3.1774 \(\int \frac{(a+\frac{b}{x})^{5/2}}{\sqrt{x}} \, dx\)

Optimal. Leaf size=97 \[ -\frac{15}{4} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )+2 \sqrt{x} \left (a+\frac{b}{x}\right )^{5/2}-\frac{5 b \left (a+\frac{b}{x}\right )^{3/2}}{2 \sqrt{x}}-\frac{15 a b \sqrt{a+\frac{b}{x}}}{4 \sqrt{x}} \]

[Out]

(-15*a*b*Sqrt[a + b/x])/(4*Sqrt[x]) - (5*b*(a + b/x)^(3/2))/(2*Sqrt[x]) + 2*(a + b/x)^(5/2)*Sqrt[x] - (15*a^2*
Sqrt[b]*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/4

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Rubi [A]  time = 0.0457083, antiderivative size = 97, 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} \[ -\frac{15}{4} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )+2 \sqrt{x} \left (a+\frac{b}{x}\right )^{5/2}-\frac{5 b \left (a+\frac{b}{x}\right )^{3/2}}{2 \sqrt{x}}-\frac{15 a b \sqrt{a+\frac{b}{x}}}{4 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)/Sqrt[x],x]

[Out]

(-15*a*b*Sqrt[a + b/x])/(4*Sqrt[x]) - (5*b*(a + b/x)^(3/2))/(2*Sqrt[x]) + 2*(a + b/x)^(5/2)*Sqrt[x] - (15*a^2*
Sqrt[b]*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/4

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 \frac{\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^2} \, dx,x,\frac{1}{\sqrt{x}}\right )\right )\\ &=2 \left (a+\frac{b}{x}\right )^{5/2} \sqrt{x}-(10 b) \operatorname{Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,\frac{1}{\sqrt{x}}\right )\\ &=-\frac{5 b \left (a+\frac{b}{x}\right )^{3/2}}{2 \sqrt{x}}+2 \left (a+\frac{b}{x}\right )^{5/2} \sqrt{x}-\frac{1}{2} (15 a b) \operatorname{Subst}\left (\int \sqrt{a+b x^2} \, dx,x,\frac{1}{\sqrt{x}}\right )\\ &=-\frac{15 a b \sqrt{a+\frac{b}{x}}}{4 \sqrt{x}}-\frac{5 b \left (a+\frac{b}{x}\right )^{3/2}}{2 \sqrt{x}}+2 \left (a+\frac{b}{x}\right )^{5/2} \sqrt{x}-\frac{1}{4} \left (15 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{\sqrt{x}}\right )\\ &=-\frac{15 a b \sqrt{a+\frac{b}{x}}}{4 \sqrt{x}}-\frac{5 b \left (a+\frac{b}{x}\right )^{3/2}}{2 \sqrt{x}}+2 \left (a+\frac{b}{x}\right )^{5/2} \sqrt{x}-\frac{1}{4} \left (15 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{1}{\sqrt{a+\frac{b}{x}} \sqrt{x}}\right )\\ &=-\frac{15 a b \sqrt{a+\frac{b}{x}}}{4 \sqrt{x}}-\frac{5 b \left (a+\frac{b}{x}\right )^{3/2}}{2 \sqrt{x}}+2 \left (a+\frac{b}{x}\right )^{5/2} \sqrt{x}-\frac{15}{4} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x}} \sqrt{x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0163347, size = 54, normalized size = 0.56 \[ \frac{2 a^2 \sqrt{x} \sqrt{a+\frac{b}{x}} \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};-\frac{b}{a x}\right )}{\sqrt{\frac{b}{a x}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(5/2)/Sqrt[x],x]

[Out]

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

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Maple [A]  time = 0.014, size = 93, normalized size = 1. \begin{align*} -{\frac{1}{4}\sqrt{{\frac{ax+b}{x}}} \left ( 15\,{\it Artanh} \left ({\frac{\sqrt{ax+b}}{\sqrt{b}}} \right ){a}^{2}b{x}^{2}-8\,{x}^{2}{a}^{2}\sqrt{b}\sqrt{ax+b}+9\,xa{b}^{3/2}\sqrt{ax+b}+2\,{b}^{5/2}\sqrt{ax+b} \right ){x}^{-{\frac{3}{2}}}{\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]

-1/4*((a*x+b)/x)^(1/2)/x^(3/2)*(15*arctanh((a*x+b)^(1/2)/b^(1/2))*a^2*b*x^2-8*x^2*a^2*b^(1/2)*(a*x+b)^(1/2)+9*
x*a*b^(3/2)*(a*x+b)^(1/2)+2*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 = 1.50287, size = 385, normalized size = 3.97 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{b} x^{2} \log \left (\frac{a x - 2 \, \sqrt{b} \sqrt{x} \sqrt{\frac{a x + b}{x}} + 2 \, b}{x}\right ) + 2 \,{\left (8 \, a^{2} x^{2} - 9 \, a b x - 2 \, b^{2}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{8 \, x^{2}}, \frac{15 \, a^{2} \sqrt{-b} x^{2} \arctan \left (\frac{\sqrt{-b} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{b}\right ) +{\left (8 \, a^{2} x^{2} - 9 \, a b x - 2 \, b^{2}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{4 \, x^{2}}\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]

[1/8*(15*a^2*sqrt(b)*x^2*log((a*x - 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) + 2*(8*a^2*x^2 - 9*a*b*x - 2
*b^2)*sqrt(x)*sqrt((a*x + b)/x))/x^2, 1/4*(15*a^2*sqrt(-b)*x^2*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x + b)/x)/b) +
(8*a^2*x^2 - 9*a*b*x - 2*b^2)*sqrt(x)*sqrt((a*x + b)/x))/x^2]

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Sympy [A]  time = 76.0237, size = 126, normalized size = 1.3 \begin{align*} \frac{2 a^{\frac{5}{2}} \sqrt{x}}{\sqrt{1 + \frac{b}{a x}}} - \frac{a^{\frac{3}{2}} b}{4 \sqrt{x} \sqrt{1 + \frac{b}{a x}}} - \frac{11 \sqrt{a} b^{2}}{4 x^{\frac{3}{2}} \sqrt{1 + \frac{b}{a x}}} - \frac{15 a^{2} \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b}}{\sqrt{a} \sqrt{x}} \right )}}{4} - \frac{b^{3}}{2 \sqrt{a} x^{\frac{5}{2}} \sqrt{1 + \frac{b}{a 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**(5/2)*sqrt(x)/sqrt(1 + b/(a*x)) - a**(3/2)*b/(4*sqrt(x)*sqrt(1 + b/(a*x))) - 11*sqrt(a)*b**2/(4*x**(3/2)*
sqrt(1 + b/(a*x))) - 15*a**2*sqrt(b)*asinh(sqrt(b)/(sqrt(a)*sqrt(x)))/4 - b**3/(2*sqrt(a)*x**(5/2)*sqrt(1 + b/
(a*x)))

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Giac [A]  time = 1.36085, size = 92, normalized size = 0.95 \begin{align*} \frac{1}{4} \,{\left (\frac{15 \, b \arctan \left (\frac{\sqrt{a x + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} + 8 \, \sqrt{a x + b} - \frac{9 \,{\left (a x + b\right )}^{\frac{3}{2}} b - 7 \, \sqrt{a x + b} b^{2}}{a^{2} x^{2}}\right )} a^{2} \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]

1/4*(15*b*arctan(sqrt(a*x + b)/sqrt(-b))/sqrt(-b) + 8*sqrt(a*x + b) - (9*(a*x + b)^(3/2)*b - 7*sqrt(a*x + b)*b
^2)/(a^2*x^2))*a^2