 3*b^2)*sqrt(x)*sqrt((a*x + b)/x))/x]

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Sympy [A]  time = 108.268, size = 99, normalized size = 1.05 \begin{align*} \frac{2 a^{2} \sqrt{b} x \sqrt{\frac{a x}{b} + 1}}{3} + \frac{14 a b^{\frac{3}{2}} \sqrt{\frac{a x}{b} + 1}}{3} + \frac{5 a b^{\frac{3}{2}} \log{\left (\frac{a x}{b} \right )}}{2} - 5 a b^{\frac{3}{2}} \log{\left (\sqrt{\frac{a x}{b} + 1} + 1 \right )} - \frac{b^{\frac{5}{2}} \sqrt{\frac{a x}{b} + 1}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**(5/2)*x**(1/2),x)

[Out]

2*a**2*sqrt(b)*x*sqrt(a*x/b + 1)/3 + 14*a*b**(3/2)*sqrt(a*x/b + 1)/3 + 5*a*b**(3/2)*log(a*x/b)/2 - 5*a*b**(3/2
)*log(sqrt(a*x/b + 1) + 1) - b**(5/2)*sqrt(a*x/b + 1)/x

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Giac [A]  time = 1.27723, size = 88, normalized size = 0.94 \begin{align*} \frac{1}{3} \,{\left (\frac{15 \, b^{2} \arctan \left (\frac{\sqrt{a x + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} + 2 \,{\left (a x + b\right )}^{\frac{3}{2}} + 12 \, \sqrt{a x + b} b - \frac{3 \, \sqrt{a x + b} b^{2}}{a x}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)*x^(1/2),x, algorithm="giac")

[Out]

1/3*(15*b^2*arctan(sqrt(a*x + b)/sqrt(-b))/sqrt(-b) + 2*(a*x + b)^(3/2) + 12*sqrt(a*x + b)*b - 3*sqrt(a*x + b)
*b^2/(a*x))*a